CHM 345 - PHYSICAL CHEMISTRY I - Calculation of
energy levels Practice Material - Set 1
1. Question:
Consider a particle of mass mtrapped in a 1D potential well defined by the potential energy
function V(x) = 1
2kx2for −a≤x≤a, where k > 0is a constant. Determine the energy levels
of the particle in this potential well.
Ans. Let’s solve this problem step by step:
1. The Schrödinger equation for a particle in a 1D potential well is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
where ˉhis the reduced Planck constant, mis the mass of the particle, V(x)is the potential
energy function, Eis the energy of the particle, and ψis the wave function.
2. In the region −a≤x≤a, the potential energy function is V(x) = 1
2kx2. Thus, the
Schrödinger equation in this region becomes:
−ˉh2
2m
d2ψ
dx2+1
2kx2ψ=Eψ
3. To simplify the equation, we introduce a dimensionless variable ξsuch that ξ=ax. Thus,
dx =dξ
aand d2
dx2=1
a2
d2
dξ2.
4. Substituting ξ=ax into the Schrödinger equation, we get:
−ˉh2
2ma2
d2ψ
dξ2+1
2ka2ξ2ψ=Eψ
5. Multiplying through by 2ma2
ˉh2and simplifying, we obtain:
−d2ψ
dξ2+2ma2k
ˉh2(1
2ξ2)ψ=2ma2
ˉh2Eψ
6. Define a new constant α2=2ma2k
ˉh2and a new energy parameter ϵ=2ma2
ˉh2E. Then the
equation becomes:
−d2ψ
dξ2+α2ξ2ψ=ϵψ
7. This is the dimensionless form of the Schrödinger equation for an oscillator. The energy
levels are quantized, and the solutions are known as Hermite polynomials multiplied by the
Gaussian function.
8. The energy levels for this system are given by:
En=ˉhω(n+1
2)
where ω=√k
mis the angular frequency of the oscillator, and n= 0,1,2, ... are the quantum
numbers corresponding to the energy levels.
2. Question: Determine the energy levels of a particle in a one-dimensional box of length L
where nis an odd integer.
Ans. Step-by-step solution: 1. The energy levels of a particle in a one-dimensional box are
given by the formula:
En=n2ˉh2π2
2mL2
where nis a positive integer, ˉhis the reduced Planck constant, mis the mass of the particle,
and Lis the length of the box. 2. Given that nis an odd integer, let’s substitute n= 2k+ 1
into the formula for energy levels:
E2k+1 =(2k+ 1)2ˉh2π2
2mL2
E2k+1 =(4k2+ 4k+ 1)ˉh2π2
2mL2
E2k+1 =4k2ˉh2π2+ 4kˉh2π2+ˉh2π2
2mL2
E2k+1 =4k2ˉh2π2
2mL2+4kˉh2π2
2mL2+ˉh2π2
2mL2
E2k+1 = 2k2ˉh2π2
mL2+ 2kˉh2π2
mL2+ˉh2π2
2mL2
E2k+1 = 2k2E1+ 2kE1+E1
E2k+1 = (2k2+ 2k+ 1)E1
where E1=ˉh2π2
2mL2is the energy level for n= 1. 3. Therefore, the energy levels for nodd are
given by E2k+1 = (2k2+ 2k+ 1)E1, where k= 0,1,2, ....
3. Question: Determine the energy levels of an electron in a one-dimensional box of length 5
nm.
Ans. Let’s denote the length of the box as L= 5 nm. The energy levels of an electron in a
one-dimensional box are given by the equation:
En=n2h2
8mL2
where nis the quantum number, his the Planck constant (6.626 ×10−34 J·s), mis the mass
of the electron (9.11 ×10−31 kg), and Lis the length of the box.
1. Calculate the energy of the electron in the ground state, n= 1. Plugging in the values,
we get:
E1=(1)2(6.626 ×10−34)2
8(9.11 ×10−31)(5 ×10−9)2
E1=1(4.389 ×10−67)
3.648 ×10−40
E1≈0.529 ×10−17J
2. Calculate the energy of the electron in the first excited state, n= 2. Plugging in the
values, we get:
E2=(2)2(6.626 ×10−34)2
8(9.11 ×10−31)(5 ×10−9)2
E2=4(4.389 ×10−67)
3.648 ×10−40
E2≈2.116 ×10−17J
Therefore, the energy levels of the electron in the one-dimensional box are approximately
0.529 ×10−17 J for the ground state and 2.116 ×10−17 J for the first excited state.
4. Question: Consider a particle of mass min a one-dimensional infinite potential well of width
L. Calculate the energy levels for this system.
Ans. Step-by-step solution: 1. In an infinite potential well, the potential energy is zero within
the well and infinite outside of it. Therefore, the Hamiltonian operator for the particle in the well
is given by
ˆ
H=−ˉh2
2m
d2
dx2+ 0
2. The Schrödinger equation for this system is:
ˆ
Hψ(x) = Eψ(x)
3. Inside the well, the wave function ψ(x)is described by the time-independent Schrödinger
equation:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
4. Rearranging the equation gives:
d2ψ(x)
dx2=−2mE
ˉh2ψ(x)
5. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh
6. Applying the boundary conditions implies the wave function should be zero at x= 0 and
x=L. This leads to the requirement that k=nπ/L, where nis a positive integer.
7. Therefore, the allowed values of energy are quantized as:
En=n2π2ˉh2
2mL2
where nis a positive integer.
8. Thus, the energy levels for the particle in a one-dimensional infinite potential well of width
Lare given by the expression above.
5. Question: Consider a particle confined to a one-dimensional box of length L. Calculate the
energy levels of the particle using quantum mechanics.
Ans. Let’s denote the energy levels of the particle as En, where nis a positive integer
representing the quantum number.
1. Setting up the problem: The energy of the particle in a one-dimensional box is given
by the equation:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck constant, mis the mass of the particle, and Lis the length of the
box.
2. Finding the energy levels: Substitute the given values into the equation:
En=n2π2ˉh2
2mL2
3. Simplifying the equation: Since ˉh=h
2π, where his the Planck constant, we can rewrite
the equation as:
En=n2h2
8mL2
Therefore, the energy levels of the particle confined to a one-dimensional box of length Lare
given by:
En=n2h2
8mL2
6. Question: Consider an electron in a one-dimensional infinite potential well of width L. Find
the energy levels of the electron in terms of ˉh,m, and L.
Ans. Let’s denote the energy levels of the electron in the potential well as En, where nis a
positive integer representing the quantum number.
1. The allowed energies of the electron in the infinite potential well are given by the equation:
En=n2π2ˉh2
2mL2,
where mis the mass of the electron and ˉhis the reduced Planck constant.
2. Substituting the given values for mand L, we can rewrite the equation as:
En=n2π2ˉh2
2mL2=n2π2ˉh2
2·(9.11 ×10−31 kg)·L2.
3. Therefore, the energy levels of the electron in the one-dimensional infinite potential well
are given by:
En=n2π2ˉh2
2·(9.11 ×10−31 kg)·L2.
7. Question 7: An electron is in a one-dimensional box with a length of 2 nm. Determine the
energy of the electron in the first excited state.
Ans. Let’s denote the length of the box as L= 2 nm. The energy levels of an electron in a
one-dimensional box are given by the formula:
En=n2π2ˉh2
2mL2
where nis the quantum number, ˉhis the reduced Planck’s constant (1.0545718 ×10−34 J·s),
and mis the mass of the electron (9.11 ×10−31 kg).
1. Determine the quantum number for the first excited state: For the first excited
state, n= 2.
2. Calculate the energy of the electron in the first excited state: Plugging in the
values into the formula, we have:
E2=(2)2π2(1.0545718 ×10−34)2
2(9.11 ×10−31)(2 ×10−9)2
E2=4×π2×(1.0545718 ×10−34)2
2×9.11 ×(2 ×10−9)2
E2=4×(π)2×(1.113 ×10−68)
3.6464 ×10−28
E2=4.396 ×10−68
3.6464 ×10−28
E2= 1.204 ×10−40 J
Therefore, the energy of the electron in the first excited state is 1.204 ×10−40 J.
8. Question: Find the energy levels of a particle in a one-dimensional infinite square well of
width L.
Ans. Let’s denote the energy levels as Enwhere nis a positive integer. The energy levels are
given by the formula:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck’s constant, and mis the mass of the particle.
Step 1. Start with the time-independent Schrödinger equation for a particle in a one-
dimensional infinite square well:
ˆ
Hψ(x) = Eψ(x)
Step 2. The Hamiltonian operator ˆ
Hfor a particle with no potential energy in the well is:
ˆ
H=−ˉh2
2m
d2
dx2
Step 3. Substitute the Hamiltonian operator into the Schrödinger equation:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
Step 4. Rearrange the equation to separate variables:
d2ψ(x)
dx2=−2mE
ˉh2ψ(x)
Step 5. The general solution to this differential equation is a linear combination of sine and
cosine functions:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
Step 6. Apply the boundary conditions ψ(0) = 0 and ψ(L)=0for a particle in an infinite
square well. This leads to the quantization condition for k:
kn=nπ
L
Step 7. Substitute the quantized values of knback into the energy formula:
En=ˉh2k2
n
2m=n2π2ˉh2
2mL2
Therefore, the energy levels of a particle in a one-dimensional infinite square well are given
by En=n2π2ˉh2
2mL2, where nis a positive integer.
9. Question: Consider a particle of mass min a one-dimensional infinite potential well of width
L. Calculate the energy levels of the particle in terms of Planck’s constant h, the mass m, and
the width of the well L.
Ans. Let’s denote the energy levels of the particle in the potential well as En, where nis a
positive integer corresponding to the quantum number. We can find the energy levels by solving
the time-independent Schrödinger equation for the particle in the potential well and applying the
boundary conditions.
1. Set up the Schrödinger equation: The time-independent Schrödinger equation for the
particle in the one-dimensional infinite potential well is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
where ψis the wave function of the particle, Eis the energy of the particle, ˉhis the reduced
Planck’s constant, mis the mass of the particle, and xis the position coordinate.
2. Solve the Schrödinger equation in the well: Inside the well, the potential energy is
zero. Thus, the Schrödinger equation simplifies to:
−ˉh2
2m
d2ψ
dx2=Eψ
3. Apply the boundary conditions: The wave function ψ(x)must be zero at both ends
of the well, i.e., ψ(0) = 0 and ψ(L) = 0. This implies that the solutions to the Schrödinger
equation are of the form:
ψn(x) = √2
Lsin (nπx
L)
where nis a positive integer.
4. Find the energy levels: Substitute the wave function ψn(x)into the Schrödinger
equation to solve for the energy levels En. Using the energy operator ˆ
H=−ˉh2
2m
d2
dx2, we have:
ˆ
Hψn(x) = Enψn(x)
En=n2π2ˉh2
2mL2
Therefore, the energy levels of the particle in the one-dimensional infinite potential well are
given by En=n2π2ˉh2
2mL2, where nis a positive integer corresponding to the quantum number.
10. Question: Let’s consider a particle trapped in a one-dimensional potential well defined by
the potential energy function:
V(x) = {0if 0≤x≤a
∞otherwise
where ais a positive constant.
Calculate the energy levels of the particle in this potential well.
Ans. To calculate the energy levels of a particle trapped in the potential well defined by the
given potential energy function, we need to solve the time-independent Schrödinger equation for
the system and apply the appropriate boundary conditions.
1. The time-independent Schrödinger equation is given by:
−ˉh2
2m
d2ψ(x)
dx2+V(x)ψ(x) = Eψ(x)
where ˉhis the reduced Planck constant, mis the mass of the particle, Eis the total energy of
the particle, and ψ(x)is the wave function.
2. Since the potential energy is infinite outside the region 0≤x≤a, the wave function must
be zero outside this region. Therefore, the wave function ψ(x)is non-zero only in the region
0≤x≤a.
3. In the region 0≤x≤a, the Schrödinger equation simplifies to:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
This is the time-independent Schrödinger equation for a particle inside a potential well.
4. Let’s further simplify the equation by dividing through by −ˉh2
2m:
d2ψ(x)
dx2=−2mE
ˉh2ψ(x)
5. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh, and Aand Bare constants to be determined.
6. Apply the boundary conditions: ψ(0) = 0 and ψ(a) = 0.
7. From ψ(0) = 0, we have:
B= 0
Therefore, the wave function becomes:
ψ(x) = Asin(kx)
8. From ψ(a) = 0, we have:
Asin(ka) = 0
Since sin(ka)cannot be zero for all k(except for k= 0), we must have:
ka =nπ
where nis a positive integer.
9. This gives us the quantized values for E:
En=n2π2ˉh2
2ma2
where n= 1,2,3, ... are the allowed energy levels of the particle in the potential well.
Therefore, the energy levels of the particle in this one-dimensional potential well are quantized
with values given by En=n2π2ˉh2
2ma2, where n= 1,2,3, ....
11. Question: Determine the energy levels for a particle trapped in a one-dimensional infinite
potential well of width L.
Ans. Let Endenote the nth energy level of the particle in the infinite potential well. The energy
levels are given by the formula:
En=n2π2ˉh2
2mL2,
where n= 1,2,3, . . . corresponds to the quantum number representing different energy levels.
Solution: 1. The Schrödinger equation for the wave function inside the well is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x).
2. Since the potential is infinite at 0and L, the wave function must be zero at these points.
Therefore, the wave function ψ(x)can be written as:
ψ(x) = Asin (nπx
L),
where Ais the normalization constant.
3. Applying the boundary conditions ψ(0) = ψ(L) = 0, we get:
ψ(0) = Asin(0) = 0 =⇒A= 0,
and
ψ(L) = Asin(nπ) = 0 =⇒nπ =mπ,
where mis an integer (but not zero).
4. The energy levels Enare then given by:
En=n2π2ˉh2
2mL2.
Therefore, the particle in the infinite potential well has discrete energy levels determined by
the quantum number n.
12. Question 12:
Consider a particle in a one-dimensional infinite square well potential with width L.
(a) Calculate the ground state energy of the particle in terms of Land the particle’s mass m.
(b) Find the energy of the first excited state in terms of Land m.
Ans. (a)
1. The ground state energy of a particle in a one-dimensional infinite square well potential is
given by the formula:
E1=π2ˉh2
2mL2
where mis the mass of the particle and Lis the width of the potential well.
(b)
1. The energy of the nth excited state in a one-dimensional infinite square well potential is
given by:
En=n2π2ˉh2
2mL2
where nis the quantum number of the state.
2. For the first excited state, n= 2, so the energy is:
E2=4π2ˉh2
2mL2=2π2ˉh2
mL2
13. Question: Determine the energy levels of a particle confined to a one-dimensional box of
length L. The particle has a mass mand the potential energy inside the box is zero.
Ans. To find the energy levels of a particle confined to a one-dimensional box, we can use
the time-independent Schrödinger equation. The general form of the Schrödinger equation for a
particle in one dimension is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
where Eis the total energy of the particle, ψis the wave function, ˉhis the reduced Planck
constant, and mis the mass of the particle.
1. Define the potential energy function: Since the potential energy inside the box is
zero, the potential energy function can be represented as V(x) = 0.
2. Set up the Schrödinger equation: Substitute V(x) = 0 into the Schrödinger equation
to obtain:
−ˉh2
2m
d2ψ
dx2=Eψ
3. Solve the differential equation: Let’s solve this differential equation to find the possible
energy levels. The general solution to the differential equation is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where Aand Bare constants to be determined, and k=√2mE
ˉh2.
4. Apply the boundary conditions: The wave function must satisfy the following boundary
conditions for a particle in a box of length L:
a) ψ(0) = 0 b) ψ(L) = 0
Applying the boundary condition ψ(0) = 0 yields:
0 = B
So, the wave function becomes:
ψ(x) = Asin(kx)
Applying the boundary condition ψ(L) = 0, we get:
0 = Asin(kL)
5. Quantization condition: For the wave function to satisfy the boundary condition ψ(L) =
0, the argument of the sine function must be an integer multiple of π:
kL =nπ
where nis a positive integer.
6. Find the energy levels: Substitute the expression for kinto the quantization condition
and solve for Eto find the energy levels:
√2mE
ˉh2·L=nπ
En=n2π2ˉh2
2mL2
Therefore, the energy levels of the particle confined to a one-dimensional box are given by
En=n2π2ˉh2
2mL2, where nis a positive integer.
14. Question: Determine the energy levels of an electron confined in a one-dimensional box of
length L= 2 nm. The mass of the electron is m= 9.11 ×10−31 kg and Planck’s constant is
h= 6.63 ×10−34 J·s.
Ans. Let’s denote the energy levels as En, where nis a positive integer representing the
quantum number.
1. The energy levels of an electron in a one-dimensional box are given by the formula:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck’s constant, ˉh=h
2π.
2. Substituting the given values, we have:
En=n2π2(h
2π)2
2mL2
En=h2n2
8mL2
3. Plugging in the values for h,m, and L, we get:
En=(6.63 ×10−34 J·s)2n2
8×9.11 ×10−31 kg ×(2 ×10−9m)2
4. Simplifying the expression:
En=43.9×10−68 J·s2n2
7.28 ×10−59 kg ·m2
En= 6.04 ×10−9J·n2
Therefore, the energy levels of the electron confined in the one-dimensional box are given by
En= 6.04 ×10−9J·n2.
15. Question: Determine the energy levels of an electron in a one-dimensional harmonic
oscillator potential given by V(x) = 1
2kx2, where kis the force constant.
Ans. Let’s solve this step-by-step:
1. The Schrödinger equation for a one-dimensional harmonic oscillator potential is given by:
−ˉh2
2m
d2ψ(x)
dx2+1
2kx2ψ(x) = Eψ(x)
where ˉhis the reduced Planck constant, mis the mass of the electron, Eis the energy of the
electron, and ψ(x)is the wave function.
2. We can simplify the Schrödinger equation by making the following substitutions:
ξ=√mω
ˉhxand λ=2E
ˉhω
where ω=√k
mis the angular frequency.
3. With these substitutions, the Schrödinger equation becomes:
−d2ψ(ξ)
dξ2+ξ2ψ(ξ) = λψ(ξ)
4. We now apply the ladder operators aand a†defined as:
a=1
√2(ξ+d
dξ )and a†=1
√2(ξ−d
dξ )
5. Using these ladder operators, we can rewrite the Schrödinger equation as:
aa†ψ(ξ) = (λ+ 1)ψ(ξ)
6. We see that aa†=1
2(ξ2−d2
dξ2−1), hence the eigenvalues of aa†are 1
2(2n+ 1), where n
is a non-negative integer.
7. Therefore, the energy levels are given by:
En=ˉhω (n+1
2)
where ntakes values 0,1,2, ....
16. Question: Determine the first four energy levels of a particle in a one-dimensional box of
length L.
Ans. Let’s begin by using the formula for the energy levels of a particle in a one-dimensional
box:
1. The formula for the energy levels of a particle in a one-dimensional box is given by:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number, ˉhis the reduced Planck’s
constant, mis the mass of the particle, and Lis the length of the box.
2. For the first energy level (n= 1), the energy is:
E1=(1)2π2ˉh2
2mL2=π2ˉh2
2mL2
3. For the second energy level (n= 2), the energy is:
E2=(2)2π2ˉh2
2mL2=4π2ˉh2
2mL2= 2 π2ˉh2
mL2
4. For the third energy level (n= 3), the energy is:
E3=(3)2π2ˉh2
2mL2=9π2ˉh2
2mL2=9
2
π2ˉh2
mL2
5. For the fourth energy level (n= 4), the energy is:
E4=(4)2π2ˉh2
2mL2=16π2ˉh2
2mL2= 8 π2ˉh2
mL2
Therefore, the first four energy levels of a particle in a one-dimensional box of length Lare:
E1=π2ˉh2
2mL2, E2= 2 π2ˉh2
mL2, E3=9
2
π2ˉh2
mL2, E4= 8 π2ˉh2
mL2
17. Let’s consider an electron confined in a one-dimensional infinite potential well of width L.
The potential energy inside the well is zero, while the potential energy outside the well is infinite.
Determine the possible energy levels for the electron in this system.
Ans. To solve the problem, we need to consider the Schrödinger equation for the infinite
potential well and apply the boundary conditions.
1. Set up the Schrödinger equation: Inside the well, the potential energy V(x)is zero,
so the Schrödinger equation in one dimension is:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
2. Solve the Schrödinger equation: The general solution to the Schrödinger equation in
this case is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
3. Apply the boundary conditions: Since the potential energy outside the well is infi-
nite, the wave function must vanish at the boundaries, i.e., ψ(0) = ψ(L)=0. This leads to
quantization of the energy levels:
kn=nπ
L
where n= 1,2,3, . . ..
4. Calculate the energy levels: Substitute the quantized values of knback into the
expression for energy:
En=ˉh2k2
n
2m=n2π2ˉh2
2mL2
Thus, the energy levels are quantized and given by:
En=n2π2ˉh2
2mL2where n= 1,2,3, . . .
These discrete energy levels correspond to the allowed energy states of the electron in the infinite
potential well.
18. Question: Consider a particle of mass mmoving in a one-dimensional box of length L.
Determine the energy levels of the particle in terms of the quantum number n.
Ans. Step-by-step solution: 1. The energy of the particle in the box is given by the formula:
En=n2π2ˉ
h2
2mL2
where nis the quantum number, ˉhis the reduced Planck’s constant, and Lis the length of the
box. 2. Substituting the given values into the formula, we get:
En=n2π2ˉh2
2mL2
3. Therefore, the energy levels of the particle in the box are quantized, with En=π2ˉh2
2mL2,4π2ˉh2
2mL2,9π2ˉh2
2mL2, ...
for n= 1,2,3, ....
19. Question: Consider an electron in a one-dimensional box of length L. At t= 0, the electron
is in the ground state of the box. If an external uniform electric field Eis suddenly turned on
along the length of the box at t= 0, find the energy of the electron at t=L2
2πeEˉh.
Ans. Let’s first determine the initial ground state of the electron in the box and then find its
energy at t=L2
2πeEˉhafter turning on the external electric field E.
1. Initial Ground State Energy: The energy levels of an electron in a one-dimensional box
of length Lare given by En=n2π2ˉh2
2mL2, where n= 1,2,3, ... represents the quantum number.
The ground state corresponds to n= 1, so the initial ground state energy E1is:
E1=π2ˉh2
2mL2
2. Energy at t=L22πeEˉh: The acceleration of the electron due to the electric field is
a=eE
m, where −eis the charge of an electron and mis the mass of the electron.
The electron’s position at time tunder constant acceleration starting from rest is given by
x=1
2at2. Substituting aand t=L2
2πeEˉh, we get:
x=1
2·eE
m·(L2
2πeEˉh)2
The final energy of the electron is given by the energy level formula with the length of the
box now equal to the distance traveled by the electron:
Ef=n2π2ˉh2
2m(x)
Substitute n= 1 and x=eEL2
4π2m2e2ˉhinto the equation to find the energy of the electron at
t=L2
2πeEˉh.
Ef=π2ˉh2m
2(eEL2
4π2m2e2ˉh)=2π2ˉh2eEˉh
mL2
Therefore, the energy of the electron at t=L2
2πeEˉhis Ef=2π2ˉh2eEˉh
mL2.
20. Question: Find the energy levels of a particle in a one-dimensional box of length Lwith
infinitely high walls if the particle has mass mand the potential energy inside the box is given by
V(x) = λx4, where λis a positive constant.
Ans. Let’s denote the energy levels of the particle as Enfor n= 1,2,3, . . .. We know that
the Schrödinger equation for a particle in a one-dimensional box with a potential energy function
V(x)is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
We are given that V(x) = λx4. To solve for the energy levels, we need to find the eigenvalues
Enfor which there exist nontrivial solutions to the Schrödinger equation.
1. We start by writing down the Schrödinger equation with the given potential energy
function:
−ˉh2
2m
d2ψ
dx2+λx4ψ=Eψ
2. To simplify the equation, let’s substitute ψ(x) = Aeikx into the equation, where Ais a
constant to be determined and kis the wave number.
3. Substitute ψ(x)into the Schrödinger equation and simplify to get a differential equation
for k:
−ˉh2
2m(−k2)eikx +λx4Aeikx =EAeikx
4. Divide the equation by Aeikx to simplify:
ˉh2k2
2m+λx4=E
5. This is a separable equation, so let’s separate the variables xand E:
ˉh2k2
2m=E−λx4
6. Now, let’s solve the separated differential equation by assuming En=αnλ1/4 and k=βn,
where αnand βnare constants.
7. Substitute these into the separated equation and solve for αnand βnto find the energy
levels En.
21. Question 21: Consider a particle of mass min a one-dimensional potential well given by
V(x) = 1
2kx2for −a≤x≤a, where kis a positive constant. Determine the allowed energy
levels of the particle in this potential well.
Ans. To determine the allowed energy levels of the particle in the potential well, we need to
solve the time-independent Schrödinger equation for the given potential and apply appropriate
boundary conditions.
1. Write the time-independent Schrödinger equation: The time-independent Schrödinger
equation for a one-dimensional potential is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
Substitute V(x) = 1
2kx2into the equation.
2. Solve the Schrödinger equation: The Schrödinger equation becomes:
−ˉh2
2m
d2ψ
dx2+1
2kx2ψ=Eψ
Simplify the equation and rearrange it in a standard form.
3. Apply the boundary conditions: The boundary conditions for a particle in a potential
well require that the wavefunction ψ(x)must be continuous and the first derivative dψ
dx must be
continuous at x=−aand x=a.
4. Solve for the energy levels: By solving the Schrödinger equation with the appropriate
boundary conditions, we can find the allowed energy levels of the particle in the potential well.
The energy levels will depend on the mass m, the constant k, and the width of the potential
well 2a.
22. What is the energy of a photon with a frequency of 6.0×1018 Hz?
Ans. To calculate the energy of a photon, we can use the equation:
E=hf
where: - Eis the energy of the photon, - his the Planck constant (6.63 ×10−34 J·s), and -
fis the frequency of the photon.
Step 1. Plug in the values into the equation:
E= (6.63 ×10−34 J·s)×(6.0×1018 Hz)
Step 2. Calculate the energy of the photon:
E= 3.978 ×10−15J
Therefore, the energy of a photon with a frequency of 6.0×1018 Hz is 3.978 ×10−15 J.
23. Question: Determine the energy levels of an electron in a one-dimensional box of length
L= 5 nm.
Ans. Step-by-step solution: 1. The energy levels for an electron in a one-dimensional box are
given by the equation
En=n2h2
8mL2
where nis the quantum number, his the Planck constant (6.626 ×10−34 J s), mis the mass of
the electron (9.11 ×10−31 kg), and Lis the length of the box.
2. Substituting the given values into the equation, we have
En=n2(6.626 ×10−34)2
8(9.11 ×10−31)(5 ×10−9)2
3. Simplifying the expression, we get
En=n2×4.3908 ×10−67
1.822 ×10−19
En=n2×2.41 ×10−48
4. Therefore, the energy levels for the electron in the one-dimensional box with L= 5 nm
are multiples of 2.41 ×10−48 J−.
24. Question: Consider a particle in a one-dimensional potential well of width L. The potential
energy inside the well is zero, while the potential energy outside the well is infinite. Determine
the energy levels of the particle in terms of ˉh,L, and the particle’s mass m.
Ans. Let’s denote the energy levels of the particle inside the well as En, where nis a positive
integer. We can solve for the energy levels by solving the time-independent Schrödinger equation
for the particle in this potential well.
1. The general form of the time-independent Schrödinger equation for this system is:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
This equation can be rewritten as:
d2ψ(x)
dx2+2mE
ˉh2ψ(x) = 0
2. Inside the well, the potential energy V(x) = 0, so the Schrödinger equation simplifies to:
d2ψ(x)
dx2+2mE
ˉh2ψ(x) = 0
3. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
4. Since the potential energy outside the well is infinite, the wavefunction ψ(x)must go to
zero at x= 0 and x=L. Hence, ψ(0) = ψ(L) = 0. This leads to the boundary conditions:
B= 0
(since cos(0) = 1)
and
Asin(kL) = 0
5. The boundary condition Asin(kL) = 0 implies that kL =nπ for nbeing a positive
integer. Therefore, we have:
k=nπ
L
6. Substituting k=nπ
Lback into the expression k2=2mEn
ˉh2yields the energy levels En:
En=ˉh2π2n2
2mL2
Therefore, the energy levels of the particle in the potential well are quantized and given by
En=ˉh2π2n2
2mL2, where nis a positive integer.
25. Question: A particle with mass mis confined within a one-dimensional potential well
defined by V(x) = 1
2kx2for |x|< a, where kis a positive constant and ais the width of the
well. Calculate the allowed energy levels of the particle in terms of m,k, and a.
Ans. Let’s denote the allowed energy levels of the particle by En, where n= 1,2,3, . . .. The
energy levels are given by the equation:
En=n2π2ˉh2
2ma2
where ˉhis the reduced Planck constant.
26. Question 26: Consider a particle in a one-dimensional infinite potential well of width a.
This particle can occupy states labeled by n= 1,2,3, ... with corresponding energy levels given
by En=n2π2ˉh2
2ma2, where ˉhis the reduced Planck constant, mis the mass of the particle, and ais
the width of the potential well.
Calculate the difference in energy levels between the n= 3 and n= 4 states in terms of ˉh,
m, and a.
Ans. To find the difference in energy levels between the n= 3 and n= 4 states, we can
subtract E4from E3:
1. Calculate the energy of the n= 3 state:
E3=32π2ˉh2
2ma2=9π2ˉh2
2ma2
2. Calculate the energy of the n= 4 state:
E4=42π2ˉh2
2ma2=16π2ˉh2
2ma2
3. Find the difference in energy levels:
∆E=E4−E3=16π2ˉh2
2ma2−9π2ˉh2
2ma2
∆E=16π2ˉh2−9π2ˉh2
2ma2=7π2ˉh2
2ma2=7
2
π2ˉh2
ma2
Therefore, the difference in energy levels between the n= 3 and n= 4 states is 7
2
π2ˉh2
ma2.
27. Question 27: A particle of mass mis in a one-dimensional box of length L. The wavefunction
of the particle is given by ψ(x) = Asin(kx), where Ais a normalization constant.
Determine the possible energy levels for this system.
Ans. To determine the possible energy levels for the system, we need to solve the time-
independent Schrödinger equation.
1. The general form of the time-independent Schrödinger equation for a one-dimensional box
is:
−ˉh2
2m
d2ψ
dx2=Eψ
where Eis the energy of the particle.
2. Given the wavefunction ψ(x) = Asin(kx), we can substitute this into the Schrödinger
equation:
−ˉh2
2mAk2sin(kx) = EA sin(kx)
3. Simplifying the equation by dividing by Asin(kx), we get:
−ˉh2k2
2m=E
4. Solving for E, we find that the energy levels are quantized and given by:
En=ˉh2k2
2m=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number.
Therefore, the possible energy levels for this system are:
En=n2π2ˉh2
2mL2
28. **Question:** Find the energy eigenvalues and corresponding eigenfunctions of the particle
in a one-dimensional box of length L, when a potential barrier of height V0and width ais inserted
at the center of the box.
**Step-by-step Solution:** Let’s denote the total energy of the system as E. Inside the
barrier (0< x < a or L−a<x<L), the potential energy is V0and outside the barrier
(a<x<L−a), the potential energy is 0.
1. **Finding the wave function inside the barrier:** The wave function inside the barrier can
be represented as:
ψ(x) = {Aeikx +Be−ikx for 0< x < a
Ceikx +De−ikx for L−a<x<L
where k=√2mE
ˉh2.
2. **Applying boundary conditions at x= 0 and x=a:** At x= 0, the wave function and
its derivative must be continuous:
ψ(0) = ψ(a)and dψ(0)
dx =dψ(a)
dx
3. **Applying boundary conditions at x=aand x=L−a:** Similarly, at x=aand
x=L−a, the wave function and its derivative must be continuous:
ψ(a) = ψ(L−a)and dψ(a)
dx =dψ(L−a)
dx
4. **Solving the system of equations:** Solving the system of equations obtained from the
boundary conditions will give the values of constants A,B,C, and D. This, in turn, will provide
the energy eigenvalues and eigenfunctions of the system.
Therefore, the energy eigenvalues and corresponding eigenfunctions of the particle in a one-
dimensional box with a potential barrier at the center can be determined by solving the Schrödinger
equation with appropriate boundary conditions.
29. Let’s consider an electron confined to a one-dimensional box with a length of L. The
potential energy V(x)inside the box is zero, but infinite outside the box. We are asked to find
the expression for the energy levels of this system.
30. A quantum mechanical particle is confined to a one-dimensional box of length L. Calculate
the energy levels for this system.
Ans. To find the energy levels of the particle in a one-dimensional box, we can start by solving
the time-independent Schrödinger equation for this system.
1. Set up the Schrödinger equation: The time-independent Schrödinger equation for the
particle in a one-dimensional box is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)for 0< x < L
with the boundary conditions ψ(0) = 0 and ψ(L) = 0.
2. Solve the differential equation: The general solution to the differential equation is of
the form:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
Applying the boundary conditions:
ψ(0) = 0 =⇒B= 0
ψ(L) = 0 =⇒Asin(kL) = 0
3. Determine the quantized values of k:From the boundary condition sin(kL) = 0, we
get kL =nπ, where nis a positive integer.
So, the allowed values of kare:
kn=nπ
L
4. Calculate the energy levels: The energy levels are given by:
En=ˉh2k2
n
2m=ˉh2π2n2
2mL2
Therefore, the energy levels for the particle in a one-dimensional box are quantized and given
by En=ˉh2π2n2
2mL2where n= 1,2,3, ....
31. Question 30: An electron is confined to move in one dimension in an infinite potential well
of width L. Determine the energy of the electron in the n= 4 energy level.
Ans. Let’s start by determining the general formula for the energy levels of an electron in an
infinite potential well. The energy levels for a particle in a 1D infinite potential well are given by:
En=n2π2ˉh2
2mL2
where nis the quantum number, ˉhis the reduced Planck’s constant, mis the mass of the
electron, and Lis the width of the well.
1. Substitute n= 4 into the formula to find the energy of the electron in the n= 4 energy
level:
E4=42π2ˉh2
2mL2
E4=16π2ˉh2
2mL2
32. Question 31: An electron in a one-dimensional quantum well with a width of 3 nm has an
energy level of -2.5 eV. Calculate the energy of the next higher energy level in the well.
Ans. To solve this problem, we can use the equation for the energy levels in a one-dimensional
quantum well:
En=n2π2ˉh2
2mL2
Where: - Enis the energy level, - nis the quantum number of the energy level, - ˉhis the
reduced Planck’s constant (1.0545718 ×10−34 Js), - mis the mass of the electron (9.11 ×10−31
kg), - Lis the width of the quantum well.
1. Calculate the energy of the next higher energy level:
Given: - E1=−2.5eV, - L= 3 nm.
First, let’s convert the values to SI units: - E1=−2.5×1.6×10−19 J (1 eV = 1.6×10−19
J), - L= 3 ×10−9m.
Now, we can solve for E2using the equation:
E2=(2)2π2ˉh2
2mL2
E2=(2)2π2·(1.0545718 ×10−34)2
2·9.11 ×10−31 ·(3 ×10−9)2
E2=4·π2·(1.0545718 ×10−34)2
2·9.11 ×10−31 ·9×10−18
E2≈ −1.107 ×10−19 J
Finally, convert the energy back to electron volts:
E2≈−1.107 ×10−19
1.6×10−19 ≈ −0.692 eV
Therefore, the energy of the next higher energy level in the quantum well is approximately
-0.692 eV.
33. Question 32: Consider a quantum mechanical system with a Hamiltonian operator given by
ˆ
H=−ˉh2
2m
d2
dx2+V(x), where V(x)is a potential function. Suppose that the potential function
is described by V(x) = 1
2mω2x2+λ
4x4, where ω, λ > 0.
Find the expression for the energy eigenvalues Enof this system.
Ans. Let us consider the time-independent Schrödinger equation for this system: ˆ
Hψ(x) =
Eψ(x), where ψ(x)is the wave function and Eis the energy eigenvalue.
1. Substituting the given Hamiltonian operator into the Schrödinger equation, we have:
−ˉh2
2m
d2ψ
dx2+(1
2mω2x2+λ
4x4)ψ=Eψ.
2. Rearranging the terms and defining a new parameter α=mω
ˉh, the equation becomes:
−ˉh2
2m
d2ψ
dx2+1
2mω2x2ψ+λ
4x4ψ−Eψ = 0.
3. Dividing through by ˉhω to simplify, and defining the dimensionless variable ξ=√mω
ˉhx,
the equation transforms into:
−1
2
d2ψ
dξ2+1
2ξ2ψ+λ
4ˉhω x4ψ−E
ˉhω ψ= 0.
4. Writing the equation in dimensionless form, we have the new Hamiltonian operator ˆ
H=
−1
2
d2
dξ2+1
2ξ2+λ
4ˉhω ξ4.
5. We can solve this dimensionless form of the Schrödinger equation to find the energy
eigenvalues Enand corresponding wave functions ψn(ξ).
6. The energy eigenvalues for this system are given by:
En=(n+1
2)ˉhω, n = 0,1,2, ...
Therefore, the expression for the energy eigenvalues Enof this quantum mechanical system
with the specified potential function is En=(n+1
2)ˉhω.
34. Question 33:
A particle of mass mis confined in a one-dimensional infinite potential well of width L.
Calculate the energy levels of the particle in this potential well.
Ans. To calculate the energy levels of the particle in the one-dimensional infinite potential well,
we can use the time-independent Schrödinger equation and the boundary conditions.
1. Write down the time-independent Schrödinger equation:
The time-independent Schrödinger equation for a particle of mass min one dimension po-
tential well is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
2. Write down the boundary conditions: The boundary conditions for an infinite potential
well of width Lare that the wave function ψ(x)must be continuous and the derivative of ψ(x)
must be discontinuous at x= 0 and x=L, since the potential is infinite at these points.
3. Solve the differential equation: The general solution to the differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
Applying the boundary condition that ψ(0) = ψ(L) = 0 gives:
ψ(0) = B= 0
ψ(L) = Asin(kL) = 0
This implies that kL =nπ for n= 1,2,3, .... Therefore, the energy levels Enare given by:
En=n2π2ˉh2
2mL2for n= 1,2,3, ...
So, the energy levels of the particle in the one-dimensional infinite potential well are quantized
and given by the formula above.
35. Find the energy levels of a particle confined to a one-dimensional box of length L, where
the potential energy inside the box is zero and infinitely large outside the box.
Ans. To find the energy levels of a particle in a one-dimensional box, we need to solve the
time-independent Schrödinger equation and apply the boundary conditions. The general form of
the wave function for a particle in a box of length Lis ψ(x) = Asin(kx) + Bcos(kx), where
k=nπ
Lfor positive integer values of n.
1. Applying the boundary conditions: Inside the box where the potential energy is zero,
the wave function should be continuous. So, at x= 0 and x=L, the wave function should be
continuous. Therefore, ψ(0) = ψ(L) = 0.
2. Applying the boundary condition at x= 0:Plugging in x= 0 into the wave function
ψ(x)gives: ψ(0) = Asin(0) + Bcos(0) = B= 0, since sin(0) = 0 and cos(0) = 1.
So, B= 0.
3. Applying the boundary condition at x=L:Plugging in x=Linto the wave function
ψ(x)gives: ψ(L) = Asin(kL) = 0.
For the wave function to be nonzero, sin(kL)must be zero, which means kL =nπ for some
positive integer n.
4. Solving for the energy levels: Since k=nπ
L, the energy levels are given by the relation
En=ˉh2k2
2m=n2π2ˉh2
2mL2.
Therefore, the energy levels of the particle in the one-dimensional box are quantized and given
by En=n2π2ˉh2
2mL2, where nis a positive integer.
36. Question 35:
Consider a particle in a one-dimensional box of length L. Determine the energy levels for the
particle confined within the box.
Ans. To find the energy levels of a particle in a one-dimensional box, we can use the Schrödinger
equation and apply the boundary conditions. The Schrödinger equation for a particle in a box is:
−ˉh2
2m
d2ψ
dx2=Eψ
where mis the mass of the particle, Eis the energy, ψis the wave function, and ˉhis the
reduced Planck’s constant.
1. Setting up the Schrödinger equation: Since the particle is confined in a one-
dimensional box of length L, the wave function ψshould satisfy the boundary conditions that
ψ(0) = ψ(L) = 0. Thus, the general form of the wave function is:
ψ(x) = Asin(kx)
where kis the wave number, and Ais a normalization constant.
2. Find the wave number k:We can find the wave number kby using the boundary
conditions:
ψ(0) = Asin(0) = 0 =⇒A= 0
ψ(L) = Asin(kL) = 0 =⇒kL =nπ
where nis a positive integer.
3. Expressing the energy levels in terms of n:Once we have the possible values of k,
we can express the energy levels:
kL =nπ =⇒k=nπ
L
The energy levels are given by:
En=ˉh2k2
2m=ˉh2(nπ
L)2
2m=n2π2ˉh2
2mL2
Therefore, the energy levels for a particle in a one-dimensional box of length Lare given by
En=n2π2ˉh2
2mL2, where nis a positive integer.
3. Question: Determine the energy levels of an electron in a one-dimensional box of length 5
nm.
Ans. Let’s denote the length of the box as L= 5 nm. The energy levels of an electron in a
one-dimensional box are given by the equation:
En=n2h2
8mL2
where nis the quantum number, his the Planck constant (6.626 ×10−34 J·s), mis the mass
of the electron (9.11 ×10−31 kg), and Lis the length of the box.
1. Calculate the energy of the electron in the ground state, n= 1. Plugging in the values,
we get:
E1=(1)2(6.626 ×10−34)2
8(9.11 ×10−31)(5 ×10−9)2
E1=1(4.389 ×10−67)
3.648 ×10−40
E1≈0.529 ×10−17J
2. Calculate the energy of the electron in the first excited state, n= 2. Plugging in the
values, we get:
E2=(2)2(6.626 ×10−34)2
8(9.11 ×10−31)(5 ×10−9)2
E2=4(4.389 ×10−67)
3.648 ×10−40
E2≈2.116 ×10−17J
Therefore, the energy levels of the electron in the one-dimensional box are approximately
0.529 ×10−17 J for the ground state and 2.116 ×10−17 J for the first excited state.
4. Question: Consider a particle of mass min a one-dimensional infinite potential well of width
L. Calculate the energy levels for this system.
Ans. Step-by-step solution: 1. In an infinite potential well, the potential energy is zero within
the well and infinite outside of it. Therefore, the Hamiltonian operator for the particle in the well
is given by
ˆ
H=−ˉh2
2m
d2
dx2+ 0
2. The Schrödinger equation for this system is:
ˆ
Hψ(x) = Eψ(x)
3. Inside the well, the wave function ψ(x)is described by the time-independent Schrödinger
equation:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
4. Rearranging the equation gives:
d2ψ(x)
dx2=−2mE
ˉh2ψ(x)
5. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh
6. Applying the boundary conditions implies the wave function should be zero at x= 0 and
x=L. This leads to the requirement that k=nπ/L, where nis a positive integer.
7. Therefore, the allowed values of energy are quantized as:
En=n2π2ˉh2
2mL2
where nis a positive integer.
8. Thus, the energy levels for the particle in a one-dimensional infinite potential well of width
Lare given by the expression above.
5. Question: Consider a particle confined to a one-dimensional box of length L. Calculate the
energy levels of the particle using quantum mechanics.
Ans. Let’s denote the energy levels of the particle as En, where nis a positive integer
representing the quantum number.
1. Setting up the problem: The energy of the particle in a one-dimensional box is given
by the equation:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck constant, mis the mass of the particle, and Lis the length of the
box.
2. Finding the energy levels: Substitute the given values into the equation:
En=n2π2ˉh2
2mL2
3. Simplifying the equation: Since ˉh=h
2π, where his the Planck constant, we can rewrite
the equation as:
En=n2h2
8mL2
Therefore, the energy levels of the particle confined to a one-dimensional box of length Lare
given by:
En=n2h2
8mL2
6. Question: Consider an electron in a one-dimensional infinite potential well of width L. Find
the energy levels of the electron in terms of ˉh,m, and L.
Ans. Let’s denote the energy levels of the electron in the potential well as En, where nis a
positive integer representing the quantum number.
1. The allowed energies of the electron in the infinite potential well are given by the equation:
En=n2π2ˉh2
2mL2,
where mis the mass of the electron and ˉhis the reduced Planck constant.
2. Substituting the given values for mand L, we can rewrite the equation as:
En=n2π2ˉh2
2mL2=n2π2ˉh2
2·(9.11 ×10−31 kg)·L2.
3. Therefore, the energy levels of the electron in the one-dimensional infinite potential well
are given by:
En=n2π2ˉh2
2·(9.11 ×10−31 kg)·L2.
7. Question 7: An electron is in a one-dimensional box with a length of 2 nm. Determine the
energy of the electron in the first excited state.
Ans. Let’s denote the length of the box as L= 2 nm. The energy levels of an electron in a
one-dimensional box are given by the formula:
En=n2π2ˉh2
2mL2
where nis the quantum number, ˉhis the reduced Planck’s constant (1.0545718 ×10−34 J·s),
and mis the mass of the electron (9.11 ×10−31 kg).
1. Determine the quantum number for the first excited state: For the first excited
state, n= 2.
2. Calculate the energy of the electron in the first excited state: Plugging in the
values into the formula, we have:
E2=(2)2π2(1.0545718 ×10−34)2
2(9.11 ×10−31)(2 ×10−9)2
E2=4×π2×(1.0545718 ×10−34)2
2×9.11 ×(2 ×10−9)2
E2=4×(π)2×(1.113 ×10−68)
3.6464 ×10−28
E2=4.396 ×10−68
3.6464 ×10−28
E2= 1.204 ×10−40 J
Therefore, the energy of the electron in the first excited state is 1.204 ×10−40 J.
8. Question: Find the energy levels of a particle in a one-dimensional infinite square well of
width L.
Ans. Let’s denote the energy levels as Enwhere nis a positive integer. The energy levels are
given by the formula:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck’s constant, and mis the mass of the particle.
Step 1. Start with the time-independent Schrödinger equation for a particle in a one-
dimensional infinite square well:
ˆ
Hψ(x) = Eψ(x)
Step 2. The Hamiltonian operator ˆ
Hfor a particle with no potential energy in the well is:
ˆ
H=−ˉh2
2m
d2
dx2
Step 3. Substitute the Hamiltonian operator into the Schrödinger equation:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
Step 4. Rearrange the equation to separate variables:
d2ψ(x)
dx2=−2mE
ˉh2ψ(x)
Step 5. The general solution to this differential equation is a linear combination of sine and
cosine functions:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
Step 6. Apply the boundary conditions ψ(0) = 0 and ψ(L)=0for a particle in an infinite
square well. This leads to the quantization condition for k:
kn=nπ
L
Step 7. Substitute the quantized values of knback into the energy formula:
En=ˉh2k2
n
2m=n2π2ˉh2
2mL2
Therefore, the energy levels of a particle in a one-dimensional infinite square well are given
by En=n2π2ˉh2
2mL2, where nis a positive integer.
9. Question: Consider a particle of mass min a one-dimensional infinite potential well of width
L. Calculate the energy levels of the particle in terms of Planck’s constant h, the mass m, and
the width of the well L.
Ans. Let’s denote the energy levels of the particle in the potential well as En, where nis a
positive integer corresponding to the quantum number. We can find the energy levels by solving
the time-independent Schrödinger equation for the particle in the potential well and applying the
boundary conditions.
1. Set up the Schrödinger equation: The time-independent Schrödinger equation for the
particle in the one-dimensional infinite potential well is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
where ψis the wave function of the particle, Eis the energy of the particle, ˉhis the reduced
Planck’s constant, mis the mass of the particle, and xis the position coordinate.
2. Solve the Schrödinger equation in the well: Inside the well, the potential energy is
zero. Thus, the Schrödinger equation simplifies to:
−ˉh2
2m
d2ψ
dx2=Eψ
3. Apply the boundary conditions: The wave function ψ(x)must be zero at both ends
of the well, i.e., ψ(0) = 0 and ψ(L) = 0. This implies that the solutions to the Schrödinger
equation are of the form:
ψn(x) = √2
Lsin (nπx
L)
where nis a positive integer.
4. Find the energy levels: Substitute the wave function ψn(x)into the Schrödinger
equation to solve for the energy levels En. Using the energy operator ˆ
H=−ˉh2
2m
d2
dx2, we have:
ˆ
Hψn(x) = Enψn(x)
En=n2π2ˉh2
2mL2
Therefore, the energy levels of the particle in the one-dimensional infinite potential well are
given by En=n2π2ˉh2
2mL2, where nis a positive integer corresponding to the quantum number.
10. Question: Let’s consider a particle trapped in a one-dimensional potential well defined by
the potential energy function:
V(x) = {0if 0≤x≤a
∞otherwise
where ais a positive constant.
Calculate the energy levels of the particle in this potential well.
Ans. To calculate the energy levels of a particle trapped in the potential well defined by the
given potential energy function, we need to solve the time-independent Schrödinger equation for
the system and apply the appropriate boundary conditions.
1. The time-independent Schrödinger equation is given by:
−ˉh2
2m
d2ψ(x)
dx2+V(x)ψ(x) = Eψ(x)
where ˉhis the reduced Planck constant, mis the mass of the particle, Eis the total energy of
the particle, and ψ(x)is the wave function.
2. Since the potential energy is infinite outside the region 0≤x≤a, the wave function must
be zero outside this region. Therefore, the wave function ψ(x)is non-zero only in the region
0≤x≤a.
3. In the region 0≤x≤a, the Schrödinger equation simplifies to:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
This is the time-independent Schrödinger equation for a particle inside a potential well.
4. Let’s further simplify the equation by dividing through by −ˉh2
2m:
d2ψ(x)
dx2=−2mE
ˉh2ψ(x)
5. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh, and Aand Bare constants to be determined.
6. Apply the boundary conditions: ψ(0) = 0 and ψ(a) = 0.
7. From ψ(0) = 0, we have:
B= 0
Therefore, the wave function becomes:
ψ(x) = Asin(kx)
8. From ψ(a) = 0, we have:
Asin(ka) = 0
Since sin(ka)cannot be zero for all k(except for k= 0), we must have:
ka =nπ
where nis a positive integer.
9. This gives us the quantized values for E:
En=n2π2ˉh2
2ma2
where n= 1,2,3, ... are the allowed energy levels of the particle in the potential well.
Therefore, the energy levels of the particle in this one-dimensional potential well are quantized
with values given by En=n2π2ˉh2
2ma2, where n= 1,2,3, ....
11. Question: Determine the energy levels for a particle trapped in a one-dimensional infinite
potential well of width L.
Ans. Let Endenote the nth energy level of the particle in the infinite potential well. The energy
levels are given by the formula:
En=n2π2ˉh2
2mL2,
where n= 1,2,3, . . . corresponds to the quantum number representing different energy levels.
Solution: 1. The Schrödinger equation for the wave function inside the well is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x).
2. Since the potential is infinite at 0and L, the wave function must be zero at these points.
Therefore, the wave function ψ(x)can be written as:
ψ(x) = Asin (nπx
L),
where Ais the normalization constant.
3. Applying the boundary conditions ψ(0) = ψ(L) = 0, we get:
ψ(0) = Asin(0) = 0 =⇒A= 0,
and
ψ(L) = Asin(nπ) = 0 =⇒nπ =mπ,
where mis an integer (but not zero).
4. The energy levels Enare then given by:
En=n2π2ˉh2
2mL2.
Therefore, the particle in the infinite potential well has discrete energy levels determined by
the quantum number n.
12. Question 12:
Consider a particle in a one-dimensional infinite square well potential with width L.
(a) Calculate the ground state energy of the particle in terms of Land the particle’s mass m.
(b) Find the energy of the first excited state in terms of Land m.
Ans. (a)
1. The ground state energy of a particle in a one-dimensional infinite square well potential is
given by the formula:
E1=π2ˉh2
2mL2
where mis the mass of the particle and Lis the width of the potential well.
(b)
1. The energy of the nth excited state in a one-dimensional infinite square well potential is
given by:
En=n2π2ˉh2
2mL2
where nis the quantum number of the state.
2. For the first excited state, n= 2, so the energy is:
E2=4π2ˉh2
2mL2=2π2ˉh2
mL2
13. Question: Determine the energy levels of a particle confined to a one-dimensional box of
length L. The particle has a mass mand the potential energy inside the box is zero.
Ans. To find the energy levels of a particle confined to a one-dimensional box, we can use
the time-independent Schrödinger equation. The general form of the Schrödinger equation for a
particle in one dimension is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
where Eis the total energy of the particle, ψis the wave function, ˉhis the reduced Planck
constant, and mis the mass of the particle.
1. Define the potential energy function: Since the potential energy inside the box is
zero, the potential energy function can be represented as V(x) = 0.
2. Set up the Schrödinger equation: Substitute V(x) = 0 into the Schrödinger equation
to obtain:
−ˉh2
2m
d2ψ
dx2=Eψ
3. Solve the differential equation: Let’s solve this differential equation to find the possible
energy levels. The general solution to the differential equation is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where Aand Bare constants to be determined, and k=√2mE
ˉh2.
4. Apply the boundary conditions: The wave function must satisfy the following boundary
conditions for a particle in a box of length L:
a) ψ(0) = 0 b) ψ(L) = 0
Applying the boundary condition ψ(0) = 0 yields:
0 = B
So, the wave function becomes:
ψ(x) = Asin(kx)
Applying the boundary condition ψ(L) = 0, we get:
0 = Asin(kL)
5. Quantization condition: For the wave function to satisfy the boundary condition ψ(L) =
0, the argument of the sine function must be an integer multiple of π:
kL =nπ
where nis a positive integer.
6. Find the energy levels: Substitute the expression for kinto the quantization condition
and solve for Eto find the energy levels:
√2mE
ˉh2·L=nπ
En=n2π2ˉh2
2mL2
Therefore, the energy levels of the particle confined to a one-dimensional box are given by
En=n2π2ˉh2
2mL2, where nis a positive integer.
14. Question: Determine the energy levels of an electron confined in a one-dimensional box of
length L= 2 nm. The mass of the electron is m= 9.11 ×10−31 kg and Planck’s constant is
h= 6.63 ×10−34 J·s.
Ans. Let’s denote the energy levels as En, where nis a positive integer representing the
quantum number.
1. The energy levels of an electron in a one-dimensional box are given by the formula:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck’s constant, ˉh=h
2π.
2. Substituting the given values, we have:
En=n2π2(h
2π)2
2mL2
En=h2n2
8mL2
3. Plugging in the values for h,m, and L, we get:
En=(6.63 ×10−34 J·s)2n2
8×9.11 ×10−31 kg ×(2 ×10−9m)2
4. Simplifying the expression:
En=43.9×10−68 J·s2n2
7.28 ×10−59 kg ·m2
En= 6.04 ×10−9J·n2
Therefore, the energy levels of the electron confined in the one-dimensional box are given by
En= 6.04 ×10−9J·n2.
15. Question: Determine the energy levels of an electron in a one-dimensional harmonic
oscillator potential given by V(x) = 1
2kx2, where kis the force constant.
Ans. Let’s solve this step-by-step:
1. The Schrödinger equation for a one-dimensional harmonic oscillator potential is given by:
−ˉh2
2m
d2ψ(x)
dx2+1
2kx2ψ(x) = Eψ(x)
where ˉhis the reduced Planck constant, mis the mass of the electron, Eis the energy of the
electron, and ψ(x)is the wave function.
2. We can simplify the Schrödinger equation by making the following substitutions:
ξ=√mω
ˉhxand λ=2E
ˉhω
where ω=√k
mis the angular frequency.
3. With these substitutions, the Schrödinger equation becomes:
−d2ψ(ξ)
dξ2+ξ2ψ(ξ) = λψ(ξ)
4. We now apply the ladder operators aand a†defined as:
a=1
√2(ξ+d
dξ )and a†=1
√2(ξ−d
dξ )
5. Using these ladder operators, we can rewrite the Schrödinger equation as:
aa†ψ(ξ) = (λ+ 1)ψ(ξ)
6. We see that aa†=1
2(ξ2−d2
dξ2−1), hence the eigenvalues of aa†are 1
2(2n+ 1), where n
is a non-negative integer.
7. Therefore, the energy levels are given by:
En=ˉhω (n+1
2)
where ntakes values 0,1,2, ....
16. Question: Determine the first four energy levels of a particle in a one-dimensional box of
length L.
Ans. Let’s begin by using the formula for the energy levels of a particle in a one-dimensional
box:
1. The formula for the energy levels of a particle in a one-dimensional box is given by:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number, ˉhis the reduced Planck’s
constant, mis the mass of the particle, and Lis the length of the box.
2. For the first energy level (n= 1), the energy is:
E1=(1)2π2ˉh2
2mL2=π2ˉh2
2mL2
3. For the second energy level (n= 2), the energy is:
E2=(2)2π2ˉh2
2mL2=4π2ˉh2
2mL2= 2 π2ˉh2
mL2
4. For the third energy level (n= 3), the energy is:
E3=(3)2π2ˉh2
2mL2=9π2ˉh2
2mL2=9
2
π2ˉh2
mL2
5. For the fourth energy level (n= 4), the energy is:
E4=(4)2π2ˉh2
2mL2=16π2ˉh2
2mL2= 8 π2ˉh2
mL2
Therefore, the first four energy levels of a particle in a one-dimensional box of length Lare:
E1=π2ˉh2
2mL2, E2= 2 π2ˉh2
mL2, E3=9
2
π2ˉh2
mL2, E4= 8 π2ˉh2
mL2
17. Let’s consider an electron confined in a one-dimensional infinite potential well of width L.
The potential energy inside the well is zero, while the potential energy outside the well is infinite.
Determine the possible energy levels for the electron in this system.
Ans. To solve the problem, we need to consider the Schrödinger equation for the infinite
potential well and apply the boundary conditions.
1. Set up the Schrödinger equation: Inside the well, the potential energy V(x)is zero,
so the Schrödinger equation in one dimension is:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
2. Solve the Schrödinger equation: The general solution to the Schrödinger equation in
this case is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
3. Apply the boundary conditions: Since the potential energy outside the well is infi-
nite, the wave function must vanish at the boundaries, i.e., ψ(0) = ψ(L)=0. This leads to
quantization of the energy levels:
kn=nπ
L
where n= 1,2,3, . . ..
4. Calculate the energy levels: Substitute the quantized values of knback into the
expression for energy:
En=ˉh2k2
n
2m=n2π2ˉh2
2mL2
Thus, the energy levels are quantized and given by:
En=n2π2ˉh2
2mL2where n= 1,2,3, . . .
These discrete energy levels correspond to the allowed energy states of the electron in the infinite
potential well.
18. Question: Consider a particle of mass mmoving in a one-dimensional box of length L.
Determine the energy levels of the particle in terms of the quantum number n.
Ans. Step-by-step solution: 1. The energy of the particle in the box is given by the formula:
En=n2π2ˉ
h2
2mL2
where nis the quantum number, ˉhis the reduced Planck’s constant, and Lis the length of the
box. 2. Substituting the given values into the formula, we get:
En=n2π2ˉh2
2mL2
3. Therefore, the energy levels of the particle in the box are quantized, with En=π2ˉh2
2mL2,4π2ˉh2
2mL2,9π2ˉh2
2mL2, ...
for n= 1,2,3, ....
19. Question: Consider an electron in a one-dimensional box of length L. At t= 0, the electron
is in the ground state of the box. If an external uniform electric field Eis suddenly turned on
along the length of the box at t= 0, find the energy of the electron at t=L2
2πeEˉh.
Ans. Let’s first determine the initial ground state of the electron in the box and then find its
energy at t=L2
2πeEˉhafter turning on the external electric field E.
1. Initial Ground State Energy: The energy levels of an electron in a one-dimensional box
of length Lare given by En=n2π2ˉh2
2mL2, where n= 1,2,3, ... represents the quantum number.
The ground state corresponds to n= 1, so the initial ground state energy E1is:
E1=π2ˉh2
2mL2
2. Energy at t=L22πeEˉh: The acceleration of the electron due to the electric field is
a=eE
m, where −eis the charge of an electron and mis the mass of the electron.
The electron’s position at time tunder constant acceleration starting from rest is given by
x=1
2at2. Substituting aand t=L2
2πeEˉh, we get:
x=1
2·eE
m·(L2
2πeEˉh)2
The final energy of the electron is given by the energy level formula with the length of the
box now equal to the distance traveled by the electron:
Ef=n2π2ˉh2
2m(x)
Substitute n= 1 and x=eEL2
4π2m2e2ˉhinto the equation to find the energy of the electron at
t=L2
2πeEˉh.
Ef=π2ˉh2m
2(eEL2
4π2m2e2ˉh)=2π2ˉh2eEˉh
mL2
Therefore, the energy of the electron at t=L2
2πeEˉhis Ef=2π2ˉh2eEˉh
mL2.
20. Question: Find the energy levels of a particle in a one-dimensional box of length Lwith
infinitely high walls if the particle has mass mand the potential energy inside the box is given by
V(x) = λx4, where λis a positive constant.
Ans. Let’s denote the energy levels of the particle as Enfor n= 1,2,3, . . .. We know that
the Schrödinger equation for a particle in a one-dimensional box with a potential energy function
V(x)is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
We are given that V(x) = λx4. To solve for the energy levels, we need to find the eigenvalues
Enfor which there exist nontrivial solutions to the Schrödinger equation.
1. We start by writing down the Schrödinger equation with the given potential energy
function:
−ˉh2
2m
d2ψ
dx2+λx4ψ=Eψ
2. To simplify the equation, let’s substitute ψ(x) = Aeikx into the equation, where Ais a
constant to be determined and kis the wave number.
3. Substitute ψ(x)into the Schrödinger equation and simplify to get a differential equation
for k:
−ˉh2
2m(−k2)eikx +λx4Aeikx =EAeikx
4. Divide the equation by Aeikx to simplify:
ˉh2k2
2m+λx4=E
5. This is a separable equation, so let’s separate the variables xand E:
ˉh2k2
2m=E−λx4
6. Now, let’s solve the separated differential equation by assuming En=αnλ1/4 and k=βn,
where αnand βnare constants.
7. Substitute these into the separated equation and solve for αnand βnto find the energy
levels En.
21. Question 21: Consider a particle of mass min a one-dimensional potential well given by
V(x) = 1
2kx2for −a≤x≤a, where kis a positive constant. Determine the allowed energy
levels of the particle in this potential well.
Ans. To determine the allowed energy levels of the particle in the potential well, we need to
solve the time-independent Schrödinger equation for the given potential and apply appropriate
boundary conditions.
1. Write the time-independent Schrödinger equation: The time-independent Schrödinger
equation for a one-dimensional potential is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
Substitute V(x) = 1
2kx2into the equation.
2. Solve the Schrödinger equation: The Schrödinger equation becomes:
−ˉh2
2m
d2ψ
dx2+1
2kx2ψ=Eψ
Simplify the equation and rearrange it in a standard form.
3. Apply the boundary conditions: The boundary conditions for a particle in a potential
well require that the wavefunction ψ(x)must be continuous and the first derivative dψ
dx must be
continuous at x=−aand x=a.
4. Solve for the energy levels: By solving the Schrödinger equation with the appropriate
boundary conditions, we can find the allowed energy levels of the particle in the potential well.
The energy levels will depend on the mass m, the constant k, and the width of the potential
well 2a.
22. What is the energy of a photon with a frequency of 6.0×1018 Hz?
Ans. To calculate the energy of a photon, we can use the equation:
E=hf
where: - Eis the energy of the photon, - his the Planck constant (6.63 ×10−34 J·s), and -
fis the frequency of the photon.
Step 1. Plug in the values into the equation:
E= (6.63 ×10−34 J·s)×(6.0×1018 Hz)
Step 2. Calculate the energy of the photon:
E= 3.978 ×10−15J
Therefore, the energy of a photon with a frequency of 6.0×1018 Hz is 3.978 ×10−15 J.
23. Question: Determine the energy levels of an electron in a one-dimensional box of length
L= 5 nm.
Ans. Step-by-step solution: 1. The energy levels for an electron in a one-dimensional box are
given by the equation
En=n2h2
8mL2
where nis the quantum number, his the Planck constant (6.626 ×10−34 J s), mis the mass of
the electron (9.11 ×10−31 kg), and Lis the length of the box.
2. Substituting the given values into the equation, we have
En=n2(6.626 ×10−34)2
8(9.11 ×10−31)(5 ×10−9)2
3. Simplifying the expression, we get
En=n2×4.3908 ×10−67
1.822 ×10−19
En=n2×2.41 ×10−48
4. Therefore, the energy levels for the electron in the one-dimensional box with L= 5 nm
are multiples of 2.41 ×10−48 J−.
24. Question: Consider a particle in a one-dimensional potential well of width L. The potential
energy inside the well is zero, while the potential energy outside the well is infinite. Determine
the energy levels of the particle in terms of ˉh,L, and the particle’s mass m.
Ans. Let’s denote the energy levels of the particle inside the well as En, where nis a positive
integer. We can solve for the energy levels by solving the time-independent Schrödinger equation
for the particle in this potential well.
1. The general form of the time-independent Schrödinger equation for this system is:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
This equation can be rewritten as:
d2ψ(x)
dx2+2mE
ˉh2ψ(x) = 0
2. Inside the well, the potential energy V(x) = 0, so the Schrödinger equation simplifies to:
d2ψ(x)
dx2+2mE
ˉh2ψ(x) = 0
3. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
4. Since the potential energy outside the well is infinite, the wavefunction ψ(x)must go to
zero at x= 0 and x=L. Hence, ψ(0) = ψ(L) = 0. This leads to the boundary conditions:
B= 0
(since cos(0) = 1)
and
Asin(kL) = 0
5. The boundary condition Asin(kL) = 0 implies that kL =nπ for nbeing a positive
integer. Therefore, we have:
k=nπ
L
6. Substituting k=nπ
Lback into the expression k2=2mEn
ˉh2yields the energy levels En:
En=ˉh2π2n2
2mL2
Therefore, the energy levels of the particle in the potential well are quantized and given by
En=ˉh2π2n2
2mL2, where nis a positive integer.
25. Question: A particle with mass mis confined within a one-dimensional potential well
defined by V(x) = 1
2kx2for |x|< a, where kis a positive constant and ais the width of the
well. Calculate the allowed energy levels of the particle in terms of m,k, and a.
Ans. Let’s denote the allowed energy levels of the particle by En, where n= 1,2,3, . . .. The
energy levels are given by the equation:
En=n2π2ˉh2
2ma2
where ˉhis the reduced Planck constant.
26. Question 26: Consider a particle in a one-dimensional infinite potential well of width a.
This particle can occupy states labeled by n= 1,2,3, ... with corresponding energy levels given
by En=n2π2ˉh2
2ma2, where ˉhis the reduced Planck constant, mis the mass of the particle, and ais
the width of the potential well.
Calculate the difference in energy levels between the n= 3 and n= 4 states in terms of ˉh,
m, and a.
Ans. To find the difference in energy levels between the n= 3 and n= 4 states, we can
subtract E4from E3:
1. Calculate the energy of the n= 3 state:
E3=32π2ˉh2
2ma2=9π2ˉh2
2ma2
2. Calculate the energy of the n= 4 state:
E4=42π2ˉh2
2ma2=16π2ˉh2
2ma2
3. Find the difference in energy levels:
∆E=E4−E3=16π2ˉh2
2ma2−9π2ˉh2
2ma2
∆E=16π2ˉh2−9π2ˉh2
2ma2=7π2ˉh2
2ma2=7
2
π2ˉh2
ma2
Therefore, the difference in energy levels between the n= 3 and n= 4 states is 7
2
π2ˉh2
ma2.
27. Question 27: A particle of mass mis in a one-dimensional box of length L. The wavefunction
of the particle is given by ψ(x) = Asin(kx), where Ais a normalization constant.
Determine the possible energy levels for this system.
Ans. To determine the possible energy levels for the system, we need to solve the time-
independent Schrödinger equation.
1. The general form of the time-independent Schrödinger equation for a one-dimensional box
is:
−ˉh2
2m
d2ψ
dx2=Eψ
where Eis the energy of the particle.
2. Given the wavefunction ψ(x) = Asin(kx), we can substitute this into the Schrödinger
equation:
−ˉh2
2mAk2sin(kx) = EA sin(kx)
3. Simplifying the equation by dividing by Asin(kx), we get:
−ˉh2k2
2m=E
4. Solving for E, we find that the energy levels are quantized and given by:
En=ˉh2k2
2m=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number.
Therefore, the possible energy levels for this system are:
En=n2π2ˉh2
2mL2
28. **Question:** Find the energy eigenvalues and corresponding eigenfunctions of the particle
in a one-dimensional box of length L, when a potential barrier of height V0and width ais inserted
at the center of the box.
**Step-by-step Solution:** Let’s denote the total energy of the system as E. Inside the
barrier (0< x < a or L−a<x<L), the potential energy is V0and outside the barrier
(a<x<L−a), the potential energy is 0.
1. **Finding the wave function inside the barrier:** The wave function inside the barrier can
be represented as:
ψ(x) = {Aeikx +Be−ikx for 0< x < a
Ceikx +De−ikx for L−a<x<L
where k=√2mE
ˉh2.
2. **Applying boundary conditions at x= 0 and x=a:** At x= 0, the wave function and
its derivative must be continuous:
ψ(0) = ψ(a)and dψ(0)
dx =dψ(a)
dx
3. **Applying boundary conditions at x=aand x=L−a:** Similarly, at x=aand
x=L−a, the wave function and its derivative must be continuous:
ψ(a) = ψ(L−a)and dψ(a)
dx =dψ(L−a)
dx
4. **Solving the system of equations:** Solving the system of equations obtained from the
boundary conditions will give the values of constants A,B,C, and D. This, in turn, will provide
the energy eigenvalues and eigenfunctions of the system.
Therefore, the energy eigenvalues and corresponding eigenfunctions of the particle in a one-
dimensional box with a potential barrier at the center can be determined by solving the Schrödinger
equation with appropriate boundary conditions.
29. Let’s consider an electron confined to a one-dimensional box with a length of L. The
potential energy V(x)inside the box is zero, but infinite outside the box. We are asked to find
the expression for the energy levels of this system.
30. A quantum mechanical particle is confined to a one-dimensional box of length L. Calculate
the energy levels for this system.
Ans. To find the energy levels of the particle in a one-dimensional box, we can start by solving
the time-independent Schrödinger equation for this system.
1. Set up the Schrödinger equation: The time-independent Schrödinger equation for the
particle in a one-dimensional box is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)for 0< x < L
with the boundary conditions ψ(0) = 0 and ψ(L) = 0.
2. Solve the differential equation: The general solution to the differential equation is of
the form:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
Applying the boundary conditions:
ψ(0) = 0 =⇒B= 0
ψ(L) = 0 =⇒Asin(kL) = 0
3. Determine the quantized values of k:From the boundary condition sin(kL) = 0, we
get kL =nπ, where nis a positive integer.
So, the allowed values of kare:
kn=nπ
L
4. Calculate the energy levels: The energy levels are given by:
En=ˉh2k2
n
2m=ˉh2π2n2
2mL2
Therefore, the energy levels for the particle in a one-dimensional box are quantized and given
by En=ˉh2π2n2
2mL2where n= 1,2,3, ....
31. Question 30: An electron is confined to move in one dimension in an infinite potential well
of width L. Determine the energy of the electron in the n= 4 energy level.
Ans. Let’s start by determining the general formula for the energy levels of an electron in an
infinite potential well. The energy levels for a particle in a 1D infinite potential well are given by:
En=n2π2ˉh2
2mL2
where nis the quantum number, ˉhis the reduced Planck’s constant, mis the mass of the
electron, and Lis the width of the well.
1. Substitute n= 4 into the formula to find the energy of the electron in the n= 4 energy
level:
E4=42π2ˉh2
2mL2
E4=16π2ˉh2
2mL2
32. Question 31: An electron in a one-dimensional quantum well with a width of 3 nm has an
energy level of -2.5 eV. Calculate the energy of the next higher energy level in the well.
Ans. To solve this problem, we can use the equation for the energy levels in a one-dimensional
quantum well:
En=n2π2ˉh2
2mL2
Where: - Enis the energy level, - nis the quantum number of the energy level, - ˉhis the
reduced Planck’s constant (1.0545718 ×10−34 Js), - mis the mass of the electron (9.11 ×10−31
kg), - Lis the width of the quantum well.
1. Calculate the energy of the next higher energy level:
Given: - E1=−2.5eV, - L= 3 nm.
First, let’s convert the values to SI units: - E1=−2.5×1.6×10−19 J (1 eV = 1.6×10−19
J), - L= 3 ×10−9m.
Now, we can solve for E2using the equation:
E2=(2)2π2ˉh2
2mL2
E2=(2)2π2·(1.0545718 ×10−34)2
2·9.11 ×10−31 ·(3 ×10−9)2
E2=4·π2·(1.0545718 ×10−34)2
2·9.11 ×10−31 ·9×10−18
E2≈ −1.107 ×10−19 J
Finally, convert the energy back to electron volts:
E2≈−1.107 ×10−19
1.6×10−19 ≈ −0.692 eV
Therefore, the energy of the next higher energy level in the quantum well is approximately
-0.692 eV.
33. Question 32: Consider a quantum mechanical system with a Hamiltonian operator given by
ˆ
H=−ˉh2
2m
d2
dx2+V(x), where V(x)is a potential function. Suppose that the potential function
is described by V(x) = 1
2mω2x2+λ
4x4, where ω, λ > 0.
Find the expression for the energy eigenvalues Enof this system.
Ans. Let us consider the time-independent Schrödinger equation for this system: ˆ
Hψ(x) =
Eψ(x), where ψ(x)is the wave function and Eis the energy eigenvalue.
1. Substituting the given Hamiltonian operator into the Schrödinger equation, we have:
−ˉh2
2m
d2ψ
dx2+(1
2mω2x2+λ
4x4)ψ=Eψ.
2. Rearranging the terms and defining a new parameter α=mω
ˉh, the equation becomes:
−ˉh2
2m
d2ψ
dx2+1
2mω2x2ψ+λ
4x4ψ−Eψ = 0.
3. Dividing through by ˉhω to simplify, and defining the dimensionless variable ξ=√mω
ˉhx,
the equation transforms into:
−1
2
d2ψ
dξ2+1
2ξ2ψ+λ
4ˉhω x4ψ−E
ˉhω ψ= 0.
4. Writing the equation in dimensionless form, we have the new Hamiltonian operator ˆ
H=
−1
2
d2
dξ2+1
2ξ2+λ
4ˉhω ξ4.
5. We can solve this dimensionless form of the Schrödinger equation to find the energy
eigenvalues Enand corresponding wave functions ψn(ξ).
6. The energy eigenvalues for this system are given by:
En=(n+1
2)ˉhω, n = 0,1,2, ...
Therefore, the expression for the energy eigenvalues Enof this quantum mechanical system
with the specified potential function is En=(n+1
2)ˉhω.
34. Question 33:
A particle of mass mis confined in a one-dimensional infinite potential well of width L.
Calculate the energy levels of the particle in this potential well.
Ans. To calculate the energy levels of the particle in the one-dimensional infinite potential well,
we can use the time-independent Schrödinger equation and the boundary conditions.
1. Write down the time-independent Schrödinger equation:
The time-independent Schrödinger equation for a particle of mass min one dimension po-
tential well is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
2. Write down the boundary conditions: The boundary conditions for an infinite potential
well of width Lare that the wave function ψ(x)must be continuous and the derivative of ψ(x)
must be discontinuous at x= 0 and x=L, since the potential is infinite at these points.
3. Solve the differential equation: The general solution to the differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
Applying the boundary condition that ψ(0) = ψ(L) = 0 gives:
ψ(0) = B= 0
ψ(L) = Asin(kL) = 0
This implies that kL =nπ for n= 1,2,3, .... Therefore, the energy levels Enare given by:
En=n2π2ˉh2
2mL2for n= 1,2,3, ...
So, the energy levels of the particle in the one-dimensional infinite potential well are quantized
and given by the formula above.
35. Find the energy levels of a particle confined to a one-dimensional box of length L, where
the potential energy inside the box is zero and infinitely large outside the box.
Ans. To find the energy levels of a particle in a one-dimensional box, we need to solve the
time-independent Schrödinger equation and apply the boundary conditions. The general form of
the wave function for a particle in a box of length Lis ψ(x) = Asin(kx) + Bcos(kx), where
k=nπ
Lfor positive integer values of n.
1. Applying the boundary conditions: Inside the box where the potential energy is zero,
the wave function should be continuous. So, at x= 0 and x=L, the wave function should be
continuous. Therefore, ψ(0) = ψ(L) = 0.
2. Applying the boundary condition at x= 0:Plugging in x= 0 into the wave function
ψ(x)gives: ψ(0) = Asin(0) + Bcos(0) = B= 0, since sin(0) = 0 and cos(0) = 1.
So, B= 0.
3. Applying the boundary condition at x=L:Plugging in x=Linto the wave function
ψ(x)gives: ψ(L) = Asin(kL) = 0.
For the wave function to be nonzero, sin(kL)must be zero, which means kL =nπ for some
positive integer n.
4. Solving for the energy levels: Since k=nπ
L, the energy levels are given by the relation
En=ˉh2k2
2m=n2π2ˉh2
2mL2.
Therefore, the energy levels of the particle in the one-dimensional box are quantized and given
by En=n2π2ˉh2
2mL2, where nis a positive integer.
36. Question 35:
Consider a particle in a one-dimensional box of length L. Determine the energy levels for the
particle confined within the box.
Ans. To find the energy levels of a particle in a one-dimensional box, we can use the Schrödinger
equation and apply the boundary conditions. The Schrödinger equation for a particle in a box is:
−ˉh2
2m
d2ψ
dx2=Eψ
where mis the mass of the particle, Eis the energy, ψis the wave function, and ˉhis the
reduced Planck’s constant.
1. Setting up the Schrödinger equation: Since the particle is confined in a one-
dimensional box of length L, the wave function ψshould satisfy the boundary conditions that
ψ(0) = ψ(L) = 0. Thus, the general form of the wave function is:
ψ(x) = Asin(kx)
where kis the wave number, and Ais a normalization constant.
2. Find the wave number k:We can find the wave number kby using the boundary
conditions:
ψ(0) = Asin(0) = 0 =⇒A= 0
ψ(L) = Asin(kL) = 0 =⇒kL =nπ
where nis a positive integer.
3. Expressing the energy levels in terms of n:Once we have the possible values of k,
we can express the energy levels:
kL =nπ =⇒k=nπ
L
The energy levels are given by:
En=ˉh2k2
2m=ˉh2(nπ
L)2
2m=n2π2ˉh2
2mL2
Therefore, the energy levels for a particle in a one-dimensional box of length Lare given by
En=n2π2ˉh2
2mL2, where nis a positive integer.
