NUCLEAR PHYSICS AND MANY-BODY SYSTEMS
1 1. NEUTRON CAPTURE IN NUCLEAR REACTORS
Problem 1. A neutron flux of 1.5×1013 neutrons/cm2/s is incident on a target material with a
neutron capture cross section of 3×10−24 cm2. Calculate the neutron capture rate in captures per
second for this target material.
Solution 1.
Given: Neutron flux = 1.5×1013 neutrons/cm2/s, Neutron capture cross section = 3×10−24
cm2.
The neutron capture rate is given by the formula:
Rate =Neutron flux ×Neutron capture cross section
Rate = (1.5×1013 neutrons/cm2/s)×(3 ×10−24 cm2)
Rate = 4.5×10−11 captures/s/cm2
Therefore, the neutron capture rate for this target material is 4.5×10−11 captures/second per
cm2.
2 2. QUANTUM MECHANICAL DESCRIPTION OF NUCLEAR FUSION
Problem 2. Consider a nuclear fusion reaction where two deuterium (2
1H) nuclei combine to
form a helium-4 (4
2He) nucleus. The energy released in this fusion reaction can be calculated using
the equation:
Q= (m1+m2−m3−m4)c2
where m1, m2, m3, m4are the rest masses of the deuterium nuclei and helium-4 nucleus in
atomic mass units, cis the speed of light, and Qis the energy released in MeV.
Given that the rest masses of the particles are:
m(2
1H)=2.014102 u, m(4
2He)=4.001506 u
Calculate the energy released in the fusion reaction and express the result in MeV.
Solution 2.
Given data:
m(2
1H) = 2.014102 u, m(4
2He)=4.001506 u
We can substitute these values into the equation for the energy released in the nuclear fusion
reaction:
Q= (2.014102 + 2.014102 −4.001506)c2
Q= 0.026698c2
To convert from atomic mass units (u) to energy (MeV), we use the conversion factor 1u=
931.5MeV/c2.
Substitute this conversion factor into the equation to get the energy in MeV:
Q= 0.026698 ×931.5
Q24.9MeV
Therefore, the energy released in the fusion reaction is approximately 24.9 MeV.
3 3. NUCLEAR DECAY PROCESSES
Problem 3. The half-life of a certain radioactive isotope is 10 days. If you start with a sample
containing 100 grams of the isotope, how much of the isotope will remain after 30 days?
Additional context: The decay of a radioactive substance follows an exponential decay model
given by the equation N(t) = N0e−λt, where: - N(t)is the amount of the substance remaining at
time t, - N0is the initial amount of the substance, - λis the decay constant, - tis the time elapsed.
a) How much of the isotope will remain after 10 days?
b) What is the decay constant, λ, for this isotope?
c) How much of the isotope will remain after 30 days?
Solution 3.
a) After 10 days:
N(10) = 100 ·e−λ·10
Given that 10 days is the half-life, N(10) = 100
2= 50, so:
50 = 100 ·e−λ·10
0.5 = e−10λ
ln(0.5) = −10λ
λ=−ln(0.5)
10
λ≈0.0693 per day
b) Decay constant, λ≈0.0693 per day.
c) After 30 days:
N(30) = 100 ·e−λ·30
= 100 ·e−0.0693·30
≈17.07 grams remaining
Therefore, after 30 days, approximately 17.07 grams of the isotope will remain.
4 4. ISOSPIN SYMMETRY IN NUCLEAR STRUCTURE
Problem 4. Consider a system of two protons and two neutrons in a nucleus. The isospin
quantum number for protons and neutrons is 1
2.
a) Determine the isospin values for the four particles in this system.
b) If we allow for isospin mixing, what are the possible values for the total isospin of the system?
c) Suppose the total isospin of the system is measured to be T= 1. What are the possible
values of the third component, Tz, of the isospin?
Solution 4. a) The isospin quantum number Tzcan take values ranging from −Tto T. In this
case, each proton and neutron has isospin T=1
2, so their Tzvalues can be −1
2or 1
2.
Therefore, the isospin values for the four particles in the system are:
Proton 1: T=1
2,Tz=1
2
Proton 2: T=1
2,Tz=1
2
Neutron 1: T=1
2,Tz=1
2
Neutron 2: T=1
2,Tz=1
2
b) When we allow for isospin mixing, the possible values for the total isospin of the system
range from |T1−T2|to T1+T2. In this case, T1=T2=1
2.
So, the possible values for the total isospin of the system are 0 and 1.
c) If the total isospin of the system is measured to be T= 1, the possible values for the third
component of isospin, Tz, are −1, 0, and 1.
5 5. ALPHA DECAY AND NUCLEAR STABILITY
Problem 5. The half-life of a certain radioactive isotope is 100 years. If there are initially 1000
atoms of this isotope, how many atoms will remain after 300 years?
Given:
•Half-life of the isotope = 100 years
•Initial number of atoms = 1000
•Time elapsed = 300 years
Solution 5. a) To calculate the decay constant, we use the formula:
N(t) = N0×e−λt
where: N(t)= number of atoms at time t,N0= initial number of atoms, λ= decay constant, t=
time.
First, we find the decay constant using the half-life formula:
T1/2=ln(2)
λ
Substitute T1/2= 100 years:
100 = ln(2)
λ
λ=ln(2)
100
λ≈0.00693 years−1
b) Next, we find the number of atoms remaining after 300 years:
N(300) = 1000 ×e−0.00693×300
N(300) = 1000 ×e−2.079
N(300) ≈1000 ×0.125
N(300) ≈125 atoms
Therefore, after 300 years, approximately 125 atoms of this isotope will remain.
6 6. NUCLEAR FISSION CHAIN REACTIONS
Problem 6. Consider a nuclear fission chain reaction in which a neutron induces the fission of
a uranium-235 nucleus, releasing on average 2.5 neutrons per fission event. The probability of a
neutron causing a subsequent fission event is 0.8.
a) If 100 neutrons are released in the initial fission event, how many total neutrons will be
released in the subsequent chain reaction?
b) Determine the number of fission events that will occur in this chain reaction.
c) If the energy released per fission event is 200 MeV, calculate the total energy released in this
chain reaction.
Solution 6.
a) In the initial fission event, 100 neutrons are released. Since on average 2.5 neutrons are
released per fission event, these 100 neutrons will lead to 100 ×2.5 = 250 neutrons being released
in the subsequent chain reaction.
b) The probability of a neutron causing a subsequent fission event is 0.8. This means 0.8neu-
trons out of the 2.5neutrons on average released per fission event will cause a subsequent fission
event. Therefore, the number of fission events that will occur in this chain reaction is 0.8×250
2.5= 80
fission events.
c) The total energy released per fission event is 200 MeV. Since there are 80 fission events, the
total energy released in this chain reaction is 80 ×200 = 16000 MeV or 16 GeV.
7 7. HYPERSPHERICAL COORDINATES IN MANY-BODY SYSTEMS
Problem 7. Consider a system of three identical particles confined in a 3-dimensional harmonic
oscillator potential. The Hamiltonian for this system can be written in terms of hyperspherical co-
ordinates as:
H=−¯h2
2m∂2
∂R2+5
R
∂
∂R +3
2mω2R2,
where Ris a collective coordinate. Calculate the ground state energy of this system in terms of the
oscillator frequency ω.
Solution 7. To find the ground state energy of the system, we need to solve the Schrödinger
equation
HΨ(R,Ω) = EΨ(R,Ω),
where Ψ(R,Ω)is the wave function of the system and Ris the hyperradius.
For the ground state energy, the wave function can be assumed to be separable as Ψ(R,Ω) =
F(R)Y(Ω).
Substituting this into the Schrödinger equation and dividing by F(R)Y(Ω), we get
−¯h2
2mF′′
F+5
R
F′
F+3
2mω2R2=E.
Separating variables, we have
F′′
F=−5
R
F′
F+2mE
¯h2−3mω2R2
¯h2.
To simplify this equation, we make the substitution u=√mωR, which transforms the equation
into
−F′′ +5
uF′= (λu2−k)F,
where λ=2mE
¯hω and k=3
2.
The solution to this differential equation can be written in terms of Laguerre polynomials Lα
k(u)
as
F(R) = Au5/2e−u2/2L5
0(u2),
where Ais a normalization constant.
Now, the ground state energy is given by E0=3
2¯hω.
Therefore, the ground state energy of the system in terms of the oscillator frequency ωis E0=
3
2¯hω.
8 8. INTRINSIC AND COLLECTIVE NUCLEAR EXCITATIONS
Problem 8. Consider a nucleus with a mass number A= 240 and a charge number Z= 92.
a) Calculate the binding energy of the nucleus in MeV given that the atomic mass unit uis
equivalent to 931.5MeV/c2.
b) Determine the energy needed to completely remove a neutron from the nucleus.
c) Find the Q-value of the alpha decay process where the nucleus decays into a daughter
nucleus with A= 236 and Z= 90, emitting an alpha particle.
Solution 8.
a) The binding energy of a nucleus can be calculated using the formula:
B.E. = (Zmp+N mn−M)c2
Where: - Zis the number of protons, - Nis the number of neutrons, - mpis the mass of a proton,
-mnis the mass of a neutron, - Mis the mass of the nucleus, and - cis the speed of light.
Given that mp=mn= 1u= 931.5MeV/c2, we first need to calculate the total number of
neutrons N=A−Z:
N= 240 −92 = 148
The mass of the nucleus Mcan be calculated by:
M=Amu= 240 ×1u= 240 ×931.5MeV/c2
Therefore, the binding energy B.E. is:
B.E. = (92 ×931.5 + 148 ×931.5−240 ×931.5) MeV = 16900 MeV
So, the binding energy of the nucleus is 16900 MeV.
b) To remove a neutron, we need to consider the mass of the neutron itself, hence the energy
required is the mass of the neutron:
E=mnc2= 1 ×931.5MeV = 931.5MeV
Hence, the energy needed to remove a neutron is 931.5MeV.
c) The Q-value of the alpha decay process is given by:
Q= (Mi−Mf−mα)c2
Where: - Miis the initial mass, - Mfis the final mass, and - mαis the mass of the alpha particle.
Given that Mi= 240×931.5MeV/c2and Mf= 236×931.5MeV/c2, and mα= 4×931.5MeV/c2,
we can calculate the Q-value:
Q= (240 ×931.5−236 ×931.5−4×931.5) MeV = 0 MeV
Therefore, the Q-value of the alpha decay process is 0MeV.
9 9. NUCLEAR FORCES AND BINDING ENERGIES
Problem 9. Consider a nucleus with atomic number Z= 6. Given that the mass of a proton is
mp= 1.67 ×10−27 kg and the mass of a neutron is mn= 1.675 ×10−27 kg, and the mass of the
nucleus is M= 12.000 amu, calculate the binding energy per nucleon in MeV.
Solution 9. a) The total number of protons and neutrons in the nucleus can be calculated using
the atomic number Z= 6.
N=Z+ (A−Z)
= 6 + (12 −6)
= 12 nucleons
b) The total mass of the nucleus can be calculated using the masses of protons and neutrons:
Mnucleus =Z·mp+ (A−Z)·mn
= 6 ·1.67 ×10−27 + 6 ·1.675 ×10−27
= 9.99 ×10−26 + 10.05 ×10−26
= 20.04 ×10−26 kg
= 20.04 ×10−26 kg ×1amu
1.66 ×10−27 kg
= 12.09 amu
The binding energy of the nucleus can be approximated using the mass defect:
∆m= 12.000 amu −12.09 amu
=−0.09 amu
c) The binding energy per nucleon can be calculated as:
BE/A =∆E
A·931.5MeV/amu
=0.09 amu
12 ·931.5MeV/amu
= 6.9375 MeV
Therefore, the binding energy per nucleon is 6.9375 MeV.
10 10. SCATTERING THEORY IN NUCLEAR PHYSICS
Problem 10. Consider a neutron scattering off a nucleus with a potential given by V(r) =
−V0e−r2/a2, where ris the distance from the center of the nucleus, V0>0, and ais a constant with
units of length.
A neutron with energy Escatters from this potential. The scattering cross section is given by
σ=4π
k2P∞
l=0(2l+ 1)|fl|2, where k=q2mE
¯h2is the wave number, flis the scattering amplitude for
orbital angular momentum l, and the sum runs over all possible values of l.
a) Show that the differential cross section for neutron-nucleus scattering, dσ/dΩ, can be written
as |f(θ)|2, where θis the scattering angle.
b) Assuming f(θ) = 1
ksin(δ0)eiα, where αis a real constant, show that dσ/dΩ = |f(θ)|2=
sin2(δ0)
k2.
Solution 10.
a) The differential cross section is defined as dσ
dΩ=|f(θ)|2, where θis the scattering angle.
This comes from the fact that the differential cross section is related to the scattering amplitude by
dσ
dΩ=|f(θ)|2. Thus, dσ/dΩ = |f(θ)|2.
b) Given f(θ) = 1
ksin(δ0)eiα, where αis real, we have |f(θ)|2=1
ksin(δ0)eiα
2=sin2(δ0)
k2. This
expression gives the differential cross section for neutron-nucleus scattering.
11 11. NUCLEAR SHAPE TRANSITIONS
Problem 11. Consider a nucleus undergoing a shape transition from a deformed shape to a
spherical shape as the excitation energy increases. The potential energy of the nucleus can be
described by the formula:
V(r) = V0
1 + e(r−r0
a)
where: V(r)is the potential energy at distance rfrom the center of the nucleus, V0= 50 MeV is
the depth of the potential well, r0= 1.2fm is the equilibrium radius of the nucleus in the deformed
state, a= 0.1fm characterizes the sharpness of the potential well.
a) Calculate the potential energy of the nucleus at r= 1.5fm in the deformed state.
b) At what distance from the center of the nucleus does the potential energy equal half of V0in
the deformed state?
c) Determine the critical excitation energy in MeV needed for the nucleus to transition from a
deformed shape to a spherical shape.
Solution 11. a) To calculate the potential energy at r= 1.5fm in the deformed state, we
substitute r= 1.5fm into the potential energy formula:
V(1.5) = 50
1 + e(1.5−1.2
0.1)=50
1 + e(3)
V(1.5) = 50
1 + e3≈50
1 + 20.0855 ≈50
21.0855 ≈2.37 MeV
Therefore, the potential energy of the nucleus at r= 1.5fm in the deformed state is approxi-
mately 2.37 MeV.
b) To find the distance at which the potential energy equals half of V0, we set V(r) = V0
2and
solve for r:
50
1 + e(r−1.2
0.1)=50
2
1 + e(r−1.2
0.1)= 2
e(r−1.2
0.1)= 1
r−1.2
0.1= 0
r= 1.2fm
Therefore, the potential energy equals half of V0at a distance of 1.2 fm from the center of the
nucleus in the deformed state.
c) The critical excitation energy ∆Eneeded for the nucleus to transition from a deformed shape
to a spherical shape occurs when the nucleus reaches the equilibrium radius of a spherical shape,
i.e., when r=r0for the potential energy formula. Thus:
V(r0) = 50
1 + e(r0−1.2
0.1)= 50
1 + e(r0−1.2
0.1)= 1
e(r0−1.2
0.1)= 0
r0= 1.2fm
Therefore, the critical excitation energy needed for the nucleus to transition from a deformed
shape to a spherical shape is 50 MeV.
12 12. QUANTUM MONTE CARLO METHODS FOR MANY-BODY SYSTEMS
Problem 12. Consider a quantum many-body system of Nparticles confined in a one-dimensional
infinite square well potential given by V(x) = 0 for 0< x < L and V(x) = ∞otherwise. Sup-
pose that the ground state wave function of this system can be approximated by a product of N
single-particle wave functions ψi(x), where each single-particle wave function can be expressed
as ψi(x) = q2
Lsin iπx
Lfor i= 1,2, . . . , N.
a) Calculate the normalization constant for the single-particle wave functions ψi(x).
b) Determine the ground state energy of the many-body system with these single-particle wave
functions.
c) Suppose the particles are non-interacting fermions. What is the ground state energy of the
system in this case?
Solution 12. a) The normalization condition for a wave function ψ(x)is R∞
−∞ |ψ(x)|2dx = 1. For
the given single-particle wave function ψi(x) = q2
Lsin iπx
L, we have:
ZL
0|ψi(x)|2dx =ZL
0 r2
Lsin iπx
L!2
dx
ZL
0|ψi(x)|2dx =2
LZL
0
sin2iπx
Ldx
ZL
0|ψi(x)|2dx =2
LZL
0
1−cos(2iπx/L)
2dx
Using trigonometric identity sin2(θ) = 1−cos(2θ)
2, we can simplify the integral as:
ZL
0|ψi(x)|2dx =1
Lx−L
2iπ sin 2iπx
LL
0
ZL
0|ψi(x)|2dx =1
L[L−0]
ZL
0|ψi(x)|2dx = 1
Therefore, the single-particle wave function ψi(x)is normalized.
b) The total ground state wave function for the many-body system is given by Ψ(x1, x2, . . . , xN) =
QN
i=1 ψi(xi). The ground state energy Efor this system is given by the sum of the single-particle
energies:
E=
N
X
i=1
¯h2π2i2
2mL2
c) Since the particles are non-interacting fermions, the ground state energy of the system is
given by the sum of the single-particle energies up to the Fermi level. The Fermi level is the highest
occupied single-particle energy level, which is the Nth level for Nnon-interacting fermions.
Efermions =
N
X
i=1
¯h2π2i2
2mL2
13 13. ELECTRON-NUCLEUS INTERACTIONS
Problem 13. An electron is scattered off a gold nucleus (197
79 Au) with a charge of +79e. The
electron has an initial kinetic energy of 5 MeV and scatters backwards at an angle of 180◦. Calculate
the recoil energy of the gold nucleus.
Given:
•Charge of the gold nucleus, Z= +79e
•Initial kinetic energy of the electron, Ei= 5 MeV
•Scattered angle, θ= 180◦
•Mass of the gold nucleus, mAu = 197 mproton
Solution 13.
a) The recoil energy of the gold nucleus can be calculated using the conservation of momentum
and energy. Since the electron scatters backwards, the final momentum of the system should be
zero.
Initial momentum of the system:
pi=p2meEi=p2×9.11 ×10−31 ×5×106≈1.74 ×10−22 kg m/s
The final kinetic energy of the gold nucleus is given by:
Ef=p2
i
2mAu
=(1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈7.62 ×10−11 J
Converting this energy to MeV:
Ef=7.62 ×10−11
1.6×10−13 ≈476.25 MeV
b) The final momentum of the gold nucleus can be calculated using the law of conservation of
momentum:
pAu =−pi=−1.74 ×10−22 kg m/s
c) The recoil energy of the gold nucleus can be calculated using the final momentum:
Erecoil =p2
Au
2mAu
=(−1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈3.81 ×10−10 J
Converting this energy to MeV:
Erecoil =3.81 ×10−10
1.6×10−13 ≈2381.25 MeV
14 14. CHIRAL EFFECTIVE FIELD THEORY IN NUCLEAR PHYSICS
Problem 14. Consider a two-neutron system in a harmonic oscillator potential with a frequency
ω= 5 MeV.
a) Calculate the energy of the ground state of the two-neutron system in MeV.
b) If the interaction potential between the neutrons is given by a contact interaction with a
strength of V0=−20 MeV, compute the energy of the first excited state of the system in MeV.
Solution 14.
a) The energy of the ground state of a harmonic oscillator potential is given by Eground =3
2¯hω.
Substituting ¯h= 197 MeV fm and ω= 5 MeV into the formula:
Eground =3
2·197 MeV fm ·5MeV =3
2·197 ·5 = 1477.5MeV
Therefore, the energy of the ground state of the two-neutron system is 1477.5 MeV.
b) The energy of the first excited state can be calculated by considering the ground state energy
plus the interaction energy. The interaction energy for a contact interaction is given by Vint =
⟨ϕ2|V0|ϕ1⟩where ϕ1and ϕ2are the ground and first excited state wave functions, respectively.
For a harmonic oscillator potential, the wave functions are known and the matrix element can
be calculated. For a two-particle system, the interaction energy is:
Vint =V01−15
8=V0−7
8=−20 MeV −7
8= 17.5MeV
Therefore, the energy of the first excited state is 1477.5 + 17.5 = 1495 MeV.
15 15. SUPERFLUIDITY IN NUCLEAR MATTER
Problem 15. Consider a system of nucleons in a superfluid state with a pairing energy gap of
∆=2MeV. Calculate the critical temperature Tcfor this superfluid system.
Given:
•The transition temperature Tc= 0.57∆
Solution 15.
•We are given the formula for Tcin terms of the pairing energy gap ∆:Tc= 0.57∆
•Substituting the given value of ∆=2MeV into the formula:
Tc= 0.57 ×2MeV
Tc= 1.14 MeV
Therefore, the critical temperature for this superfluid system is Tc= 1.14 MeV.
16 16. NUCLEAR SPIN-ORBIT COUPLING
Problem 16. Consider a nucleus with a spin-orbit coupling constant equal to a= 20 MeV.
a) Calculate the energy splitting between the l= 1 states for a nucleus with N= 50 neutrons
and Z= 40 protons.
b) Calculate the energy splitting between the l= 1 states for a nucleus with N= 70 neutrons
and Z= 40 protons.
Solution 16.
a) The energy splitting between l= 1 states can be calculated using the formula:
∆E=a·(N−Z)
Plugging in the values a= 20 MeV, N= 50, and Z= 40, we find:
∆E= 20 MeV ·(50 −40) = 200 MeV
So, the energy splitting between the l= 1 states for this nucleus is 200 MeV.
b) Similarly, for the nucleus with N= 70 and Z= 40, the energy splitting between l= 1 states
is given by:
∆E= 20 MeV ·(70 −40) = 20 MeV ·30 = 600 MeV
Therefore, for a nucleus with N= 70 neutrons and Z= 40 protons, the energy splitting between
the l= 1 states is 600 MeV.
17 17. DEUTERON SCATTERING AND BOUND STATES
Problem 17. Consider a deuteron, a bound state of a proton and a neutron. The total energy
of the deuteron is given by E=−2.224 MeV. The radius of the deuteron is r= 2 fm.
a) Calculate the reduced mass of the proton-neutron system.
b) Calculate the binding energy of the deuteron.
c) If the deuteron is in a state of higher angular momentum, how would its radius compare to
r= 2 fm?
Solution 17.
a) The reduced mass µof the proton-neutron system can be calculated as:
µ=mp×mn
mp+mn
where mpis the mass of the proton and mnis the mass of the neutron. Given that mp≈
1.67 ×10−27 kg and mn≈1.675 ×10−27 kg, we have:
µ=(1.67 ×10−27 kg)×(1.675 ×10−27 kg)
1.67 ×10−27 kg + 1.675 ×10−27 kg ≈1.673 ×10−27 kg
Therefore, the reduced mass of the proton-neutron system is µ≈1.673 ×10−27 kg.
b) The binding energy Bof the deuteron is related to the total energy Eas:
B=−E
Therefore, the binding energy of the deuteron is B=−(−2.224) = 2.224 MeV.
c) The radius r′of the deuteron in a state of higher angular momentum can be related to the
initial radius rand the new angular momentum l′as:
r′=r×l+ 1
l′+ 11/3
Since the state has higher angular momentum, l′> l, which implies that r′< r. Therefore, the
radius of the deuteron in a state of higher angular momentum would be smaller than r= 2 fm.
I. Problem:
Problem 18. Consider a particle of mass mand energy Eapproaching a one-dimensional
potential barrier of height V0and width a. The potential inside the barrier is zero. Assuming E > V0,
calculate the transmission coefficient Tfor the particle through the barrier, where T=|t|2, with t
being the transmission amplitude.
Additional Information: The potential barrier is given by:
V(x) = (0,for 0< x < a
V0,for x < 0and x>a
Solution 18: a) To find the transmission coefficient T, we first need to calculate the transmission
amplitude t. The transmission coefficient is then given by T=|t|2.
b) The transmission coefficient can be expressed in terms of the wave numbers k1and k2inside
and outside the barrier, respectively. We have:
T=k2
k1
c) The wave numbers k1and k2can be related to the energy Eand potential V0using the
following equations:
k1=p2m(E−V0)
¯h
k2=√2mE
¯h
Now, let’s substitute k1and k2into the expression for the transmission coefficient to find T.
c) Detailed Solution: a) The transmission amplitude tis given by:
t=e−ika
b) Substituting the expressions for k1and k2into the formula for T:
T=e−ika
2=e−2ika
c) Substituting the values of k1and k2into the expression, we get:
T= √2mE
¯h! p2m(E−V0)
¯h!
Thus, the transmission coefficient Tin this scenario is determined by the given particle’s energy
E, the potential barrier height V0, and the barrier width a.
18 19. NUCLEAR DENSITY FUNCTIONAL THEORY
Problem 19. Consider a system of nucleons with a total binding energy described by the
Skyrme energy density functional in the form:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
where ρ(r)is the nucleon density, τ(r)is the kinetic energy density, and C0,C3are constants.
Given that the nucleon density is ρ(r) = 3
4πR3e−3r
2R, where R= 1.2fm, and the constants are
C0=−400 MeV, C3= 100 MeV, calculate the total binding energy of the system.
Solution 19. The total binding energy E[ρ]can be calculated by evaluating the integral over
the entire volume:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
Substitute the given nucleon density ρ(r)into the expression for E[ρ]:
E[ρ] = Zd3r ¯h2
2mτ(r) + C03
4πR3e−3r
2R2
+C33
4πR3e−3r
2R3!
Since the kinetic energy density term τ(r)is not given, let’s assume it to be zero for simplicity.
We can now integrate the potential energy terms over the entire volume by performing the following
steps:
a) Calculate ρ2(r) = 3
4πR3e−3r
2R2and ρ3(r) = 3
4πR3e−3r
2R3.
b) Substitute these expressions into the integral and solve for the total binding energy.
c) Evaluate the integral numerically using appropriate techniques to find the total binding energy
of the system.
19 20. LIGHT NUCLEI AND FEW-BODY SYSTEMS
Problem 20. Consider a helium-4 nucleus (alpha particle) consisting of 2 protons and 2 neu-
trons.
Given that the rest masses of a proton and neutron are mp= 1.6726219 ×10−27 kg and mn=
1.6749275 ×10−27 kg, respectively, and the speed of light c= 3.00 ×108m/s, calculate the binding
energy of a helium-4 nucleus. Assume the atomic mass unit (u) is defined as 1u= 1.660539×10−27
kg.
Solution 20. The binding energy of a nucleus is the energy required to completely disassemble
a nucleus into its constituent protons and neutrons. It is defined as the difference between the total
rest mass energy of the separate protons and neutrons and the rest mass energy of the nucleus.
a) The total rest mass energy of a helium-4 nucleus:
mnucleus = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Enucleus =mnucleusc2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
b) The total rest mass energy of the separate protons and neutrons:
mseparate = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Eseparate =mseparatec2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
c) The binding energy of the helium-4 nucleus is:
Therefore, the binding energy of the helium-4 nucleus is 0 Joules, indicating that the nucleus
is stable.
20 21. GAMOW-TELLER TRANSITIONS IN NUCLEI
Problem 21. Consider a nucleus undergoing a Gamow-Teller transition, where the initial state
has quantum numbers Jπ= 1+and the final state has quantum numbers Jπ= 0−. If the energy
difference between the initial and final states is 2 MeV, and the reduced transition probability B(GT )
is 3×10−3units, calculate the half-life of this transition.
Solution 21.
a) The decay rate λof a nuclear transition is related to the half-life T1
2
by the equation:
T1
2=ln(2)
λ
b) The decay rate of a Gamow-Teller transition can be calculated using the formula:
λ=1
T1
2
B(GT )
2Ji+1 2
where T1
2
is the half-life, B(GT )is the reduced transition probability, and Jiis the total angular
momentum of the initial state.
c) Substituting the given values into the formula, we have:
λ=1
3×10−3
3×10−3
2×1+1 2
λ=1
3×10−3×9
(3 ×10−3)2
λ=1
3×10−3×9
9×10−6
λ=1
3×10−3×106= 103sec−1
Using the relation between decay rate and half-life, we find:
T1
2=ln(2)
λ=ln(2)
103≈0.693 ms
Therefore, the half-life of this Gamow-Teller transition is approximately 0.693 milliseconds.
21 22. CLUSTERING PHENOMENA IN NUCLEAR STRUCTURE
Problem 22. Consider a nucleus with mass number A= 12, consisting of two alpha clus-
ters and 4 neutrons. The energy of an alpha particle is 28.3 MeV and the mass of a neutron is
approximately 939.6 MeV/c2.
a) Calculate the binding energy per nucleon of this nucleus.
b) Determine the total energy of this nucleus.
c) If the neutrons interact only through the nuclear force, what is the total nuclear force energy
for this nucleus?
Solution 22. a) The total binding energy of the nucleus is given by the sum of the binding
energy of the two alpha clusters and the binding energy of the 4 neutrons. The binding energy per
nucleon is then the total binding energy divided by the number of nucleons.
The binding energy of an alpha particle is Eα= 28.3MeV and the mass of an alpha particle
is approximately 4 times the mass of a nucleon. So, the binding energy per nucleon for the alpha
clusters would be Eα/4 = 28.3/4 = 7.075 MeV.
The binding energy of a neutron is zero (since it is a free particle). Therefore, the binding energy
per nucleon for the 4 neutrons is 0/4=0MeV.
The total binding energy for the nucleus is 2×Eα+ 4 ×0=2×28.3 = 56.6MeV.
Therefore, the binding energy per nucleon is 56.6/12 = 4.717 MeV.
b) The total energy of the nucleus is the sum of the rest mass energy and the binding energy.
Since the rest mass energy is not given in the problem, we will neglect it for this specific calculation.
Therefore, the total energy of the nucleus is just the binding energy, which is 56.6 MeV.
c) The total nuclear force energy for this nucleus would be zero since the neutrons are not
interacting with each other.
22 Nuclear Physics and Many-Body Systems
Problem: Consider a nucleus with a total of 20 protons and 30 neutrons. Assume that each
proton can pair with a neutron in the ground state of the nucleus. Determine the total angular
momentum and total spin of the nucleus.
Solution:
a) To find the total angular momentum of the nucleus, we first need to calculate the individual
angular momenta of the protons and neutrons in the ground state. Since each proton can pair with
a neutron, we have a total of 20 pairs. The individual angular momentum of a proton is l=1
2and
for a neutron is also l=1
2.
The total angular momentum of a pair is given by L=Lp+Ln, where Lpis the angular mo-
mentum of the proton and Lnis the angular momentum of the neutron. Therefore, the total angular
momentum of a pair is 1
2+1
2= 1.
Since we have 20 pairs, the total angular momentum of the nucleus is Ltotal = 20 ×1 = 20.
b) Next, to find the total spin of the nucleus, we know that the total spin of a nucleus is the sum
of the spins of each nucleon. Both protons and neutrons have spin s=1
2.
Since we have 20 protons and 30 neutrons, the total spin of the nucleus is Stotal = (20 ×1
2) +
(30 ×1
2) = 25.
Therefore, the total angular momentum of the nucleus is 20 and the total spin is 25.
I. Problem 1:
Consider a heavy nucleus with a mass number A= 240 and a proton number Z= 92. The
ground state of this nucleus has a binding energy of B= 7.6MeV per nucleon. Calculate the total
binding energy of this nucleus.
Solution 1: The total binding energy of the nucleus can be calculated using the formula:
Ebinding =B×A
Substitute the given values into the formula:
Ebinding = 7.6MeV/nucleon ×240 nucleons
Ebinding = 1824 MeV
Therefore, the total binding energy of this nucleus is 1824 MeV.
II. Problem 2:
A heavy nucleus in an excited state decays by emitting a beta particle (electron) and a neutrino.
If the energy released in this decay is 2 MeV, calculate the energy of the excited state before the
decay.
Solution 2: The energy released in the decay is related to the energy of the excited state by
the equation:
Ereleased =Ei−Ef
Where Eiis the energy of the excited state and Efis the energy of the final state after decay.
Given that Ereleased = 2 MeV, and the final state consists of the beta particle and the neutrino
which have negligible energy, the energy of the final state Efcan be considered as nearly 0.
Hence, Ei=Ereleased = 2 MeV.
Therefore, the energy of the excited state before the decay is 2 MeV.
23 25. NUCLEAR PHYSICS APPLICATIONS IN ASTROPHYSICS
Problem 25. Consider the nuclear reaction 4
2He +3
1H→7
3Li.
Given the rest mass of:
•Helium-4 atom 4
2He is 4.0026 u
•Hydrogen-3 atom 3
1His 3.0155 u
•Lithium-7 atom 7
3Li is 7.0160 u
Calculate the energy released in the reaction in MeV.
Solution 25. The energy released in the nuclear reaction can be calculated using Einstein’s
mass-energy equivalence E= ∆mc2.
The change in mass (∆m) is the total rest mass of the reactants minus the total rest mass of
the products:
∆m= (4.0026 + 3.0155) −7.0160
= 7.0181 −7.0160
= 0.0021 u
Using the conversion factor 1u= 931.5MeV/c2, we can calculate the energy released:
E= (0.0021 u)×(931.5MeV/c2)
= 1.95915 MeV
Therefore, the energy released in the reaction is 1.95915 M eV .
Q= 0.026698c2
To convert from atomic mass units (u) to energy (MeV), we use the conversion factor 1u=
931.5MeV/c2.
Substitute this conversion factor into the equation to get the energy in MeV:
Q= 0.026698 ×931.5
Q24.9MeV
Therefore, the energy released in the fusion reaction is approximately 24.9 MeV.
3 3. NUCLEAR DECAY PROCESSES
Problem 3. The half-life of a certain radioactive isotope is 10 days. If you start with a sample
containing 100 grams of the isotope, how much of the isotope will remain after 30 days?
Additional context: The decay of a radioactive substance follows an exponential decay model
given by the equation N(t) = N0e−λt, where: - N(t)is the amount of the substance remaining at
time t, - N0is the initial amount of the substance, - λis the decay constant, - tis the time elapsed.
a) How much of the isotope will remain after 10 days?
b) What is the decay constant, λ, for this isotope?
c) How much of the isotope will remain after 30 days?
Solution 3.
a) After 10 days:
N(10) = 100 ·e−λ·10
Given that 10 days is the half-life, N(10) = 100
2= 50, so:
50 = 100 ·e−λ·10
0.5 = e−10λ
ln(0.5) = −10λ
λ=−ln(0.5)
10
λ≈0.0693 per day
b) Decay constant, λ≈0.0693 per day.
c) After 30 days:
N(30) = 100 ·e−λ·30
= 100 ·e−0.0693·30
≈17.07 grams remaining
Therefore, after 30 days, approximately 17.07 grams of the isotope will remain.
4 4. ISOSPIN SYMMETRY IN NUCLEAR STRUCTURE
Problem 4. Consider a system of two protons and two neutrons in a nucleus. The isospin
quantum number for protons and neutrons is 1
2.
a) Determine the isospin values for the four particles in this system.
b) If we allow for isospin mixing, what are the possible values for the total isospin of the system?
c) Suppose the total isospin of the system is measured to be T= 1. What are the possible
values of the third component, Tz, of the isospin?
Solution 4. a) The isospin quantum number Tzcan take values ranging from −Tto T. In this
case, each proton and neutron has isospin T=1
2, so their Tzvalues can be −1
2or 1
2.
Therefore, the isospin values for the four particles in the system are:
Proton 1: T=1
2,Tz=1
2
Proton 2: T=1
2,Tz=1
2
Neutron 1: T=1
2,Tz=1
2
Neutron 2: T=1
2,Tz=1
2
b) When we allow for isospin mixing, the possible values for the total isospin of the system
range from |T1−T2|to T1+T2. In this case, T1=T2=1
2.
So, the possible values for the total isospin of the system are 0 and 1.
c) If the total isospin of the system is measured to be T= 1, the possible values for the third
component of isospin, Tz, are −1, 0, and 1.
5 5. ALPHA DECAY AND NUCLEAR STABILITY
Problem 5. The half-life of a certain radioactive isotope is 100 years. If there are initially 1000
atoms of this isotope, how many atoms will remain after 300 years?
Given:
•Half-life of the isotope = 100 years
•Initial number of atoms = 1000
•Time elapsed = 300 years
Solution 5. a) To calculate the decay constant, we use the formula:
N(t) = N0×e−λt
where: N(t)= number of atoms at time t,N0= initial number of atoms, λ= decay constant, t=
time.
First, we find the decay constant using the half-life formula:
T1/2=ln(2)
λ
Substitute T1/2= 100 years:
100 = ln(2)
λ
λ=ln(2)
100
λ≈0.00693 years−1
b) Next, we find the number of atoms remaining after 300 years:
N(300) = 1000 ×e−0.00693×300
N(300) = 1000 ×e−2.079
N(300) ≈1000 ×0.125
N(300) ≈125 atoms
Therefore, after 300 years, approximately 125 atoms of this isotope will remain.
6 6. NUCLEAR FISSION CHAIN REACTIONS
Problem 6. Consider a nuclear fission chain reaction in which a neutron induces the fission of
a uranium-235 nucleus, releasing on average 2.5 neutrons per fission event. The probability of a
neutron causing a subsequent fission event is 0.8.
a) If 100 neutrons are released in the initial fission event, how many total neutrons will be
released in the subsequent chain reaction?
b) Determine the number of fission events that will occur in this chain reaction.
c) If the energy released per fission event is 200 MeV, calculate the total energy released in this
chain reaction.
Solution 6.
a) In the initial fission event, 100 neutrons are released. Since on average 2.5 neutrons are
released per fission event, these 100 neutrons will lead to 100 ×2.5 = 250 neutrons being released
in the subsequent chain reaction.
b) The probability of a neutron causing a subsequent fission event is 0.8. This means 0.8neu-
trons out of the 2.5neutrons on average released per fission event will cause a subsequent fission
event. Therefore, the number of fission events that will occur in this chain reaction is 0.8×250
2.5= 80
fission events.
c) The total energy released per fission event is 200 MeV. Since there are 80 fission events, the
total energy released in this chain reaction is 80 ×200 = 16000 MeV or 16 GeV.
7 7. HYPERSPHERICAL COORDINATES IN MANY-BODY SYSTEMS
Problem 7. Consider a system of three identical particles confined in a 3-dimensional harmonic
oscillator potential. The Hamiltonian for this system can be written in terms of hyperspherical co-
ordinates as:
H=−¯h2
2m∂2
∂R2+5
R
∂
∂R +3
2mω2R2,
where Ris a collective coordinate. Calculate the ground state energy of this system in terms of the
oscillator frequency ω.
Solution 7. To find the ground state energy of the system, we need to solve the Schrödinger
equation
HΨ(R,Ω) = EΨ(R,Ω),
where Ψ(R,Ω)is the wave function of the system and Ris the hyperradius.
For the ground state energy, the wave function can be assumed to be separable as Ψ(R,Ω) =
F(R)Y(Ω).
Substituting this into the Schrödinger equation and dividing by F(R)Y(Ω), we get
−¯h2
2mF′′
F+5
R
F′
F+3
2mω2R2=E.
Separating variables, we have
F′′
F=−5
R
F′
F+2mE
¯h2−3mω2R2
¯h2.
To simplify this equation, we make the substitution u=√mωR, which transforms the equation
into
−F′′ +5
uF′= (λu2−k)F,
where λ=2mE
¯hω and k=3
2.
The solution to this differential equation can be written in terms of Laguerre polynomials Lα
k(u)
as
F(R) = Au5/2e−u2/2L5
0(u2),
where Ais a normalization constant.
Now, the ground state energy is given by E0=3
2¯hω.
Therefore, the ground state energy of the system in terms of the oscillator frequency ωis E0=
3
2¯hω.
8 8. INTRINSIC AND COLLECTIVE NUCLEAR EXCITATIONS
Problem 8. Consider a nucleus with a mass number A= 240 and a charge number Z= 92.
a) Calculate the binding energy of the nucleus in MeV given that the atomic mass unit uis
equivalent to 931.5MeV/c2.
b) Determine the energy needed to completely remove a neutron from the nucleus.
c) Find the Q-value of the alpha decay process where the nucleus decays into a daughter
nucleus with A= 236 and Z= 90, emitting an alpha particle.
Solution 8.
a) The binding energy of a nucleus can be calculated using the formula:
B.E. = (Zmp+N mn−M)c2
Where: - Zis the number of protons, - Nis the number of neutrons, - mpis the mass of a proton,
-mnis the mass of a neutron, - Mis the mass of the nucleus, and - cis the speed of light.
Given that mp=mn= 1u= 931.5MeV/c2, we first need to calculate the total number of
neutrons N=A−Z:
N= 240 −92 = 148
The mass of the nucleus Mcan be calculated by:
M=Amu= 240 ×1u= 240 ×931.5MeV/c2
Therefore, the binding energy B.E. is:
B.E. = (92 ×931.5 + 148 ×931.5−240 ×931.5) MeV = 16900 MeV
So, the binding energy of the nucleus is 16900 MeV.
b) To remove a neutron, we need to consider the mass of the neutron itself, hence the energy
required is the mass of the neutron:
E=mnc2= 1 ×931.5MeV = 931.5MeV
Hence, the energy needed to remove a neutron is 931.5MeV.
c) The Q-value of the alpha decay process is given by:
Q= (Mi−Mf−mα)c2
Where: - Miis the initial mass, - Mfis the final mass, and - mαis the mass of the alpha particle.
Given that Mi= 240×931.5MeV/c2and Mf= 236×931.5MeV/c2, and mα= 4×931.5MeV/c2,
we can calculate the Q-value:
Q= (240 ×931.5−236 ×931.5−4×931.5) MeV = 0 MeV
Therefore, the Q-value of the alpha decay process is 0MeV.
9 9. NUCLEAR FORCES AND BINDING ENERGIES
Problem 9. Consider a nucleus with atomic number Z= 6. Given that the mass of a proton is
mp= 1.67 ×10−27 kg and the mass of a neutron is mn= 1.675 ×10−27 kg, and the mass of the
nucleus is M= 12.000 amu, calculate the binding energy per nucleon in MeV.
Solution 9. a) The total number of protons and neutrons in the nucleus can be calculated using
the atomic number Z= 6.
N=Z+ (A−Z)
= 6 + (12 −6)
= 12 nucleons
b) The total mass of the nucleus can be calculated using the masses of protons and neutrons:
Mnucleus =Z·mp+ (A−Z)·mn
= 6 ·1.67 ×10−27 + 6 ·1.675 ×10−27
= 9.99 ×10−26 + 10.05 ×10−26
= 20.04 ×10−26 kg
= 20.04 ×10−26 kg ×1amu
1.66 ×10−27 kg
= 12.09 amu
The binding energy of the nucleus can be approximated using the mass defect:
∆m= 12.000 amu −12.09 amu
=−0.09 amu
c) The binding energy per nucleon can be calculated as:
BE/A =∆E
A·931.5MeV/amu
=0.09 amu
12 ·931.5MeV/amu
= 6.9375 MeV
Therefore, the binding energy per nucleon is 6.9375 MeV.
10 10. SCATTERING THEORY IN NUCLEAR PHYSICS
Problem 10. Consider a neutron scattering off a nucleus with a potential given by V(r) =
−V0e−r2/a2, where ris the distance from the center of the nucleus, V0>0, and ais a constant with
units of length.
A neutron with energy Escatters from this potential. The scattering cross section is given by
σ=4π
k2P∞
l=0(2l+ 1)|fl|2, where k=q2mE
¯h2is the wave number, flis the scattering amplitude for
orbital angular momentum l, and the sum runs over all possible values of l.
a) Show that the differential cross section for neutron-nucleus scattering, dσ/dΩ, can be written
as |f(θ)|2, where θis the scattering angle.
b) Assuming f(θ) = 1
ksin(δ0)eiα, where αis a real constant, show that dσ/dΩ = |f(θ)|2=
sin2(δ0)
k2.
Solution 10.
a) The differential cross section is defined as dσ
dΩ=|f(θ)|2, where θis the scattering angle.
This comes from the fact that the differential cross section is related to the scattering amplitude by
dσ
dΩ=|f(θ)|2. Thus, dσ/dΩ = |f(θ)|2.
b) Given f(θ) = 1
ksin(δ0)eiα, where αis real, we have |f(θ)|2=1
ksin(δ0)eiα
2=sin2(δ0)
k2. This
expression gives the differential cross section for neutron-nucleus scattering.
11 11. NUCLEAR SHAPE TRANSITIONS
Problem 11. Consider a nucleus undergoing a shape transition from a deformed shape to a
spherical shape as the excitation energy increases. The potential energy of the nucleus can be
described by the formula:
V(r) = V0
1 + e(r−r0
a)
where: V(r)is the potential energy at distance rfrom the center of the nucleus, V0= 50 MeV is
the depth of the potential well, r0= 1.2fm is the equilibrium radius of the nucleus in the deformed
state, a= 0.1fm characterizes the sharpness of the potential well.
a) Calculate the potential energy of the nucleus at r= 1.5fm in the deformed state.
b) At what distance from the center of the nucleus does the potential energy equal half of V0in
the deformed state?
c) Determine the critical excitation energy in MeV needed for the nucleus to transition from a
deformed shape to a spherical shape.
Solution 11. a) To calculate the potential energy at r= 1.5fm in the deformed state, we
substitute r= 1.5fm into the potential energy formula:
V(1.5) = 50
1 + e(1.5−1.2
0.1)=50
1 + e(3)
V(1.5) = 50
1 + e3≈50
1 + 20.0855 ≈50
21.0855 ≈2.37 MeV
Therefore, the potential energy of the nucleus at r= 1.5fm in the deformed state is approxi-
mately 2.37 MeV.
b) To find the distance at which the potential energy equals half of V0, we set V(r) = V0
2and
solve for r:
50
1 + e(r−1.2
0.1)=50
2
1 + e(r−1.2
0.1)= 2
e(r−1.2
0.1)= 1
r−1.2
0.1= 0
r= 1.2fm
Therefore, the potential energy equals half of V0at a distance of 1.2 fm from the center of the
nucleus in the deformed state.
c) The critical excitation energy ∆Eneeded for the nucleus to transition from a deformed shape
to a spherical shape occurs when the nucleus reaches the equilibrium radius of a spherical shape,
i.e., when r=r0for the potential energy formula. Thus:
V(r0) = 50
1 + e(r0−1.2
0.1)= 50
1 + e(r0−1.2
0.1)= 1
e(r0−1.2
0.1)= 0
r0= 1.2fm
Therefore, the critical excitation energy needed for the nucleus to transition from a deformed
shape to a spherical shape is 50 MeV.
12 12. QUANTUM MONTE CARLO METHODS FOR MANY-BODY SYSTEMS
Problem 12. Consider a quantum many-body system of Nparticles confined in a one-dimensional
infinite square well potential given by V(x) = 0 for 0< x < L and V(x) = ∞otherwise. Sup-
pose that the ground state wave function of this system can be approximated by a product of N
single-particle wave functions ψi(x), where each single-particle wave function can be expressed
as ψi(x) = q2
Lsin iπx
Lfor i= 1,2, . . . , N.
a) Calculate the normalization constant for the single-particle wave functions ψi(x).
b) Determine the ground state energy of the many-body system with these single-particle wave
functions.
c) Suppose the particles are non-interacting fermions. What is the ground state energy of the
system in this case?
Solution 12. a) The normalization condition for a wave function ψ(x)is R∞
−∞ |ψ(x)|2dx = 1. For
the given single-particle wave function ψi(x) = q2
Lsin iπx
L, we have:
ZL
0|ψi(x)|2dx =ZL
0 r2
Lsin iπx
L!2
dx
ZL
0|ψi(x)|2dx =2
LZL
0
sin2iπx
Ldx
ZL
0|ψi(x)|2dx =2
LZL
0
1−cos(2iπx/L)
2dx
Using trigonometric identity sin2(θ) = 1−cos(2θ)
2, we can simplify the integral as:
ZL
0|ψi(x)|2dx =1
Lx−L
2iπ sin 2iπx
LL
0
ZL
0|ψi(x)|2dx =1
L[L−0]
ZL
0|ψi(x)|2dx = 1
Therefore, the single-particle wave function ψi(x)is normalized.
b) The total ground state wave function for the many-body system is given by Ψ(x1, x2, . . . , xN) =
QN
i=1 ψi(xi). The ground state energy Efor this system is given by the sum of the single-particle
energies:
E=
N
X
i=1
¯h2π2i2
2mL2
c) Since the particles are non-interacting fermions, the ground state energy of the system is
given by the sum of the single-particle energies up to the Fermi level. The Fermi level is the highest
occupied single-particle energy level, which is the Nth level for Nnon-interacting fermions.
Efermions =
N
X
i=1
¯h2π2i2
2mL2
13 13. ELECTRON-NUCLEUS INTERACTIONS
Problem 13. An electron is scattered off a gold nucleus (197
79 Au) with a charge of +79e. The
electron has an initial kinetic energy of 5 MeV and scatters backwards at an angle of 180◦. Calculate
the recoil energy of the gold nucleus.
Given:
•Charge of the gold nucleus, Z= +79e
•Initial kinetic energy of the electron, Ei= 5 MeV
•Scattered angle, θ= 180◦
•Mass of the gold nucleus, mAu = 197 mproton
Solution 13.
a) The recoil energy of the gold nucleus can be calculated using the conservation of momentum
and energy. Since the electron scatters backwards, the final momentum of the system should be
zero.
Initial momentum of the system:
pi=p2meEi=p2×9.11 ×10−31 ×5×106≈1.74 ×10−22 kg m/s
The final kinetic energy of the gold nucleus is given by:
Ef=p2
i
2mAu
=(1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈7.62 ×10−11 J
Converting this energy to MeV:
Ef=7.62 ×10−11
1.6×10−13 ≈476.25 MeV
b) The final momentum of the gold nucleus can be calculated using the law of conservation of
momentum:
pAu =−pi=−1.74 ×10−22 kg m/s
c) The recoil energy of the gold nucleus can be calculated using the final momentum:
Erecoil =p2
Au
2mAu
=(−1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈3.81 ×10−10 J
Converting this energy to MeV:
Erecoil =3.81 ×10−10
1.6×10−13 ≈2381.25 MeV
14 14. CHIRAL EFFECTIVE FIELD THEORY IN NUCLEAR PHYSICS
Problem 14. Consider a two-neutron system in a harmonic oscillator potential with a frequency
ω= 5 MeV.
a) Calculate the energy of the ground state of the two-neutron system in MeV.
b) If the interaction potential between the neutrons is given by a contact interaction with a
strength of V0=−20 MeV, compute the energy of the first excited state of the system in MeV.
Solution 14.
a) The energy of the ground state of a harmonic oscillator potential is given by Eground =3
2¯hω.
Substituting ¯h= 197 MeV fm and ω= 5 MeV into the formula:
Eground =3
2·197 MeV fm ·5MeV =3
2·197 ·5 = 1477.5MeV
Therefore, the energy of the ground state of the two-neutron system is 1477.5 MeV.
b) The energy of the first excited state can be calculated by considering the ground state energy
plus the interaction energy. The interaction energy for a contact interaction is given by Vint =
⟨ϕ2|V0|ϕ1⟩where ϕ1and ϕ2are the ground and first excited state wave functions, respectively.
For a harmonic oscillator potential, the wave functions are known and the matrix element can
be calculated. For a two-particle system, the interaction energy is:
Vint =V01−15
8=V0−7
8=−20 MeV −7
8= 17.5MeV
Therefore, the energy of the first excited state is 1477.5 + 17.5 = 1495 MeV.
15 15. SUPERFLUIDITY IN NUCLEAR MATTER
Problem 15. Consider a system of nucleons in a superfluid state with a pairing energy gap of
∆=2MeV. Calculate the critical temperature Tcfor this superfluid system.
Given:
•The transition temperature Tc= 0.57∆
Solution 15.
•We are given the formula for Tcin terms of the pairing energy gap ∆:Tc= 0.57∆
•Substituting the given value of ∆=2MeV into the formula:
Tc= 0.57 ×2MeV
Tc= 1.14 MeV
Therefore, the critical temperature for this superfluid system is Tc= 1.14 MeV.
16 16. NUCLEAR SPIN-ORBIT COUPLING
Problem 16. Consider a nucleus with a spin-orbit coupling constant equal to a= 20 MeV.
a) Calculate the energy splitting between the l= 1 states for a nucleus with N= 50 neutrons
and Z= 40 protons.
b) Calculate the energy splitting between the l= 1 states for a nucleus with N= 70 neutrons
and Z= 40 protons.
Solution 16.
a) The energy splitting between l= 1 states can be calculated using the formula:
∆E=a·(N−Z)
Plugging in the values a= 20 MeV, N= 50, and Z= 40, we find:
∆E= 20 MeV ·(50 −40) = 200 MeV
So, the energy splitting between the l= 1 states for this nucleus is 200 MeV.
b) Similarly, for the nucleus with N= 70 and Z= 40, the energy splitting between l= 1 states
is given by:
∆E= 20 MeV ·(70 −40) = 20 MeV ·30 = 600 MeV
Therefore, for a nucleus with N= 70 neutrons and Z= 40 protons, the energy splitting between
the l= 1 states is 600 MeV.
17 17. DEUTERON SCATTERING AND BOUND STATES
Problem 17. Consider a deuteron, a bound state of a proton and a neutron. The total energy
of the deuteron is given by E=−2.224 MeV. The radius of the deuteron is r= 2 fm.
a) Calculate the reduced mass of the proton-neutron system.
b) Calculate the binding energy of the deuteron.
c) If the deuteron is in a state of higher angular momentum, how would its radius compare to
r= 2 fm?
Solution 17.
a) The reduced mass µof the proton-neutron system can be calculated as:
µ=mp×mn
mp+mn
where mpis the mass of the proton and mnis the mass of the neutron. Given that mp≈
1.67 ×10−27 kg and mn≈1.675 ×10−27 kg, we have:
µ=(1.67 ×10−27 kg)×(1.675 ×10−27 kg)
1.67 ×10−27 kg + 1.675 ×10−27 kg ≈1.673 ×10−27 kg
Therefore, the reduced mass of the proton-neutron system is µ≈1.673 ×10−27 kg.
b) The binding energy Bof the deuteron is related to the total energy Eas:
B=−E
Therefore, the binding energy of the deuteron is B=−(−2.224) = 2.224 MeV.
c) The radius r′of the deuteron in a state of higher angular momentum can be related to the
initial radius rand the new angular momentum l′as:
r′=r×l+ 1
l′+ 11/3
Since the state has higher angular momentum, l′> l, which implies that r′< r. Therefore, the
radius of the deuteron in a state of higher angular momentum would be smaller than r= 2 fm.
I. Problem:
Problem 18. Consider a particle of mass mand energy Eapproaching a one-dimensional
potential barrier of height V0and width a. The potential inside the barrier is zero. Assuming E > V0,
calculate the transmission coefficient Tfor the particle through the barrier, where T=|t|2, with t
being the transmission amplitude.
Additional Information: The potential barrier is given by:
V(x) = (0,for 0< x < a
V0,for x < 0and x>a
Solution 18: a) To find the transmission coefficient T, we first need to calculate the transmission
amplitude t. The transmission coefficient is then given by T=|t|2.
b) The transmission coefficient can be expressed in terms of the wave numbers k1and k2inside
and outside the barrier, respectively. We have:
T=k2
k1
c) The wave numbers k1and k2can be related to the energy Eand potential V0using the
following equations:
k1=p2m(E−V0)
¯h
k2=√2mE
¯h
Now, let’s substitute k1and k2into the expression for the transmission coefficient to find T.
c) Detailed Solution: a) The transmission amplitude tis given by:
t=e−ika
b) Substituting the expressions for k1and k2into the formula for T:
T=e−ika
2=e−2ika
c) Substituting the values of k1and k2into the expression, we get:
T= √2mE
¯h! p2m(E−V0)
¯h!
Thus, the transmission coefficient Tin this scenario is determined by the given particle’s energy
E, the potential barrier height V0, and the barrier width a.
18 19. NUCLEAR DENSITY FUNCTIONAL THEORY
Problem 19. Consider a system of nucleons with a total binding energy described by the
Skyrme energy density functional in the form:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
where ρ(r)is the nucleon density, τ(r)is the kinetic energy density, and C0,C3are constants.
Given that the nucleon density is ρ(r) = 3
4πR3e−3r
2R, where R= 1.2fm, and the constants are
C0=−400 MeV, C3= 100 MeV, calculate the total binding energy of the system.
Solution 19. The total binding energy E[ρ]can be calculated by evaluating the integral over
the entire volume:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
Substitute the given nucleon density ρ(r)into the expression for E[ρ]:
E[ρ] = Zd3r ¯h2
2mτ(r) + C03
4πR3e−3r
2R2
+C33
4πR3e−3r
2R3!
Since the kinetic energy density term τ(r)is not given, let’s assume it to be zero for simplicity.
We can now integrate the potential energy terms over the entire volume by performing the following
steps:
a) Calculate ρ2(r) = 3
4πR3e−3r
2R2and ρ3(r) = 3
4πR3e−3r
2R3.
b) Substitute these expressions into the integral and solve for the total binding energy.
c) Evaluate the integral numerically using appropriate techniques to find the total binding energy
of the system.
19 20. LIGHT NUCLEI AND FEW-BODY SYSTEMS
Problem 20. Consider a helium-4 nucleus (alpha particle) consisting of 2 protons and 2 neu-
trons.
Given that the rest masses of a proton and neutron are mp= 1.6726219 ×10−27 kg and mn=
1.6749275 ×10−27 kg, respectively, and the speed of light c= 3.00 ×108m/s, calculate the binding
energy of a helium-4 nucleus. Assume the atomic mass unit (u) is defined as 1u= 1.660539×10−27
kg.
Solution 20. The binding energy of a nucleus is the energy required to completely disassemble
a nucleus into its constituent protons and neutrons. It is defined as the difference between the total
rest mass energy of the separate protons and neutrons and the rest mass energy of the nucleus.
a) The total rest mass energy of a helium-4 nucleus:
mnucleus = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Enucleus =mnucleusc2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
b) The total rest mass energy of the separate protons and neutrons:
mseparate = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Eseparate =mseparatec2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
c) The binding energy of the helium-4 nucleus is:
Therefore, the binding energy of the helium-4 nucleus is 0 Joules, indicating that the nucleus
is stable.
20 21. GAMOW-TELLER TRANSITIONS IN NUCLEI
Problem 21. Consider a nucleus undergoing a Gamow-Teller transition, where the initial state
has quantum numbers Jπ= 1+and the final state has quantum numbers Jπ= 0−. If the energy
difference between the initial and final states is 2 MeV, and the reduced transition probability B(GT )
is 3×10−3units, calculate the half-life of this transition.
Solution 21.
a) The decay rate λof a nuclear transition is related to the half-life T1
2
by the equation:
T1
2=ln(2)
λ
b) The decay rate of a Gamow-Teller transition can be calculated using the formula:
λ=1
T1
2
B(GT )
2Ji+1 2
where T1
2
is the half-life, B(GT )is the reduced transition probability, and Jiis the total angular
momentum of the initial state.
c) Substituting the given values into the formula, we have:
λ=1
3×10−3
3×10−3
2×1+1 2
λ=1
3×10−3×9
(3 ×10−3)2
λ=1
3×10−3×9
9×10−6
λ=1
3×10−3×106= 103sec−1
Using the relation between decay rate and half-life, we find:
T1
2=ln(2)
λ=ln(2)
103≈0.693 ms
Therefore, the half-life of this Gamow-Teller transition is approximately 0.693 milliseconds.
21 22. CLUSTERING PHENOMENA IN NUCLEAR STRUCTURE
Problem 22. Consider a nucleus with mass number A= 12, consisting of two alpha clus-
ters and 4 neutrons. The energy of an alpha particle is 28.3 MeV and the mass of a neutron is
approximately 939.6 MeV/c2.
a) Calculate the binding energy per nucleon of this nucleus.
b) Determine the total energy of this nucleus.
c) If the neutrons interact only through the nuclear force, what is the total nuclear force energy
for this nucleus?
Solution 22. a) The total binding energy of the nucleus is given by the sum of the binding
energy of the two alpha clusters and the binding energy of the 4 neutrons. The binding energy per
nucleon is then the total binding energy divided by the number of nucleons.
The binding energy of an alpha particle is Eα= 28.3MeV and the mass of an alpha particle
is approximately 4 times the mass of a nucleon. So, the binding energy per nucleon for the alpha
clusters would be Eα/4 = 28.3/4 = 7.075 MeV.
The binding energy of a neutron is zero (since it is a free particle). Therefore, the binding energy
per nucleon for the 4 neutrons is 0/4=0MeV.
The total binding energy for the nucleus is 2×Eα+ 4 ×0=2×28.3 = 56.6MeV.
Therefore, the binding energy per nucleon is 56.6/12 = 4.717 MeV.
b) The total energy of the nucleus is the sum of the rest mass energy and the binding energy.
Since the rest mass energy is not given in the problem, we will neglect it for this specific calculation.
Therefore, the total energy of the nucleus is just the binding energy, which is 56.6 MeV.
c) The total nuclear force energy for this nucleus would be zero since the neutrons are not
interacting with each other.
22 Nuclear Physics and Many-Body Systems
Problem: Consider a nucleus with a total of 20 protons and 30 neutrons. Assume that each
proton can pair with a neutron in the ground state of the nucleus. Determine the total angular
momentum and total spin of the nucleus.
Solution:
a) To find the total angular momentum of the nucleus, we first need to calculate the individual
angular momenta of the protons and neutrons in the ground state. Since each proton can pair with
a neutron, we have a total of 20 pairs. The individual angular momentum of a proton is l=1
2and
for a neutron is also l=1
2.
The total angular momentum of a pair is given by L=Lp+Ln, where Lpis the angular mo-
mentum of the proton and Lnis the angular momentum of the neutron. Therefore, the total angular
momentum of a pair is 1
2+1
2= 1.
Since we have 20 pairs, the total angular momentum of the nucleus is Ltotal = 20 ×1 = 20.
b) Next, to find the total spin of the nucleus, we know that the total spin of a nucleus is the sum
of the spins of each nucleon. Both protons and neutrons have spin s=1
2.
Since we have 20 protons and 30 neutrons, the total spin of the nucleus is Stotal = (20 ×1
2) +
(30 ×1
2) = 25.
Therefore, the total angular momentum of the nucleus is 20 and the total spin is 25.
I. Problem 1:
Consider a heavy nucleus with a mass number A= 240 and a proton number Z= 92. The
ground state of this nucleus has a binding energy of B= 7.6MeV per nucleon. Calculate the total
binding energy of this nucleus.
Solution 1: The total binding energy of the nucleus can be calculated using the formula:
Ebinding =B×A
Substitute the given values into the formula:
Ebinding = 7.6MeV/nucleon ×240 nucleons
Ebinding = 1824 MeV
Therefore, the total binding energy of this nucleus is 1824 MeV.
II. Problem 2:
A heavy nucleus in an excited state decays by emitting a beta particle (electron) and a neutrino.
If the energy released in this decay is 2 MeV, calculate the energy of the excited state before the
decay.
Solution 2: The energy released in the decay is related to the energy of the excited state by
the equation:
Ereleased =Ei−Ef
Where Eiis the energy of the excited state and Efis the energy of the final state after decay.
Given that Ereleased = 2 MeV, and the final state consists of the beta particle and the neutrino
which have negligible energy, the energy of the final state Efcan be considered as nearly 0.
Hence, Ei=Ereleased = 2 MeV.
Therefore, the energy of the excited state before the decay is 2 MeV.
23 25. NUCLEAR PHYSICS APPLICATIONS IN ASTROPHYSICS
Problem 25. Consider the nuclear reaction 4
2He +3
1H→7
3Li.
Given the rest mass of:
•Helium-4 atom 4
2He is 4.0026 u
•Hydrogen-3 atom 3
1His 3.0155 u
•Lithium-7 atom 7
3Li is 7.0160 u
Calculate the energy released in the reaction in MeV.
Solution 25. The energy released in the nuclear reaction can be calculated using Einstein’s
mass-energy equivalence E= ∆mc2.
The change in mass (∆m) is the total rest mass of the reactants minus the total rest mass of
the products:
∆m= (4.0026 + 3.0155) −7.0160
= 7.0181 −7.0160
= 0.0021 u
Using the conversion factor 1u= 931.5MeV/c2, we can calculate the energy released:
E= (0.0021 u)×(931.5MeV/c2)
= 1.95915 MeV
Therefore, the energy released in the reaction is 1.95915 M eV .
Q= 0.026698c2
To convert from atomic mass units (u) to energy (MeV), we use the conversion factor 1u=
931.5MeV/c2.
Substitute this conversion factor into the equation to get the energy in MeV:
Q= 0.026698 ×931.5
Q24.9MeV
Therefore, the energy released in the fusion reaction is approximately 24.9 MeV.
3 3. NUCLEAR DECAY PROCESSES
Problem 3. The half-life of a certain radioactive isotope is 10 days. If you start with a sample
containing 100 grams of the isotope, how much of the isotope will remain after 30 days?
Additional context: The decay of a radioactive substance follows an exponential decay model
given by the equation N(t) = N0e−λt, where: - N(t)is the amount of the substance remaining at
time t, - N0is the initial amount of the substance, - λis the decay constant, - tis the time elapsed.
a) How much of the isotope will remain after 10 days?
b) What is the decay constant, λ, for this isotope?
c) How much of the isotope will remain after 30 days?
Solution 3.
a) After 10 days:
N(10) = 100 ·e−λ·10
Given that 10 days is the half-life, N(10) = 100
2= 50, so:
50 = 100 ·e−λ·10
0.5 = e−10λ
ln(0.5) = −10λ
λ=−ln(0.5)
10
λ≈0.0693 per day
b) Decay constant, λ≈0.0693 per day.
c) After 30 days:
N(30) = 100 ·e−λ·30
= 100 ·e−0.0693·30
≈17.07 grams remaining
Therefore, after 30 days, approximately 17.07 grams of the isotope will remain.
4 4. ISOSPIN SYMMETRY IN NUCLEAR STRUCTURE
Problem 4. Consider a system of two protons and two neutrons in a nucleus. The isospin
quantum number for protons and neutrons is 1
2.
a) Determine the isospin values for the four particles in this system.
b) If we allow for isospin mixing, what are the possible values for the total isospin of the system?
c) Suppose the total isospin of the system is measured to be T= 1. What are the possible
values of the third component, Tz, of the isospin?
Solution 4. a) The isospin quantum number Tzcan take values ranging from −Tto T. In this
case, each proton and neutron has isospin T=1
2, so their Tzvalues can be −1
2or 1
2.
Therefore, the isospin values for the four particles in the system are:
Proton 1: T=1
2,Tz=1
2
Proton 2: T=1
2,Tz=1
2
Neutron 1: T=1
2,Tz=1
2
Neutron 2: T=1
2,Tz=1
2
b) When we allow for isospin mixing, the possible values for the total isospin of the system
range from |T1−T2|to T1+T2. In this case, T1=T2=1
2.
So, the possible values for the total isospin of the system are 0 and 1.
c) If the total isospin of the system is measured to be T= 1, the possible values for the third
component of isospin, Tz, are −1, 0, and 1.
5 5. ALPHA DECAY AND NUCLEAR STABILITY
Problem 5. The half-life of a certain radioactive isotope is 100 years. If there are initially 1000
atoms of this isotope, how many atoms will remain after 300 years?
Given:
•Half-life of the isotope = 100 years
•Initial number of atoms = 1000
•Time elapsed = 300 years
Solution 5. a) To calculate the decay constant, we use the formula:
N(t) = N0×e−λt
where: N(t)= number of atoms at time t,N0= initial number of atoms, λ= decay constant, t=
time.
First, we find the decay constant using the half-life formula:
T1/2=ln(2)
λ
Substitute T1/2= 100 years:
100 = ln(2)
λ
λ=ln(2)
100
λ≈0.00693 years−1
b) Next, we find the number of atoms remaining after 300 years:
N(300) = 1000 ×e−0.00693×300
N(300) = 1000 ×e−2.079
N(300) ≈1000 ×0.125
N(300) ≈125 atoms
Therefore, after 300 years, approximately 125 atoms of this isotope will remain.
6 6. NUCLEAR FISSION CHAIN REACTIONS
Problem 6. Consider a nuclear fission chain reaction in which a neutron induces the fission of
a uranium-235 nucleus, releasing on average 2.5 neutrons per fission event. The probability of a
neutron causing a subsequent fission event is 0.8.
a) If 100 neutrons are released in the initial fission event, how many total neutrons will be
released in the subsequent chain reaction?
b) Determine the number of fission events that will occur in this chain reaction.
c) If the energy released per fission event is 200 MeV, calculate the total energy released in this
chain reaction.
Solution 6.
a) In the initial fission event, 100 neutrons are released. Since on average 2.5 neutrons are
released per fission event, these 100 neutrons will lead to 100 ×2.5 = 250 neutrons being released
in the subsequent chain reaction.
b) The probability of a neutron causing a subsequent fission event is 0.8. This means 0.8neu-
trons out of the 2.5neutrons on average released per fission event will cause a subsequent fission
event. Therefore, the number of fission events that will occur in this chain reaction is 0.8×250
2.5= 80
fission events.
c) The total energy released per fission event is 200 MeV. Since there are 80 fission events, the
total energy released in this chain reaction is 80 ×200 = 16000 MeV or 16 GeV.
7 7. HYPERSPHERICAL COORDINATES IN MANY-BODY SYSTEMS
Problem 7. Consider a system of three identical particles confined in a 3-dimensional harmonic
oscillator potential. The Hamiltonian for this system can be written in terms of hyperspherical co-
ordinates as:
H=−¯h2
2m∂2
∂R2+5
R
∂
∂R +3
2mω2R2,
where Ris a collective coordinate. Calculate the ground state energy of this system in terms of the
oscillator frequency ω.
Solution 7. To find the ground state energy of the system, we need to solve the Schrödinger
equation
HΨ(R,Ω) = EΨ(R,Ω),
where Ψ(R,Ω)is the wave function of the system and Ris the hyperradius.
For the ground state energy, the wave function can be assumed to be separable as Ψ(R,Ω) =
F(R)Y(Ω).
Substituting this into the Schrödinger equation and dividing by F(R)Y(Ω), we get
−¯h2
2mF′′
F+5
R
F′
F+3
2mω2R2=E.
Separating variables, we have
F′′
F=−5
R
F′
F+2mE
¯h2−3mω2R2
¯h2.
To simplify this equation, we make the substitution u=√mωR, which transforms the equation
into
−F′′ +5
uF′= (λu2−k)F,
where λ=2mE
¯hω and k=3
2.
The solution to this differential equation can be written in terms of Laguerre polynomials Lα
k(u)
as
F(R) = Au5/2e−u2/2L5
0(u2),
where Ais a normalization constant.
Now, the ground state energy is given by E0=3
2¯hω.
Therefore, the ground state energy of the system in terms of the oscillator frequency ωis E0=
3
2¯hω.
8 8. INTRINSIC AND COLLECTIVE NUCLEAR EXCITATIONS
Problem 8. Consider a nucleus with a mass number A= 240 and a charge number Z= 92.
a) Calculate the binding energy of the nucleus in MeV given that the atomic mass unit uis
equivalent to 931.5MeV/c2.
b) Determine the energy needed to completely remove a neutron from the nucleus.
c) Find the Q-value of the alpha decay process where the nucleus decays into a daughter
nucleus with A= 236 and Z= 90, emitting an alpha particle.
Solution 8.
a) The binding energy of a nucleus can be calculated using the formula:
B.E. = (Zmp+N mn−M)c2
Where: - Zis the number of protons, - Nis the number of neutrons, - mpis the mass of a proton,
-mnis the mass of a neutron, - Mis the mass of the nucleus, and - cis the speed of light.
Given that mp=mn= 1u= 931.5MeV/c2, we first need to calculate the total number of
neutrons N=A−Z:
N= 240 −92 = 148
The mass of the nucleus Mcan be calculated by:
M=Amu= 240 ×1u= 240 ×931.5MeV/c2
Therefore, the binding energy B.E. is:
B.E. = (92 ×931.5 + 148 ×931.5−240 ×931.5) MeV = 16900 MeV
So, the binding energy of the nucleus is 16900 MeV.
b) To remove a neutron, we need to consider the mass of the neutron itself, hence the energy
required is the mass of the neutron:
E=mnc2= 1 ×931.5MeV = 931.5MeV
Hence, the energy needed to remove a neutron is 931.5MeV.
c) The Q-value of the alpha decay process is given by:
Q= (Mi−Mf−mα)c2
Where: - Miis the initial mass, - Mfis the final mass, and - mαis the mass of the alpha particle.
Given that Mi= 240×931.5MeV/c2and Mf= 236×931.5MeV/c2, and mα= 4×931.5MeV/c2,
we can calculate the Q-value:
Q= (240 ×931.5−236 ×931.5−4×931.5) MeV = 0 MeV
Therefore, the Q-value of the alpha decay process is 0MeV.
9 9. NUCLEAR FORCES AND BINDING ENERGIES
Problem 9. Consider a nucleus with atomic number Z= 6. Given that the mass of a proton is
mp= 1.67 ×10−27 kg and the mass of a neutron is mn= 1.675 ×10−27 kg, and the mass of the
nucleus is M= 12.000 amu, calculate the binding energy per nucleon in MeV.
Solution 9. a) The total number of protons and neutrons in the nucleus can be calculated using
the atomic number Z= 6.
N=Z+ (A−Z)
= 6 + (12 −6)
= 12 nucleons
b) The total mass of the nucleus can be calculated using the masses of protons and neutrons:
Mnucleus =Z·mp+ (A−Z)·mn
= 6 ·1.67 ×10−27 + 6 ·1.675 ×10−27
= 9.99 ×10−26 + 10.05 ×10−26
= 20.04 ×10−26 kg
= 20.04 ×10−26 kg ×1amu
1.66 ×10−27 kg
= 12.09 amu
The binding energy of the nucleus can be approximated using the mass defect:
∆m= 12.000 amu −12.09 amu
=−0.09 amu
c) The binding energy per nucleon can be calculated as:
BE/A =∆E
A·931.5MeV/amu
=0.09 amu
12 ·931.5MeV/amu
= 6.9375 MeV
Therefore, the binding energy per nucleon is 6.9375 MeV.
10 10. SCATTERING THEORY IN NUCLEAR PHYSICS
Problem 10. Consider a neutron scattering off a nucleus with a potential given by V(r) =
−V0e−r2/a2, where ris the distance from the center of the nucleus, V0>0, and ais a constant with
units of length.
A neutron with energy Escatters from this potential. The scattering cross section is given by
σ=4π
k2P∞
l=0(2l+ 1)|fl|2, where k=q2mE
¯h2is the wave number, flis the scattering amplitude for
orbital angular momentum l, and the sum runs over all possible values of l.
a) Show that the differential cross section for neutron-nucleus scattering, dσ/dΩ, can be written
as |f(θ)|2, where θis the scattering angle.
b) Assuming f(θ) = 1
ksin(δ0)eiα, where αis a real constant, show that dσ/dΩ = |f(θ)|2=
sin2(δ0)
k2.
Solution 10.
a) The differential cross section is defined as dσ
dΩ=|f(θ)|2, where θis the scattering angle.
This comes from the fact that the differential cross section is related to the scattering amplitude by
dσ
dΩ=|f(θ)|2. Thus, dσ/dΩ = |f(θ)|2.
b) Given f(θ) = 1
ksin(δ0)eiα, where αis real, we have |f(θ)|2=1
ksin(δ0)eiα
2=sin2(δ0)
k2. This
expression gives the differential cross section for neutron-nucleus scattering.
11 11. NUCLEAR SHAPE TRANSITIONS
Problem 11. Consider a nucleus undergoing a shape transition from a deformed shape to a
spherical shape as the excitation energy increases. The potential energy of the nucleus can be
described by the formula:
V(r) = V0
1 + e(r−r0
a)
where: V(r)is the potential energy at distance rfrom the center of the nucleus, V0= 50 MeV is
the depth of the potential well, r0= 1.2fm is the equilibrium radius of the nucleus in the deformed
state, a= 0.1fm characterizes the sharpness of the potential well.
a) Calculate the potential energy of the nucleus at r= 1.5fm in the deformed state.
b) At what distance from the center of the nucleus does the potential energy equal half of V0in
the deformed state?
c) Determine the critical excitation energy in MeV needed for the nucleus to transition from a
deformed shape to a spherical shape.
Solution 11. a) To calculate the potential energy at r= 1.5fm in the deformed state, we
substitute r= 1.5fm into the potential energy formula:
V(1.5) = 50
1 + e(1.5−1.2
0.1)=50
1 + e(3)
V(1.5) = 50
1 + e3≈50
1 + 20.0855 ≈50
21.0855 ≈2.37 MeV
Therefore, the potential energy of the nucleus at r= 1.5fm in the deformed state is approxi-
mately 2.37 MeV.
b) To find the distance at which the potential energy equals half of V0, we set V(r) = V0
2and
solve for r:
50
1 + e(r−1.2
0.1)=50
2
1 + e(r−1.2
0.1)= 2
e(r−1.2
0.1)= 1
r−1.2
0.1= 0
r= 1.2fm
Therefore, the potential energy equals half of V0at a distance of 1.2 fm from the center of the
nucleus in the deformed state.
c) The critical excitation energy ∆Eneeded for the nucleus to transition from a deformed shape
to a spherical shape occurs when the nucleus reaches the equilibrium radius of a spherical shape,
i.e., when r=r0for the potential energy formula. Thus:
V(r0) = 50
1 + e(r0−1.2
0.1)= 50
1 + e(r0−1.2
0.1)= 1
e(r0−1.2
0.1)= 0
r0= 1.2fm
Therefore, the critical excitation energy needed for the nucleus to transition from a deformed
shape to a spherical shape is 50 MeV.
12 12. QUANTUM MONTE CARLO METHODS FOR MANY-BODY SYSTEMS
Problem 12. Consider a quantum many-body system of Nparticles confined in a one-dimensional
infinite square well potential given by V(x) = 0 for 0< x < L and V(x) = ∞otherwise. Sup-
pose that the ground state wave function of this system can be approximated by a product of N
single-particle wave functions ψi(x), where each single-particle wave function can be expressed
as ψi(x) = q2
Lsin iπx
Lfor i= 1,2, . . . , N.
a) Calculate the normalization constant for the single-particle wave functions ψi(x).
b) Determine the ground state energy of the many-body system with these single-particle wave
functions.
c) Suppose the particles are non-interacting fermions. What is the ground state energy of the
system in this case?
Solution 12. a) The normalization condition for a wave function ψ(x)is R∞
−∞ |ψ(x)|2dx = 1. For
the given single-particle wave function ψi(x) = q2
Lsin iπx
L, we have:
ZL
0|ψi(x)|2dx =ZL
0 r2
Lsin iπx
L!2
dx
ZL
0|ψi(x)|2dx =2
LZL
0
sin2iπx
Ldx
ZL
0|ψi(x)|2dx =2
LZL
0
1−cos(2iπx/L)
2dx
Using trigonometric identity sin2(θ) = 1−cos(2θ)
2, we can simplify the integral as:
ZL
0|ψi(x)|2dx =1
Lx−L
2iπ sin 2iπx
LL
0
ZL
0|ψi(x)|2dx =1
L[L−0]
ZL
0|ψi(x)|2dx = 1
Therefore, the single-particle wave function ψi(x)is normalized.
b) The total ground state wave function for the many-body system is given by Ψ(x1, x2, . . . , xN) =
QN
i=1 ψi(xi). The ground state energy Efor this system is given by the sum of the single-particle
energies:
E=
N
X
i=1
¯h2π2i2
2mL2
c) Since the particles are non-interacting fermions, the ground state energy of the system is
given by the sum of the single-particle energies up to the Fermi level. The Fermi level is the highest
occupied single-particle energy level, which is the Nth level for Nnon-interacting fermions.
Efermions =
N
X
i=1
¯h2π2i2
2mL2
13 13. ELECTRON-NUCLEUS INTERACTIONS
Problem 13. An electron is scattered off a gold nucleus (197
79 Au) with a charge of +79e. The
electron has an initial kinetic energy of 5 MeV and scatters backwards at an angle of 180◦. Calculate
the recoil energy of the gold nucleus.
Given:
•Charge of the gold nucleus, Z= +79e
•Initial kinetic energy of the electron, Ei= 5 MeV
•Scattered angle, θ= 180◦
•Mass of the gold nucleus, mAu = 197 mproton
Solution 13.
a) The recoil energy of the gold nucleus can be calculated using the conservation of momentum
and energy. Since the electron scatters backwards, the final momentum of the system should be
zero.
Initial momentum of the system:
pi=p2meEi=p2×9.11 ×10−31 ×5×106≈1.74 ×10−22 kg m/s
The final kinetic energy of the gold nucleus is given by:
Ef=p2
i
2mAu
=(1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈7.62 ×10−11 J
Converting this energy to MeV:
Ef=7.62 ×10−11
1.6×10−13 ≈476.25 MeV
b) The final momentum of the gold nucleus can be calculated using the law of conservation of
momentum:
pAu =−pi=−1.74 ×10−22 kg m/s
c) The recoil energy of the gold nucleus can be calculated using the final momentum:
Erecoil =p2
Au
2mAu
=(−1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈3.81 ×10−10 J
Converting this energy to MeV:
Erecoil =3.81 ×10−10
1.6×10−13 ≈2381.25 MeV
14 14. CHIRAL EFFECTIVE FIELD THEORY IN NUCLEAR PHYSICS
Problem 14. Consider a two-neutron system in a harmonic oscillator potential with a frequency
ω= 5 MeV.
a) Calculate the energy of the ground state of the two-neutron system in MeV.
b) If the interaction potential between the neutrons is given by a contact interaction with a
strength of V0=−20 MeV, compute the energy of the first excited state of the system in MeV.
Solution 14.
a) The energy of the ground state of a harmonic oscillator potential is given by Eground =3
2¯hω.
Substituting ¯h= 197 MeV fm and ω= 5 MeV into the formula:
Eground =3
2·197 MeV fm ·5MeV =3
2·197 ·5 = 1477.5MeV
Therefore, the energy of the ground state of the two-neutron system is 1477.5 MeV.
b) The energy of the first excited state can be calculated by considering the ground state energy
plus the interaction energy. The interaction energy for a contact interaction is given by Vint =
⟨ϕ2|V0|ϕ1⟩where ϕ1and ϕ2are the ground and first excited state wave functions, respectively.
For a harmonic oscillator potential, the wave functions are known and the matrix element can
be calculated. For a two-particle system, the interaction energy is:
Vint =V01−15
8=V0−7
8=−20 MeV −7
8= 17.5MeV
Therefore, the energy of the first excited state is 1477.5 + 17.5 = 1495 MeV.
15 15. SUPERFLUIDITY IN NUCLEAR MATTER
Problem 15. Consider a system of nucleons in a superfluid state with a pairing energy gap of
∆=2MeV. Calculate the critical temperature Tcfor this superfluid system.
Given:
•The transition temperature Tc= 0.57∆
Solution 15.
•We are given the formula for Tcin terms of the pairing energy gap ∆:Tc= 0.57∆
•Substituting the given value of ∆=2MeV into the formula:
Tc= 0.57 ×2MeV
Tc= 1.14 MeV
Therefore, the critical temperature for this superfluid system is Tc= 1.14 MeV.
16 16. NUCLEAR SPIN-ORBIT COUPLING
Problem 16. Consider a nucleus with a spin-orbit coupling constant equal to a= 20 MeV.
a) Calculate the energy splitting between the l= 1 states for a nucleus with N= 50 neutrons
and Z= 40 protons.
b) Calculate the energy splitting between the l= 1 states for a nucleus with N= 70 neutrons
and Z= 40 protons.
Solution 16.
a) The energy splitting between l= 1 states can be calculated using the formula:
∆E=a·(N−Z)
Plugging in the values a= 20 MeV, N= 50, and Z= 40, we find:
∆E= 20 MeV ·(50 −40) = 200 MeV
So, the energy splitting between the l= 1 states for this nucleus is 200 MeV.
b) Similarly, for the nucleus with N= 70 and Z= 40, the energy splitting between l= 1 states
is given by:
∆E= 20 MeV ·(70 −40) = 20 MeV ·30 = 600 MeV
Therefore, for a nucleus with N= 70 neutrons and Z= 40 protons, the energy splitting between
the l= 1 states is 600 MeV.
17 17. DEUTERON SCATTERING AND BOUND STATES
Problem 17. Consider a deuteron, a bound state of a proton and a neutron. The total energy
of the deuteron is given by E=−2.224 MeV. The radius of the deuteron is r= 2 fm.
a) Calculate the reduced mass of the proton-neutron system.
b) Calculate the binding energy of the deuteron.
c) If the deuteron is in a state of higher angular momentum, how would its radius compare to
r= 2 fm?
Solution 17.
a) The reduced mass µof the proton-neutron system can be calculated as:
µ=mp×mn
mp+mn
where mpis the mass of the proton and mnis the mass of the neutron. Given that mp≈
1.67 ×10−27 kg and mn≈1.675 ×10−27 kg, we have:
µ=(1.67 ×10−27 kg)×(1.675 ×10−27 kg)
1.67 ×10−27 kg + 1.675 ×10−27 kg ≈1.673 ×10−27 kg
Therefore, the reduced mass of the proton-neutron system is µ≈1.673 ×10−27 kg.
b) The binding energy Bof the deuteron is related to the total energy Eas:
B=−E
Therefore, the binding energy of the deuteron is B=−(−2.224) = 2.224 MeV.
c) The radius r′of the deuteron in a state of higher angular momentum can be related to the
initial radius rand the new angular momentum l′as:
r′=r×l+ 1
l′+ 11/3
Since the state has higher angular momentum, l′> l, which implies that r′< r. Therefore, the
radius of the deuteron in a state of higher angular momentum would be smaller than r= 2 fm.
I. Problem:
Problem 18. Consider a particle of mass mand energy Eapproaching a one-dimensional
potential barrier of height V0and width a. The potential inside the barrier is zero. Assuming E > V0,
calculate the transmission coefficient Tfor the particle through the barrier, where T=|t|2, with t
being the transmission amplitude.
Additional Information: The potential barrier is given by:
V(x) = (0,for 0< x < a
V0,for x < 0and x>a
Solution 18: a) To find the transmission coefficient T, we first need to calculate the transmission
amplitude t. The transmission coefficient is then given by T=|t|2.
b) The transmission coefficient can be expressed in terms of the wave numbers k1and k2inside
and outside the barrier, respectively. We have:
T=k2
k1
c) The wave numbers k1and k2can be related to the energy Eand potential V0using the
following equations:
k1=p2m(E−V0)
¯h
k2=√2mE
¯h
Now, let’s substitute k1and k2into the expression for the transmission coefficient to find T.
c) Detailed Solution: a) The transmission amplitude tis given by:
t=e−ika
b) Substituting the expressions for k1and k2into the formula for T:
T=e−ika
2=e−2ika
c) Substituting the values of k1and k2into the expression, we get:
T= √2mE
¯h! p2m(E−V0)
¯h!
Thus, the transmission coefficient Tin this scenario is determined by the given particle’s energy
E, the potential barrier height V0, and the barrier width a.
18 19. NUCLEAR DENSITY FUNCTIONAL THEORY
Problem 19. Consider a system of nucleons with a total binding energy described by the
Skyrme energy density functional in the form:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
where ρ(r)is the nucleon density, τ(r)is the kinetic energy density, and C0,C3are constants.
Given that the nucleon density is ρ(r) = 3
4πR3e−3r
2R, where R= 1.2fm, and the constants are
C0=−400 MeV, C3= 100 MeV, calculate the total binding energy of the system.
Solution 19. The total binding energy E[ρ]can be calculated by evaluating the integral over
the entire volume:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
Substitute the given nucleon density ρ(r)into the expression for E[ρ]:
E[ρ] = Zd3r ¯h2
2mτ(r) + C03
4πR3e−3r
2R2
+C33
4πR3e−3r
2R3!
Since the kinetic energy density term τ(r)is not given, let’s assume it to be zero for simplicity.
We can now integrate the potential energy terms over the entire volume by performing the following
steps:
a) Calculate ρ2(r) = 3
4πR3e−3r
2R2and ρ3(r) = 3
4πR3e−3r
2R3.
b) Substitute these expressions into the integral and solve for the total binding energy.
c) Evaluate the integral numerically using appropriate techniques to find the total binding energy
of the system.
19 20. LIGHT NUCLEI AND FEW-BODY SYSTEMS
Problem 20. Consider a helium-4 nucleus (alpha particle) consisting of 2 protons and 2 neu-
trons.
Given that the rest masses of a proton and neutron are mp= 1.6726219 ×10−27 kg and mn=
1.6749275 ×10−27 kg, respectively, and the speed of light c= 3.00 ×108m/s, calculate the binding
energy of a helium-4 nucleus. Assume the atomic mass unit (u) is defined as 1u= 1.660539×10−27
kg.
Solution 20. The binding energy of a nucleus is the energy required to completely disassemble
a nucleus into its constituent protons and neutrons. It is defined as the difference between the total
rest mass energy of the separate protons and neutrons and the rest mass energy of the nucleus.
a) The total rest mass energy of a helium-4 nucleus:
mnucleus = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Enucleus =mnucleusc2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
b) The total rest mass energy of the separate protons and neutrons:
mseparate = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Eseparate =mseparatec2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
c) The binding energy of the helium-4 nucleus is:
Therefore, the binding energy of the helium-4 nucleus is 0 Joules, indicating that the nucleus
is stable.
20 21. GAMOW-TELLER TRANSITIONS IN NUCLEI
Problem 21. Consider a nucleus undergoing a Gamow-Teller transition, where the initial state
has quantum numbers Jπ= 1+and the final state has quantum numbers Jπ= 0−. If the energy
difference between the initial and final states is 2 MeV, and the reduced transition probability B(GT )
is 3×10−3units, calculate the half-life of this transition.
Solution 21.
a) The decay rate λof a nuclear transition is related to the half-life T1
2
by the equation:
T1
2=ln(2)
λ
b) The decay rate of a Gamow-Teller transition can be calculated using the formula:
λ=1
T1
2
B(GT )
2Ji+1 2
where T1
2
is the half-life, B(GT )is the reduced transition probability, and Jiis the total angular
momentum of the initial state.
c) Substituting the given values into the formula, we have:
λ=1
3×10−3
3×10−3
2×1+1 2
λ=1
3×10−3×9
(3 ×10−3)2
λ=1
3×10−3×9
9×10−6
λ=1
3×10−3×106= 103sec−1
Using the relation between decay rate and half-life, we find:
T1
2=ln(2)
λ=ln(2)
103≈0.693 ms
Therefore, the half-life of this Gamow-Teller transition is approximately 0.693 milliseconds.
21 22. CLUSTERING PHENOMENA IN NUCLEAR STRUCTURE
Problem 22. Consider a nucleus with mass number A= 12, consisting of two alpha clus-
ters and 4 neutrons. The energy of an alpha particle is 28.3 MeV and the mass of a neutron is
approximately 939.6 MeV/c2.
a) Calculate the binding energy per nucleon of this nucleus.
b) Determine the total energy of this nucleus.
c) If the neutrons interact only through the nuclear force, what is the total nuclear force energy
for this nucleus?
Solution 22. a) The total binding energy of the nucleus is given by the sum of the binding
energy of the two alpha clusters and the binding energy of the 4 neutrons. The binding energy per
nucleon is then the total binding energy divided by the number of nucleons.
The binding energy of an alpha particle is Eα= 28.3MeV and the mass of an alpha particle
is approximately 4 times the mass of a nucleon. So, the binding energy per nucleon for the alpha
clusters would be Eα/4 = 28.3/4 = 7.075 MeV.
The binding energy of a neutron is zero (since it is a free particle). Therefore, the binding energy
per nucleon for the 4 neutrons is 0/4=0MeV.
The total binding energy for the nucleus is 2×Eα+ 4 ×0=2×28.3 = 56.6MeV.
Therefore, the binding energy per nucleon is 56.6/12 = 4.717 MeV.
b) The total energy of the nucleus is the sum of the rest mass energy and the binding energy.
Since the rest mass energy is not given in the problem, we will neglect it for this specific calculation.
Therefore, the total energy of the nucleus is just the binding energy, which is 56.6 MeV.
c) The total nuclear force energy for this nucleus would be zero since the neutrons are not
interacting with each other.
22 Nuclear Physics and Many-Body Systems
Problem: Consider a nucleus with a total of 20 protons and 30 neutrons. Assume that each
proton can pair with a neutron in the ground state of the nucleus. Determine the total angular
momentum and total spin of the nucleus.
Solution:
a) To find the total angular momentum of the nucleus, we first need to calculate the individual
angular momenta of the protons and neutrons in the ground state. Since each proton can pair with
a neutron, we have a total of 20 pairs. The individual angular momentum of a proton is l=1
2and
for a neutron is also l=1
2.
The total angular momentum of a pair is given by L=Lp+Ln, where Lpis the angular mo-
mentum of the proton and Lnis the angular momentum of the neutron. Therefore, the total angular
momentum of a pair is 1
2+1
2= 1.
Since we have 20 pairs, the total angular momentum of the nucleus is Ltotal = 20 ×1 = 20.
b) Next, to find the total spin of the nucleus, we know that the total spin of a nucleus is the sum
of the spins of each nucleon. Both protons and neutrons have spin s=1
2.
Since we have 20 protons and 30 neutrons, the total spin of the nucleus is Stotal = (20 ×1
2) +
(30 ×1
2) = 25.
Therefore, the total angular momentum of the nucleus is 20 and the total spin is 25.
I. Problem 1:
Consider a heavy nucleus with a mass number A= 240 and a proton number Z= 92. The
ground state of this nucleus has a binding energy of B= 7.6MeV per nucleon. Calculate the total
binding energy of this nucleus.
Solution 1: The total binding energy of the nucleus can be calculated using the formula:
Ebinding =B×A
Substitute the given values into the formula:
Ebinding = 7.6MeV/nucleon ×240 nucleons
Ebinding = 1824 MeV
Therefore, the total binding energy of this nucleus is 1824 MeV.
II. Problem 2:
A heavy nucleus in an excited state decays by emitting a beta particle (electron) and a neutrino.
If the energy released in this decay is 2 MeV, calculate the energy of the excited state before the
decay.
Solution 2: The energy released in the decay is related to the energy of the excited state by
the equation:
Ereleased =Ei−Ef
Where Eiis the energy of the excited state and Efis the energy of the final state after decay.
Given that Ereleased = 2 MeV, and the final state consists of the beta particle and the neutrino
which have negligible energy, the energy of the final state Efcan be considered as nearly 0.
Hence, Ei=Ereleased = 2 MeV.
Therefore, the energy of the excited state before the decay is 2 MeV.
23 25. NUCLEAR PHYSICS APPLICATIONS IN ASTROPHYSICS
Problem 25. Consider the nuclear reaction 4
2He +3
1H→7
3Li.
Given the rest mass of:
•Helium-4 atom 4
2He is 4.0026 u
•Hydrogen-3 atom 3
1His 3.0155 u
•Lithium-7 atom 7
3Li is 7.0160 u
Calculate the energy released in the reaction in MeV.
Solution 25. The energy released in the nuclear reaction can be calculated using Einstein’s
mass-energy equivalence E= ∆mc2.
The change in mass (∆m) is the total rest mass of the reactants minus the total rest mass of
the products:
∆m= (4.0026 + 3.0155) −7.0160
= 7.0181 −7.0160
= 0.0021 u
Using the conversion factor 1u= 931.5MeV/c2, we can calculate the energy released:
E= (0.0021 u)×(931.5MeV/c2)
= 1.95915 MeV
Therefore, the energy released in the reaction is 1.95915 M eV .
Q= 0.026698c2
To convert from atomic mass units (u) to energy (MeV), we use the conversion factor 1u=
931.5MeV/c2.
Substitute this conversion factor into the equation to get the energy in MeV:
Q= 0.026698 ×931.5
Q24.9MeV
Therefore, the energy released in the fusion reaction is approximately 24.9 MeV.
3 3. NUCLEAR DECAY PROCESSES
Problem 3. The half-life of a certain radioactive isotope is 10 days. If you start with a sample
containing 100 grams of the isotope, how much of the isotope will remain after 30 days?
Additional context: The decay of a radioactive substance follows an exponential decay model
given by the equation N(t) = N0e−λt, where: - N(t)is the amount of the substance remaining at
time t, - N0is the initial amount of the substance, - λis the decay constant, - tis the time elapsed.
a) How much of the isotope will remain after 10 days?
b) What is the decay constant, λ, for this isotope?
c) How much of the isotope will remain after 30 days?
Solution 3.
a) After 10 days:
N(10) = 100 ·e−λ·10
Given that 10 days is the half-life, N(10) = 100
2= 50, so:
50 = 100 ·e−λ·10
0.5 = e−10λ
ln(0.5) = −10λ
λ=−ln(0.5)
10
λ≈0.0693 per day
b) Decay constant, λ≈0.0693 per day.
c) After 30 days:
N(30) = 100 ·e−λ·30
= 100 ·e−0.0693·30
≈17.07 grams remaining
Therefore, after 30 days, approximately 17.07 grams of the isotope will remain.
4 4. ISOSPIN SYMMETRY IN NUCLEAR STRUCTURE
Problem 4. Consider a system of two protons and two neutrons in a nucleus. The isospin
quantum number for protons and neutrons is 1
2.
a) Determine the isospin values for the four particles in this system.
b) If we allow for isospin mixing, what are the possible values for the total isospin of the system?
c) Suppose the total isospin of the system is measured to be T= 1. What are the possible
values of the third component, Tz, of the isospin?
Solution 4. a) The isospin quantum number Tzcan take values ranging from −Tto T. In this
case, each proton and neutron has isospin T=1
2, so their Tzvalues can be −1
2or 1
2.
Therefore, the isospin values for the four particles in the system are:
Proton 1: T=1
2,Tz=1
2
Proton 2: T=1
2,Tz=1
2
Neutron 1: T=1
2,Tz=1
2
Neutron 2: T=1
2,Tz=1
2
b) When we allow for isospin mixing, the possible values for the total isospin of the system
range from |T1−T2|to T1+T2. In this case, T1=T2=1
2.
So, the possible values for the total isospin of the system are 0 and 1.
c) If the total isospin of the system is measured to be T= 1, the possible values for the third
component of isospin, Tz, are −1, 0, and 1.
5 5. ALPHA DECAY AND NUCLEAR STABILITY
Problem 5. The half-life of a certain radioactive isotope is 100 years. If there are initially 1000
atoms of this isotope, how many atoms will remain after 300 years?
Given:
•Half-life of the isotope = 100 years
•Initial number of atoms = 1000
•Time elapsed = 300 years
Solution 5. a) To calculate the decay constant, we use the formula:
N(t) = N0×e−λt
where: N(t)= number of atoms at time t,N0= initial number of atoms, λ= decay constant, t=
time.
First, we find the decay constant using the half-life formula:
T1/2=ln(2)
λ
Substitute T1/2= 100 years:
100 = ln(2)
λ
λ=ln(2)
100
λ≈0.00693 years−1
b) Next, we find the number of atoms remaining after 300 years:
N(300) = 1000 ×e−0.00693×300
N(300) = 1000 ×e−2.079
N(300) ≈1000 ×0.125
N(300) ≈125 atoms
Therefore, after 300 years, approximately 125 atoms of this isotope will remain.
6 6. NUCLEAR FISSION CHAIN REACTIONS
Problem 6. Consider a nuclear fission chain reaction in which a neutron induces the fission of
a uranium-235 nucleus, releasing on average 2.5 neutrons per fission event. The probability of a
neutron causing a subsequent fission event is 0.8.
a) If 100 neutrons are released in the initial fission event, how many total neutrons will be
released in the subsequent chain reaction?
b) Determine the number of fission events that will occur in this chain reaction.
c) If the energy released per fission event is 200 MeV, calculate the total energy released in this
chain reaction.
Solution 6.
a) In the initial fission event, 100 neutrons are released. Since on average 2.5 neutrons are
released per fission event, these 100 neutrons will lead to 100 ×2.5 = 250 neutrons being released
in the subsequent chain reaction.
b) The probability of a neutron causing a subsequent fission event is 0.8. This means 0.8neu-
trons out of the 2.5neutrons on average released per fission event will cause a subsequent fission
event. Therefore, the number of fission events that will occur in this chain reaction is 0.8×250
2.5= 80
fission events.
c) The total energy released per fission event is 200 MeV. Since there are 80 fission events, the
total energy released in this chain reaction is 80 ×200 = 16000 MeV or 16 GeV.
7 7. HYPERSPHERICAL COORDINATES IN MANY-BODY SYSTEMS
Problem 7. Consider a system of three identical particles confined in a 3-dimensional harmonic
oscillator potential. The Hamiltonian for this system can be written in terms of hyperspherical co-
ordinates as:
H=−¯h2
2m∂2
∂R2+5
R
∂
∂R +3
2mω2R2,
where Ris a collective coordinate. Calculate the ground state energy of this system in terms of the
oscillator frequency ω.
Solution 7. To find the ground state energy of the system, we need to solve the Schrödinger
equation
HΨ(R,Ω) = EΨ(R,Ω),
where Ψ(R,Ω)is the wave function of the system and Ris the hyperradius.
For the ground state energy, the wave function can be assumed to be separable as Ψ(R,Ω) =
F(R)Y(Ω).
Substituting this into the Schrödinger equation and dividing by F(R)Y(Ω), we get
−¯h2
2mF′′
F+5
R
F′
F+3
2mω2R2=E.
Separating variables, we have
F′′
F=−5
R
F′
F+2mE
¯h2−3mω2R2
¯h2.
To simplify this equation, we make the substitution u=√mωR, which transforms the equation
into
−F′′ +5
uF′= (λu2−k)F,
where λ=2mE
¯hω and k=3
2.
The solution to this differential equation can be written in terms of Laguerre polynomials Lα
k(u)
as
F(R) = Au5/2e−u2/2L5
0(u2),
where Ais a normalization constant.
Now, the ground state energy is given by E0=3
2¯hω.
Therefore, the ground state energy of the system in terms of the oscillator frequency ωis E0=
3
2¯hω.
8 8. INTRINSIC AND COLLECTIVE NUCLEAR EXCITATIONS
Problem 8. Consider a nucleus with a mass number A= 240 and a charge number Z= 92.
a) Calculate the binding energy of the nucleus in MeV given that the atomic mass unit uis
equivalent to 931.5MeV/c2.
b) Determine the energy needed to completely remove a neutron from the nucleus.
c) Find the Q-value of the alpha decay process where the nucleus decays into a daughter
nucleus with A= 236 and Z= 90, emitting an alpha particle.
Solution 8.
a) The binding energy of a nucleus can be calculated using the formula:
B.E. = (Zmp+N mn−M)c2
Where: - Zis the number of protons, - Nis the number of neutrons, - mpis the mass of a proton,
-mnis the mass of a neutron, - Mis the mass of the nucleus, and - cis the speed of light.
Given that mp=mn= 1u= 931.5MeV/c2, we first need to calculate the total number of
neutrons N=A−Z:
N= 240 −92 = 148
The mass of the nucleus Mcan be calculated by:
M=Amu= 240 ×1u= 240 ×931.5MeV/c2
Therefore, the binding energy B.E. is:
B.E. = (92 ×931.5 + 148 ×931.5−240 ×931.5) MeV = 16900 MeV
So, the binding energy of the nucleus is 16900 MeV.
b) To remove a neutron, we need to consider the mass of the neutron itself, hence the energy
required is the mass of the neutron:
E=mnc2= 1 ×931.5MeV = 931.5MeV
Hence, the energy needed to remove a neutron is 931.5MeV.
c) The Q-value of the alpha decay process is given by:
Q= (Mi−Mf−mα)c2
Where: - Miis the initial mass, - Mfis the final mass, and - mαis the mass of the alpha particle.
Given that Mi= 240×931.5MeV/c2and Mf= 236×931.5MeV/c2, and mα= 4×931.5MeV/c2,
we can calculate the Q-value:
Q= (240 ×931.5−236 ×931.5−4×931.5) MeV = 0 MeV
Therefore, the Q-value of the alpha decay process is 0MeV.
9 9. NUCLEAR FORCES AND BINDING ENERGIES
Problem 9. Consider a nucleus with atomic number Z= 6. Given that the mass of a proton is
mp= 1.67 ×10−27 kg and the mass of a neutron is mn= 1.675 ×10−27 kg, and the mass of the
nucleus is M= 12.000 amu, calculate the binding energy per nucleon in MeV.
Solution 9. a) The total number of protons and neutrons in the nucleus can be calculated using
the atomic number Z= 6.
N=Z+ (A−Z)
= 6 + (12 −6)
= 12 nucleons
b) The total mass of the nucleus can be calculated using the masses of protons and neutrons:
Mnucleus =Z·mp+ (A−Z)·mn
= 6 ·1.67 ×10−27 + 6 ·1.675 ×10−27
= 9.99 ×10−26 + 10.05 ×10−26
= 20.04 ×10−26 kg
= 20.04 ×10−26 kg ×1amu
1.66 ×10−27 kg
= 12.09 amu
The binding energy of the nucleus can be approximated using the mass defect:
∆m= 12.000 amu −12.09 amu
=−0.09 amu
c) The binding energy per nucleon can be calculated as:
BE/A =∆E
A·931.5MeV/amu
=0.09 amu
12 ·931.5MeV/amu
= 6.9375 MeV
Therefore, the binding energy per nucleon is 6.9375 MeV.
10 10. SCATTERING THEORY IN NUCLEAR PHYSICS
Problem 10. Consider a neutron scattering off a nucleus with a potential given by V(r) =
−V0e−r2/a2, where ris the distance from the center of the nucleus, V0>0, and ais a constant with
units of length.
A neutron with energy Escatters from this potential. The scattering cross section is given by
σ=4π
k2P∞
l=0(2l+ 1)|fl|2, where k=q2mE
¯h2is the wave number, flis the scattering amplitude for
orbital angular momentum l, and the sum runs over all possible values of l.
a) Show that the differential cross section for neutron-nucleus scattering, dσ/dΩ, can be written
as |f(θ)|2, where θis the scattering angle.
b) Assuming f(θ) = 1
ksin(δ0)eiα, where αis a real constant, show that dσ/dΩ = |f(θ)|2=
sin2(δ0)
k2.
Solution 10.
a) The differential cross section is defined as dσ
dΩ=|f(θ)|2, where θis the scattering angle.
This comes from the fact that the differential cross section is related to the scattering amplitude by
dσ
dΩ=|f(θ)|2. Thus, dσ/dΩ = |f(θ)|2.
b) Given f(θ) = 1
ksin(δ0)eiα, where αis real, we have |f(θ)|2=1
ksin(δ0)eiα
2=sin2(δ0)
k2. This
expression gives the differential cross section for neutron-nucleus scattering.
11 11. NUCLEAR SHAPE TRANSITIONS
Problem 11. Consider a nucleus undergoing a shape transition from a deformed shape to a
spherical shape as the excitation energy increases. The potential energy of the nucleus can be
described by the formula:
V(r) = V0
1 + e(r−r0
a)
where: V(r)is the potential energy at distance rfrom the center of the nucleus, V0= 50 MeV is
the depth of the potential well, r0= 1.2fm is the equilibrium radius of the nucleus in the deformed
state, a= 0.1fm characterizes the sharpness of the potential well.
a) Calculate the potential energy of the nucleus at r= 1.5fm in the deformed state.
b) At what distance from the center of the nucleus does the potential energy equal half of V0in
the deformed state?
c) Determine the critical excitation energy in MeV needed for the nucleus to transition from a
deformed shape to a spherical shape.
Solution 11. a) To calculate the potential energy at r= 1.5fm in the deformed state, we
substitute r= 1.5fm into the potential energy formula:
V(1.5) = 50
1 + e(1.5−1.2
0.1)=50
1 + e(3)
V(1.5) = 50
1 + e3≈50
1 + 20.0855 ≈50
21.0855 ≈2.37 MeV
Therefore, the potential energy of the nucleus at r= 1.5fm in the deformed state is approxi-
mately 2.37 MeV.
b) To find the distance at which the potential energy equals half of V0, we set V(r) = V0
2and
solve for r:
50
1 + e(r−1.2
0.1)=50
2
1 + e(r−1.2
0.1)= 2
e(r−1.2
0.1)= 1
r−1.2
0.1= 0
r= 1.2fm
Therefore, the potential energy equals half of V0at a distance of 1.2 fm from the center of the
nucleus in the deformed state.
c) The critical excitation energy ∆Eneeded for the nucleus to transition from a deformed shape
to a spherical shape occurs when the nucleus reaches the equilibrium radius of a spherical shape,
i.e., when r=r0for the potential energy formula. Thus:
V(r0) = 50
1 + e(r0−1.2
0.1)= 50
1 + e(r0−1.2
0.1)= 1
e(r0−1.2
0.1)= 0
r0= 1.2fm
Therefore, the critical excitation energy needed for the nucleus to transition from a deformed
shape to a spherical shape is 50 MeV.
12 12. QUANTUM MONTE CARLO METHODS FOR MANY-BODY SYSTEMS
Problem 12. Consider a quantum many-body system of Nparticles confined in a one-dimensional
infinite square well potential given by V(x) = 0 for 0< x < L and V(x) = ∞otherwise. Sup-
pose that the ground state wave function of this system can be approximated by a product of N
single-particle wave functions ψi(x), where each single-particle wave function can be expressed
as ψi(x) = q2
Lsin iπx
Lfor i= 1,2, . . . , N.
a) Calculate the normalization constant for the single-particle wave functions ψi(x).
b) Determine the ground state energy of the many-body system with these single-particle wave
functions.
c) Suppose the particles are non-interacting fermions. What is the ground state energy of the
system in this case?
Solution 12. a) The normalization condition for a wave function ψ(x)is R∞
−∞ |ψ(x)|2dx = 1. For
the given single-particle wave function ψi(x) = q2
Lsin iπx
L, we have:
ZL
0|ψi(x)|2dx =ZL
0 r2
Lsin iπx
L!2
dx
ZL
0|ψi(x)|2dx =2
LZL
0
sin2iπx
Ldx
ZL
0|ψi(x)|2dx =2
LZL
0
1−cos(2iπx/L)
2dx
Using trigonometric identity sin2(θ) = 1−cos(2θ)
2, we can simplify the integral as:
ZL
0|ψi(x)|2dx =1
Lx−L
2iπ sin 2iπx
LL
0
ZL
0|ψi(x)|2dx =1
L[L−0]
ZL
0|ψi(x)|2dx = 1
Therefore, the single-particle wave function ψi(x)is normalized.
b) The total ground state wave function for the many-body system is given by Ψ(x1, x2, . . . , xN) =
QN
i=1 ψi(xi). The ground state energy Efor this system is given by the sum of the single-particle
energies:
E=
N
X
i=1
¯h2π2i2
2mL2
c) Since the particles are non-interacting fermions, the ground state energy of the system is
given by the sum of the single-particle energies up to the Fermi level. The Fermi level is the highest
occupied single-particle energy level, which is the Nth level for Nnon-interacting fermions.
Efermions =
N
X
i=1
¯h2π2i2
2mL2
13 13. ELECTRON-NUCLEUS INTERACTIONS
Problem 13. An electron is scattered off a gold nucleus (197
79 Au) with a charge of +79e. The
electron has an initial kinetic energy of 5 MeV and scatters backwards at an angle of 180◦. Calculate
the recoil energy of the gold nucleus.
Given:
•Charge of the gold nucleus, Z= +79e
•Initial kinetic energy of the electron, Ei= 5 MeV
•Scattered angle, θ= 180◦
•Mass of the gold nucleus, mAu = 197 mproton
Solution 13.
a) The recoil energy of the gold nucleus can be calculated using the conservation of momentum
and energy. Since the electron scatters backwards, the final momentum of the system should be
zero.
Initial momentum of the system:
pi=p2meEi=p2×9.11 ×10−31 ×5×106≈1.74 ×10−22 kg m/s
The final kinetic energy of the gold nucleus is given by:
Ef=p2
i
2mAu
=(1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈7.62 ×10−11 J
Converting this energy to MeV:
Ef=7.62 ×10−11
1.6×10−13 ≈476.25 MeV
b) The final momentum of the gold nucleus can be calculated using the law of conservation of
momentum:
pAu =−pi=−1.74 ×10−22 kg m/s
c) The recoil energy of the gold nucleus can be calculated using the final momentum:
Erecoil =p2
Au
2mAu
=(−1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈3.81 ×10−10 J
Converting this energy to MeV:
Erecoil =3.81 ×10−10
1.6×10−13 ≈2381.25 MeV
14 14. CHIRAL EFFECTIVE FIELD THEORY IN NUCLEAR PHYSICS
Problem 14. Consider a two-neutron system in a harmonic oscillator potential with a frequency
ω= 5 MeV.
a) Calculate the energy of the ground state of the two-neutron system in MeV.
b) If the interaction potential between the neutrons is given by a contact interaction with a
strength of V0=−20 MeV, compute the energy of the first excited state of the system in MeV.
Solution 14.
a) The energy of the ground state of a harmonic oscillator potential is given by Eground =3
2¯hω.
Substituting ¯h= 197 MeV fm and ω= 5 MeV into the formula:
Eground =3
2·197 MeV fm ·5MeV =3
2·197 ·5 = 1477.5MeV
Therefore, the energy of the ground state of the two-neutron system is 1477.5 MeV.
b) The energy of the first excited state can be calculated by considering the ground state energy
plus the interaction energy. The interaction energy for a contact interaction is given by Vint =
⟨ϕ2|V0|ϕ1⟩where ϕ1and ϕ2are the ground and first excited state wave functions, respectively.
For a harmonic oscillator potential, the wave functions are known and the matrix element can
be calculated. For a two-particle system, the interaction energy is:
Vint =V01−15
8=V0−7
8=−20 MeV −7
8= 17.5MeV
Therefore, the energy of the first excited state is 1477.5 + 17.5 = 1495 MeV.
15 15. SUPERFLUIDITY IN NUCLEAR MATTER
Problem 15. Consider a system of nucleons in a superfluid state with a pairing energy gap of
∆=2MeV. Calculate the critical temperature Tcfor this superfluid system.
Given:
•The transition temperature Tc= 0.57∆
Solution 15.
•We are given the formula for Tcin terms of the pairing energy gap ∆:Tc= 0.57∆
•Substituting the given value of ∆=2MeV into the formula:
Tc= 0.57 ×2MeV
Tc= 1.14 MeV
Therefore, the critical temperature for this superfluid system is Tc= 1.14 MeV.
16 16. NUCLEAR SPIN-ORBIT COUPLING
Problem 16. Consider a nucleus with a spin-orbit coupling constant equal to a= 20 MeV.
a) Calculate the energy splitting between the l= 1 states for a nucleus with N= 50 neutrons
and Z= 40 protons.
b) Calculate the energy splitting between the l= 1 states for a nucleus with N= 70 neutrons
and Z= 40 protons.
Solution 16.
a) The energy splitting between l= 1 states can be calculated using the formula:
∆E=a·(N−Z)
Plugging in the values a= 20 MeV, N= 50, and Z= 40, we find:
∆E= 20 MeV ·(50 −40) = 200 MeV
So, the energy splitting between the l= 1 states for this nucleus is 200 MeV.
b) Similarly, for the nucleus with N= 70 and Z= 40, the energy splitting between l= 1 states
is given by:
∆E= 20 MeV ·(70 −40) = 20 MeV ·30 = 600 MeV
Therefore, for a nucleus with N= 70 neutrons and Z= 40 protons, the energy splitting between
the l= 1 states is 600 MeV.
17 17. DEUTERON SCATTERING AND BOUND STATES
Problem 17. Consider a deuteron, a bound state of a proton and a neutron. The total energy
of the deuteron is given by E=−2.224 MeV. The radius of the deuteron is r= 2 fm.
a) Calculate the reduced mass of the proton-neutron system.
b) Calculate the binding energy of the deuteron.
c) If the deuteron is in a state of higher angular momentum, how would its radius compare to
r= 2 fm?
Solution 17.
a) The reduced mass µof the proton-neutron system can be calculated as:
µ=mp×mn
mp+mn
where mpis the mass of the proton and mnis the mass of the neutron. Given that mp≈
1.67 ×10−27 kg and mn≈1.675 ×10−27 kg, we have:
µ=(1.67 ×10−27 kg)×(1.675 ×10−27 kg)
1.67 ×10−27 kg + 1.675 ×10−27 kg ≈1.673 ×10−27 kg
Therefore, the reduced mass of the proton-neutron system is µ≈1.673 ×10−27 kg.
b) The binding energy Bof the deuteron is related to the total energy Eas:
B=−E
Therefore, the binding energy of the deuteron is B=−(−2.224) = 2.224 MeV.
c) The radius r′of the deuteron in a state of higher angular momentum can be related to the
initial radius rand the new angular momentum l′as:
r′=r×l+ 1
l′+ 11/3
Since the state has higher angular momentum, l′> l, which implies that r′< r. Therefore, the
radius of the deuteron in a state of higher angular momentum would be smaller than r= 2 fm.
I. Problem:
Problem 18. Consider a particle of mass mand energy Eapproaching a one-dimensional
potential barrier of height V0and width a. The potential inside the barrier is zero. Assuming E > V0,
calculate the transmission coefficient Tfor the particle through the barrier, where T=|t|2, with t
being the transmission amplitude.
Additional Information: The potential barrier is given by:
V(x) = (0,for 0< x < a
V0,for x < 0and x>a
Solution 18: a) To find the transmission coefficient T, we first need to calculate the transmission
amplitude t. The transmission coefficient is then given by T=|t|2.
b) The transmission coefficient can be expressed in terms of the wave numbers k1and k2inside
and outside the barrier, respectively. We have:
T=k2
k1
c) The wave numbers k1and k2can be related to the energy Eand potential V0using the
following equations:
k1=p2m(E−V0)
¯h
k2=√2mE
¯h
Now, let’s substitute k1and k2into the expression for the transmission coefficient to find T.
c) Detailed Solution: a) The transmission amplitude tis given by:
t=e−ika
b) Substituting the expressions for k1and k2into the formula for T:
T=e−ika
2=e−2ika
c) Substituting the values of k1and k2into the expression, we get:
T= √2mE
¯h! p2m(E−V0)
¯h!
Thus, the transmission coefficient Tin this scenario is determined by the given particle’s energy
E, the potential barrier height V0, and the barrier width a.
18 19. NUCLEAR DENSITY FUNCTIONAL THEORY
Problem 19. Consider a system of nucleons with a total binding energy described by the
Skyrme energy density functional in the form:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
where ρ(r)is the nucleon density, τ(r)is the kinetic energy density, and C0,C3are constants.
Given that the nucleon density is ρ(r) = 3
4πR3e−3r
2R, where R= 1.2fm, and the constants are
C0=−400 MeV, C3= 100 MeV, calculate the total binding energy of the system.
Solution 19. The total binding energy E[ρ]can be calculated by evaluating the integral over
the entire volume:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
Substitute the given nucleon density ρ(r)into the expression for E[ρ]:
E[ρ] = Zd3r ¯h2
2mτ(r) + C03
4πR3e−3r
2R2
+C33
4πR3e−3r
2R3!
Since the kinetic energy density term τ(r)is not given, let’s assume it to be zero for simplicity.
We can now integrate the potential energy terms over the entire volume by performing the following
steps:
a) Calculate ρ2(r) = 3
4πR3e−3r
2R2and ρ3(r) = 3
4πR3e−3r
2R3.
b) Substitute these expressions into the integral and solve for the total binding energy.
c) Evaluate the integral numerically using appropriate techniques to find the total binding energy
of the system.
19 20. LIGHT NUCLEI AND FEW-BODY SYSTEMS
Problem 20. Consider a helium-4 nucleus (alpha particle) consisting of 2 protons and 2 neu-
trons.
Given that the rest masses of a proton and neutron are mp= 1.6726219 ×10−27 kg and mn=
1.6749275 ×10−27 kg, respectively, and the speed of light c= 3.00 ×108m/s, calculate the binding
energy of a helium-4 nucleus. Assume the atomic mass unit (u) is defined as 1u= 1.660539×10−27
kg.
Solution 20. The binding energy of a nucleus is the energy required to completely disassemble
a nucleus into its constituent protons and neutrons. It is defined as the difference between the total
rest mass energy of the separate protons and neutrons and the rest mass energy of the nucleus.
a) The total rest mass energy of a helium-4 nucleus:
mnucleus = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Enucleus =mnucleusc2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
b) The total rest mass energy of the separate protons and neutrons:
mseparate = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Eseparate =mseparatec2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
c) The binding energy of the helium-4 nucleus is:
Therefore, the binding energy of the helium-4 nucleus is 0 Joules, indicating that the nucleus
is stable.
20 21. GAMOW-TELLER TRANSITIONS IN NUCLEI
Problem 21. Consider a nucleus undergoing a Gamow-Teller transition, where the initial state
has quantum numbers Jπ= 1+and the final state has quantum numbers Jπ= 0−. If the energy
difference between the initial and final states is 2 MeV, and the reduced transition probability B(GT )
is 3×10−3units, calculate the half-life of this transition.
Solution 21.
a) The decay rate λof a nuclear transition is related to the half-life T1
2
by the equation:
T1
2=ln(2)
λ
b) The decay rate of a Gamow-Teller transition can be calculated using the formula:
λ=1
T1
2
B(GT )
2Ji+1 2
where T1
2
is the half-life, B(GT )is the reduced transition probability, and Jiis the total angular
momentum of the initial state.
c) Substituting the given values into the formula, we have:
λ=1
3×10−3
3×10−3
2×1+1 2
λ=1
3×10−3×9
(3 ×10−3)2
λ=1
3×10−3×9
9×10−6
λ=1
3×10−3×106= 103sec−1
Using the relation between decay rate and half-life, we find:
T1
2=ln(2)
λ=ln(2)
103≈0.693 ms
Therefore, the half-life of this Gamow-Teller transition is approximately 0.693 milliseconds.
21 22. CLUSTERING PHENOMENA IN NUCLEAR STRUCTURE
Problem 22. Consider a nucleus with mass number A= 12, consisting of two alpha clus-
ters and 4 neutrons. The energy of an alpha particle is 28.3 MeV and the mass of a neutron is
approximately 939.6 MeV/c2.
a) Calculate the binding energy per nucleon of this nucleus.
b) Determine the total energy of this nucleus.
c) If the neutrons interact only through the nuclear force, what is the total nuclear force energy
for this nucleus?
Solution 22. a) The total binding energy of the nucleus is given by the sum of the binding
energy of the two alpha clusters and the binding energy of the 4 neutrons. The binding energy per
nucleon is then the total binding energy divided by the number of nucleons.
The binding energy of an alpha particle is Eα= 28.3MeV and the mass of an alpha particle
is approximately 4 times the mass of a nucleon. So, the binding energy per nucleon for the alpha
clusters would be Eα/4 = 28.3/4 = 7.075 MeV.
The binding energy of a neutron is zero (since it is a free particle). Therefore, the binding energy
per nucleon for the 4 neutrons is 0/4=0MeV.
The total binding energy for the nucleus is 2×Eα+ 4 ×0=2×28.3 = 56.6MeV.
Therefore, the binding energy per nucleon is 56.6/12 = 4.717 MeV.
b) The total energy of the nucleus is the sum of the rest mass energy and the binding energy.
Since the rest mass energy is not given in the problem, we will neglect it for this specific calculation.
Therefore, the total energy of the nucleus is just the binding energy, which is 56.6 MeV.
c) The total nuclear force energy for this nucleus would be zero since the neutrons are not
interacting with each other.
22 Nuclear Physics and Many-Body Systems
Problem: Consider a nucleus with a total of 20 protons and 30 neutrons. Assume that each
proton can pair with a neutron in the ground state of the nucleus. Determine the total angular
momentum and total spin of the nucleus.
Solution:
a) To find the total angular momentum of the nucleus, we first need to calculate the individual
angular momenta of the protons and neutrons in the ground state. Since each proton can pair with
a neutron, we have a total of 20 pairs. The individual angular momentum of a proton is l=1
2and
for a neutron is also l=1
2.
The total angular momentum of a pair is given by L=Lp+Ln, where Lpis the angular mo-
mentum of the proton and Lnis the angular momentum of the neutron. Therefore, the total angular
momentum of a pair is 1
2+1
2= 1.
Since we have 20 pairs, the total angular momentum of the nucleus is Ltotal = 20 ×1 = 20.
b) Next, to find the total spin of the nucleus, we know that the total spin of a nucleus is the sum
of the spins of each nucleon. Both protons and neutrons have spin s=1
2.
Since we have 20 protons and 30 neutrons, the total spin of the nucleus is Stotal = (20 ×1
2) +
(30 ×1
2) = 25.
Therefore, the total angular momentum of the nucleus is 20 and the total spin is 25.
I. Problem 1:
Consider a heavy nucleus with a mass number A= 240 and a proton number Z= 92. The
ground state of this nucleus has a binding energy of B= 7.6MeV per nucleon. Calculate the total
binding energy of this nucleus.
Solution 1: The total binding energy of the nucleus can be calculated using the formula:
Ebinding =B×A
Substitute the given values into the formula:
Ebinding = 7.6MeV/nucleon ×240 nucleons
Ebinding = 1824 MeV
Therefore, the total binding energy of this nucleus is 1824 MeV.
II. Problem 2:
A heavy nucleus in an excited state decays by emitting a beta particle (electron) and a neutrino.
If the energy released in this decay is 2 MeV, calculate the energy of the excited state before the
decay.
Solution 2: The energy released in the decay is related to the energy of the excited state by
the equation:
Ereleased =Ei−Ef
Where Eiis the energy of the excited state and Efis the energy of the final state after decay.
Given that Ereleased = 2 MeV, and the final state consists of the beta particle and the neutrino
which have negligible energy, the energy of the final state Efcan be considered as nearly 0.
Hence, Ei=Ereleased = 2 MeV.
Therefore, the energy of the excited state before the decay is 2 MeV.
23 25. NUCLEAR PHYSICS APPLICATIONS IN ASTROPHYSICS
Problem 25. Consider the nuclear reaction 4
2He +3
1H→7
3Li.
Given the rest mass of:
•Helium-4 atom 4
2He is 4.0026 u
•Hydrogen-3 atom 3
1His 3.0155 u
•Lithium-7 atom 7
3Li is 7.0160 u
Calculate the energy released in the reaction in MeV.
Solution 25. The energy released in the nuclear reaction can be calculated using Einstein’s
mass-energy equivalence E= ∆mc2.
The change in mass (∆m) is the total rest mass of the reactants minus the total rest mass of
the products:
∆m= (4.0026 + 3.0155) −7.0160
= 7.0181 −7.0160
= 0.0021 u
Using the conversion factor 1u= 931.5MeV/c2, we can calculate the energy released:
E= (0.0021 u)×(931.5MeV/c2)
= 1.95915 MeV
Therefore, the energy released in the reaction is 1.95915 M eV .
Q= 0.026698c2
To convert from atomic mass units (u) to energy (MeV), we use the conversion factor 1u=
931.5MeV/c2.
Substitute this conversion factor into the equation to get the energy in MeV:
Q= 0.026698 ×931.5
Q24.9MeV
Therefore, the energy released in the fusion reaction is approximately 24.9 MeV.
3 3. NUCLEAR DECAY PROCESSES
Problem 3. The half-life of a certain radioactive isotope is 10 days. If you start with a sample
containing 100 grams of the isotope, how much of the isotope will remain after 30 days?
Additional context: The decay of a radioactive substance follows an exponential decay model
given by the equation N(t) = N0e−λt, where: - N(t)is the amount of the substance remaining at
time t, - N0is the initial amount of the substance, - λis the decay constant, - tis the time elapsed.
a) How much of the isotope will remain after 10 days?
b) What is the decay constant, λ, for this isotope?
c) How much of the isotope will remain after 30 days?
Solution 3.
a) After 10 days:
N(10) = 100 ·e−λ·10
Given that 10 days is the half-life, N(10) = 100
2= 50, so:
50 = 100 ·e−λ·10
0.5 = e−10λ
ln(0.5) = −10λ
λ=−ln(0.5)
10
λ≈0.0693 per day
b) Decay constant, λ≈0.0693 per day.
c) After 30 days:
N(30) = 100 ·e−λ·30
= 100 ·e−0.0693·30
≈17.07 grams remaining
Therefore, after 30 days, approximately 17.07 grams of the isotope will remain.
4 4. ISOSPIN SYMMETRY IN NUCLEAR STRUCTURE
Problem 4. Consider a system of two protons and two neutrons in a nucleus. The isospin
quantum number for protons and neutrons is 1
2.
a) Determine the isospin values for the four particles in this system.
b) If we allow for isospin mixing, what are the possible values for the total isospin of the system?
c) Suppose the total isospin of the system is measured to be T= 1. What are the possible
values of the third component, Tz, of the isospin?
Solution 4. a) The isospin quantum number Tzcan take values ranging from −Tto T. In this
case, each proton and neutron has isospin T=1
2, so their Tzvalues can be −1
2or 1
2.
Therefore, the isospin values for the four particles in the system are:
Proton 1: T=1
2,Tz=1
2
Proton 2: T=1
2,Tz=1
2
Neutron 1: T=1
2,Tz=1
2
Neutron 2: T=1
2,Tz=1
2
b) When we allow for isospin mixing, the possible values for the total isospin of the system
range from |T1−T2|to T1+T2. In this case, T1=T2=1
2.
So, the possible values for the total isospin of the system are 0 and 1.
c) If the total isospin of the system is measured to be T= 1, the possible values for the third
component of isospin, Tz, are −1, 0, and 1.
5 5. ALPHA DECAY AND NUCLEAR STABILITY
Problem 5. The half-life of a certain radioactive isotope is 100 years. If there are initially 1000
atoms of this isotope, how many atoms will remain after 300 years?
Given:
•Half-life of the isotope = 100 years
•Initial number of atoms = 1000
•Time elapsed = 300 years
Solution 5. a) To calculate the decay constant, we use the formula:
N(t) = N0×e−λt
where: N(t)= number of atoms at time t,N0= initial number of atoms, λ= decay constant, t=
time.
First, we find the decay constant using the half-life formula:
T1/2=ln(2)
λ
Substitute T1/2= 100 years:
100 = ln(2)
λ
λ=ln(2)
100
λ≈0.00693 years−1
b) Next, we find the number of atoms remaining after 300 years:
N(300) = 1000 ×e−0.00693×300
N(300) = 1000 ×e−2.079
N(300) ≈1000 ×0.125
N(300) ≈125 atoms
Therefore, after 300 years, approximately 125 atoms of this isotope will remain.
6 6. NUCLEAR FISSION CHAIN REACTIONS
Problem 6. Consider a nuclear fission chain reaction in which a neutron induces the fission of
a uranium-235 nucleus, releasing on average 2.5 neutrons per fission event. The probability of a
neutron causing a subsequent fission event is 0.8.
a) If 100 neutrons are released in the initial fission event, how many total neutrons will be
released in the subsequent chain reaction?
b) Determine the number of fission events that will occur in this chain reaction.
c) If the energy released per fission event is 200 MeV, calculate the total energy released in this
chain reaction.
Solution 6.
a) In the initial fission event, 100 neutrons are released. Since on average 2.5 neutrons are
released per fission event, these 100 neutrons will lead to 100 ×2.5 = 250 neutrons being released
in the subsequent chain reaction.
b) The probability of a neutron causing a subsequent fission event is 0.8. This means 0.8neu-
trons out of the 2.5neutrons on average released per fission event will cause a subsequent fission
event. Therefore, the number of fission events that will occur in this chain reaction is 0.8×250
2.5= 80
fission events.
c) The total energy released per fission event is 200 MeV. Since there are 80 fission events, the
total energy released in this chain reaction is 80 ×200 = 16000 MeV or 16 GeV.
7 7. HYPERSPHERICAL COORDINATES IN MANY-BODY SYSTEMS
Problem 7. Consider a system of three identical particles confined in a 3-dimensional harmonic
oscillator potential. The Hamiltonian for this system can be written in terms of hyperspherical co-
ordinates as:
H=−¯h2
2m∂2
∂R2+5
R
∂
∂R +3
2mω2R2,
where Ris a collective coordinate. Calculate the ground state energy of this system in terms of the
oscillator frequency ω.
Solution 7. To find the ground state energy of the system, we need to solve the Schrödinger
equation
HΨ(R,Ω) = EΨ(R,Ω),
where Ψ(R,Ω)is the wave function of the system and Ris the hyperradius.
For the ground state energy, the wave function can be assumed to be separable as Ψ(R,Ω) =
F(R)Y(Ω).
Substituting this into the Schrödinger equation and dividing by F(R)Y(Ω), we get
−¯h2
2mF′′
F+5
R
F′
F+3
2mω2R2=E.
Separating variables, we have
F′′
F=−5
R
F′
F+2mE
¯h2−3mω2R2
¯h2.
To simplify this equation, we make the substitution u=√mωR, which transforms the equation
into
−F′′ +5
uF′= (λu2−k)F,
where λ=2mE
¯hω and k=3
2.
The solution to this differential equation can be written in terms of Laguerre polynomials Lα
k(u)
as
F(R) = Au5/2e−u2/2L5
0(u2),
where Ais a normalization constant.
Now, the ground state energy is given by E0=3
2¯hω.
Therefore, the ground state energy of the system in terms of the oscillator frequency ωis E0=
3
2¯hω.
8 8. INTRINSIC AND COLLECTIVE NUCLEAR EXCITATIONS
Problem 8. Consider a nucleus with a mass number A= 240 and a charge number Z= 92.
a) Calculate the binding energy of the nucleus in MeV given that the atomic mass unit uis
equivalent to 931.5MeV/c2.
b) Determine the energy needed to completely remove a neutron from the nucleus.
c) Find the Q-value of the alpha decay process where the nucleus decays into a daughter
nucleus with A= 236 and Z= 90, emitting an alpha particle.
Solution 8.
a) The binding energy of a nucleus can be calculated using the formula:
B.E. = (Zmp+N mn−M)c2
Where: - Zis the number of protons, - Nis the number of neutrons, - mpis the mass of a proton,
-mnis the mass of a neutron, - Mis the mass of the nucleus, and - cis the speed of light.
Given that mp=mn= 1u= 931.5MeV/c2, we first need to calculate the total number of
neutrons N=A−Z:
N= 240 −92 = 148
The mass of the nucleus Mcan be calculated by:
M=Amu= 240 ×1u= 240 ×931.5MeV/c2
Therefore, the binding energy B.E. is:
B.E. = (92 ×931.5 + 148 ×931.5−240 ×931.5) MeV = 16900 MeV
So, the binding energy of the nucleus is 16900 MeV.
b) To remove a neutron, we need to consider the mass of the neutron itself, hence the energy
required is the mass of the neutron:
E=mnc2= 1 ×931.5MeV = 931.5MeV
Hence, the energy needed to remove a neutron is 931.5MeV.
c) The Q-value of the alpha decay process is given by:
Q= (Mi−Mf−mα)c2
Where: - Miis the initial mass, - Mfis the final mass, and - mαis the mass of the alpha particle.
Given that Mi= 240×931.5MeV/c2and Mf= 236×931.5MeV/c2, and mα= 4×931.5MeV/c2,
we can calculate the Q-value:
Q= (240 ×931.5−236 ×931.5−4×931.5) MeV = 0 MeV
Therefore, the Q-value of the alpha decay process is 0MeV.
9 9. NUCLEAR FORCES AND BINDING ENERGIES
Problem 9. Consider a nucleus with atomic number Z= 6. Given that the mass of a proton is
mp= 1.67 ×10−27 kg and the mass of a neutron is mn= 1.675 ×10−27 kg, and the mass of the
nucleus is M= 12.000 amu, calculate the binding energy per nucleon in MeV.
Solution 9. a) The total number of protons and neutrons in the nucleus can be calculated using
the atomic number Z= 6.
N=Z+ (A−Z)
= 6 + (12 −6)
= 12 nucleons
b) The total mass of the nucleus can be calculated using the masses of protons and neutrons:
Mnucleus =Z·mp+ (A−Z)·mn
= 6 ·1.67 ×10−27 + 6 ·1.675 ×10−27
= 9.99 ×10−26 + 10.05 ×10−26
= 20.04 ×10−26 kg
= 20.04 ×10−26 kg ×1amu
1.66 ×10−27 kg
= 12.09 amu
The binding energy of the nucleus can be approximated using the mass defect:
∆m= 12.000 amu −12.09 amu
=−0.09 amu
c) The binding energy per nucleon can be calculated as:
BE/A =∆E
A·931.5MeV/amu
=0.09 amu
12 ·931.5MeV/amu
= 6.9375 MeV
Therefore, the binding energy per nucleon is 6.9375 MeV.
10 10. SCATTERING THEORY IN NUCLEAR PHYSICS
Problem 10. Consider a neutron scattering off a nucleus with a potential given by V(r) =
−V0e−r2/a2, where ris the distance from the center of the nucleus, V0>0, and ais a constant with
units of length.
A neutron with energy Escatters from this potential. The scattering cross section is given by
σ=4π
k2P∞
l=0(2l+ 1)|fl|2, where k=q2mE
¯h2is the wave number, flis the scattering amplitude for
orbital angular momentum l, and the sum runs over all possible values of l.
a) Show that the differential cross section for neutron-nucleus scattering, dσ/dΩ, can be written
as |f(θ)|2, where θis the scattering angle.
b) Assuming f(θ) = 1
ksin(δ0)eiα, where αis a real constant, show that dσ/dΩ = |f(θ)|2=
sin2(δ0)
k2.
Solution 10.
a) The differential cross section is defined as dσ
dΩ=|f(θ)|2, where θis the scattering angle.
This comes from the fact that the differential cross section is related to the scattering amplitude by
dσ
dΩ=|f(θ)|2. Thus, dσ/dΩ = |f(θ)|2.
b) Given f(θ) = 1
ksin(δ0)eiα, where αis real, we have |f(θ)|2=1
ksin(δ0)eiα
2=sin2(δ0)
k2. This
expression gives the differential cross section for neutron-nucleus scattering.
11 11. NUCLEAR SHAPE TRANSITIONS
Problem 11. Consider a nucleus undergoing a shape transition from a deformed shape to a
spherical shape as the excitation energy increases. The potential energy of the nucleus can be
described by the formula:
V(r) = V0
1 + e(r−r0
a)
where: V(r)is the potential energy at distance rfrom the center of the nucleus, V0= 50 MeV is
the depth of the potential well, r0= 1.2fm is the equilibrium radius of the nucleus in the deformed
state, a= 0.1fm characterizes the sharpness of the potential well.
a) Calculate the potential energy of the nucleus at r= 1.5fm in the deformed state.
b) At what distance from the center of the nucleus does the potential energy equal half of V0in
the deformed state?
c) Determine the critical excitation energy in MeV needed for the nucleus to transition from a
deformed shape to a spherical shape.
Solution 11. a) To calculate the potential energy at r= 1.5fm in the deformed state, we
substitute r= 1.5fm into the potential energy formula:
V(1.5) = 50
1 + e(1.5−1.2
0.1)=50
1 + e(3)
V(1.5) = 50
1 + e3≈50
1 + 20.0855 ≈50
21.0855 ≈2.37 MeV
Therefore, the potential energy of the nucleus at r= 1.5fm in the deformed state is approxi-
mately 2.37 MeV.
b) To find the distance at which the potential energy equals half of V0, we set V(r) = V0
2and
solve for r:
50
1 + e(r−1.2
0.1)=50
2
1 + e(r−1.2
0.1)= 2
e(r−1.2
0.1)= 1
r−1.2
0.1= 0
r= 1.2fm
Therefore, the potential energy equals half of V0at a distance of 1.2 fm from the center of the
nucleus in the deformed state.
c) The critical excitation energy ∆Eneeded for the nucleus to transition from a deformed shape
to a spherical shape occurs when the nucleus reaches the equilibrium radius of a spherical shape,
i.e., when r=r0for the potential energy formula. Thus:
V(r0) = 50
1 + e(r0−1.2
0.1)= 50
1 + e(r0−1.2
0.1)= 1
e(r0−1.2
0.1)= 0
r0= 1.2fm
Therefore, the critical excitation energy needed for the nucleus to transition from a deformed
shape to a spherical shape is 50 MeV.
12 12. QUANTUM MONTE CARLO METHODS FOR MANY-BODY SYSTEMS
Problem 12. Consider a quantum many-body system of Nparticles confined in a one-dimensional
infinite square well potential given by V(x) = 0 for 0< x < L and V(x) = ∞otherwise. Sup-
pose that the ground state wave function of this system can be approximated by a product of N
single-particle wave functions ψi(x), where each single-particle wave function can be expressed
as ψi(x) = q2
Lsin iπx
Lfor i= 1,2, . . . , N.
a) Calculate the normalization constant for the single-particle wave functions ψi(x).
b) Determine the ground state energy of the many-body system with these single-particle wave
functions.
c) Suppose the particles are non-interacting fermions. What is the ground state energy of the
system in this case?
Solution 12. a) The normalization condition for a wave function ψ(x)is R∞
−∞ |ψ(x)|2dx = 1. For
the given single-particle wave function ψi(x) = q2
Lsin iπx
L, we have:
ZL
0|ψi(x)|2dx =ZL
0 r2
Lsin iπx
L!2
dx
ZL
0|ψi(x)|2dx =2
LZL
0
sin2iπx
Ldx
ZL
0|ψi(x)|2dx =2
LZL
0
1−cos(2iπx/L)
2dx
Using trigonometric identity sin2(θ) = 1−cos(2θ)
2, we can simplify the integral as:
ZL
0|ψi(x)|2dx =1
Lx−L
2iπ sin 2iπx
LL
0
ZL
0|ψi(x)|2dx =1
L[L−0]
ZL
0|ψi(x)|2dx = 1
Therefore, the single-particle wave function ψi(x)is normalized.
b) The total ground state wave function for the many-body system is given by Ψ(x1, x2, . . . , xN) =
QN
i=1 ψi(xi). The ground state energy Efor this system is given by the sum of the single-particle
energies:
E=
N
X
i=1
¯h2π2i2
2mL2
c) Since the particles are non-interacting fermions, the ground state energy of the system is
given by the sum of the single-particle energies up to the Fermi level. The Fermi level is the highest
occupied single-particle energy level, which is the Nth level for Nnon-interacting fermions.
Efermions =
N
X
i=1
¯h2π2i2
2mL2
13 13. ELECTRON-NUCLEUS INTERACTIONS
Problem 13. An electron is scattered off a gold nucleus (197
79 Au) with a charge of +79e. The
electron has an initial kinetic energy of 5 MeV and scatters backwards at an angle of 180◦. Calculate
the recoil energy of the gold nucleus.
Given:
•Charge of the gold nucleus, Z= +79e
•Initial kinetic energy of the electron, Ei= 5 MeV
•Scattered angle, θ= 180◦
•Mass of the gold nucleus, mAu = 197 mproton
Solution 13.
a) The recoil energy of the gold nucleus can be calculated using the conservation of momentum
and energy. Since the electron scatters backwards, the final momentum of the system should be
zero.
Initial momentum of the system:
pi=p2meEi=p2×9.11 ×10−31 ×5×106≈1.74 ×10−22 kg m/s
The final kinetic energy of the gold nucleus is given by:
Ef=p2
i
2mAu
=(1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈7.62 ×10−11 J
Converting this energy to MeV:
Ef=7.62 ×10−11
1.6×10−13 ≈476.25 MeV
b) The final momentum of the gold nucleus can be calculated using the law of conservation of
momentum:
pAu =−pi=−1.74 ×10−22 kg m/s
c) The recoil energy of the gold nucleus can be calculated using the final momentum:
Erecoil =p2
Au
2mAu
=(−1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈3.81 ×10−10 J
Converting this energy to MeV:
Erecoil =3.81 ×10−10
1.6×10−13 ≈2381.25 MeV
14 14. CHIRAL EFFECTIVE FIELD THEORY IN NUCLEAR PHYSICS
Problem 14. Consider a two-neutron system in a harmonic oscillator potential with a frequency
ω= 5 MeV.
a) Calculate the energy of the ground state of the two-neutron system in MeV.
b) If the interaction potential between the neutrons is given by a contact interaction with a
strength of V0=−20 MeV, compute the energy of the first excited state of the system in MeV.
Solution 14.
a) The energy of the ground state of a harmonic oscillator potential is given by Eground =3
2¯hω.
Substituting ¯h= 197 MeV fm and ω= 5 MeV into the formula:
Eground =3
2·197 MeV fm ·5MeV =3
2·197 ·5 = 1477.5MeV
Therefore, the energy of the ground state of the two-neutron system is 1477.5 MeV.
b) The energy of the first excited state can be calculated by considering the ground state energy
plus the interaction energy. The interaction energy for a contact interaction is given by Vint =
⟨ϕ2|V0|ϕ1⟩where ϕ1and ϕ2are the ground and first excited state wave functions, respectively.
For a harmonic oscillator potential, the wave functions are known and the matrix element can
be calculated. For a two-particle system, the interaction energy is:
Vint =V01−15
8=V0−7
8=−20 MeV −7
8= 17.5MeV
Therefore, the energy of the first excited state is 1477.5 + 17.5 = 1495 MeV.
15 15. SUPERFLUIDITY IN NUCLEAR MATTER
Problem 15. Consider a system of nucleons in a superfluid state with a pairing energy gap of
∆=2MeV. Calculate the critical temperature Tcfor this superfluid system.
Given:
•The transition temperature Tc= 0.57∆
Solution 15.
•We are given the formula for Tcin terms of the pairing energy gap ∆:Tc= 0.57∆
•Substituting the given value of ∆=2MeV into the formula:
Tc= 0.57 ×2MeV
Tc= 1.14 MeV
Therefore, the critical temperature for this superfluid system is Tc= 1.14 MeV.
16 16. NUCLEAR SPIN-ORBIT COUPLING
Problem 16. Consider a nucleus with a spin-orbit coupling constant equal to a= 20 MeV.
a) Calculate the energy splitting between the l= 1 states for a nucleus with N= 50 neutrons
and Z= 40 protons.
b) Calculate the energy splitting between the l= 1 states for a nucleus with N= 70 neutrons
and Z= 40 protons.
Solution 16.
a) The energy splitting between l= 1 states can be calculated using the formula:
∆E=a·(N−Z)
Plugging in the values a= 20 MeV, N= 50, and Z= 40, we find:
∆E= 20 MeV ·(50 −40) = 200 MeV
So, the energy splitting between the l= 1 states for this nucleus is 200 MeV.
b) Similarly, for the nucleus with N= 70 and Z= 40, the energy splitting between l= 1 states
is given by:
∆E= 20 MeV ·(70 −40) = 20 MeV ·30 = 600 MeV
Therefore, for a nucleus with N= 70 neutrons and Z= 40 protons, the energy splitting between
the l= 1 states is 600 MeV.
17 17. DEUTERON SCATTERING AND BOUND STATES
Problem 17. Consider a deuteron, a bound state of a proton and a neutron. The total energy
of the deuteron is given by E=−2.224 MeV. The radius of the deuteron is r= 2 fm.
a) Calculate the reduced mass of the proton-neutron system.
b) Calculate the binding energy of the deuteron.
c) If the deuteron is in a state of higher angular momentum, how would its radius compare to
r= 2 fm?
Solution 17.
a) The reduced mass µof the proton-neutron system can be calculated as:
µ=mp×mn
mp+mn
where mpis the mass of the proton and mnis the mass of the neutron. Given that mp≈
1.67 ×10−27 kg and mn≈1.675 ×10−27 kg, we have:
µ=(1.67 ×10−27 kg)×(1.675 ×10−27 kg)
1.67 ×10−27 kg + 1.675 ×10−27 kg ≈1.673 ×10−27 kg
Therefore, the reduced mass of the proton-neutron system is µ≈1.673 ×10−27 kg.
b) The binding energy Bof the deuteron is related to the total energy Eas:
B=−E
Therefore, the binding energy of the deuteron is B=−(−2.224) = 2.224 MeV.
c) The radius r′of the deuteron in a state of higher angular momentum can be related to the
initial radius rand the new angular momentum l′as:
r′=r×l+ 1
l′+ 11/3
Since the state has higher angular momentum, l′> l, which implies that r′< r. Therefore, the
radius of the deuteron in a state of higher angular momentum would be smaller than r= 2 fm.
I. Problem:
Problem 18. Consider a particle of mass mand energy Eapproaching a one-dimensional
potential barrier of height V0and width a. The potential inside the barrier is zero. Assuming E > V0,
calculate the transmission coefficient Tfor the particle through the barrier, where T=|t|2, with t
being the transmission amplitude.
Additional Information: The potential barrier is given by:
V(x) = (0,for 0< x < a
V0,for x < 0and x>a
Solution 18: a) To find the transmission coefficient T, we first need to calculate the transmission
amplitude t. The transmission coefficient is then given by T=|t|2.
b) The transmission coefficient can be expressed in terms of the wave numbers k1and k2inside
and outside the barrier, respectively. We have:
T=k2
k1
c) The wave numbers k1and k2can be related to the energy Eand potential V0using the
following equations:
k1=p2m(E−V0)
¯h
k2=√2mE
¯h
Now, let’s substitute k1and k2into the expression for the transmission coefficient to find T.
c) Detailed Solution: a) The transmission amplitude tis given by:
t=e−ika
b) Substituting the expressions for k1and k2into the formula for T:
T=e−ika
2=e−2ika
c) Substituting the values of k1and k2into the expression, we get:
T= √2mE
¯h! p2m(E−V0)
¯h!
Thus, the transmission coefficient Tin this scenario is determined by the given particle’s energy
E, the potential barrier height V0, and the barrier width a.
18 19. NUCLEAR DENSITY FUNCTIONAL THEORY
Problem 19. Consider a system of nucleons with a total binding energy described by the
Skyrme energy density functional in the form:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
where ρ(r)is the nucleon density, τ(r)is the kinetic energy density, and C0,C3are constants.
Given that the nucleon density is ρ(r) = 3
4πR3e−3r
2R, where R= 1.2fm, and the constants are
C0=−400 MeV, C3= 100 MeV, calculate the total binding energy of the system.
Solution 19. The total binding energy E[ρ]can be calculated by evaluating the integral over
the entire volume:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
Substitute the given nucleon density ρ(r)into the expression for E[ρ]:
E[ρ] = Zd3r ¯h2
2mτ(r) + C03
4πR3e−3r
2R2
+C33
4πR3e−3r
2R3!
Since the kinetic energy density term τ(r)is not given, let’s assume it to be zero for simplicity.
We can now integrate the potential energy terms over the entire volume by performing the following
steps:
a) Calculate ρ2(r) = 3
4πR3e−3r
2R2and ρ3(r) = 3
4πR3e−3r
2R3.
b) Substitute these expressions into the integral and solve for the total binding energy.
c) Evaluate the integral numerically using appropriate techniques to find the total binding energy
of the system.
19 20. LIGHT NUCLEI AND FEW-BODY SYSTEMS
Problem 20. Consider a helium-4 nucleus (alpha particle) consisting of 2 protons and 2 neu-
trons.
Given that the rest masses of a proton and neutron are mp= 1.6726219 ×10−27 kg and mn=
1.6749275 ×10−27 kg, respectively, and the speed of light c= 3.00 ×108m/s, calculate the binding
energy of a helium-4 nucleus. Assume the atomic mass unit (u) is defined as 1u= 1.660539×10−27
kg.
Solution 20. The binding energy of a nucleus is the energy required to completely disassemble
a nucleus into its constituent protons and neutrons. It is defined as the difference between the total
rest mass energy of the separate protons and neutrons and the rest mass energy of the nucleus.
a) The total rest mass energy of a helium-4 nucleus:
mnucleus = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Enucleus =mnucleusc2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
b) The total rest mass energy of the separate protons and neutrons:
mseparate = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Eseparate =mseparatec2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
c) The binding energy of the helium-4 nucleus is:
Therefore, the binding energy of the helium-4 nucleus is 0 Joules, indicating that the nucleus
is stable.
20 21. GAMOW-TELLER TRANSITIONS IN NUCLEI
Problem 21. Consider a nucleus undergoing a Gamow-Teller transition, where the initial state
has quantum numbers Jπ= 1+and the final state has quantum numbers Jπ= 0−. If the energy
difference between the initial and final states is 2 MeV, and the reduced transition probability B(GT )
is 3×10−3units, calculate the half-life of this transition.
Solution 21.
a) The decay rate λof a nuclear transition is related to the half-life T1
2
by the equation:
T1
2=ln(2)
λ
b) The decay rate of a Gamow-Teller transition can be calculated using the formula:
λ=1
T1
2
B(GT )
2Ji+1 2
where T1
2
is the half-life, B(GT )is the reduced transition probability, and Jiis the total angular
momentum of the initial state.
c) Substituting the given values into the formula, we have:
λ=1
3×10−3
3×10−3
2×1+1 2
λ=1
3×10−3×9
(3 ×10−3)2
λ=1
3×10−3×9
9×10−6
λ=1
3×10−3×106= 103sec−1
Using the relation between decay rate and half-life, we find:
T1
2=ln(2)
λ=ln(2)
103≈0.693 ms
Therefore, the half-life of this Gamow-Teller transition is approximately 0.693 milliseconds.
21 22. CLUSTERING PHENOMENA IN NUCLEAR STRUCTURE
Problem 22. Consider a nucleus with mass number A= 12, consisting of two alpha clus-
ters and 4 neutrons. The energy of an alpha particle is 28.3 MeV and the mass of a neutron is
approximately 939.6 MeV/c2.
a) Calculate the binding energy per nucleon of this nucleus.
b) Determine the total energy of this nucleus.
c) If the neutrons interact only through the nuclear force, what is the total nuclear force energy
for this nucleus?
Solution 22. a) The total binding energy of the nucleus is given by the sum of the binding
energy of the two alpha clusters and the binding energy of the 4 neutrons. The binding energy per
nucleon is then the total binding energy divided by the number of nucleons.
The binding energy of an alpha particle is Eα= 28.3MeV and the mass of an alpha particle
is approximately 4 times the mass of a nucleon. So, the binding energy per nucleon for the alpha
clusters would be Eα/4 = 28.3/4 = 7.075 MeV.
The binding energy of a neutron is zero (since it is a free particle). Therefore, the binding energy
per nucleon for the 4 neutrons is 0/4=0MeV.
The total binding energy for the nucleus is 2×Eα+ 4 ×0=2×28.3 = 56.6MeV.
Therefore, the binding energy per nucleon is 56.6/12 = 4.717 MeV.
b) The total energy of the nucleus is the sum of the rest mass energy and the binding energy.
Since the rest mass energy is not given in the problem, we will neglect it for this specific calculation.
Therefore, the total energy of the nucleus is just the binding energy, which is 56.6 MeV.
c) The total nuclear force energy for this nucleus would be zero since the neutrons are not
interacting with each other.
22 Nuclear Physics and Many-Body Systems
Problem: Consider a nucleus with a total of 20 protons and 30 neutrons. Assume that each
proton can pair with a neutron in the ground state of the nucleus. Determine the total angular
momentum and total spin of the nucleus.
Solution:
a) To find the total angular momentum of the nucleus, we first need to calculate the individual
angular momenta of the protons and neutrons in the ground state. Since each proton can pair with
a neutron, we have a total of 20 pairs. The individual angular momentum of a proton is l=1
2and
for a neutron is also l=1
2.
The total angular momentum of a pair is given by L=Lp+Ln, where Lpis the angular mo-
mentum of the proton and Lnis the angular momentum of the neutron. Therefore, the total angular
momentum of a pair is 1
2+1
2= 1.
Since we have 20 pairs, the total angular momentum of the nucleus is Ltotal = 20 ×1 = 20.
b) Next, to find the total spin of the nucleus, we know that the total spin of a nucleus is the sum
of the spins of each nucleon. Both protons and neutrons have spin s=1
2.
Since we have 20 protons and 30 neutrons, the total spin of the nucleus is Stotal = (20 ×1
2) +
(30 ×1
2) = 25.
Therefore, the total angular momentum of the nucleus is 20 and the total spin is 25.
I. Problem 1:
Consider a heavy nucleus with a mass number A= 240 and a proton number Z= 92. The
ground state of this nucleus has a binding energy of B= 7.6MeV per nucleon. Calculate the total
binding energy of this nucleus.
Solution 1: The total binding energy of the nucleus can be calculated using the formula:
Ebinding =B×A
Substitute the given values into the formula:
Ebinding = 7.6MeV/nucleon ×240 nucleons
Ebinding = 1824 MeV
Therefore, the total binding energy of this nucleus is 1824 MeV.
II. Problem 2:
A heavy nucleus in an excited state decays by emitting a beta particle (electron) and a neutrino.
If the energy released in this decay is 2 MeV, calculate the energy of the excited state before the
decay.
Solution 2: The energy released in the decay is related to the energy of the excited state by
the equation:
Ereleased =Ei−Ef
Where Eiis the energy of the excited state and Efis the energy of the final state after decay.
Given that Ereleased = 2 MeV, and the final state consists of the beta particle and the neutrino
which have negligible energy, the energy of the final state Efcan be considered as nearly 0.
Hence, Ei=Ereleased = 2 MeV.
Therefore, the energy of the excited state before the decay is 2 MeV.
23 25. NUCLEAR PHYSICS APPLICATIONS IN ASTROPHYSICS
Problem 25. Consider the nuclear reaction 4
2He +3
1H→7
3Li.
Given the rest mass of:
•Helium-4 atom 4
2He is 4.0026 u
•Hydrogen-3 atom 3
1His 3.0155 u
•Lithium-7 atom 7
3Li is 7.0160 u
Calculate the energy released in the reaction in MeV.
Solution 25. The energy released in the nuclear reaction can be calculated using Einstein’s
mass-energy equivalence E= ∆mc2.
The change in mass (∆m) is the total rest mass of the reactants minus the total rest mass of
the products:
∆m= (4.0026 + 3.0155) −7.0160
= 7.0181 −7.0160
= 0.0021 u
Using the conversion factor 1u= 931.5MeV/c2, we can calculate the energy released:
E= (0.0021 u)×(931.5MeV/c2)
= 1.95915 MeV
Therefore, the energy released in the reaction is 1.95915 M eV .
Q= 0.026698c2
To convert from atomic mass units (u) to energy (MeV), we use the conversion factor 1u=
931.5MeV/c2.
Substitute this conversion factor into the equation to get the energy in MeV:
Q= 0.026698 ×931.5
Q24.9MeV
Therefore, the energy released in the fusion reaction is approximately 24.9 MeV.
3 3. NUCLEAR DECAY PROCESSES
Problem 3. The half-life of a certain radioactive isotope is 10 days. If you start with a sample
containing 100 grams of the isotope, how much of the isotope will remain after 30 days?
Additional context: The decay of a radioactive substance follows an exponential decay model
given by the equation N(t) = N0e−λt, where: - N(t)is the amount of the substance remaining at
time t, - N0is the initial amount of the substance, - λis the decay constant, - tis the time elapsed.
a) How much of the isotope will remain after 10 days?
b) What is the decay constant, λ, for this isotope?
c) How much of the isotope will remain after 30 days?
Solution 3.
a) After 10 days:
N(10) = 100 ·e−λ·10
Given that 10 days is the half-life, N(10) = 100
2= 50, so:
50 = 100 ·e−λ·10
0.5 = e−10λ
ln(0.5) = −10λ
λ=−ln(0.5)
10
λ≈0.0693 per day
b) Decay constant, λ≈0.0693 per day.
c) After 30 days:
N(30) = 100 ·e−λ·30
= 100 ·e−0.0693·30
≈17.07 grams remaining
Therefore, after 30 days, approximately 17.07 grams of the isotope will remain.
4 4. ISOSPIN SYMMETRY IN NUCLEAR STRUCTURE
Problem 4. Consider a system of two protons and two neutrons in a nucleus. The isospin
quantum number for protons and neutrons is 1
2.
a) Determine the isospin values for the four particles in this system.
b) If we allow for isospin mixing, what are the possible values for the total isospin of the system?
c) Suppose the total isospin of the system is measured to be T= 1. What are the possible
values of the third component, Tz, of the isospin?
Solution 4. a) The isospin quantum number Tzcan take values ranging from −Tto T. In this
case, each proton and neutron has isospin T=1
2, so their Tzvalues can be −1
2or 1
2.
Therefore, the isospin values for the four particles in the system are:
Proton 1: T=1
2,Tz=1
2
Proton 2: T=1
2,Tz=1
2
Neutron 1: T=1
2,Tz=1
2
Neutron 2: T=1
2,Tz=1
2
b) When we allow for isospin mixing, the possible values for the total isospin of the system
range from |T1−T2|to T1+T2. In this case, T1=T2=1
2.
So, the possible values for the total isospin of the system are 0 and 1.
c) If the total isospin of the system is measured to be T= 1, the possible values for the third
component of isospin, Tz, are −1, 0, and 1.
5 5. ALPHA DECAY AND NUCLEAR STABILITY
Problem 5. The half-life of a certain radioactive isotope is 100 years. If there are initially 1000
atoms of this isotope, how many atoms will remain after 300 years?
Given:
•Half-life of the isotope = 100 years
•Initial number of atoms = 1000
•Time elapsed = 300 years
Solution 5. a) To calculate the decay constant, we use the formula:
N(t) = N0×e−λt
where: N(t)= number of atoms at time t,N0= initial number of atoms, λ= decay constant, t=
time.
First, we find the decay constant using the half-life formula:
T1/2=ln(2)
λ
Substitute T1/2= 100 years:
100 = ln(2)
λ
λ=ln(2)
100
λ≈0.00693 years−1
b) Next, we find the number of atoms remaining after 300 years:
N(300) = 1000 ×e−0.00693×300
N(300) = 1000 ×e−2.079
N(300) ≈1000 ×0.125
N(300) ≈125 atoms
Therefore, after 300 years, approximately 125 atoms of this isotope will remain.
6 6. NUCLEAR FISSION CHAIN REACTIONS
Problem 6. Consider a nuclear fission chain reaction in which a neutron induces the fission of
a uranium-235 nucleus, releasing on average 2.5 neutrons per fission event. The probability of a
neutron causing a subsequent fission event is 0.8.
a) If 100 neutrons are released in the initial fission event, how many total neutrons will be
released in the subsequent chain reaction?
b) Determine the number of fission events that will occur in this chain reaction.
c) If the energy released per fission event is 200 MeV, calculate the total energy released in this
chain reaction.
Solution 6.
a) In the initial fission event, 100 neutrons are released. Since on average 2.5 neutrons are
released per fission event, these 100 neutrons will lead to 100 ×2.5 = 250 neutrons being released
in the subsequent chain reaction.
b) The probability of a neutron causing a subsequent fission event is 0.8. This means 0.8neu-
trons out of the 2.5neutrons on average released per fission event will cause a subsequent fission
event. Therefore, the number of fission events that will occur in this chain reaction is 0.8×250
2.5= 80
fission events.
c) The total energy released per fission event is 200 MeV. Since there are 80 fission events, the
total energy released in this chain reaction is 80 ×200 = 16000 MeV or 16 GeV.
7 7. HYPERSPHERICAL COORDINATES IN MANY-BODY SYSTEMS
Problem 7. Consider a system of three identical particles confined in a 3-dimensional harmonic
oscillator potential. The Hamiltonian for this system can be written in terms of hyperspherical co-
ordinates as:
H=−¯h2
2m∂2
∂R2+5
R
∂
∂R +3
2mω2R2,
where Ris a collective coordinate. Calculate the ground state energy of this system in terms of the
oscillator frequency ω.
Solution 7. To find the ground state energy of the system, we need to solve the Schrödinger
equation
HΨ(R,Ω) = EΨ(R,Ω),
where Ψ(R,Ω)is the wave function of the system and Ris the hyperradius.
For the ground state energy, the wave function can be assumed to be separable as Ψ(R,Ω) =
F(R)Y(Ω).
Substituting this into the Schrödinger equation and dividing by F(R)Y(Ω), we get
−¯h2
2mF′′
F+5
R
F′
F+3
2mω2R2=E.
Separating variables, we have
F′′
F=−5
R
F′
F+2mE
¯h2−3mω2R2
¯h2.
To simplify this equation, we make the substitution u=√mωR, which transforms the equation
into
−F′′ +5
uF′= (λu2−k)F,
where λ=2mE
¯hω and k=3
2.
The solution to this differential equation can be written in terms of Laguerre polynomials Lα
k(u)
as
F(R) = Au5/2e−u2/2L5
0(u2),
where Ais a normalization constant.
Now, the ground state energy is given by E0=3
2¯hω.
Therefore, the ground state energy of the system in terms of the oscillator frequency ωis E0=
3
2¯hω.
8 8. INTRINSIC AND COLLECTIVE NUCLEAR EXCITATIONS
Problem 8. Consider a nucleus with a mass number A= 240 and a charge number Z= 92.
a) Calculate the binding energy of the nucleus in MeV given that the atomic mass unit uis
equivalent to 931.5MeV/c2.
b) Determine the energy needed to completely remove a neutron from the nucleus.
c) Find the Q-value of the alpha decay process where the nucleus decays into a daughter
nucleus with A= 236 and Z= 90, emitting an alpha particle.
Solution 8.
a) The binding energy of a nucleus can be calculated using the formula:
B.E. = (Zmp+N mn−M)c2
Where: - Zis the number of protons, - Nis the number of neutrons, - mpis the mass of a proton,
-mnis the mass of a neutron, - Mis the mass of the nucleus, and - cis the speed of light.
Given that mp=mn= 1u= 931.5MeV/c2, we first need to calculate the total number of
neutrons N=A−Z:
N= 240 −92 = 148
The mass of the nucleus Mcan be calculated by:
M=Amu= 240 ×1u= 240 ×931.5MeV/c2
Therefore, the binding energy B.E. is:
B.E. = (92 ×931.5 + 148 ×931.5−240 ×931.5) MeV = 16900 MeV
So, the binding energy of the nucleus is 16900 MeV.
b) To remove a neutron, we need to consider the mass of the neutron itself, hence the energy
required is the mass of the neutron:
E=mnc2= 1 ×931.5MeV = 931.5MeV
Hence, the energy needed to remove a neutron is 931.5MeV.
c) The Q-value of the alpha decay process is given by:
Q= (Mi−Mf−mα)c2
Where: - Miis the initial mass, - Mfis the final mass, and - mαis the mass of the alpha particle.
Given that Mi= 240×931.5MeV/c2and Mf= 236×931.5MeV/c2, and mα= 4×931.5MeV/c2,
we can calculate the Q-value:
Q= (240 ×931.5−236 ×931.5−4×931.5) MeV = 0 MeV
Therefore, the Q-value of the alpha decay process is 0MeV.
9 9. NUCLEAR FORCES AND BINDING ENERGIES
Problem 9. Consider a nucleus with atomic number Z= 6. Given that the mass of a proton is
mp= 1.67 ×10−27 kg and the mass of a neutron is mn= 1.675 ×10−27 kg, and the mass of the
nucleus is M= 12.000 amu, calculate the binding energy per nucleon in MeV.
Solution 9. a) The total number of protons and neutrons in the nucleus can be calculated using
the atomic number Z= 6.
N=Z+ (A−Z)
= 6 + (12 −6)
= 12 nucleons
b) The total mass of the nucleus can be calculated using the masses of protons and neutrons:
Mnucleus =Z·mp+ (A−Z)·mn
= 6 ·1.67 ×10−27 + 6 ·1.675 ×10−27
= 9.99 ×10−26 + 10.05 ×10−26
= 20.04 ×10−26 kg
= 20.04 ×10−26 kg ×1amu
1.66 ×10−27 kg
= 12.09 amu
The binding energy of the nucleus can be approximated using the mass defect:
∆m= 12.000 amu −12.09 amu
=−0.09 amu
c) The binding energy per nucleon can be calculated as:
BE/A =∆E
A·931.5MeV/amu
=0.09 amu
12 ·931.5MeV/amu
= 6.9375 MeV
Therefore, the binding energy per nucleon is 6.9375 MeV.
10 10. SCATTERING THEORY IN NUCLEAR PHYSICS
Problem 10. Consider a neutron scattering off a nucleus with a potential given by V(r) =
−V0e−r2/a2, where ris the distance from the center of the nucleus, V0>0, and ais a constant with
units of length.
A neutron with energy Escatters from this potential. The scattering cross section is given by
σ=4π
k2P∞
l=0(2l+ 1)|fl|2, where k=q2mE
¯h2is the wave number, flis the scattering amplitude for
orbital angular momentum l, and the sum runs over all possible values of l.
a) Show that the differential cross section for neutron-nucleus scattering, dσ/dΩ, can be written
as |f(θ)|2, where θis the scattering angle.
b) Assuming f(θ) = 1
ksin(δ0)eiα, where αis a real constant, show that dσ/dΩ = |f(θ)|2=
sin2(δ0)
k2.
Solution 10.
a) The differential cross section is defined as dσ
dΩ=|f(θ)|2, where θis the scattering angle.
This comes from the fact that the differential cross section is related to the scattering amplitude by
dσ
dΩ=|f(θ)|2. Thus, dσ/dΩ = |f(θ)|2.
b) Given f(θ) = 1
ksin(δ0)eiα, where αis real, we have |f(θ)|2=1
ksin(δ0)eiα
2=sin2(δ0)
k2. This
expression gives the differential cross section for neutron-nucleus scattering.
11 11. NUCLEAR SHAPE TRANSITIONS
Problem 11. Consider a nucleus undergoing a shape transition from a deformed shape to a
spherical shape as the excitation energy increases. The potential energy of the nucleus can be
described by the formula:
V(r) = V0
1 + e(r−r0
a)
where: V(r)is the potential energy at distance rfrom the center of the nucleus, V0= 50 MeV is
the depth of the potential well, r0= 1.2fm is the equilibrium radius of the nucleus in the deformed
state, a= 0.1fm characterizes the sharpness of the potential well.
a) Calculate the potential energy of the nucleus at r= 1.5fm in the deformed state.
b) At what distance from the center of the nucleus does the potential energy equal half of V0in
the deformed state?
c) Determine the critical excitation energy in MeV needed for the nucleus to transition from a
deformed shape to a spherical shape.
Solution 11. a) To calculate the potential energy at r= 1.5fm in the deformed state, we
substitute r= 1.5fm into the potential energy formula:
V(1.5) = 50
1 + e(1.5−1.2
0.1)=50
1 + e(3)
V(1.5) = 50
1 + e3≈50
1 + 20.0855 ≈50
21.0855 ≈2.37 MeV
Therefore, the potential energy of the nucleus at r= 1.5fm in the deformed state is approxi-
mately 2.37 MeV.
b) To find the distance at which the potential energy equals half of V0, we set V(r) = V0
2and
solve for r:
50
1 + e(r−1.2
0.1)=50
2
1 + e(r−1.2
0.1)= 2
e(r−1.2
0.1)= 1
r−1.2
0.1= 0
r= 1.2fm
Therefore, the potential energy equals half of V0at a distance of 1.2 fm from the center of the
nucleus in the deformed state.
c) The critical excitation energy ∆Eneeded for the nucleus to transition from a deformed shape
to a spherical shape occurs when the nucleus reaches the equilibrium radius of a spherical shape,
i.e., when r=r0for the potential energy formula. Thus:
V(r0) = 50
1 + e(r0−1.2
0.1)= 50
1 + e(r0−1.2
0.1)= 1
e(r0−1.2
0.1)= 0
r0= 1.2fm
Therefore, the critical excitation energy needed for the nucleus to transition from a deformed
shape to a spherical shape is 50 MeV.
12 12. QUANTUM MONTE CARLO METHODS FOR MANY-BODY SYSTEMS
Problem 12. Consider a quantum many-body system of Nparticles confined in a one-dimensional
infinite square well potential given by V(x) = 0 for 0< x < L and V(x) = ∞otherwise. Sup-
pose that the ground state wave function of this system can be approximated by a product of N
single-particle wave functions ψi(x), where each single-particle wave function can be expressed
as ψi(x) = q2
Lsin iπx
Lfor i= 1,2, . . . , N.
a) Calculate the normalization constant for the single-particle wave functions ψi(x).
b) Determine the ground state energy of the many-body system with these single-particle wave
functions.
c) Suppose the particles are non-interacting fermions. What is the ground state energy of the
system in this case?
Solution 12. a) The normalization condition for a wave function ψ(x)is R∞
−∞ |ψ(x)|2dx = 1. For
the given single-particle wave function ψi(x) = q2
Lsin iπx
L, we have:
ZL
0|ψi(x)|2dx =ZL
0 r2
Lsin iπx
L!2
dx
ZL
0|ψi(x)|2dx =2
LZL
0
sin2iπx
Ldx
ZL
0|ψi(x)|2dx =2
LZL
0
1−cos(2iπx/L)
2dx
Using trigonometric identity sin2(θ) = 1−cos(2θ)
2, we can simplify the integral as:
ZL
0|ψi(x)|2dx =1
Lx−L
2iπ sin 2iπx
LL
0
ZL
0|ψi(x)|2dx =1
L[L−0]
ZL
0|ψi(x)|2dx = 1
Therefore, the single-particle wave function ψi(x)is normalized.
b) The total ground state wave function for the many-body system is given by Ψ(x1, x2, . . . , xN) =
QN
i=1 ψi(xi). The ground state energy Efor this system is given by the sum of the single-particle
energies:
E=
N
X
i=1
¯h2π2i2
2mL2
c) Since the particles are non-interacting fermions, the ground state energy of the system is
given by the sum of the single-particle energies up to the Fermi level. The Fermi level is the highest
occupied single-particle energy level, which is the Nth level for Nnon-interacting fermions.
Efermions =
N
X
i=1
¯h2π2i2
2mL2
13 13. ELECTRON-NUCLEUS INTERACTIONS
Problem 13. An electron is scattered off a gold nucleus (197
79 Au) with a charge of +79e. The
electron has an initial kinetic energy of 5 MeV and scatters backwards at an angle of 180◦. Calculate
the recoil energy of the gold nucleus.
Given:
•Charge of the gold nucleus, Z= +79e
•Initial kinetic energy of the electron, Ei= 5 MeV
•Scattered angle, θ= 180◦
•Mass of the gold nucleus, mAu = 197 mproton
Solution 13.
a) The recoil energy of the gold nucleus can be calculated using the conservation of momentum
and energy. Since the electron scatters backwards, the final momentum of the system should be
zero.
Initial momentum of the system:
pi=p2meEi=p2×9.11 ×10−31 ×5×106≈1.74 ×10−22 kg m/s
The final kinetic energy of the gold nucleus is given by:
Ef=p2
i
2mAu
=(1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈7.62 ×10−11 J
Converting this energy to MeV:
Ef=7.62 ×10−11
1.6×10−13 ≈476.25 MeV
b) The final momentum of the gold nucleus can be calculated using the law of conservation of
momentum:
pAu =−pi=−1.74 ×10−22 kg m/s
c) The recoil energy of the gold nucleus can be calculated using the final momentum:
Erecoil =p2
Au
2mAu
=(−1.74 ×10−22)2
2×197 ×1.67 ×10−27 ≈3.81 ×10−10 J
Converting this energy to MeV:
Erecoil =3.81 ×10−10
1.6×10−13 ≈2381.25 MeV
14 14. CHIRAL EFFECTIVE FIELD THEORY IN NUCLEAR PHYSICS
Problem 14. Consider a two-neutron system in a harmonic oscillator potential with a frequency
ω= 5 MeV.
a) Calculate the energy of the ground state of the two-neutron system in MeV.
b) If the interaction potential between the neutrons is given by a contact interaction with a
strength of V0=−20 MeV, compute the energy of the first excited state of the system in MeV.
Solution 14.
a) The energy of the ground state of a harmonic oscillator potential is given by Eground =3
2¯hω.
Substituting ¯h= 197 MeV fm and ω= 5 MeV into the formula:
Eground =3
2·197 MeV fm ·5MeV =3
2·197 ·5 = 1477.5MeV
Therefore, the energy of the ground state of the two-neutron system is 1477.5 MeV.
b) The energy of the first excited state can be calculated by considering the ground state energy
plus the interaction energy. The interaction energy for a contact interaction is given by Vint =
⟨ϕ2|V0|ϕ1⟩where ϕ1and ϕ2are the ground and first excited state wave functions, respectively.
For a harmonic oscillator potential, the wave functions are known and the matrix element can
be calculated. For a two-particle system, the interaction energy is:
Vint =V01−15
8=V0−7
8=−20 MeV −7
8= 17.5MeV
Therefore, the energy of the first excited state is 1477.5 + 17.5 = 1495 MeV.
15 15. SUPERFLUIDITY IN NUCLEAR MATTER
Problem 15. Consider a system of nucleons in a superfluid state with a pairing energy gap of
∆=2MeV. Calculate the critical temperature Tcfor this superfluid system.
Given:
•The transition temperature Tc= 0.57∆
Solution 15.
•We are given the formula for Tcin terms of the pairing energy gap ∆:Tc= 0.57∆
•Substituting the given value of ∆=2MeV into the formula:
Tc= 0.57 ×2MeV
Tc= 1.14 MeV
Therefore, the critical temperature for this superfluid system is Tc= 1.14 MeV.
16 16. NUCLEAR SPIN-ORBIT COUPLING
Problem 16. Consider a nucleus with a spin-orbit coupling constant equal to a= 20 MeV.
a) Calculate the energy splitting between the l= 1 states for a nucleus with N= 50 neutrons
and Z= 40 protons.
b) Calculate the energy splitting between the l= 1 states for a nucleus with N= 70 neutrons
and Z= 40 protons.
Solution 16.
a) The energy splitting between l= 1 states can be calculated using the formula:
∆E=a·(N−Z)
Plugging in the values a= 20 MeV, N= 50, and Z= 40, we find:
∆E= 20 MeV ·(50 −40) = 200 MeV
So, the energy splitting between the l= 1 states for this nucleus is 200 MeV.
b) Similarly, for the nucleus with N= 70 and Z= 40, the energy splitting between l= 1 states
is given by:
∆E= 20 MeV ·(70 −40) = 20 MeV ·30 = 600 MeV
Therefore, for a nucleus with N= 70 neutrons and Z= 40 protons, the energy splitting between
the l= 1 states is 600 MeV.
17 17. DEUTERON SCATTERING AND BOUND STATES
Problem 17. Consider a deuteron, a bound state of a proton and a neutron. The total energy
of the deuteron is given by E=−2.224 MeV. The radius of the deuteron is r= 2 fm.
a) Calculate the reduced mass of the proton-neutron system.
b) Calculate the binding energy of the deuteron.
c) If the deuteron is in a state of higher angular momentum, how would its radius compare to
r= 2 fm?
Solution 17.
a) The reduced mass µof the proton-neutron system can be calculated as:
µ=mp×mn
mp+mn
where mpis the mass of the proton and mnis the mass of the neutron. Given that mp≈
1.67 ×10−27 kg and mn≈1.675 ×10−27 kg, we have:
µ=(1.67 ×10−27 kg)×(1.675 ×10−27 kg)
1.67 ×10−27 kg + 1.675 ×10−27 kg ≈1.673 ×10−27 kg
Therefore, the reduced mass of the proton-neutron system is µ≈1.673 ×10−27 kg.
b) The binding energy Bof the deuteron is related to the total energy Eas:
B=−E
Therefore, the binding energy of the deuteron is B=−(−2.224) = 2.224 MeV.
c) The radius r′of the deuteron in a state of higher angular momentum can be related to the
initial radius rand the new angular momentum l′as:
r′=r×l+ 1
l′+ 11/3
Since the state has higher angular momentum, l′> l, which implies that r′< r. Therefore, the
radius of the deuteron in a state of higher angular momentum would be smaller than r= 2 fm.
I. Problem:
Problem 18. Consider a particle of mass mand energy Eapproaching a one-dimensional
potential barrier of height V0and width a. The potential inside the barrier is zero. Assuming E > V0,
calculate the transmission coefficient Tfor the particle through the barrier, where T=|t|2, with t
being the transmission amplitude.
Additional Information: The potential barrier is given by:
V(x) = (0,for 0< x < a
V0,for x < 0and x>a
Solution 18: a) To find the transmission coefficient T, we first need to calculate the transmission
amplitude t. The transmission coefficient is then given by T=|t|2.
b) The transmission coefficient can be expressed in terms of the wave numbers k1and k2inside
and outside the barrier, respectively. We have:
T=k2
k1
c) The wave numbers k1and k2can be related to the energy Eand potential V0using the
following equations:
k1=p2m(E−V0)
¯h
k2=√2mE
¯h
Now, let’s substitute k1and k2into the expression for the transmission coefficient to find T.
c) Detailed Solution: a) The transmission amplitude tis given by:
t=e−ika
b) Substituting the expressions for k1and k2into the formula for T:
T=e−ika
2=e−2ika
c) Substituting the values of k1and k2into the expression, we get:
T= √2mE
¯h! p2m(E−V0)
¯h!
Thus, the transmission coefficient Tin this scenario is determined by the given particle’s energy
E, the potential barrier height V0, and the barrier width a.
18 19. NUCLEAR DENSITY FUNCTIONAL THEORY
Problem 19. Consider a system of nucleons with a total binding energy described by the
Skyrme energy density functional in the form:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
where ρ(r)is the nucleon density, τ(r)is the kinetic energy density, and C0,C3are constants.
Given that the nucleon density is ρ(r) = 3
4πR3e−3r
2R, where R= 1.2fm, and the constants are
C0=−400 MeV, C3= 100 MeV, calculate the total binding energy of the system.
Solution 19. The total binding energy E[ρ]can be calculated by evaluating the integral over
the entire volume:
E[ρ] = Zd3r¯h2
2mτ(r) + C0ρ2(r) + C3ρ3(r)
Substitute the given nucleon density ρ(r)into the expression for E[ρ]:
E[ρ] = Zd3r ¯h2
2mτ(r) + C03
4πR3e−3r
2R2
+C33
4πR3e−3r
2R3!
Since the kinetic energy density term τ(r)is not given, let’s assume it to be zero for simplicity.
We can now integrate the potential energy terms over the entire volume by performing the following
steps:
a) Calculate ρ2(r) = 3
4πR3e−3r
2R2and ρ3(r) = 3
4πR3e−3r
2R3.
b) Substitute these expressions into the integral and solve for the total binding energy.
c) Evaluate the integral numerically using appropriate techniques to find the total binding energy
of the system.
19 20. LIGHT NUCLEI AND FEW-BODY SYSTEMS
Problem 20. Consider a helium-4 nucleus (alpha particle) consisting of 2 protons and 2 neu-
trons.
Given that the rest masses of a proton and neutron are mp= 1.6726219 ×10−27 kg and mn=
1.6749275 ×10−27 kg, respectively, and the speed of light c= 3.00 ×108m/s, calculate the binding
energy of a helium-4 nucleus. Assume the atomic mass unit (u) is defined as 1u= 1.660539×10−27
kg.
Solution 20. The binding energy of a nucleus is the energy required to completely disassemble
a nucleus into its constituent protons and neutrons. It is defined as the difference between the total
rest mass energy of the separate protons and neutrons and the rest mass energy of the nucleus.
a) The total rest mass energy of a helium-4 nucleus:
mnucleus = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Enucleus =mnucleusc2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
b) The total rest mass energy of the separate protons and neutrons:
mseparate = 2mp+ 2mn
= 2(1.6726219 ×10−27 kg) + 2(1.6749275 ×10−27 kg)
= 6.6950988 ×10−27 kg
The rest mass energy E=mc2can be calculated as:
Eseparate =mseparatec2
= 6.6950988 ×10−27 kg ×(3.00 ×108m/s)2
= 6.0255899 ×10−11 J
c) The binding energy of the helium-4 nucleus is:
Therefore, the binding energy of the helium-4 nucleus is 0 Joules, indicating that the nucleus
is stable.
20 21. GAMOW-TELLER TRANSITIONS IN NUCLEI
Problem 21. Consider a nucleus undergoing a Gamow-Teller transition, where the initial state
has quantum numbers Jπ= 1+and the final state has quantum numbers Jπ= 0−. If the energy
difference between the initial and final states is 2 MeV, and the reduced transition probability B(GT )
is 3×10−3units, calculate the half-life of this transition.
Solution 21.
a) The decay rate λof a nuclear transition is related to the half-life T1
2
by the equation:
T1
2=ln(2)
λ
b) The decay rate of a Gamow-Teller transition can be calculated using the formula:
λ=1
T1
2
B(GT )
2Ji+1 2
where T1
2
is the half-life, B(GT )is the reduced transition probability, and Jiis the total angular
momentum of the initial state.
c) Substituting the given values into the formula, we have:
λ=1
3×10−3
3×10−3
2×1+1 2
λ=1
3×10−3×9
(3 ×10−3)2
λ=1
3×10−3×9
9×10−6
λ=1
3×10−3×106= 103sec−1
Using the relation between decay rate and half-life, we find:
T1
2=ln(2)
λ=ln(2)
103≈0.693 ms
Therefore, the half-life of this Gamow-Teller transition is approximately 0.693 milliseconds.
21 22. CLUSTERING PHENOMENA IN NUCLEAR STRUCTURE
Problem 22. Consider a nucleus with mass number A= 12, consisting of two alpha clus-
ters and 4 neutrons. The energy of an alpha particle is 28.3 MeV and the mass of a neutron is
approximately 939.6 MeV/c2.
a) Calculate the binding energy per nucleon of this nucleus.
b) Determine the total energy of this nucleus.
c) If the neutrons interact only through the nuclear force, what is the total nuclear force energy
for this nucleus?
Solution 22. a) The total binding energy of the nucleus is given by the sum of the binding
energy of the two alpha clusters and the binding energy of the 4 neutrons. The binding energy per
nucleon is then the total binding energy divided by the number of nucleons.
The binding energy of an alpha particle is Eα= 28.3MeV and the mass of an alpha particle
is approximately 4 times the mass of a nucleon. So, the binding energy per nucleon for the alpha
clusters would be Eα/4 = 28.3/4 = 7.075 MeV.
The binding energy of a neutron is zero (since it is a free particle). Therefore, the binding energy
per nucleon for the 4 neutrons is 0/4=0MeV.
The total binding energy for the nucleus is 2×Eα+ 4 ×0=2×28.3 = 56.6MeV.
Therefore, the binding energy per nucleon is 56.6/12 = 4.717 MeV.
b) The total energy of the nucleus is the sum of the rest mass energy and the binding energy.
Since the rest mass energy is not given in the problem, we will neglect it for this specific calculation.
Therefore, the total energy of the nucleus is just the binding energy, which is 56.6 MeV.
c) The total nuclear force energy for this nucleus would be zero since the neutrons are not
interacting with each other.
22 Nuclear Physics and Many-Body Systems
Problem: Consider a nucleus with a total of 20 protons and 30 neutrons. Assume that each
proton can pair with a neutron in the ground state of the nucleus. Determine the total angular
momentum and total spin of the nucleus.
Solution:
a) To find the total angular momentum of the nucleus, we first need to calculate the individual
angular momenta of the protons and neutrons in the ground state. Since each proton can pair with
a neutron, we have a total of 20 pairs. The individual angular momentum of a proton is l=1
2and
for a neutron is also l=1
2.
The total angular momentum of a pair is given by L=Lp+Ln, where Lpis the angular mo-
mentum of the proton and Lnis the angular momentum of the neutron. Therefore, the total angular
momentum of a pair is 1
2+1
2= 1.
Since we have 20 pairs, the total angular momentum of the nucleus is Ltotal = 20 ×1 = 20.
b) Next, to find the total spin of the nucleus, we know that the total spin of a nucleus is the sum
of the spins of each nucleon. Both protons and neutrons have spin s=1
2.
Since we have 20 protons and 30 neutrons, the total spin of the nucleus is Stotal = (20 ×1
2) +
(30 ×1
2) = 25.
Therefore, the total angular momentum of the nucleus is 20 and the total spin is 25.
I. Problem 1:
Consider a heavy nucleus with a mass number A= 240 and a proton number Z= 92. The
ground state of this nucleus has a binding energy of B= 7.6MeV per nucleon. Calculate the total
binding energy of this nucleus.
Solution 1: The total binding energy of the nucleus can be calculated using the formula:
Ebinding =B×A
Substitute the given values into the formula:
Ebinding = 7.6MeV/nucleon ×240 nucleons
Ebinding = 1824 MeV
Therefore, the total binding energy of this nucleus is 1824 MeV.
II. Problem 2:
A heavy nucleus in an excited state decays by emitting a beta particle (electron) and a neutrino.
If the energy released in this decay is 2 MeV, calculate the energy of the excited state before the
decay.
Solution 2: The energy released in the decay is related to the energy of the excited state by
the equation:
Ereleased =Ei−Ef
Where Eiis the energy of the excited state and Efis the energy of the final state after decay.
Given that Ereleased = 2 MeV, and the final state consists of the beta particle and the neutrino
which have negligible energy, the energy of the final state Efcan be considered as nearly 0.
Hence, Ei=Ereleased = 2 MeV.
Therefore, the energy of the excited state before the decay is 2 MeV.
23 25. NUCLEAR PHYSICS APPLICATIONS IN ASTROPHYSICS
Problem 25. Consider the nuclear reaction 4
2He +3
1H→7
3Li.
Given the rest mass of:
•Helium-4 atom 4
2He is 4.0026 u
•Hydrogen-3 atom 3
1His 3.0155 u
•Lithium-7 atom 7
3Li is 7.0160 u
Calculate the energy released in the reaction in MeV.
Solution 25. The energy released in the nuclear reaction can be calculated using Einstein’s
mass-energy equivalence E= ∆mc2.
The change in mass (∆m) is the total rest mass of the reactants minus the total rest mass of
the products:
∆m= (4.0026 + 3.0155) −7.0160
= 7.0181 −7.0160
= 0.0021 u
Using the conversion factor 1u= 931.5MeV/c2, we can calculate the energy released:
E= (0.0021 u)×(931.5MeV/c2)
= 1.95915 MeV
Therefore, the energy released in the reaction is 1.95915 M eV .
