CHM 345 - PHYSICAL CHEMISTRY I - Calculation of
energy levels Practice Material - Set 5
1. What are the energy levels of an electron confined within a one-dimensional box of length
L? Express the energy levels in terms of Planck’s constant h, electron mass m, and the length
L.
Ans. The energy levels of an electron confined within a one-dimensional box of length Lare
given by:
En=n2h2
8mL2,
where n= 1,2,3, ... is the quantum number representing the nth energy level.
2. Question: Find the energy levels of an electron in a one-dimensional infinite square well
potential of width a.
Ans. Step-by-step solution: 1. The Schrödinger equation for a particle in a one-dimensional
infinite square well potential is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function, mis the mass of the particle, Eis the energy of the particle,
and ˉhis the reduced Planck constant. 2. The general solutions to the Schrödinger equation in
the regions 0≤x≤aand a≤x≤2aare:
ψ(x) = Asin(kx) + Bcos(kx)
ψ(x) = Csin(k′(x−a)) + Dcos(k′(x−a))
respectively, where k=√2mE
ˉh2and k′=√2m(E−V)
ˉh2. 3. Applying the boundary conditions
ψ(0) = ψ(2a) = 0 leads to:
B= 0
Csin(k′a) = 0
4. Since Ccannot be zero for non-trivial solutions, we have k′a=nπ, where nis a positive
integer. 5. Therefore, the energy levels Enare given by:
En=n2π2ˉh2
2ma2
where n= 1,2,3, ... represents the quantum number corresponding to the energy levels. 6. This
quantization of energy levels is a consequence of the particle being confined within the potential
well, leading to discrete allowed energy values.
3. Question: Determine the energy levels of a particle in a one-dimensional box of length L,
where the particle has a mass mand is subject to a potential energy V(x) = 0 inside the box
and V(x) = ∞outside the box.
Ans. Let’s solve this using the time-independent Schrödinger equation and boundary conditions.
1. The time-independent Schrödinger equation for this system 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, and mis the mass of the particle.
2. Inside the box (0<x<L), the potential energy V(x)=0, so the Schrödinger equation
becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
3. The general solution to the Schrödinger equation inside the box is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉhand Aand Bare constants to be determined.
4. Applying the boundary conditions: ψ(0) = 0 and ψ(L) = 0, we find:
{B= 0
kL =nπ where n= 1,2,3, . . .
5. From the second boundary condition, we get the quantization condition for the energy
levels:
knL=nπ =⇒√2mEnL
ˉh=nπ
6. Solving for the energy levels En, we find:
En=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.
4. Find the energy levels of a particle confined to a one-dimensional infinite potential well with
width L.
Ans. To find the energy levels of a particle in a one-dimensional infinite potential well, we can
use the Schrödinger equation and the boundary conditions to solve for the allowed energy levels.
1. The general form of the wave function inside the well is Ψ(x) = Asin(kx) + Bcos(kx),
where k=√2mE
ˉh2.
2. Applying the boundary conditions, Ψ(0) = 0 and Ψ(L) = 0, we find that k=nπ
L, where
n= 1,2,3, ... (quantization condition).
3. The energy levels Enare then given by En=n2π2ˉh2
2mL2.
Therefore, the energy levels of a particle confined to a one-dimensional infinite potential well with
width Lare En=n2π2ˉh2
2mL2, where n= 1,2,3, ....
5. Question: Determine the energy levels for a particle in a one-dimensional box of length L
with infinitely high walls.
Ans. Let’s denote the allowed energy levels as En. The energy levels can be calculated using
the formula:
En=n2h2
8mL2
where nis a positive integer representing the quantum number of the energy level, his the
Planck constant, and mis the mass of the particle.
Step 1. Given that the walls are infinitely high, the particle’s wavefunction must go to zero
at the walls. Thus, we can write the wavefunction as:
ψ(x) = Asin (nπx
L)
where Ais a normalization constant.
Step 2. The probability density function, |ψ(x)|2, must be normalized. Therefore, we find
Ain such a way that:
∫L
0|ψ(x)|2dx = 1
It follows that A=√2
L.
Step 3. To determine the energy levels, we must find the expectation value of the energy,
⟨E⟩, for an arbitrary wavefunction ψ(x). This can be expressed as:
⟨E⟩=∫ψ∗(x)(−ˉh2
2m
d2
dx2)ψ(x)dx
∫|ψ(x)|2dx
Substitute ψ(x)into the equation and simplify.
Step 4. Applying the kinetic and potential energy operators to ψ(x), we obtain the Schrödinger
equation. Solve the resulting differential equation to find the energy levels En.
Step 5. Finally, use the formula En=n2h2
8mL2to determine the values of the energy levels En
for the particle in the box.
6. Question 6:
Consider a particle in a one-dimensional box of length L. Calculate the first four energy levels
of the particle in terms of the particle’s mass mand the width of the box L.
Ans. To solve for the energy levels of a particle in a one-dimensional box, we can use the
Schrödinger equation and apply the boundary conditions.
1. Setting up the Schrödinger equation: The time-independent Schrödinger equation
for the particle in a box is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
2. Applying boundary conditions: The boundary conditions for a particle in a box are
ψ(0) = ψ(L) = 0. This implies that the wave function must be zero at x= 0 and x=L.
3. Solving for the energy levels: Let’s assume that the wave function is in the form
ψ(x) = Asin(kx). Substitute ψ(x) = Asin(kx)into the Schrödinger equation and apply the
boundary conditions to solve for kand E.
At x= 0:ψ(0) = Asin(0) = 0 This implies that A= 0 is not a permissible value. Therefore,
for non-trivial solutions, we have sin(k·0) = 0, which gives k=nπ/L, where nis a positive
integer.
The energy levels are given by En=ˉh2k2
2m=ˉh2π2n2
2mL2.
4. Calculating the first four energy levels: For n= 1,E1=ˉh2π2
2mL2. For n= 2,
E2=4ˉh2π2
2mL2. For n= 3,E3=9ˉh2π2
2mL2. For n= 4,E4=16ˉh2π2
2mL2.
Therefore, the first four energy levels of the particle in the box are: E1=ˉh2π2
2mL2,E2=4ˉh2π2
2mL2,
E3=9ˉh2π2
2mL2,E4=16ˉh2π2
2mL2.
7. Question: Find the energy levels of a particle in a one-dimensional box of length Lif the
particle has mass mand the potential energy inside the box is zero.
Ans. Let nbe the quantum number representing the energy level. The energy levels Enare
given by:
En=n2π2ˉh2
2mL2
where nis a positive integer.
Step 1. Write down the time-independent Schrödinger equation for the particle in a one-
dimensional box:
−ˉh2
2m
d2ψ
dx2=Eψ
Step 2. The general solution to this differential equation is:
ψ(x) = Asin (nπx
L)+Bcos (nπx
L)
where Aand Bare constants to be determined.
Step 3. Apply the boundary conditions to determine the constants Aand B. The boundary
conditions are:
ψ(0) = 0 and ψ(L) = 0
Step 4. The boundary condition ψ(0) = 0 gives:
0 = Asin(0) + Bcos(0) = B
Step 5. Substituting B= 0 back into the general solution gives:
ψ(x) = Asin (nπx
L)
Step 6. Applying the boundary condition ψ(L) = 0:
0 = Asin(nπ)
Since sin(nπ) = 0 for integer n, we have n= 1,2,3, . . .
Step 7. Therefore, the energy levels are given by:
En=n2π2ˉh2
2mL2
where nis a positive integer.
8. Question 8:
Consider a particle confined to a one-dimensional box of length L. The potential in the box
is given by V(x) = 0 for 0< x < L and infinite elsewhere. Determine the energy levels of the
particle in terms of n.
Ans. To determine the energy levels of the particle in the one-dimensional box, we need to
solve the time-independent Schrödinger equation for the system:
1. The time-independent Schrödinger equation is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
2. For 0< x < L, the potential is zero, so the Schrödinger equation becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
3. This is a second-order ordinary differential equation. The general solution is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2
4. To satisfy the boundary conditions at x= 0 and x=L,ψ(0) = 0 and ψ(L)=0, the
boundary conditions ψ(0) = 0 gives B= 0.
5. The boundary condition at x=Limplies sin(kL)=0, which gives kL =nπ for
n= 1,2,3, ....
6. Solving kL =nπ for E, we get the energy levels:
En=n2π2ˉh2
2mL2
where n= 1,2,3, ...
7. Therefore, the energy levels of the particle in the one-dimensional box are given by En=
n2π2ˉh2
2mL2for n= 1,2,3, ....
9. Question: Determine the energy levels of a particle in a one-dimensional box of length
Lif the potential energy function is given by V(x) = V0for 0< x < L/3,V(x)=0for
L/3 < x < 2L/3, and V(x) = 2V0for 2L/3 < x < L, where V0is a constant.
Ans. Let’s denote the energy levels as Enand the corresponding wavefunctions as ψn(x). The
general form of the wavefunction inside the box is ψn(x) = Asin(kx) + Bcos(kx), where A,B,
and kare constants.
1. Find the wavefunction and energy levels for 0<x<L/3:Inside this region,
V(x) = V0. The Schrödinger equation is −ˉh2
2m
d2ψ
dx2+V0ψ=Eψ. Solving this differential equation
gives ψ1(x) = A1sin(kx) + B1cos(kx)and E1=ˉh2k2
2m+V0.
2. Find the wavefunction and energy levels for L/3 <x<2L/3:Inside this region,
V(x) = 0. The Schrödinger equation is −ˉh2
2m
d2ψ
dx2=Eψ. Solving this differential equation gives
ψ2(x) = A2sin(kx) + B2cos(kx)and E2=ˉh2k2
2m.
3. Find the wavefunction and energy levels for 2L/3 < x < L:Inside this region,
V(x)=2V0. The Schrödinger equation is −ˉh2
2m
d2ψ
dx2+ 2V0ψ=Eψ. Solving this differential
equation gives ψ3(x) = A3sin(kx) + B3cos(kx)and E3=ˉh2k2
2m+ 2V0.
We need to apply continuity conditions on the wavefunction and its derivative at the bound-
aries x=L/3 and x= 2L/3 to determine the values of kand E.
10. Question: Determine the energy levels of an electron in a one-dimensional infinite square
well of width L. Express your answer in terms of the quantum number n.
Ans. Let’s denote the energy levels of an electron in a one-dimensional infinite square well as
En. The general formula for the energy levels in an infinite square well is given by:
En=n2π2ˉh2
2mL2
where: En= energy level of the electron in the n-th quantum state, n= quantum number
(positive integer), ˉh= reduced Planck’s constant (1.0545718 ×10−34 J s), m= mass of the
electron (9.10938356 ×10−31 kg), L= width of the infinite square well.
Step 1. Start with the general formula for energy levels:
En=n2π2ˉh2
2mL2
Step 2. Substitute the given values into the formula:
En=n2π2(1.0545718 ×10−34)2
2(9.10938356 ×10−31)L2
Step 3. Simplify the expression to get the energy level in terms of n:
En=n2π2×(1.1100599 ×10−68)
2(9.10938356 ×10−31)L2
Step 4. Finally, express the energy levels in terms of the quantum number n:
En=n2π2×(1.1100599 ×10−68)
2mL2
11. Question: Consider an electron in a one-dimensional infinite potential well of width L.
Determine the energy levels for the electron in terms of n, where nis a positive integer representing
the quantum number.
Ans. Step-by-step solution: 1. In a one-dimensional infinite potential well, the Schrödinger
equation for the electron can be written as:
−ˉh2
2m
d2ψ
dx2=Eψ
Where ψ(x)is the wave function, Eis the energy of the electron, ˉhis the reduced Planck
constant, mis the mass of the electron, and x∈[0, L]within the well. 2. The general solution
to this differential equation is given by:
ψn(x) = Asin (nπx
L)
Where nis a positive integer representing the quantum number and Ais a normalization constant.
3. To determine the energy levels for the electron, we substitute the wave function into the
Schrödinger equation:
−ˉh2
2m
d2
dx2(Asin (nπx
L))=EA sin (nπx
L)
4. Simplifying the differential equation:
(nπˉh
L)2
=2mE
ˉh2
E=n2π2ˉh2
2mL2
5. Therefore, the energy levels for the electron in the one-dimensional infinite potential well are
given by:
En=n2π2ˉh2
2mL2
Where nis a positive integer representing the quantum number.
12. Question 12: Consider a particle in a one-dimensional infinite potential well of width L.
The wave function of the particle in this potential well is given by
ψ(x) = Asin (2πx
L)+Bsin (3πx
L),
where Aand Bare normalization constants. Find the energies corresponding to the particle in
this potential well.
Ans. To find the energies of the particle in the potential well, we need to solve the time-
independent Schrödinger equation for the system. The general form of the time-independent
Schrödinger equation for a one-dimensional system is given by
ˆ
Hψ(x) = Eψ(x),
where ˆ
His the Hamiltonian operator, Eis the energy, and ψ(x)is the wave function.
1. Normalize the wave function:
To normalize the wave function ψ(x), we need to ensure that the integral of the square of
the wave function over the entire range is equal to 1:
∫L
0|ψ(x)|2dx = 1.
Given the form of the wave function ψ(x), we have
∫L
0|ψ(x)|2dx =∫L
0|Asin (2πx
L)+Bsin (3πx
L)|2dx = 1.
2. Apply the normalization condition:
Using the orthogonality of sine functions, we can simplify the integral. The integral of a
product of sines over a full period is zero unless they are the same function. Therefore, the
normalization condition would require that the two terms in ψ(x)are orthogonal:
∫L
0
sin (2πx
L)sin (3πx
L)dx.
3. Solve for the normalization constants:
From the orthogonality condition above, we get
∫L
0
sin (2πx
L)sin (3πx
L)dx = 0.
This implies that the two terms in the wave function are orthogonal, so the term involving A
will not contribute to the normalization condition, and we only need to normalize the term with
coefficient B.
4. Calculate the energies:
After normalizing the wave function, we can then solve the time-independent Schrödinger
equation to find the energies of the particle in the potential well. The general form of the kinetic
energy operator in one dimension is ˆ
K=−ˉh2
2m
d2
dx2.
By solving the Schrödinger equation ˆ
Hψ(x) = Eψ(x)with the given potential well and
normalized wave function, we can find the values of Ecorresponding to the allowed energy levels.
13. Question: Determine the energy levels of a particle in a one-dimensional box of length Lif
the potential energy is given by V(x) = 1
2kx2, where kis a positive constant.
Ans. Let’s consider the time-independent Schrödinger equation for this system, where the total
energy E=T+V.
1. The Schrödinger equation for this system is given by:
−ˉh2
2m
d2ψ(x)
dx2+1
2kx2ψ(x) = Eψ(x)
2. We can rewrite this equation as:
d2ψ(x)
dx2=−2m
ˉh2(E−1
2kx2)ψ(x)
3. To solve this differential equation, we make the substitution α=√mk
ˉh2and λ= 2α√E.
4. The general solution to the differential equation is given by:
ψ(x) = Ae−αx2/2Hn(αx)
where Hn(·)is the n-th Hermite polynomial.
5. The boundary conditions for the particle in a box are that ψ(0) = ψ(L) = 0.
6. Therefore, we find that the energy levels are quantized as:
En=ˉhω (n+1
2)
with corresponding wavefunctions:
ψn(x) = √2
Lsin (nπx
L)
where nis a positive integer representing different energy levels, ωis the angular frequency given
by ω=α
L, and Ais the normalization constant.
14. Question 14: A particle is in a one-dimensional potential well given by the function
V(x) = {0,0≤x≤a
V0, x > a , where aand V0are positive constants. Determine the energy levels
of the particle.
Ans. To determine the energy levels of the particle in the given potential well, we need to solve
the time-independent Schrödinger equation and apply the appropriate boundary conditions.
1. Setup the Schrödinger equation: The time-independent Schrödinger equation in one
dimension is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
In the regions where V(x) = 0, the equation becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
d2ψ
dx2+2mE
ˉh2ψ= 0
2. Solve the Schrödinger equation in different regions: Since V(x) = 0 for 0≤x≤a,
the general solution to the Schrödinger equation in this region is:
ψ1(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2
In the region x > a where V(x) = V0, the general solution is:
ψ2(x) = Ce−κx +Deκx
where κ=√2m(V0−E)
ˉh2
3. Apply continuity conditions: At x=a, the wavefunction and its derivative must be
continuous, i.e., ψ1(a) = ψ2(a)and ψ′
1(a) = ψ′
2(a).
4. Determine the energy levels: Solve for the energy levels by solving the transcendental
equations obtained from the boundary conditions.
These steps will lead to determining the discrete energy levels of the particle in the given
potential well.
15. Question 15: An electron is in a one-dimensional infinite potential well of width L. Calculate
the energy levels for this system.
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 of the energy level.
1. Write down the Schrödinger equation for this system. The time-independent
Schrödinger equation for a one-dimensional infinite potential well is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
2. Formal solution for the wavefunction. The wavefunction ψ(x)for an electron in an
infinite potential well can be written as:
ψ(x) = Asin (nπx
L)
where Ais the normalization constant and nis the quantum number.
3. Apply boundary conditions. For an infinite potential well, the boundary conditions are
ψ(0) = 0 and ψ(L) = 0. Applying these conditions gives:
sin (nπ ·0
L)= 0 ⇒n= 0
sin (nπ ·L
L)= 0 ⇒nπ =mπ
4. Determine the allowed values of energy. The energy levels of the system can be
found by substituting the wavefunction form back into the Schrödinger equation and solving for
En. The energy levels are given by:
En=n2π2ˉh2
2mL2
where n= 1,2,3, ... is the quantum number.
Therefore, the energy levels for an electron in a one-dimensional infinite potential well of
width Lare quantized and given by En=n2π2ˉh2
2mL2.
16. Question text here An electron in a one-dimensional infinite potential well is in the n= 4
energy level. Suppose the width of the well is a. Calculate the energy of the electron in terms of
a.
Ans. Step 1. The energy levels for an electron in a one-dimensional infinite potential well are
given by the formula: En=n2h2
8ma2, where nis the quantum number, his Planck’s constant, mis
the mass of the electron, and ais the width of the well. Step 2. Substituting n= 4 into the
formula, we get: E4=42h2
8ma2=16h2
8ma2=2h2
ma2.Step 3. Therefore, the energy of the electron in
the n= 4 level in terms of ais 2h2
ma2.
17. Question: Determine the energy levels of a particle in a one-dimensional box of length L,
where Lis given and the particle has a mass m.
Ans. Let nbe the quantum number representing the energy level of the particle. The energy
levels Encan be found using the formula for the energy eigenvalues of a particle in a one-
dimensional box:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck constant and mis the mass of the particle.
Solution: 1. Determine the formula for the energy levels of a particle in a one-dimensional
box. Given that the energy eigenvalues of a particle in a one-dimensional box are given by the
formula:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck constant, mis the mass of the particle, Lis the length of the
box, and nis the quantum number representing the energy level.
2. Substitute the given values into the formula. For a particle in a one-dimensional box with
a length L, the mass m, and the quantum number n, we can substitute these values into the
formula to find the energy levels En.
3. Determine the energy levels. By substituting the given values into the formula, we find
that the energy levels Enare:
En=n2π2ˉh2
2mL2
This formula gives the quantized energy levels of the particle in the one-dimensional box.
18. Question:
An electron is confined within a one-dimensional box of length 2 nm. Determine the energy
levels of the electron in this box using the Schrödinger equation.
Ans. Let’s start by setting up the Schrödinger equation for the one-dimensional box:
1. The Schrödinger equation for a particle in a one-dimensional box is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
2. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉhis the wave number and Aand Bare coefficients to be determined.
3. Applying the boundary conditions for a particle in a box, we have ψ(0) = ψ(2 nm)=0.
This gives us:
{B= 0
k=nπ
2nm ,where n= 1,2,3, ...
4. Substituting kback into the general solution, we have:
ψn(x) = Asin (nπx
2nm)
5. To find the energy levels En, we use the relation En=ˉh2k2
n
2m:
En=ˉh2
2m(nπ
2nm)2
Thus, the energy levels are given by:
En=n2π2ˉh2
8mnm2
where n= 1,2,3, ....
19. Suppose a particle of mass mis confined to a one-dimensional box of length L. Determine
the energy levels of the particle in terms of m,L, and Planck’s constant h.
Ans. To determine the energy levels of the particle in the one-dimensional box, we must solve
the time-independent Schrödinger equation for the particle within the box and apply appropriate
boundary conditions. This will result in the quantization of the energy levels. Let’s denote the
energy levels as Enwhere nis a positive integer.
1. Writing down the time-independent Schrödinger equation:
The time-independent Schrödinger equation for a one-dimensional box is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function of the particle, Eis the energy of the particle, and ˉhis the
reduced Planck’s constant.
2. Applying boundary conditions:
In the one-dimensional box, the particle is confined between x= 0 and x=L. The boundary
conditions are: - ψ(0) = 0 (wave function at x= 0 is zero) - ψ(L) = 0 (wave function at x=L
is zero)
3. Solving the Schrödinger equation:
Let ψ(x) = Asin(kx)be the wave function, where Ais the normalization constant and kis
the wave number. Substituting this into the Schrödinger equation, we get:
k=nπ
L,where n= 1,2,3, . . .
4. Calculating the energy levels:
The energy levels Enare given by:
En=n2π2ˉh2
2mL2,where n= 1,2,3, . . .
Therefore, the energy levels of the particle in the one-dimensional box are quantized and given
by the expression above.
20. Question: Find the energy levels of a particle in a one-dimensional box of length L= 2
nm. The particle has a mass of m= 9.11 ×10−31 kg and the potential energy outside the box
is infinite.
Ans. Step-by-step solution: 1. The energy levels of a particle in a one-dimensional box are
given by the formula:
En=n2π2ˉh2
2mL2
where nis the quantum number, ˉhis the reduced Planck constant (1.05 ×10−34 J s), mis the
mass of the particle, and Lis the length of the box.
2. Substituting the given values into the formula, we have:
En=n2π2(1.05 ×10−34 J s)2
2×9.11 ×10−31 kg ×(2 ×10−9m)2
3. Simplifying the expression, we get:
En=n2π2×1.1025 ×10−68
3.6444 ×10−38
4. Further simplifying, we find:
En=1.1025π2×10−68n2
3.6444 ×10−38
5. Therefore, the energy levels are expressed as:
En≈3.0214 ×10−31n2
1
where nis the quantum number.
21. What is the energy of a particle trapped in a one-dimensional potential well of width a, if
the particle has mass m?
Ans. Let’s denote the energy of the particle as E. 1. The Schrödinger equation for a
particle in a potential well is given by −ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ, where V(x)is the potential
function. Inside the well, V(x)=0, so the equation simplifies to −ˉh2
2m
d2ψ
dx2=Eψ. 2. We can
rewrite this equation as d2ψ
dx2=−2mE
ˉh2ψ. 3. The general solution to this differential equation is
ψ(x) = Asin(kx) + Bcos(kx), where k=√2mE
ˉh2. Since the particle is trapped in a well of
width a, we must have ψ(0) = ψ(a)=0. 4. Applying the boundary conditions, we find that
B= 0 (since cos(ka) = 0) and k=nπ
a, where nis a positive integer. 5. Therefore, the energy
levels are quantized and given by En=ˉh2π2n2
2ma2.
22. Let’s consider an electron in a one-dimensional box of length L.
Question 22: Calculate the energy levels for an electron in a one-dimensional box of length
Lwith fixed endpoints.
Ans. To find the energy levels of the electron in the box, we can use the time-independent
Schrödinger equation for a particle in a box:
−ˉh2
2m
d2ψ
dx2=Eψ
where Eis the total energy of the system, ˉhis the reduced Planck’s constant, mis the mass of
the electron, xis the position, and ψis the wavefunction.
1. Boundary conditions: At the endpoints of the box (x= 0 and x=L), the wavefunction
must be zero, since the particle is confined to the box:
ψ(0) = 0 and ψ(L) = 0
2. Form of the wavefunction: The general form of the wavefunction inside the box is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉhand A,Bare constants.
3. Applying boundary conditions: Using the boundary conditions ψ(0) = 0 and ψ(L) = 0:
ψ(0) = 0 =⇒B= 0
ψ(L) = 0 =⇒Asin(kL) = 0
Since sin(kL) = 0, then kL =nπ, where nis a positive integer.
4. Determining energy levels: From kL =nπ, we can solve for E:
√2mE
ˉh·L=nπ
√2mE =nπˉh
L
2mE =(nπˉh
L)2
E=n2π2ˉh2
2mL2
Therefore, the energy levels for the electron in a one-dimensional box with fixed endpoints
are given by:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the energy level.
23. Question: Find the energy levels of a particle in a one-dimensional box of length L. The
particle has mass mand the box has infinite potential walls.
Ans. Let’s denote the energy levels as Enwhere nis a positive integer. The energy levels are
given by the formula:
En=n2ˉh2π2
2mL2
where ˉhis the reduced Planck’s constant.
Solution: 1. The time-independent Schrödinger equation for a particle in a box is:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
2. Applying the boundary conditions for this case (infinite potential walls), we have ψ(0) = 0
and ψ(L) = 0. This leads to the general solution:
ψ(x) = Asin (nπx
L)
3. Now, we apply the normalization condition:
∫L
0|ψ(x)|2dx = 1
4. Substituting the general solution into the normalization condition and solving for A, we
find:
A=√2
L
5. The energy levels are obtained by solving the time-independent Schrödinger equation with
the general solution. Thus, we get:
En=n2ˉh2π2
2mL2
Therefore, the energy levels for a particle in a one-dimensional box of length Lare given by
En=n2ˉh2π2
2mL2.
24. Question 24: Consider an electron confined within a one-dimensional infinite potential well
of width L. Determine the energy levels Enfor the electron in terms of Planck’s constant h,
electron mass meand the width Lof the well.
Ans. To find the energy levels of the electron in the one-dimensional infinite potential well,
we can use the Schrödinger equation and the boundary conditions at the edges of the well.
The general form of the time-independent Schrödinger equation for a particle of mass min a
one-dimensional region where the potential energy is zero is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
Where Eis the energy of the particle and ψis the wave function of the particle.
1. Inside the well, the potential energy is zero, so the equation simplifies to:
d2ψ
dx2=−2mE
ˉh2ψ
2. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
Where k=√2mE
ˉh2.
3. Applying the boundary conditions at x= 0 and x=L, we find that ψ(0) = ψ(L) = 0,
which gives us:
B= 0
kL =nπ
Where n= 1,2,3, ....
4. Solving for Ein terms of nand L, we have:
k=nπ
L
E=ˉh2π2n2
2mL2
Therefore, the energy levels Enfor the electron in the one-dimensional infinite potential well
are given by:
En=ˉh2π2n2
2mL2
25. Question: Consider an electron confined in a one-dimensional infinite square well potential
with width L. Calculate the energy levels of the system in terms of the fundamental parameters.
Ans. Step-by-step solution: The energy levels of an electron in an infinite square well potential
are quantized and given by the formula:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number, ˉhis the reduced Planck’s
constant, mis the mass of the electron, and Lis the width of the square well.
1. First, we need to derive the expression for the energy levels of the electron in the infinite
square well potential. In one dimension, the Schrödinger equation for the system is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
2. Inside the well, the potential energy is zero and the equation becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
3. The general solution to this differential equation is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
4. Applying boundary conditions where ψ(0) = 0 and ψ(L) = 0, we find that kL =nπ for
integer n. This gives the quantized energy levels as:
En=n2π2ˉh2
2mL2
Therefore, the energy levels of an electron in an infinite square well potential are quantized
and given by the above formula.
26. Question: Let’s consider a particle in a one-dimensional box of length L. Suppose the
particle has a mass mand the box has an infinite potential barrier at x= 0 and x=L. The
wave function of the particle is given by:
ψ(x) = Asin (nπx
L)
where Ais the normalization constant and nis a positive integer. Find the energy levels of the
particle in terms of m,L, and n.
Ans. Let’s start by finding the normalization constant A. 1. Normalize the wave function:
∫L
0|ψ(x)|2dx = 1
∫L
0|Asin(nπx
L)|2dx = 1
A2∫L
0
sin2(nπx
L)dx = 1
A2∫L
0
1−cos(2nπx
L)
2dx = 1
A2[x
2−L
2nπ sin(2nπx
L)]L
0= 1
A2[L
2−L
2nπ sin(2nπ)] = 1
Since sin(2nπ) = 0, we get:
A2·L
2= 1
A=√2
L
2. Calculate the energy levels: The energy of the particle can be calculated using the time-
independent Schrödinger equation:
Enψ(x) = −ˉh2
2m
d2ψ(x)
dx2
Substitute ψ(x)into the equation:
EnAsin (nπx
L)=−ˉh2
2mAn2π2
L2sin (nπx
L)
En=n2π2ˉh2
2mL2
Hence, the energy levels of the particle are given by:
En=n2π2ˉh2
2mL2
27. Question 27:
A particle of mass mis confined to move on a ring of radius a. Find the energy levels for this
system in terms of quantum number n.
Ans. To find the energy levels for a particle confined to move on a ring of radius a, we need
to consider the quantization of angular momentum. The angular momentum Lof the particle
is quantized as L=nˉhwhere nis a non-negative integer. The energy levels Enare given by
En=L2
2I, where Iis the moment of inertia of the particle.
1. Finding the moment of inertia: The moment of inertia Ifor a particle of mass m
moving on a ring of radius ais given by I=ma2.
2. Substituting Land Iinto the energy expression: Substitute L=nˉhand I=ma2
into the expression for energy levels, we have:
En=(nˉh)2
2ma2=n2ˉh2
2ma2
Therefore, the energy levels Enfor a particle confined to move on a ring of radius aare given
by En=n2ˉh2
2ma2, where nis a non-negative integer representing the quantum number.
28. Let’s consider a one-dimensional quantum harmonic oscillator with the Hamiltonian operator
given by ˆ
H=−ˉh2
2m
d2
dx2+1
2kx2, where ˉ
his the reduced Planck constant, mis the mass of the
particle, kis the spring constant, and xis the position operator.
Question: Find the energy levels of the quantum harmonic oscillator described by the Hamil-
tonian operator ˆ
H.
Ans. To find the energy levels of the quantum harmonic oscillator, we need to solve the time-
independent Schrödinger equation: ˆ
Hψ =Eψ, where ψis the wave function and Eis the energy
eigenvalue.
1. Form of the wave function: Let’s assume the wave function has the form ψ(x) =
Ce−αx2, where Cis the normalization constant and αis a constant to be determined.
2. Calculate the second derivative: The second derivative of the wave function with
respect to xis given by:
d2
dx2ψ(x) = −2αCe−αx2(2αx2−1)
3. Substitute into the Schrödinger equation: Substitute the wave function and its
second derivative into the Schrödinger equation. We get:
−ˉh2
2m(−2αCe−αx2(2αx2−1)) + 1
2kx2Ce−αx2=ECe−αx2
4. Simplify the equation: Simplify the equation by dividing through by Ce−αx2and
rearrange to get:
α2ˉh2−4α2ˉh2x2+ 2mE =kx2
5. Equating coefficients: In order to find the values of αand E, we compare coefficients
of x2. This gives us:
α2ˉh2=k
2m
E=ˉhα
2m
6. Energy levels: Substitute the value of αinto the expression for Eto find the energy
levels:
En=(n+1
2)ˉhω
where ω=√k
mis the angular frequency of the oscillator and n= 0,1,2, ....
29. Question: Determine the energy levels of a particle confined to a one-dimensional box of
length L. If the particle has a mass mand the box has infinitely high barriers, find the energy
levels in terms of the quantum number n.
Ans. Step-by-step solution: 1. The energy levels of a particle in a one-dimensional box are
given by the formula:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number. 2. Substituting the given values
into the formula, we have:
En=n2π2ˉh2
2mL2
3. Simplifying the expression further, we get:
En=π2ˉh2
2mL2·n2
Therefore, the energy levels of a particle confined to a one-dimensional box of length Lare
quantized and given by En=π2ˉh2
2mL2·n2, where nis a positive integer representing the quantum
number.
30. Question 30: A particle of mass mis in a one-dimensional infinite square well potential of
width L. Find the energy levels of the particle in terms of m,L, and fundamental constants.
Ans. To find the energy levels of a particle in an infinite square well potential, we solve the
time-independent Schrödinger equation for the particle within the well.
1. Setup: The time-independent Schrödinger equation for this system is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function, Eis the energy of the particle, and ˉhis the reduced Planck
constant.
2. Applying Boundary Conditions: Since the potential energy is infinite at the boundaries
of the well, the wave function must be zero at x= 0 and x=L. Thus, we have boundary
conditions ψ(0) = ψ(L) = 0.
3. Solving the Schrödinger Equation: The general solution to the Schrödinger equation
is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2and Aand Bare constants.
Applying the boundary conditions ψ(0) = ψ(L) = 0, we get:
{Bcos(0) = 0 =⇒B= 0
Asin(kL) = 0 =⇒kL =nπ (for n= 1,2,3, ...)
4. Determining Energy Levels: From the second boundary condition, we find the quantized
energy levels:
kL =nπ =⇒√2mE
ˉh2=nπ
L
Solving for Egives:
En=n2π2ˉh2
2mL2for n= 1,2,3, ...
Therefore, the energy levels of the particle in the infinite square well potential are quantized
and given by En=n2π2ˉh2
2mL2.
31. Question 31: Consider a one-dimensional infinite square well potential with width a. Find
the first three energy levels of a particle in this potential well in terms of ˉhand a.
Ans. Let’s denote the ground state energy level as E1, the first excited state as E2, and the
second excited state as E3. We can find these energy levels by solving the time-independent
Schrödinger equation for the infinite square well potential.
1. Finding the ground state energy level (E1): For the ground state, the wavefunction
of the particle is given by ψ1(x) = √2
asin (πx
a). The energy associated with this state is given
by E1=ˉh2π2
2ma2.
2. Finding the first excited state energy level (E2): For the first excited state, the
wavefunction of the particle is given by ψ2(x) = √2
asin (2πx
a). The energy associated with this
state is E2=4ˉh2π2
2ma2.
3. Finding the second excited state energy level (E3): For the second excited state,
the wavefunction of the particle is given by ψ3(x) = √2
asin (3πx
a). The energy associated with
this state is E3=9ˉh2π2
2ma2.
Therefore, the first three energy levels for a particle in a one-dimensional infinite square well
potential of width aare: E1=ˉh2π2
2ma2,E2=4ˉh2π2
2ma2, and E3=9ˉh2π2
2ma2.
32. Find the energy levels of a particle in a one-dimensional box of length Lwith an infinite
potential energy outside the box.
Ans. Let’s denote the energy levels of the particle in the box as En, with n= 1,2,3, . . .. The
energy levels are given by the formula:
En=n2π2ˉh2
2mL2,
where mis the mass of the particle and ˉhis the reduced Planck constant.
33. Question 33:
An electron is confined to move in one-dimension in an infinite potential well of width L.
Calculate the energy levels of the electron in terms of the fundamental constants.
Ans. To calculate the energy levels of the electron in an infinite potential well of width L:
1. The potential energy in the well is zero and the wave function must vanish at the walls,
so the energy levels are quantized. The allowed values of energy are given by:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number associated with the energy
level, ˉhis the reduced Planck constant, and mis the mass of the electron.
2. The lowest possible energy level corresponds to n= 1:
E1=π2ˉh2
2mL2
3. The second energy level corresponds to n= 2:
E2=4π2ˉh2
2mL2= 4E1
4. The third energy level corresponds to n= 3:
E3=9π2ˉh2
2mL2= 9E1
5. And so on, the energy levels increase with the square of the quantum number n.
34. Let’s consider a particle of mass mmoving in one dimension under the influence of a
potential V(x) = 1
2kx2, where kis a positive constant.
Question: Find the energy levels of the particle in terms of kand fundamental constants.
Ans. To find the energy levels of the particle, we can solve the time-independent Schrödinger
equation for this system. The Schrödinger equation for a one-dimensional system is given by:
−ˉh2
2m
d2ψ(x)
dx2+V(x)ψ(x) = Eψ(x)
where Eis the energy eigenvalue and ψ(x)is the wave function.
1. Writing the Schrödinger equation: Substitute the expression for V(x)into the
Schrödinger equation to obtain:
−ˉh2
2m
d2ψ(x)
dx2+1
2kx2ψ(x) = Eψ(x)
2. Simplifying the equation: Multiply through by 2m
ˉh2to simplify the equation:
−d2ψ(x)
dx2+k
ˉh2x2ψ(x) = 2m
ˉh2Eψ(x)
3. Non-dimensionalize the equation: Introduce a new dimensionless variable ξdefined
by ξ=√k
ˉhxto make the equation dimensionless. This gives:
−d2ψ(ξ)
dξ2+ξ2ψ(ξ) = λψ(ξ)
where λ=2mE
ˉh2.
4. Solving the equation: The solution to this equation can be expressed in terms of Hermite
polynomials Hn(ξ). The energy eigenvalues are quantized and given by:
En=(n+1
2)ˉh√k
2m
where n= 0,1,2, ... are the energy levels or quantum numbers.
Therefore, the energy levels of the particle in terms of kand fundamental constants are
En=(n+1
2)ˉh√k
2m.
35. Question 35:
A particle of mass mis confined to a one-dimensional box of length L. Consider the energy
levels of the particle in this box. Show that the energy levels Enare quantized according to the
equation:
En=n2ˉh2π2
2mL2
where nis a positive integer representing the quantum number of the energy level.
Ans. To derive the quantization of energy levels in a one-dimensional box, we will use the
time-independent Schrödinger equation for the particle:
1. The time-independent Schrödinger equation is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wavefunction and Eis the energy of the particle.
2. In the case of a particle in a one-dimensional box of length L, the wavefunction must
satisfy boundary conditions at x= 0 and x=L. These conditions are: ψ(0) = ψ(L) = 0.
3. The general solution to the wavefunction inside the box is given by:
ψ(x) = Asin (nπx
L)
where Ais a normalization constant and nis a positive integer.
4. Plugging the wavefunction into the Schrödinger equation, we get:
−ˉh2
2mA(nπ
L)2
sin (nπx
L)=EA sin (nπx
L)
5. Simplifying the equation above, we find the energy levels En:
En=n2ˉh2π2
2mL2
Thus, the energy levels of a particle in a one-dimensional box are quantized according to
the equation above. Each energy level is indexed by a positive integer n, representing different
quantum states of the particle.
4. Since Ccannot be zero for non-trivial solutions, we have k′a=nπ, where nis a positive
integer. 5. Therefore, the energy levels Enare given by:
En=n2π2ˉh2
2ma2
where n= 1,2,3, ... represents the quantum number corresponding to the energy levels. 6. This
quantization of energy levels is a consequence of the particle being confined within the potential
well, leading to discrete allowed energy values.
3. Question: Determine the energy levels of a particle in a one-dimensional box of length L,
where the particle has a mass mand is subject to a potential energy V(x) = 0 inside the box
and V(x) = ∞outside the box.
Ans. Let’s solve this using the time-independent Schrödinger equation and boundary conditions.
1. The time-independent Schrödinger equation for this system 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, and mis the mass of the particle.
2. Inside the box (0<x<L), the potential energy V(x)=0, so the Schrödinger equation
becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
3. The general solution to the Schrödinger equation inside the box is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉhand Aand Bare constants to be determined.
4. Applying the boundary conditions: ψ(0) = 0 and ψ(L) = 0, we find:
{B= 0
kL =nπ where n= 1,2,3, . . .
5. From the second boundary condition, we get the quantization condition for the energy
levels:
knL=nπ =⇒√2mEnL
ˉh=nπ
6. Solving for the energy levels En, we find:
En=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.
4. Find the energy levels of a particle confined to a one-dimensional infinite potential well with
width L.
Ans. To find the energy levels of a particle in a one-dimensional infinite potential well, we can
use the Schrödinger equation and the boundary conditions to solve for the allowed energy levels.
1. The general form of the wave function inside the well is Ψ(x) = Asin(kx) + Bcos(kx),
where k=√2mE
ˉh2.
2. Applying the boundary conditions, Ψ(0) = 0 and Ψ(L) = 0, we find that k=nπ
L, where
n= 1,2,3, ... (quantization condition).
3. The energy levels Enare then given by En=n2π2ˉh2
2mL2.
Therefore, the energy levels of a particle confined to a one-dimensional infinite potential well with
width Lare En=n2π2ˉh2
2mL2, where n= 1,2,3, ....
5. Question: Determine the energy levels for a particle in a one-dimensional box of length L
with infinitely high walls.
Ans. Let’s denote the allowed energy levels as En. The energy levels can be calculated using
the formula:
En=n2h2
8mL2
where nis a positive integer representing the quantum number of the energy level, his the
Planck constant, and mis the mass of the particle.
Step 1. Given that the walls are infinitely high, the particle’s wavefunction must go to zero
at the walls. Thus, we can write the wavefunction as:
ψ(x) = Asin (nπx
L)
where Ais a normalization constant.
Step 2. The probability density function, |ψ(x)|2, must be normalized. Therefore, we find
Ain such a way that:
∫L
0|ψ(x)|2dx = 1
It follows that A=√2
L.
Step 3. To determine the energy levels, we must find the expectation value of the energy,
⟨E⟩, for an arbitrary wavefunction ψ(x). This can be expressed as:
⟨E⟩=∫ψ∗(x)(−ˉh2
2m
d2
dx2)ψ(x)dx
∫|ψ(x)|2dx
Substitute ψ(x)into the equation and simplify.
Step 4. Applying the kinetic and potential energy operators to ψ(x), we obtain the Schrödinger
equation. Solve the resulting differential equation to find the energy levels En.
Step 5. Finally, use the formula En=n2h2
8mL2to determine the values of the energy levels En
for the particle in the box.
6. Question 6:
Consider a particle in a one-dimensional box of length L. Calculate the first four energy levels
of the particle in terms of the particle’s mass mand the width of the box L.
Ans. To solve for the energy levels of a particle in a one-dimensional box, we can use the
Schrödinger equation and apply the boundary conditions.
1. Setting up the Schrödinger equation: The time-independent Schrödinger equation
for the particle in a box is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
2. Applying boundary conditions: The boundary conditions for a particle in a box are
ψ(0) = ψ(L) = 0. This implies that the wave function must be zero at x= 0 and x=L.
3. Solving for the energy levels: Let’s assume that the wave function is in the form
ψ(x) = Asin(kx). Substitute ψ(x) = Asin(kx)into the Schrödinger equation and apply the
boundary conditions to solve for kand E.
At x= 0:ψ(0) = Asin(0) = 0 This implies that A= 0 is not a permissible value. Therefore,
for non-trivial solutions, we have sin(k·0) = 0, which gives k=nπ/L, where nis a positive
integer.
The energy levels are given by En=ˉh2k2
2m=ˉh2π2n2
2mL2.
4. Calculating the first four energy levels: For n= 1,E1=ˉh2π2
2mL2. For n= 2,
E2=4ˉh2π2
2mL2. For n= 3,E3=9ˉh2π2
2mL2. For n= 4,E4=16ˉh2π2
2mL2.
Therefore, the first four energy levels of the particle in the box are: E1=ˉh2π2
2mL2,E2=4ˉh2π2
2mL2,
E3=9ˉh2π2
2mL2,E4=16ˉh2π2
2mL2.
7. Question: Find the energy levels of a particle in a one-dimensional box of length Lif the
particle has mass mand the potential energy inside the box is zero.
Ans. Let nbe the quantum number representing the energy level. The energy levels Enare
given by:
En=n2π2ˉh2
2mL2
where nis a positive integer.
Step 1. Write down the time-independent Schrödinger equation for the particle in a one-
dimensional box:
−ˉh2
2m
d2ψ
dx2=Eψ
Step 2. The general solution to this differential equation is:
ψ(x) = Asin (nπx
L)+Bcos (nπx
L)
where Aand Bare constants to be determined.
Step 3. Apply the boundary conditions to determine the constants Aand B. The boundary
conditions are:
ψ(0) = 0 and ψ(L) = 0
Step 4. The boundary condition ψ(0) = 0 gives:
0 = Asin(0) + Bcos(0) = B
Step 5. Substituting B= 0 back into the general solution gives:
ψ(x) = Asin (nπx
L)
Step 6. Applying the boundary condition ψ(L) = 0:
0 = Asin(nπ)
Since sin(nπ) = 0 for integer n, we have n= 1,2,3, . . .
Step 7. Therefore, the energy levels are given by:
En=n2π2ˉh2
2mL2
where nis a positive integer.
8. Question 8:
Consider a particle confined to a one-dimensional box of length L. The potential in the box
is given by V(x) = 0 for 0< x < L and infinite elsewhere. Determine the energy levels of the
particle in terms of n.
Ans. To determine the energy levels of the particle in the one-dimensional box, we need to
solve the time-independent Schrödinger equation for the system:
1. The time-independent Schrödinger equation is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
2. For 0< x < L, the potential is zero, so the Schrödinger equation becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
3. This is a second-order ordinary differential equation. The general solution is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2
4. To satisfy the boundary conditions at x= 0 and x=L,ψ(0) = 0 and ψ(L)=0, the
boundary conditions ψ(0) = 0 gives B= 0.
5. The boundary condition at x=Limplies sin(kL)=0, which gives kL =nπ for
n= 1,2,3, ....
6. Solving kL =nπ for E, we get the energy levels:
En=n2π2ˉh2
2mL2
where n= 1,2,3, ...
7. Therefore, the energy levels of the particle in the one-dimensional box are given by En=
n2π2ˉh2
2mL2for n= 1,2,3, ....
9. Question: Determine the energy levels of a particle in a one-dimensional box of length
Lif the potential energy function is given by V(x) = V0for 0< x < L/3,V(x)=0for
L/3 < x < 2L/3, and V(x) = 2V0for 2L/3 < x < L, where V0is a constant.
Ans. Let’s denote the energy levels as Enand the corresponding wavefunctions as ψn(x). The
general form of the wavefunction inside the box is ψn(x) = Asin(kx) + Bcos(kx), where A,B,
and kare constants.
1. Find the wavefunction and energy levels for 0<x<L/3:Inside this region,
V(x) = V0. The Schrödinger equation is −ˉh2
2m
d2ψ
dx2+V0ψ=Eψ. Solving this differential equation
gives ψ1(x) = A1sin(kx) + B1cos(kx)and E1=ˉh2k2
2m+V0.
2. Find the wavefunction and energy levels for L/3 <x<2L/3:Inside this region,
V(x) = 0. The Schrödinger equation is −ˉh2
2m
d2ψ
dx2=Eψ. Solving this differential equation gives
ψ2(x) = A2sin(kx) + B2cos(kx)and E2=ˉh2k2
2m.
3. Find the wavefunction and energy levels for 2L/3 < x < L:Inside this region,
V(x)=2V0. The Schrödinger equation is −ˉh2
2m
d2ψ
dx2+ 2V0ψ=Eψ. Solving this differential
equation gives ψ3(x) = A3sin(kx) + B3cos(kx)and E3=ˉh2k2
2m+ 2V0.
We need to apply continuity conditions on the wavefunction and its derivative at the bound-
aries x=L/3 and x= 2L/3 to determine the values of kand E.
10. Question: Determine the energy levels of an electron in a one-dimensional infinite square
well of width L. Express your answer in terms of the quantum number n.
Ans. Let’s denote the energy levels of an electron in a one-dimensional infinite square well as
En. The general formula for the energy levels in an infinite square well is given by:
En=n2π2ˉh2
2mL2
where: En= energy level of the electron in the n-th quantum state, n= quantum number
(positive integer), ˉh= reduced Planck’s constant (1.0545718 ×10−34 J s), m= mass of the
electron (9.10938356 ×10−31 kg), L= width of the infinite square well.
Step 1. Start with the general formula for energy levels:
En=n2π2ˉh2
2mL2
Step 2. Substitute the given values into the formula:
En=n2π2(1.0545718 ×10−34)2
2(9.10938356 ×10−31)L2
Step 3. Simplify the expression to get the energy level in terms of n:
En=n2π2×(1.1100599 ×10−68)
2(9.10938356 ×10−31)L2
Step 4. Finally, express the energy levels in terms of the quantum number n:
En=n2π2×(1.1100599 ×10−68)
2mL2
11. Question: Consider an electron in a one-dimensional infinite potential well of width L.
Determine the energy levels for the electron in terms of n, where nis a positive integer representing
the quantum number.
Ans. Step-by-step solution: 1. In a one-dimensional infinite potential well, the Schrödinger
equation for the electron can be written as:
−ˉh2
2m
d2ψ
dx2=Eψ
Where ψ(x)is the wave function, Eis the energy of the electron, ˉhis the reduced Planck
constant, mis the mass of the electron, and x∈[0, L]within the well. 2. The general solution
to this differential equation is given by:
ψn(x) = Asin (nπx
L)
Where nis a positive integer representing the quantum number and Ais a normalization constant.
3. To determine the energy levels for the electron, we substitute the wave function into the
Schrödinger equation:
−ˉh2
2m
d2
dx2(Asin (nπx
L))=EA sin (nπx
L)
4. Simplifying the differential equation:
(nπˉh
L)2
=2mE
ˉh2
E=n2π2ˉh2
2mL2
5. Therefore, the energy levels for the electron in the one-dimensional infinite potential well are
given by:
En=n2π2ˉh2
2mL2
Where nis a positive integer representing the quantum number.
12. Question 12: Consider a particle in a one-dimensional infinite potential well of width L.
The wave function of the particle in this potential well is given by
ψ(x) = Asin (2πx
L)+Bsin (3πx
L),
where Aand Bare normalization constants. Find the energies corresponding to the particle in
this potential well.
Ans. To find the energies of the particle in the potential well, we need to solve the time-
independent Schrödinger equation for the system. The general form of the time-independent
Schrödinger equation for a one-dimensional system is given by
ˆ
Hψ(x) = Eψ(x),
where ˆ
His the Hamiltonian operator, Eis the energy, and ψ(x)is the wave function.
1. Normalize the wave function:
To normalize the wave function ψ(x), we need to ensure that the integral of the square of
the wave function over the entire range is equal to 1:
∫L
0|ψ(x)|2dx = 1.
Given the form of the wave function ψ(x), we have
∫L
0|ψ(x)|2dx =∫L
0|Asin (2πx
L)+Bsin (3πx
L)|2dx = 1.
2. Apply the normalization condition:
Using the orthogonality of sine functions, we can simplify the integral. The integral of a
product of sines over a full period is zero unless they are the same function. Therefore, the
normalization condition would require that the two terms in ψ(x)are orthogonal:
∫L
0
sin (2πx
L)sin (3πx
L)dx.
3. Solve for the normalization constants:
From the orthogonality condition above, we get
∫L
0
sin (2πx
L)sin (3πx
L)dx = 0.
This implies that the two terms in the wave function are orthogonal, so the term involving A
will not contribute to the normalization condition, and we only need to normalize the term with
coefficient B.
4. Calculate the energies:
After normalizing the wave function, we can then solve the time-independent Schrödinger
equation to find the energies of the particle in the potential well. The general form of the kinetic
energy operator in one dimension is ˆ
K=−ˉh2
2m
d2
dx2.
By solving the Schrödinger equation ˆ
Hψ(x) = Eψ(x)with the given potential well and
normalized wave function, we can find the values of Ecorresponding to the allowed energy levels.
13. Question: Determine the energy levels of a particle in a one-dimensional box of length Lif
the potential energy is given by V(x) = 1
2kx2, where kis a positive constant.
Ans. Let’s consider the time-independent Schrödinger equation for this system, where the total
energy E=T+V.
1. The Schrödinger equation for this system is given by:
−ˉh2
2m
d2ψ(x)
dx2+1
2kx2ψ(x) = Eψ(x)
2. We can rewrite this equation as:
d2ψ(x)
dx2=−2m
ˉh2(E−1
2kx2)ψ(x)
3. To solve this differential equation, we make the substitution α=√mk
ˉh2and λ= 2α√E.
4. The general solution to the differential equation is given by:
ψ(x) = Ae−αx2/2Hn(αx)
where Hn(·)is the n-th Hermite polynomial.
5. The boundary conditions for the particle in a box are that ψ(0) = ψ(L) = 0.
6. Therefore, we find that the energy levels are quantized as:
En=ˉhω (n+1
2)
with corresponding wavefunctions:
ψn(x) = √2
Lsin (nπx
L)
where nis a positive integer representing different energy levels, ωis the angular frequency given
by ω=α
L, and Ais the normalization constant.
14. Question 14: A particle is in a one-dimensional potential well given by the function
V(x) = {0,0≤x≤a
V0, x > a , where aand V0are positive constants. Determine the energy levels
of the particle.
Ans. To determine the energy levels of the particle in the given potential well, we need to solve
the time-independent Schrödinger equation and apply the appropriate boundary conditions.
1. Setup the Schrödinger equation: The time-independent Schrödinger equation in one
dimension is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
In the regions where V(x) = 0, the equation becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
d2ψ
dx2+2mE
ˉh2ψ= 0
2. Solve the Schrödinger equation in different regions: Since V(x) = 0 for 0≤x≤a,
the general solution to the Schrödinger equation in this region is:
ψ1(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2
In the region x > a where V(x) = V0, the general solution is:
ψ2(x) = Ce−κx +Deκx
where κ=√2m(V0−E)
ˉh2
3. Apply continuity conditions: At x=a, the wavefunction and its derivative must be
continuous, i.e., ψ1(a) = ψ2(a)and ψ′
1(a) = ψ′
2(a).
4. Determine the energy levels: Solve for the energy levels by solving the transcendental
equations obtained from the boundary conditions.
These steps will lead to determining the discrete energy levels of the particle in the given
potential well.
15. Question 15: An electron is in a one-dimensional infinite potential well of width L. Calculate
the energy levels for this system.
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 of the energy level.
1. Write down the Schrödinger equation for this system. The time-independent
Schrödinger equation for a one-dimensional infinite potential well is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
2. Formal solution for the wavefunction. The wavefunction ψ(x)for an electron in an
infinite potential well can be written as:
ψ(x) = Asin (nπx
L)
where Ais the normalization constant and nis the quantum number.
3. Apply boundary conditions. For an infinite potential well, the boundary conditions are
ψ(0) = 0 and ψ(L) = 0. Applying these conditions gives:
sin (nπ ·0
L)= 0 ⇒n= 0
sin (nπ ·L
L)= 0 ⇒nπ =mπ
4. Determine the allowed values of energy. The energy levels of the system can be
found by substituting the wavefunction form back into the Schrödinger equation and solving for
En. The energy levels are given by:
En=n2π2ˉh2
2mL2
where n= 1,2,3, ... is the quantum number.
Therefore, the energy levels for an electron in a one-dimensional infinite potential well of
width Lare quantized and given by En=n2π2ˉh2
2mL2.
16. Question text here An electron in a one-dimensional infinite potential well is in the n= 4
energy level. Suppose the width of the well is a. Calculate the energy of the electron in terms of
a.
Ans. Step 1. The energy levels for an electron in a one-dimensional infinite potential well are
given by the formula: En=n2h2
8ma2, where nis the quantum number, his Planck’s constant, mis
the mass of the electron, and ais the width of the well. Step 2. Substituting n= 4 into the
formula, we get: E4=42h2
8ma2=16h2
8ma2=2h2
ma2.Step 3. Therefore, the energy of the electron in
the n= 4 level in terms of ais 2h2
ma2.
17. Question: Determine the energy levels of a particle in a one-dimensional box of length L,
where Lis given and the particle has a mass m.
Ans. Let nbe the quantum number representing the energy level of the particle. The energy
levels Encan be found using the formula for the energy eigenvalues of a particle in a one-
dimensional box:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck constant and mis the mass of the particle.
Solution: 1. Determine the formula for the energy levels of a particle in a one-dimensional
box. Given that the energy eigenvalues of a particle in a one-dimensional box are given by the
formula:
En=n2π2ˉh2
2mL2
where ˉhis the reduced Planck constant, mis the mass of the particle, Lis the length of the
box, and nis the quantum number representing the energy level.
2. Substitute the given values into the formula. For a particle in a one-dimensional box with
a length L, the mass m, and the quantum number n, we can substitute these values into the
formula to find the energy levels En.
3. Determine the energy levels. By substituting the given values into the formula, we find
that the energy levels Enare:
En=n2π2ˉh2
2mL2
This formula gives the quantized energy levels of the particle in the one-dimensional box.
18. Question:
An electron is confined within a one-dimensional box of length 2 nm. Determine the energy
levels of the electron in this box using the Schrödinger equation.
Ans. Let’s start by setting up the Schrödinger equation for the one-dimensional box:
1. The Schrödinger equation for a particle in a one-dimensional box is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
2. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉhis the wave number and Aand Bare coefficients to be determined.
3. Applying the boundary conditions for a particle in a box, we have ψ(0) = ψ(2 nm)=0.
This gives us:
{B= 0
k=nπ
2nm ,where n= 1,2,3, ...
4. Substituting kback into the general solution, we have:
ψn(x) = Asin (nπx
2nm)
5. To find the energy levels En, we use the relation En=ˉh2k2
n
2m:
En=ˉh2
2m(nπ
2nm)2
Thus, the energy levels are given by:
En=n2π2ˉh2
8mnm2
where n= 1,2,3, ....
19. Suppose a particle of mass mis confined to a one-dimensional box of length L. Determine
the energy levels of the particle in terms of m,L, and Planck’s constant h.
Ans. To determine the energy levels of the particle in the one-dimensional box, we must solve
the time-independent Schrödinger equation for the particle within the box and apply appropriate
boundary conditions. This will result in the quantization of the energy levels. Let’s denote the
energy levels as Enwhere nis a positive integer.
1. Writing down the time-independent Schrödinger equation:
The time-independent Schrödinger equation for a one-dimensional box is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function of the particle, Eis the energy of the particle, and ˉhis the
reduced Planck’s constant.
2. Applying boundary conditions:
In the one-dimensional box, the particle is confined between x= 0 and x=L. The boundary
conditions are: - ψ(0) = 0 (wave function at x= 0 is zero) - ψ(L) = 0 (wave function at x=L
is zero)
3. Solving the Schrödinger equation:
Let ψ(x) = Asin(kx)be the wave function, where Ais the normalization constant and kis
the wave number. Substituting this into the Schrödinger equation, we get:
k=nπ
L,where n= 1,2,3, . . .
4. Calculating the energy levels:
The energy levels Enare given by:
En=n2π2ˉh2
2mL2,where n= 1,2,3, . . .
Therefore, the energy levels of the particle in the one-dimensional box are quantized and given
by the expression above.
20. Question: Find the energy levels of a particle in a one-dimensional box of length L= 2
nm. The particle has a mass of m= 9.11 ×10−31 kg and the potential energy outside the box
is infinite.
Ans. Step-by-step solution: 1. The energy levels of a particle in a one-dimensional box are
given by the formula:
En=n2π2ˉh2
2mL2
where nis the quantum number, ˉhis the reduced Planck constant (1.05 ×10−34 J s), mis the
mass of the particle, and Lis the length of the box.
2. Substituting the given values into the formula, we have:
En=n2π2(1.05 ×10−34 J s)2
2×9.11 ×10−31 kg ×(2 ×10−9m)2
3. Simplifying the expression, we get:
En=n2π2×1.1025 ×10−68
3.6444 ×10−38
4. Further simplifying, we find:
En=1.1025π2×10−68n2
3.6444 ×10−38
5. Therefore, the energy levels are expressed as:
En≈3.0214 ×10−31n2
1
where nis the quantum number.
21. What is the energy of a particle trapped in a one-dimensional potential well of width a, if
the particle has mass m?
Ans. Let’s denote the energy of the particle as E. 1. The Schrödinger equation for a
particle in a potential well is given by −ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ, where V(x)is the potential
function. Inside the well, V(x)=0, so the equation simplifies to −ˉh2
2m
d2ψ
dx2=Eψ. 2. We can
rewrite this equation as d2ψ
dx2=−2mE
ˉh2ψ. 3. The general solution to this differential equation is
ψ(x) = Asin(kx) + Bcos(kx), where k=√2mE
ˉh2. Since the particle is trapped in a well of
width a, we must have ψ(0) = ψ(a)=0. 4. Applying the boundary conditions, we find that
B= 0 (since cos(ka) = 0) and k=nπ
a, where nis a positive integer. 5. Therefore, the energy
levels are quantized and given by En=ˉh2π2n2
2ma2.
22. Let’s consider an electron in a one-dimensional box of length L.
Question 22: Calculate the energy levels for an electron in a one-dimensional box of length
Lwith fixed endpoints.
Ans. To find the energy levels of the electron in the box, we can use the time-independent
Schrödinger equation for a particle in a box:
−ˉh2
2m
d2ψ
dx2=Eψ
where Eis the total energy of the system, ˉhis the reduced Planck’s constant, mis the mass of
the electron, xis the position, and ψis the wavefunction.
1. Boundary conditions: At the endpoints of the box (x= 0 and x=L), the wavefunction
must be zero, since the particle is confined to the box:
ψ(0) = 0 and ψ(L) = 0
2. Form of the wavefunction: The general form of the wavefunction inside the box is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉhand A,Bare constants.
3. Applying boundary conditions: Using the boundary conditions ψ(0) = 0 and ψ(L) = 0:
ψ(0) = 0 =⇒B= 0
ψ(L) = 0 =⇒Asin(kL) = 0
Since sin(kL) = 0, then kL =nπ, where nis a positive integer.
4. Determining energy levels: From kL =nπ, we can solve for E:
√2mE
ˉh·L=nπ
√2mE =nπˉh
L
2mE =(nπˉh
L)2
E=n2π2ˉh2
2mL2
Therefore, the energy levels for the electron in a one-dimensional box with fixed endpoints
are given by:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the energy level.
23. Question: Find the energy levels of a particle in a one-dimensional box of length L. The
particle has mass mand the box has infinite potential walls.
Ans. Let’s denote the energy levels as Enwhere nis a positive integer. The energy levels are
given by the formula:
En=n2ˉh2π2
2mL2
where ˉhis the reduced Planck’s constant.
Solution: 1. The time-independent Schrödinger equation for a particle in a box is:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
2. Applying the boundary conditions for this case (infinite potential walls), we have ψ(0) = 0
and ψ(L) = 0. This leads to the general solution:
ψ(x) = Asin (nπx
L)
3. Now, we apply the normalization condition:
∫L
0|ψ(x)|2dx = 1
4. Substituting the general solution into the normalization condition and solving for A, we
find:
A=√2
L
5. The energy levels are obtained by solving the time-independent Schrödinger equation with
the general solution. Thus, we get:
En=n2ˉh2π2
2mL2
Therefore, the energy levels for a particle in a one-dimensional box of length Lare given by
En=n2ˉh2π2
2mL2.
24. Question 24: Consider an electron confined within a one-dimensional infinite potential well
of width L. Determine the energy levels Enfor the electron in terms of Planck’s constant h,
electron mass meand the width Lof the well.
Ans. To find the energy levels of the electron in the one-dimensional infinite potential well,
we can use the Schrödinger equation and the boundary conditions at the edges of the well.
The general form of the time-independent Schrödinger equation for a particle of mass min a
one-dimensional region where the potential energy is zero is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
Where Eis the energy of the particle and ψis the wave function of the particle.
1. Inside the well, the potential energy is zero, so the equation simplifies to:
d2ψ
dx2=−2mE
ˉh2ψ
2. The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
Where k=√2mE
ˉh2.
3. Applying the boundary conditions at x= 0 and x=L, we find that ψ(0) = ψ(L) = 0,
which gives us:
B= 0
kL =nπ
Where n= 1,2,3, ....
4. Solving for Ein terms of nand L, we have:
k=nπ
L
E=ˉh2π2n2
2mL2
Therefore, the energy levels Enfor the electron in the one-dimensional infinite potential well
are given by:
En=ˉh2π2n2
2mL2
25. Question: Consider an electron confined in a one-dimensional infinite square well potential
with width L. Calculate the energy levels of the system in terms of the fundamental parameters.
Ans. Step-by-step solution: The energy levels of an electron in an infinite square well potential
are quantized and given by the formula:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number, ˉhis the reduced Planck’s
constant, mis the mass of the electron, and Lis the width of the square well.
1. First, we need to derive the expression for the energy levels of the electron in the infinite
square well potential. In one dimension, the Schrödinger equation for the system is given by:
−ˉh2
2m
d2ψ
dx2=Eψ
2. Inside the well, the potential energy is zero and the equation becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
3. The general solution to this differential equation is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
4. Applying boundary conditions where ψ(0) = 0 and ψ(L) = 0, we find that kL =nπ for
integer n. This gives the quantized energy levels as:
En=n2π2ˉh2
2mL2
Therefore, the energy levels of an electron in an infinite square well potential are quantized
and given by the above formula.
26. Question: Let’s consider a particle in a one-dimensional box of length L. Suppose the
particle has a mass mand the box has an infinite potential barrier at x= 0 and x=L. The
wave function of the particle is given by:
ψ(x) = Asin (nπx
L)
where Ais the normalization constant and nis a positive integer. Find the energy levels of the
particle in terms of m,L, and n.
Ans. Let’s start by finding the normalization constant A. 1. Normalize the wave function:
∫L
0|ψ(x)|2dx = 1
∫L
0|Asin(nπx
L)|2dx = 1
A2∫L
0
sin2(nπx
L)dx = 1
A2∫L
0
1−cos(2nπx
L)
2dx = 1
A2[x
2−L
2nπ sin(2nπx
L)]L
0= 1
A2[L
2−L
2nπ sin(2nπ)] = 1
Since sin(2nπ) = 0, we get:
A2·L
2= 1
A=√2
L
2. Calculate the energy levels: The energy of the particle can be calculated using the time-
independent Schrödinger equation:
Enψ(x) = −ˉh2
2m
d2ψ(x)
dx2
Substitute ψ(x)into the equation:
EnAsin (nπx
L)=−ˉh2
2mAn2π2
L2sin (nπx
L)
En=n2π2ˉh2
2mL2
Hence, the energy levels of the particle are given by:
En=n2π2ˉh2
2mL2
27. Question 27:
A particle of mass mis confined to move on a ring of radius a. Find the energy levels for this
system in terms of quantum number n.
Ans. To find the energy levels for a particle confined to move on a ring of radius a, we need
to consider the quantization of angular momentum. The angular momentum Lof the particle
is quantized as L=nˉhwhere nis a non-negative integer. The energy levels Enare given by
En=L2
2I, where Iis the moment of inertia of the particle.
1. Finding the moment of inertia: The moment of inertia Ifor a particle of mass m
moving on a ring of radius ais given by I=ma2.
2. Substituting Land Iinto the energy expression: Substitute L=nˉhand I=ma2
into the expression for energy levels, we have:
En=(nˉh)2
2ma2=n2ˉh2
2ma2
Therefore, the energy levels Enfor a particle confined to move on a ring of radius aare given
by En=n2ˉh2
2ma2, where nis a non-negative integer representing the quantum number.
28. Let’s consider a one-dimensional quantum harmonic oscillator with the Hamiltonian operator
given by ˆ
H=−ˉh2
2m
d2
dx2+1
2kx2, where ˉ
his the reduced Planck constant, mis the mass of the
particle, kis the spring constant, and xis the position operator.
Question: Find the energy levels of the quantum harmonic oscillator described by the Hamil-
tonian operator ˆ
H.
Ans. To find the energy levels of the quantum harmonic oscillator, we need to solve the time-
independent Schrödinger equation: ˆ
Hψ =Eψ, where ψis the wave function and Eis the energy
eigenvalue.
1. Form of the wave function: Let’s assume the wave function has the form ψ(x) =
Ce−αx2, where Cis the normalization constant and αis a constant to be determined.
2. Calculate the second derivative: The second derivative of the wave function with
respect to xis given by:
d2
dx2ψ(x) = −2αCe−αx2(2αx2−1)
3. Substitute into the Schrödinger equation: Substitute the wave function and its
second derivative into the Schrödinger equation. We get:
−ˉh2
2m(−2αCe−αx2(2αx2−1)) + 1
2kx2Ce−αx2=ECe−αx2
4. Simplify the equation: Simplify the equation by dividing through by Ce−αx2and
rearrange to get:
α2ˉh2−4α2ˉh2x2+ 2mE =kx2
5. Equating coefficients: In order to find the values of αand E, we compare coefficients
of x2. This gives us:
α2ˉh2=k
2m
E=ˉhα
2m
6. Energy levels: Substitute the value of αinto the expression for Eto find the energy
levels:
En=(n+1
2)ˉhω
where ω=√k
mis the angular frequency of the oscillator and n= 0,1,2, ....
29. Question: Determine the energy levels of a particle confined to a one-dimensional box of
length L. If the particle has a mass mand the box has infinitely high barriers, find the energy
levels in terms of the quantum number n.
Ans. Step-by-step solution: 1. The energy levels of a particle in a one-dimensional box are
given by the formula:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number. 2. Substituting the given values
into the formula, we have:
En=n2π2ˉh2
2mL2
3. Simplifying the expression further, we get:
En=π2ˉh2
2mL2·n2
Therefore, the energy levels of a particle confined to a one-dimensional box of length Lare
quantized and given by En=π2ˉh2
2mL2·n2, where nis a positive integer representing the quantum
number.
30. Question 30: A particle of mass mis in a one-dimensional infinite square well potential of
width L. Find the energy levels of the particle in terms of m,L, and fundamental constants.
Ans. To find the energy levels of a particle in an infinite square well potential, we solve the
time-independent Schrödinger equation for the particle within the well.
1. Setup: The time-independent Schrödinger equation for this system is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function, Eis the energy of the particle, and ˉhis the reduced Planck
constant.
2. Applying Boundary Conditions: Since the potential energy is infinite at the boundaries
of the well, the wave function must be zero at x= 0 and x=L. Thus, we have boundary
conditions ψ(0) = ψ(L) = 0.
3. Solving the Schrödinger Equation: The general solution to the Schrödinger equation
is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2and Aand Bare constants.
Applying the boundary conditions ψ(0) = ψ(L) = 0, we get:
{Bcos(0) = 0 =⇒B= 0
Asin(kL) = 0 =⇒kL =nπ (for n= 1,2,3, ...)
4. Determining Energy Levels: From the second boundary condition, we find the quantized
energy levels:
kL =nπ =⇒√2mE
ˉh2=nπ
L
Solving for Egives:
En=n2π2ˉh2
2mL2for n= 1,2,3, ...
Therefore, the energy levels of the particle in the infinite square well potential are quantized
and given by En=n2π2ˉh2
2mL2.
31. Question 31: Consider a one-dimensional infinite square well potential with width a. Find
the first three energy levels of a particle in this potential well in terms of ˉhand a.
Ans. Let’s denote the ground state energy level as E1, the first excited state as E2, and the
second excited state as E3. We can find these energy levels by solving the time-independent
Schrödinger equation for the infinite square well potential.
1. Finding the ground state energy level (E1): For the ground state, the wavefunction
of the particle is given by ψ1(x) = √2
asin (πx
a). The energy associated with this state is given
by E1=ˉh2π2
2ma2.
2. Finding the first excited state energy level (E2): For the first excited state, the
wavefunction of the particle is given by ψ2(x) = √2
asin (2πx
a). The energy associated with this
state is E2=4ˉh2π2
2ma2.
3. Finding the second excited state energy level (E3): For the second excited state,
the wavefunction of the particle is given by ψ3(x) = √2
asin (3πx
a). The energy associated with
this state is E3=9ˉh2π2
2ma2.
Therefore, the first three energy levels for a particle in a one-dimensional infinite square well
potential of width aare: E1=ˉh2π2
2ma2,E2=4ˉh2π2
2ma2, and E3=9ˉh2π2
2ma2.
32. Find the energy levels of a particle in a one-dimensional box of length Lwith an infinite
potential energy outside the box.
Ans. Let’s denote the energy levels of the particle in the box as En, with n= 1,2,3, . . .. The
energy levels are given by the formula:
En=n2π2ˉh2
2mL2,
where mis the mass of the particle and ˉhis the reduced Planck constant.
33. Question 33:
An electron is confined to move in one-dimension in an infinite potential well of width L.
Calculate the energy levels of the electron in terms of the fundamental constants.
Ans. To calculate the energy levels of the electron in an infinite potential well of width L:
1. The potential energy in the well is zero and the wave function must vanish at the walls,
so the energy levels are quantized. The allowed values of energy are given by:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantum number associated with the energy
level, ˉhis the reduced Planck constant, and mis the mass of the electron.
2. The lowest possible energy level corresponds to n= 1:
E1=π2ˉh2
2mL2
3. The second energy level corresponds to n= 2:
E2=4π2ˉh2
2mL2= 4E1
4. The third energy level corresponds to n= 3:
E3=9π2ˉh2
2mL2= 9E1
5. And so on, the energy levels increase with the square of the quantum number n.
34. Let’s consider a particle of mass mmoving in one dimension under the influence of a
potential V(x) = 1
2kx2, where kis a positive constant.
Question: Find the energy levels of the particle in terms of kand fundamental constants.
Ans. To find the energy levels of the particle, we can solve the time-independent Schrödinger
equation for this system. The Schrödinger equation for a one-dimensional system is given by:
−ˉh2
2m
d2ψ(x)
dx2+V(x)ψ(x) = Eψ(x)
where Eis the energy eigenvalue and ψ(x)is the wave function.
1. Writing the Schrödinger equation: Substitute the expression for V(x)into the
Schrödinger equation to obtain:
−ˉh2
2m
d2ψ(x)
dx2+1
2kx2ψ(x) = Eψ(x)
2. Simplifying the equation: Multiply through by 2m
ˉh2to simplify the equation:
−d2ψ(x)
dx2+k
ˉh2x2ψ(x) = 2m
ˉh2Eψ(x)
3. Non-dimensionalize the equation: Introduce a new dimensionless variable ξdefined
by ξ=√k
ˉhxto make the equation dimensionless. This gives:
−d2ψ(ξ)
dξ2+ξ2ψ(ξ) = λψ(ξ)
where λ=2mE
ˉh2.
4. Solving the equation: The solution to this equation can be expressed in terms of Hermite
polynomials Hn(ξ). The energy eigenvalues are quantized and given by:
En=(n+1
2)ˉh√k
2m
where n= 0,1,2, ... are the energy levels or quantum numbers.
Therefore, the energy levels of the particle in terms of kand fundamental constants are
En=(n+1
2)ˉh√k
2m.
35. Question 35:
A particle of mass mis confined to a one-dimensional box of length L. Consider the energy
levels of the particle in this box. Show that the energy levels Enare quantized according to the
equation:
En=n2ˉh2π2
2mL2
where nis a positive integer representing the quantum number of the energy level.
Ans. To derive the quantization of energy levels in a one-dimensional box, we will use the
time-independent Schrödinger equation for the particle:
1. The time-independent Schrödinger equation is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wavefunction and Eis the energy of the particle.
2. In the case of a particle in a one-dimensional box of length L, the wavefunction must
satisfy boundary conditions at x= 0 and x=L. These conditions are: ψ(0) = ψ(L) = 0.
3. The general solution to the wavefunction inside the box is given by:
ψ(x) = Asin (nπx
L)
where Ais a normalization constant and nis a positive integer.
4. Plugging the wavefunction into the Schrödinger equation, we get:
−ˉh2
2mA(nπ
L)2
sin (nπx
L)=EA sin (nπx
L)
5. Simplifying the equation above, we find the energy levels En:
En=n2ˉh2π2
2mL2
Thus, the energy levels of a particle in a one-dimensional box are quantized according to
the equation above. Each energy level is indexed by a positive integer n, representing different
quantum states of the particle.
