CHM 345 - PHYSICAL CHEMISTRY I - Calculation of
energy levels Practice Material - Set 2
1. Find the energy levels of a particle in a one-dimensional box of length L.
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
given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function, Eis the energy of the particle, ˉhis the reduced Planck
constant, and mis the mass of the particle.
1. Formulating the general solution:
The general solution to the Schrödinger equation for a particle in a box is:
ψn(x) = Asin (nπx
L)
where Ais the normalization constant, nis the quantum number, and Lis the length of the
box.
2. Applying boundary conditions:
The boundary conditions for a particle in a box are ψ(0) = 0 and ψ(L)=0. Applying these
boundary conditions to the general solution, we get:
At x= 0:ψ(0) = Asin(0) = 0
This implies that A= 0 for non-trivial solutions.
At x=L:ψ(L) = Asin (nπL
L)=Asin(nπ) = 0
This condition gives us nπ =kπ where kis a non-zero integer.
3. Finding energy levels:
The energy levels Enare quantized and given by:
En=n2π2ˉh2
2mL2
where nis the quantum number.
Therefore, the energy levels of a particle in a one-dimensional box of length Lare quantized
and depend on the quantum number n.
2. Question: Consider a particle in a one-dimensional box of length L. What are the possible
energy levels for this particle 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 and represents the quantum number, ˉhis the reduced Planck’s
constant, and mis the mass of the particle. 2. Substituting the given values into the formula,
we get:
En=n2π2ˉh2
2mL2
3. Therefore, the possible energy levels for the particle in the one-dimensional box in terms of
the quantum number nare:
En=π2ˉh2
2mL2,4π2ˉh2
2mL2,9π2ˉh2
2mL2, . . .
3. Find the energy levels of a particle in a one-dimensional box of length L.
Ans. To find the energy levels, we can start by solving the time-independent Schrödinger
equation for the particle in a one-dimensional box. The general form of the Schrödinger equation
for this system is:
−ˉ
h2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function of the particle, mis the mass of the particle, Eis the energy
of the particle, xis the position coordinate, and ˉhis the reduced Planck’s constant.
1. We can start by assuming a solution of the form ψ(x) = Asin(kx) + Bcos(kx), where A
and Bare constants to be determined and kis related to the energy Eas k=√2mE
ˉh2.
2. Calculate the first derivative of the wave function and substitute it back into the Schrödinger
equation, −ˉh2
2m
d2ψ(x)
dx2=Eψ(x).
3. Solve the resulting differential equation to find possible values of k, and thus possible
energy levels E.
The energy levels for a particle in a one-dimensional box of length Lare given by:
En=n2π2ˉh2
2mL2
where n= 1,2,3, ... represents the quantum number corresponding to each energy level.
4. Question: Consider a particle of mass min a one-dimensional box of length L. The wave
function of the particle is given by ψ(x) = Asin(kx), where Ais a normalization constant and
k=nπ
Lfor n= 1,2,3, .... Determine the energy levels of the particle in terms of n.
Ans. Solution: For a particle in a one-dimensional box, the allowed energy levels are given by
En=n2π2ˉh2
2mL2. We know that the wave function of the particle is given by ψ(x) = Asin(kx),
where k=nπ
L.
1. Normalize the wave function: The normalization condition is ∫L
0|ψ(x)|2dx = 1. Thus,
∫L
0A2sin2(kx)dx = 1. Solving the integral gives ∫L
0A21−cos(2kx)
2dx = 1. This simplifies to
A2[x
2−sin(2kx)
4k]L
0= 1. Substitute in the boundaries 0and L:A2[L
2−sin(2nπ)
4nπ ]= 1. Since
sin(2nπ) = 0 for all integers n, we get A2L
2= 1. Thus, A=√2
L.
2. Calculate the energy levels: Substitute A=√2
Linto k=nπ
Lto get k=nπ
L. The energy
levels are then given by En=k2ˉh2
2m. Substitute k=nπ
Linto the expression to get En=n2π2ˉh2
2mL2.
Therefore, the energy levels of the particle in terms of nare En=n2π2ˉh2
2mL2.
5. Question: Find the energy levels of a particle in a one-dimensional box of length Lwith
infinitely high potential walls using the Schrödinger equation.
Ans. Step-by-step solution: 1. The Schrödinger equation for a particle in a one-dimensional
box with infinitely high potential walls is given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function, Eis the energy eigenvalue, ˉhis the reduced Planck’s constant,
and mis the mass of the particle. 2. The general solution to this differential equation in the
region between the walls at x= 0 and x=Lis:
ψ(x) = Asin (nπx
L)
where nis a positive integer corresponding to the energy level. 3. Applying the boundary
conditions that ψ(0) = 0 and ψ(L) = 0, we find:
Asin (0) = 0 ⇒A= 0
Asin (nπ) = 0 ⇒nπ =kL, where k∈Z
4. Therefore, the quantized energy levels are given by:
En=n2π2ˉh2
2mL2
where n= 1,2,3, . . ., representing the energy levels of the particle in the box. 5. Each energy
level corresponds to a different eigenfunction ψ(x)and energy eigenvalue En, where the particle
can exist within the box.
6. Question: Find the energy levels of an electron in a one-dimensional infinite square well
potential with width L. How does the energy change if the potential well is modified to have
width 2L?
Ans. Let’s solve this problem step by step:
1. The energy levels of an electron in a one-dimensional infinite square well potential are
given by the formula:
En=n2π2ˉh2
2mL2
where nis the quantum number, ˉhis the reduced Planck’s constant, and mis the mass of the
electron.
2. If the potential well is modified to have width 2L, the energy levels can be calculated using
the same formula but with Lreplaced by 2L:
E′
n=n2π2ˉh2
2m(2L)2=n2π2ˉh2
8mL2
3. Comparing the expressions for the energy levels in the two cases, we see that the energy
levels for the modified well are one-fourth of the original energy levels:
E′
n=1
4En
So, the energy levels decrease when the width of the potential well is doubled.
Therefore, the energy levels of the electron in a one-dimensional infinite square well potential
with width 2Lare one-fourth of the original energy levels of the well with width L.
7. Question: Consider a particle of mass mmoving in a one-dimensional potential energy
function given by V(x) = 1
2kx2, where kis a positive constant. Show that the allowed energy
levels of the particle are quantized and determine the expression for these energy levels in terms
of kand fundamental constants.
Ans. Let’s denote the total energy of the particle as E. According to the Schrödinger equation,
the time-independent equation for this system is:
−ˉh2
2m
d2ψ
dx2+ (V(x)−E)ψ= 0
1. Substituting V(x) = 1
2kx2into the Schrödinger equation, we get:
−ˉh2
2m
d2ψ
dx2+(1
2kx2−E)ψ= 0
2. We can simplify the equation by dividing through by ˉh2
2mto obtain:
−d2ψ
dx2+2m
ˉh2(1
2kx2−E)ψ= 0
3. It is common to define a new variable α=√mk
ˉh2for convenience. Substituting this into
the equation gives:
−d2ψ
dx2+ 2α2(1
2x2−E
k)ψ= 0
4. Let’s make a further substitution: y=√αx. This changes the equation into a simpler
form:
−d2ψ
dy2+ (y2−β)ψ= 0,where β=2E
ˉhω with ω=ˉh
√mk
5. The solutions to this differential equation are the Hermite polynomials, Hn(y). Therefore,
we have:
ψn(y) = NnHn(y)e−y2
2,where Nnis the normalization constant
6. The quantization condition comes from the fact that the wave function must be normal-
izable, i.e., ∫∞
−∞ |ψn(y)|2dy = 1. Applying this condition gives the quantized energy levels:
En=ˉhω (n+1
2),where n= 0,1,2, . . .
Therefore, the allowed energy levels of the particle are quantized with values determined by
the expression En=ˉhω (n+1
2).
8. Question: Consider a particle of mass min a one-dimensional box of length L. What are
the possible energy levels for this particle?
Ans. Step-by-step solution: 1. The energy levels for a particle in a one-dimensional box are
given by the formula:
En=n2π2ˉh2
2mL2
where nis a positive integer (quantum number), mis the mass of the particle, ˉhis the reduced
Planck constant, and Lis the length of the box.
2. Substituting the given values into the formula, we get:
En=n2π2ˉh2
2mL2
3. Therefore, the possible energy levels for the particle in the one-dimensional box are de-
termined by the values of n. Each permissible value of ncorresponds to a different energy
level.
4. The energy levels correspond to n= 1,2,3, ... and increase as nincreases. This means
that the particle can have energy levels E1, E2, E3, ... where En+1 > Enfor all n.
5. In summary, the energy levels of a particle in a one-dimensional box are quantized and given
by the formula En=n2π2ˉh2
2mL2where n= 1,2,3, ... represent the different energy levels available to
the particle.
9. Suppose an electron is confined within a one-dimensional infinite potential well with a width
of L= 0.1nm. Calculate the energy levels of the electron in electron volts (eV) within the
potential well.
Ans. To find the energy levels of the electron within the potential well, we can use the formula
for the energy levels of a particle in a one-dimensional infinite potential well:
En=n2π2ˉh2
2mL2
where: En= energy level of the electron, n= quantum number, ˉh= reduced Planck’s
constant, m= mass of the electron, L= width of the potential well.
1. Calculating the mass of the electron: The mass of the electron, m= 9.11 ×10−31
kg.
2. Converting the width of the potential well to meters: Given L= 0.1nm, we convert
this to meters by dividing by 109.L= 0.1×10−9= 1 ×10−10 m.
3. Calculating the energy levels for the electron: Substitute the values into the formula
to find the energy levels:
En=n2π2ˉh2
2mL2
En=n2π2(1.05 ×10−34)2
2×9.11 ×10−31 ×(1 ×10−10)2
4. Converting the energy levels to electron volts (eV): To convert the energy levels
from joules to electron volts, we use the conversion factor: 1eV = 1.602 ×10−19 J.
Now, compute the energy levels for a few values of nto find the corresponding energy levels
in eV.
10. A particle of mass mis constrained to move in a one-dimensional box of length L. Determine
the energy levels of the particle in the box.
Ans. To determine the energy levels of a particle in a one-dimensional box of length L, we need
to solve the time-independent Schrödinger equation for the particle inside the box. The general
form of the time-independent Schrödinger equation 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, ˉhis the
reduced Planck constant, mis the mass of the particle, and xis the position coordinate.
1. Applying the Schrödinger Equation:
We consider the particle inside the box, where the wave function ψ(x)must satisfy the
boundary conditions. Inside the box, the potential is zero, so the Schrödinger equation simplifies
to:
d2ψ(x)
dx2=−2mE
ˉh2ψ(x)
2. Solving the Schrödinger Equation:
The general solution to the above differential equation can be written as:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
3. Applying Boundary Conditions:
Since the particle is inside a box of length L, the wave function must satisfy the boundary
conditions ψ(0) = 0 and ψ(L) = 0.
Applying the boundary condition ψ(0) = 0, we have:
B= 0
Applying the boundary condition ψ(L) = 0, we get:
Asin(kL) = 0
This condition gives us the quantization condition for kL.
4. Quantization of Energy Levels:
The quantization condition kL =nπ for n= 1,2,3, ... allows us to determine the allowed
values of Eand subsequently the energy levels of the particle in the box.
Substituting k=nπ
Lback into the expression for E, we get:
En=n2π2ˉh2
2mL2
Therefore, the energy levels of the particle in the box are quantized and given by En=n2π2ˉh2
2mL2
for n= 1,2,3, ....
11. Find the allowed energy levels for an electron in a one-dimensional box of length L= 3 nm.
Ans. To find the allowed energy levels for an electron in a one-dimensional box, we can use the
following formula:
En=n2h2
8mL2
where: En= energy of the electron in level n,n= quantum number (1, 2, 3, ...), h=
Planck’s constant (6.626 ×10−34 J s), m= mass of the electron (9.11 ×10−31 kg), and L=
length of the box.
1. Substitute the known values into the formula:
En=n2·(6.626 ×10−34 J s)2
8·(9.11 ×10−31 kg)·(3 ×10−9m)2
2. Simplify the expression:
En=n2·4.39 ×10−34 J2s2
6.55 ×10−38 kg ·m2
En=4.39 ×10−34 J·s
6.55 ×10−38 kg ·n2
En= 6.71 ×10−3J·m−1·n2
3. The allowed energy levels are given by plugging in different values of ninto the formula.
For n= 1,2,3, ...:
n= 1 : E1= 6.71×10−3J·m−1·(1)2= 6.71×10−3Jn= 2 : E2= 6.71×10−3J·m−1·(2)2=
26.84 ×10−3Jn= 3 : E3= 6.71 ×10−3J·m−1·(3)2= 60.57 ×10−3J and so on.
Therefore, the allowed energy levels for the electron in the one-dimensional box with L= 3
nm are 6.71 ×10−3J, 26.84 ×10−3J, 60.57 ×10−3J, and so on.
12. Question 12: A particle is in a one-dimensional potential well given by V(x) = 0 for
0<x<Land V(x) = ∞otherwise. Calculate the first three energy levels, En, of the particle
in this potential well.
Ans. To find the energy levels of the particle in the potential well, we need to solve the
time-independent Schrödinger equation:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
Where ψis the wave function, Eis the energy, and mis the mass of the particle. Since the
potential energy is zero in the well, the equation becomes:
1. Inside the well:
−ˉh2
2m
d2ψ
dx2=Eψ
This is the same as the time-independent Schrödinger equation for a free particle.
2. The general solution to the above equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
3. Applying the boundary conditions, ψ(0) = ψ(L) = 0, we get:
ψ(0) = Bcos(0) = B= 0
ψ(L) = Asin(kL) = 0
This implies that kL =nπ where n= 1,2,3, ....
4. Substituting back k=nπ
Linto the expression for energy, we find the energy levels:
En=ˉh2π2n2
2mL2
Therefore, the first three energy levels are:
E1=ˉh2π2
2mL2, E2=4ˉh2π2
2mL2, E3=9ˉh2π2
2mL2
13. Let’s consider a particle in a one-dimensional potential well defined by the following potential
energy function:
V(x) = {0if 0≤x≤a
∞otherwise
The particle is described by the time-independent Schrödinger equation
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
We want to find the energy eigenvalues for this system.
Ans. The general solution to the time-independent Schrödinger equation in this case can be
written as:
ψ(x) = {Asin(kx) + Bcos(kx)if 0≤x≤a
0otherwise
where k=√2mE
ˉh.
To find the energy eigenvalues, we need to apply the boundary conditions. The wavefunction
should be continuous and its derivative should be discontinuous at the boundaries of the potential
well. Thus, we have:
1. At x= 0: The wavefunction continuity condition gives us:
B= 0
2. At x=a: The wavefunction continuity condition gives us:
Asin(ka) = 0
This condition gives us two possibilities:
a) A= 0: This corresponds to the trivial solution where the particle doesn’t exist in the well.
b) sin(ka) = 0: This condition gives us the quantization condition for the energy levels:
ka =nπ
where nis a positive integer. Solving this equation gives us the allowed values of E.
Therefore, the energy eigenvalues for this system are given by:
En=n2π2ˉh2
2ma2
where n= 1,2,3, ....
14. Suppose an electron is confined to a one-dimensional box of length L. Calculate the energy
levels of the electron in this system.
Ans. To calculate the energy levels of the electron in a one-dimensional box of length L, we
can use the formula for the energy levels of a particle in a box:
En=n2π2ˉh2
2mL2,
where nis the quantum number, ˉhis the reduced Planck constant, and mis the mass of the
electron.
1. The energy levels Enof the electron in the system can be calculated using the formula
above. The lowest energy level (n= 1) corresponds to the ground state energy:
E1=π2ˉh2
2mL2.
2. The first excited state (n= 2) energy level can be calculated as:
E2=4π2ˉh2
2mL2.
3. Continuing this pattern, the energy level for the n-th state is given by:
En=n2π2ˉh2
2mL2.
So, the energy levels of the electron in the one-dimensional box of length Lare quantized
and given by En=n2π2ˉh2
2mL2for n= 1,2,3, ...
15. Question 15:
Consider a particle of mass min a one-dimensional infinite square well potential defined by
V(x) = 0 for 0< x < a and V(x) = ∞otherwise.
Determine the first three energy levels of the particle in this potential well.
Ans. To find the energy levels of the particle in the infinite square well potential, we need to
solve the time-independent Schrödinger equation for the given potential.
1. Formulation of the Schrödinger equation: The time-independent Schrödinger equa-
tion for the particle in the potential well is given by:
−ˉh2
2m
d2ψ(x)
dx2+V(x)ψ(x) = Eψ(x)
For the infinite square well potential, V(x) = 0 for 0< x < a and ∞otherwise, which means
ψ(x) = 0 for x≤0and x≥a.
2. General solution in the well: Since V(x) = 0 in the region 0< x < a, we have:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
Applying the boundary conditions ψ(0) = 0 and ψ(a) = 0, we find that B= 0 and k=nπ
a,
where nis a positive integer.
3. Determining energy levels: The energy levels are quantized and given by:
En=n2ˉh2π2
2ma2
where nis a positive integer.
For the first three energy levels (n= 1,2,3), we have: 1. E1=ˉh2π2
2ma22. E2=4ˉh2π2
2ma2= 4E1
3. E3=9ˉh2π2
2ma2= 9E1
Therefore, the first three energy levels of the particle in the infinite square well potential are
ˉh2π2
2ma2,4ˉh2π2
2ma2, and 9ˉh2π2
2ma2respectively.
16. Question: Consider an electron confined in a one-dimensional infinite potential well of width
L. Find the energy levels of the electron in terms of the fundamental constants and the integer
n.
Ans. Let’s denote the energy levels of the electron as En, where nis a positive integer
representing the quantum number. The energy levels are quantized and can be found by solving
the time-independent Schrödinger equation for a particle in a one-dimensional infinite potential
well.
1. Setting 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(x)
dx2=Eψ(x)
where ˉhis the reduced Planck’s constant, mis the mass of the electron, ψ(x)is the wave function,
Eis the energy, and xis the position within the well.
2. Applying boundary conditions: Inside the well, the potential energy is zero, so the
Schrödinger equation becomes:
−ˉ
h2
2m
d2ψ(x)
dx2=Eψ(x)
with the boundary conditions that ψ(0) = ψ(L) = 0.
3. Solving the Schrödinger equation: The solution to the Schrödinger equation can be
separated into three regions: 0< x < L,x= 0, and x=L. By solving the differential equation
within the well and applying the boundary conditions, the solutions for ψ(x)are found to be:
ψn(x) = √2
Lsin (nπx
L)
where nis a positive integer.
4. Determining the energy levels: The energy levels can be found by substituting the
wave function solutions back into the Schrödinger equation:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantized energy levels. Hence, the energy levels
of the electron in the one-dimensional infinite potential well are given by En=n2π2ˉh2
2mL2.
17. Question 17:
Consider a particle in a one-dimensional potential well defined by the following piecewise
function:
V(x) = {0if 0≤x≤a
V0if x > a
where V0is a positive constant. Determine the expression for the energy levels of this particle
in the potential well.
Ans. To find the energy levels of the particle in the potential well described, we need to solve
the time-independent Schrödinger equation for this system and apply the appropriate boundary
conditions.
1. Set up the Schrödinger equation:
The time-independent Schrödinger equation for one-dimensional motion is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
Substitute the potential function into the equation:
−ˉh2
2m
d2ψ
dx2= (E−V(x))ψ
2. Solve the Schrödinger equation in different regions:
In region 0≤x≤a, the potential is zero (V(x) = 0), so the Schrödinger equation becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
The solutions for this region are general solutions of the form ψ(x) = Aeikx +Be−ikx, where
k=√2mE
ˉ
h.
In region x > a, the potential is V0, so the Schrödinger equation becomes:
−ˉh2
2m
d2ψ
dx2+V0ψ=Eψ
The solutions for this region are general solutions of the form ψ(x) = Ceqx +De−qx, where
q=√2m(V0−E)
ˉh.
3. Apply boundary conditions:
The wavefunction ψ(x)and its derivative must be continuous at x=a. This gives us two
conditions:
1) ψ(a−) = ψ(a+)2) dψ
dx (a−) = dψ
dx (a+)
4. Solve for the energy levels:
By applying the boundary conditions and solving the resulting equations, we can find the
expression for the energy levels of the particle in the potential well.
This process involves working through the matching conditions at the boundary, solving the
transcendental equation that arises, and finding the allowed energy levels.
The final expression for the energy levels will involve constants a,V0,ˉh, and m.
Therefore, the expression for the energy levels of the particle in the potential well described
by the given piecewise function can be obtained by solving the Schrödinger equation and applying
the appropriate boundary conditions.
18. Question: Consider a one-dimensional particle in a box with a length of 2 nm. Calculate
the energy levels of the particle in electron volts (eV) compared to its ground state energy.
Ans. Step-by-step solution: Let’s first determine the ground state energy of the particle in the
box. 1. The energy of a particle in a box is given by the equation:
En=n2π2ˉh2
2mL2
2. Substituting the given values: n= 1,ˉh= 1.054572 ×10−34 Js, m= 9.10938356 ×10−31
kg, and L= 2 ×10−9m, into the equation, we get:
E1=(1)2π2(1.054572 ×10−34)2
2(9.10938356 ×10−31)(2 ×10−9)2
3. Calculating the ground state energy E1, we find:
E1≈2.448 ×10−18 J
4. To convert the ground state energy from joules to electron volts, we use the conversion
factor:
1eV = 1.602 ×10−19 J
5. Therefore, to obtain the ground state energy in electron volts, we divide 2.448 ×10−18 J
by 1.602 ×10−19 J/eV:
E1≈15.29 eV
6. Now, let’s calculate the first excited state energy E2by plugging in n= 2 into the energy
equation:
E2=(2)2π2(1.054572 ×10−34)2
2(9.10938356 ×10−31)(2 ×10−9)2
7. Calculating E2, we get:
E2≈9.794 ×10−18 J
8. Converting E2from joules to electron volts, we find:
E2≈61.17 eV
Therefore, the energy levels of the particle in electron volts compared to its ground state
energy are approximately 15.29 eV for the ground state and 61.17 eV for the first excited state.
19. Question: Consider an electron in a one-dimensional infinite square well potential of width
L. Calculate the energy levels in terms of the well width L.
Ans. Let’s denote the energy levels of the electron in the infinite square well potential as En,
where nis a positive integer representing the quantum number.
1. Setting up the problem: Inside the square well potential, the electron obeys the time-
independent Schrödinger equation:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function for the electron at position x,mis the mass of the electron, E
is the total energy of the electron, and ˉhis the reduced Planck constant.
2. General solution of the Schrödinger equation: Since the electron is in a potential
well, the wave function ψ(x)must satisfy boundary conditions. Therefore, the general solution
for ψ(x)is:
ψn(x) = Asin(knx) + Bcos(knx)
where kn=nπ
L, and Aand Bare constants to be determined.
3. Applying boundary conditions: The boundary conditions for the infinite square well
potential are that the wave function must be zero at x= 0 and x=L. Therefore, ψ(0) =
ψ(L) = 0.
Applying the boundary condition ψ(0) = 0, we get:
B= 0
Applying the boundary condition ψ(L) = 0, we get:
Asin(knL) = 0
This implies that knL=nπ. Therefore, the allowed values of knare:
kn=nπ
L
4. Determining the energy levels: The total energy Enof the electron is related to knby
the equation:
En=ˉh2k2
n
2m
Substitute the value of kninto the expression for En, we have:
En=n2π2ˉh2
2mL2
Therefore, the energy levels of the electron in the one-dimensional infinite square well potential
are given by:
En=n2π2ˉh2
2mL2
where nis a positive integer.
20. Question: Determine the energy levels of a particle in a one-dimensional box of length Lif
the particle is subject to a potential energy function given by V(x) = V0cos2(πx
L).
Ans. To find the energy levels of the particle in the box, we need to solve the time-independent
Schrödinger equation Hψ =Eψ, where H=−ˉh2
2m
d2
dx2+V(x)and ψis the wave function.
1. First, we need to write the Schrödinger equation for the given potential:
Hψ =Eψ
(−ˉh2
2m
d2
dx2+V0cos2(πx
L))ψ=Eψ
2. Next, let’s separate the Schrödinger equation into two parts: the kinetic energy operator
and the potential operator:
−ˉh2
2m
d2ψ
dx2+V0cos2(πx
L)ψ=Eψ
3. Now, we can simplify the equation by substituting k=π
L:
−ˉh2
2m
d2ψ
dx2+V0cos2(kx)ψ=Eψ
4. The potential energy function V(x) = V0cos2(kx)can be rewritten as:
V(x) = V0
2(1 + cos(2kx))
5. We can now substitute the potential energy function back into the Schrödinger equation:
−ˉh2
2m
d2ψ
dx2+V0
2(1 + cos(2kx)) ψ=Eψ
6. Solve the Schrödinger equation to find the energy levels Eand the corresponding wave
functions ψ(x).
21. Find the first three energy levels for a particle in a one-dimensional box of length L.
Ans. Let’s denote the energy levels as Enwhere nis a positive integer. The energy levels for
a particle in a one-dimensional box are given by the formula:
En=n2π2ˉh2
2mL2
1. To find the first energy level E1, substitute n= 1 into the formula:
E1=(1)2π2ˉh2
2mL2=π2ˉh2
2mL2
2. For the second energy level E2, substitute n= 2:
E2=(2)2π2ˉh2
2mL2=4π2ˉh2
2mL2=2π2ˉh2
mL2
3. Finally, for the third energy level E3, substitute n= 3:
E3=(3)2π2ˉh2
2mL2=9π2ˉh2
2mL2=9π2ˉh2
2mL2
22. Question: Consider a potential energy function given by V(x) = 1
2kx2−1
3gx3, where k
and gare positive constants. Calculate the energy levels for a particle in this potential.
Ans. Let Ebe the total energy of the particle. The energy levels are given by the solutions to
the equation V(x) = E.
Step 1. Substitute the given potential function into the equation V(x) = E:
1
2kx2−1
3gx3=E
Step 2. Rearrange the equation to isolate x:
1
2kx2−1
3gx3−E= 0
Step 3. Multiply through by 6 to clear the fractions:
3kx2−2gx3−6E= 0
Step 4. This is a cubic equation in the form ax3+bx2+cx +d= 0, where a=−2g,
b= 0,c= 3k, and d=−6E. We are interested in finding the values of Ethat give physically
meaningful solutions for x.
Step 5. The energy levels correspond to the values of Efor which the particle is bound
within the potential. This means that there will be a finite number of real solutions for x.
Step 6. The number of energy levels can be determined by analyzing the shape of the
potential (V(x)) and considering the number of turning points in the graph.
Step 7. Once you have found the energy levels (E), you can find the corresponding wave-
functions and probabilities of finding the particle at different positions using the Schrödinger
equation and normalization conditions.
23. Suppose an electron is confined to a one-dimensional box of length L= 1 nm. Calculate
the energy levels of the system in eV.
Ans. To calculate the energy levels of the system, we can use the formula for the energy levels
of a particle in a one-dimensional box:
En=n2π2ˉh2
2mL2
where nis the quantum number (positive integer), 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. Let’s first convert the given quantities to SI units:
L= 1 nm = 1 ×10−9m
2. Now, we can substitute the values into the energy formula:
En=n2π2ˉh2
2mL2=n2π2(1.0545718 ×10−34)2
2(9.11 ×10−31)(1 ×10−9)2
3. Simplifying the expression, we get:
En=n2π2(1.0545718 ×10−34)2
2(9.11 ×10−31)(1 ×10−9)2
En=n2×9.8696 ×10−68
1.65 ×10−39
En=n2×6×10−29
4. Therefore, the energy levels of the electron in the one-dimensional box are given by
En=n2×6×10−29 J. To convert this to eV, we use the conversion factor 1eV = 1.602 ×10−19
J:
En=n2×6×10−29 ×1.602 ×10−19
En=n2×9.61 ×10−10 eV
Hence, the energy levels in eV are En=n2×9.61 ×10−10.
24. Question: Determine the energy levels of an electron in a one-dimensional harmonic
oscillator with a restoring force constant of k= 8.0N/m.
Ans. Let’s denote the energy levels as En, where nis the quantum number.
1. The energy levels of a one-dimensional harmonic oscillator are given by the formula:
En=(n+1
2)ˉhω
where ωis the angular frequency of the oscillator and is related to the force constant kas
ω=√k
m.
2. We need to find the mass mof the electron. The mass of an electron is 9.11 ×10−31 kg.
3. Calculate the angular frequency ω:
ω=√k
m=√8.0N/m
9.11 ×10−31 kg
4. Substitute the angular frequency back into the formula for energy levels:
En=(n+1
2)ˉh√k
m
5. Calculate the energy levels with n= 0,1,2.
E0=(0 + 1
2)ˉhω
E1=(1 + 1
2)ˉhω
E2=(2 + 1
2)ˉhω
25. Question 25:
An electron in a one-dimensional infinite square well of width Lis in the ground state.
Calculate the uncertainty in its kinetic energy.
Ans. Let’s denote the uncertainty in kinetic energy as ∆K. The ground state wavefunction for
an infinite square well of width Lis given by:
ψ(x) = √2
Lsin (πx
L)
The expectation value ⟨K⟩for kinetic energy can be calculated as:
⟨K⟩=∫L
0
ψ∗(−ˉh2
2m
∂2
∂x2)ψ dx
Since the electron is in the ground state, the kinetic energy is:
K=π2ˉh2
2mL2
The uncertainty in kinetic energy ∆Kcan be calculated using the uncertainty principle:
∆K∆x≥ˉh
2
Given that the electron is in the ground state, the uncertainty in position can be considered
as the width of the well L. Therefore,
∆x=L
Hence,
∆K≥ˉh
2L
Therefore, the uncertainty in kinetic energy of the electron is ∆K≥ˉh
2L.
26. What is the energy (in eV) of an electron in the third energy level of a hydrogen atom?
(Given: energy of the first energy level is -13.6 eV)
Ans. Let’s denote the energy of the electron in the third energy level as E3. We know that the
energy of an electron in the nth energy level of a hydrogen atom is given by the formula:
En=−13.6
n2eV
where nis the principal quantum number.
1. Substitute n= 3 into the formula:
E3=−13.6
32eV
E3=−13.6
9eV
E3=−1.5111 eV
Therefore, the energy of an electron in the third energy level of a hydrogen atom is −1.5111 eV.
27. Question 27:
Consider a particle in a one-dimensional potential well of width a. The potential energy is
given by
V(x) = {0 0 < x < a
∞otherwise
Determine the energy levels for this system using the time-independent Schrödinger equation.
Ans. Let’s solve for the energy levels step by step:
1. The time-independent Schrödinger equation for this system is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
2. Since the potential energy is zero within the well, the equation simplifies to:
−ˉh2
2m
d2ψ
dx2=Eψ
3. Let’s denote k=√2mE
ˉh2. The general solution to the differential equation in the well
(0< x < a) is:
ψ(x) = Asin(kx) + Bcos(kx)
4. Applying the boundary conditions at x= 0 and x=a, we have:
ψ(0) = 0 ⇒B= 0
5. This gives us:
ψ(x) = Asin(kx)
6. Applying the boundary condition at x=a, we have:
ψ(a) = 0 ⇒Asin(ka) = 0
Since sin(ka)= 0, we must have A= 0. This implies that there are no acceptable solutions
for the wavefunction, making the energy levels in this potential well undefined.
Thus, no quantized energy levels exist for the given potential well.
28. Question:
An electron in a one-dimensional infinite potential well of length Lhas an energy level of E1.
If the electron is in the ground state of another one-dimensional infinite potential well of length
2L, what is the energy level it occupies in this second well?
Ans. Step-by-step solution:
Let’s denote the energy level in the second potential well as E2.
1. The energy of the electron in the first potential well is given by the formula:
E1=n2
1h2
8mL2
where n1is the quantum number corresponding to the energy level E1,his the Planck constant,
mis the mass of the electron, and Lis the length of the first potential well.
2. The energy of the electron in the second potential well can also be expressed in terms of
the quantum number n2corresponding to the energy level E2:
E2=n2
2h2
8m(2L)2
3. Since the electron is in the ground state of the second potential well, n2= 1. Substituting
n2= 1 into the equation for E2gives:
E2=h2
32mL2
4. To relate E1and E2, we know that the energy levels are inversely proportional to the
square of the lengths of the potential wells. Thus, we can write the following relationship:
E1
E2
=(L
2L)2
=1
4
5. Substituting the expression for E2into the equation above and rearranging, we find:
E1= 4E2
6. Therefore, the energy level of the electron in the second potential well is four times the
energy level E1. Thus, the electron in the second well occupies the energy level equivalent to
4E1.
29. Find the energy levels of a particle in a one-dimensional infinite square well of width a.
Ans. Let’s first recall the formula for the energy levels of a particle in a one-dimensional infinite
square well:
En=n2π2ˉh2
2ma2
where nis a positive integer representing the quantum number associated with the energy
level, ˉhis the reduced Planck’s constant, and mis the mass of the particle.
Solution:
1. Given the formula for the energy levels, we can now plug in the values of the constants:
n= 1,2,3, ...;mis the mass of the particle; ais the width of the well.
2. The energy level for n= 1 is:
E1=(1)2π2ˉh2
2ma2
3. The energy level for n= 2 is:
E2=(2)2π2ˉh2
2ma2
4. The energy level for n= 3 is:
E3=(3)2π2ˉh2
2ma2
5. The energy level for a general nis:
En=n2π2ˉh2
2ma2
Therefore, the energy levels of a particle in a one-dimensional infinite square well are quantized
and given by the above formula for all positive integers n.
30. Question 30: Determine the energy levels of a particle confined to a one-dimensional box of
length L. Let’s denote the energy levels as En, where nis a positive integer. Find an expression
for Enin terms of nand fundamental constants.
Ans. Let’s start by using the Schrödinger equation for a particle in a one-dimensional box:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
Where ˉhis the reduced Planck constant, mis the mass of the particle, Eis the energy of the
particle, and ψ(x)is the wave function.
1. The general solution to the above differential equation is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
2. To satisfy the boundary conditions for a particle in a box of length L, we impose that
ψ(0) = ψ(L) = 0. This leads to the quantization condition:
kL =nπ
where nis a positive integer.
3. Solving for the energy levels En, we get:
k=nπ
L
En=ˉh2π2n2
2mL2
Therefore, the expression for the energy levels Enin terms of nis:
En=ˉh2π2n2
2mL2
31. Question 31: Consider a particle of mass mmoving in one dimension under the influence of
a potential energy function given by V(x) = 1
2kx2+1
4rx4, where kand rare positive constants.
Determine the energy levels of this system.
Ans. To determine the energy levels of the system, we need to solve the time-independent
Schrödinger equation, Hψ =Eψ, where His the Hamiltonian operator, ψis the wave function,
Eis the energy eigenvalue, and we have H=−ˉh2
2m
d2
dx2+V(x).
1. Writing the Hamiltonian operator: The Hamiltonian operator for this system is given
by
H=−ˉh2
2m
d2
dx2+1
2kx2+1
4rx4.
2. Writing the time-independent Schrödinger equation: The time-independent Schrödinger
equation for this system is
−ˉh2
2m
d2ψ
dx2+(1
2kx2+1
4rx4)ψ=Eψ.
3. Solving the Schrödinger equation: Let ψ(x) = u(x)e−αx2with α=√mk
ˉh2as the trial
solution.
Substitute ψ(x)into the Schrödinger equation and simplify to get
d2u
dx2= 2α(2αx2−k)u+ 4α2x2u−2rα2x4u−2mE
ˉh2u.
4. Simplifying the equation: The above differential equation simplifies to
d2u
dx2= (4α2x2−2ku + 4α(2α−rx2)u−2mE
ˉh2u.
5. Determining the energy levels: By inspection, we see that the potential term is
a harmonic oscillator (1
2kx2) and an anharmonic oscillator (1
4rx4). The energy levels for this
system will involve solving the above differential equation, which may not have a simple analytical
solution. Numerical methods or approximation techniques may be required to find the energy
levels in this case.
32. Question 32:
Consider a particle of mass mmoving in a one-dimensional potential well defined by V(x) =
V0x2, where V0is a positive constant. Determine the allowed energy levels of this system.
Ans. To find the allowed energy levels of the system, we will solve the time-independent
Schrödinger equation for this potential and apply boundary conditions to ensure the wavefunction
is well-behaved within the potential well.
1. Write down the time-independent Schrödinger equation:
The time-independent Schrödinger equation for a one-dimensional system is given by:
−ˉh2
2m
d2
dx2ψ(x) + V(x)ψ(x) = Eψ(x)
Substituting the given potential V(x) = V0x2, the equation becomes:
−ˉh2
2m
d2
dx2ψ(x) + V0x2ψ(x) = Eψ(x)
2. Solve the Schrödinger equation in different regions:
We will consider two regions: x < 0and x > 0. In these regions, the potential term simplifies
to −V0x2.
The general solution in region x < 0can be written as:
ψ(x) = Aekx +Be−kx
where k=√2mE
ˉ
h2.
The general solution in region x > 0can be written as:
ψ(x) = Ceik′x+De−ik′x
where k′=√2m(E+V0x2)
ˉh2.
3. Apply boundary conditions:
At the boundaries x= 0 and x=L, the wavefunction ψ(x)must be continuous. This means:
ψ(0−) = ψ(0+)
and
ψ(L−) = ψ(L+)
4. Find the allowed energy levels:
By solving the Schrödinger equation in each region and applying the boundary conditions, we
can find the allowed energy levels of the system for different values of V0and the length of the
potential well.
33. Question 33:
A particle of mass mmoves in a one-dimensional box of length L. What are the energy levels
for this system?
Ans. To find the energy levels for a particle in a one-dimensional box, we can use the time-
independent Schrödinger equation.
1. Set up the Schrödinger equation: The time-independent Schrödinger equation for a
particle in a one-dimensional box is given by:
−h2
8m
d2ψ
dx2=Eψ
where ψis the wavefunction of the particle, his the reduced Planck constant, mis the mass of
the particle, xis the position, and Eis the energy.
2. Solve the differential equation: The general solution to the differential equation is:
ψ(x) = Asin (nπx
L)
where Ais a normalization constant and nis a positive integer corresponding to the energy level.
3. Apply boundary conditions: At x= 0 and x=L, the wavefunction must go to zero
to satisfy the boundary conditions. Therefore, we have:
ψ(0) = Asin(0) = 0 =⇒A= 0
ψ(L) = Asin(nπ) = 0 =⇒nπ =mπ =⇒n= 1,2,3, ...
4. Determine the energy levels: The energy levels are quantized and given by:
En=n2π2ˉh2
2mL2
where nis the quantum number.
Therefore, the energy levels for the particle in a one-dimensional box are quantized and depend
on the quantum number n.
34. Question: Determine the energy levels of a particle in a one-dimensional infinite potential
well of width L.
Ans. Step-by-step solution: 1. In a one-dimensional infinite potential well, the Schrödinger
equation for a particle of mass mis given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where Eis the energy eigenvalue and ψ(x)is the wavefunction.
2. The general solution to the Schrödinger equation inside the well can be written as:
ψn(x) = Asin (nπx
L)
where Ais a normalization constant and nis a positive integer representing the quantum
number associated with the energy level.
3. To determine the energy levels, we substitute the general solution back into the Schrödinger
equation:
−ˉh2
2m
d2
dx2[Asin (nπx
L)]=EA sin (nπx
L)
4. By taking the second derivative of ψ(x)and equating it to Eψ(x), we find:
n2π2ˉh2
2mL2=E
5. The energy levels Enfor the particle in the infinite potential well are quantized and given
by:
En=n2π2ˉh2
2mL2
where n= 1,2,3, ... are the quantum numbers corresponding to the allowed energy levels.
6. Therefore, the energy levels of a particle in a one-dimensional infinite potential well of
width Lare quantized and determined by the equation En=n2π2ˉh2
2mL2.
35. Question: Consider a particle of mass mconfined to a one-dimensional box of length L.
Determine the first three energy levels of the particle in terms of the constants m,L, and h
(Planck’s constant).
Ans. Let’s denote the energy levels of the particle in the box by En, where nis the quantum
number indicating the energy level. The energy levels are given by the formula:
En=n2π2ˉh2
2mL2,
where n= 1,2,3, ....
1. To find the first energy level E1, substitute n= 1 into the formula:
E1=(1)2π2ˉh2
2mL2=π2ˉh2
2mL2.
2. Next, let’s calculate the second energy level E2:
E2=(2)2π2ˉh2
2mL2=4π2ˉh2
2mL2= 2 (π2ˉh2
2mL2)= 2E1.
3. Finally, we find the third energy level E3by substituting n= 3 into the formula:
E3=(3)2π2ˉh2
2mL2=9π2ˉh2
2mL2= 3 (π2ˉh2
2mL2)= 3E1.
Therefore, the first three energy levels of the particle in the box are:
E1=π2ˉh2
2mL2,
E2= 2 (π2ˉh2
2mL2)= 2E1,
E3= 3 (π2ˉh2
2mL2)= 3E1.
9. Suppose an electron is confined within a one-dimensional infinite potential well with a width
of L= 0.1nm. Calculate the energy levels of the electron in electron volts (eV) within the
potential well.
Ans. To find the energy levels of the electron within the potential well, we can use the formula
for the energy levels of a particle in a one-dimensional infinite potential well:
En=n2π2ˉh2
2mL2
where: En= energy level of the electron, n= quantum number, ˉh= reduced Planck’s
constant, m= mass of the electron, L= width of the potential well.
1. Calculating the mass of the electron: The mass of the electron, m= 9.11 ×10−31
kg.
2. Converting the width of the potential well to meters: Given L= 0.1nm, we convert
this to meters by dividing by 109.L= 0.1×10−9= 1 ×10−10 m.
3. Calculating the energy levels for the electron: Substitute the values into the formula
to find the energy levels:
En=n2π2ˉh2
2mL2
En=n2π2(1.05 ×10−34)2
2×9.11 ×10−31 ×(1 ×10−10)2
4. Converting the energy levels to electron volts (eV): To convert the energy levels
from joules to electron volts, we use the conversion factor: 1eV = 1.602 ×10−19 J.
Now, compute the energy levels for a few values of nto find the corresponding energy levels
in eV.
10. A particle of mass mis constrained to move in a one-dimensional box of length L. Determine
the energy levels of the particle in the box.
Ans. To determine the energy levels of a particle in a one-dimensional box of length L, we need
to solve the time-independent Schrödinger equation for the particle inside the box. The general
form of the time-independent Schrödinger equation 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, ˉhis the
reduced Planck constant, mis the mass of the particle, and xis the position coordinate.
1. Applying the Schrödinger Equation:
We consider the particle inside the box, where the wave function ψ(x)must satisfy the
boundary conditions. Inside the box, the potential is zero, so the Schrödinger equation simplifies
to:
d2ψ(x)
dx2=−2mE
ˉh2ψ(x)
2. Solving the Schrödinger Equation:
The general solution to the above differential equation can be written as:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
3. Applying Boundary Conditions:
Since the particle is inside a box of length L, the wave function must satisfy the boundary
conditions ψ(0) = 0 and ψ(L) = 0.
Applying the boundary condition ψ(0) = 0, we have:
B= 0
Applying the boundary condition ψ(L) = 0, we get:
Asin(kL) = 0
This condition gives us the quantization condition for kL.
4. Quantization of Energy Levels:
The quantization condition kL =nπ for n= 1,2,3, ... allows us to determine the allowed
values of Eand subsequently the energy levels of the particle in the box.
Substituting k=nπ
Lback into the expression for E, we get:
En=n2π2ˉh2
2mL2
Therefore, the energy levels of the particle in the box are quantized and given by En=n2π2ˉh2
2mL2
for n= 1,2,3, ....
11. Find the allowed energy levels for an electron in a one-dimensional box of length L= 3 nm.
Ans. To find the allowed energy levels for an electron in a one-dimensional box, we can use the
following formula:
En=n2h2
8mL2
where: En= energy of the electron in level n,n= quantum number (1, 2, 3, ...), h=
Planck’s constant (6.626 ×10−34 J s), m= mass of the electron (9.11 ×10−31 kg), and L=
length of the box.
1. Substitute the known values into the formula:
En=n2·(6.626 ×10−34 J s)2
8·(9.11 ×10−31 kg)·(3 ×10−9m)2
2. Simplify the expression:
En=n2·4.39 ×10−34 J2s2
6.55 ×10−38 kg ·m2
En=4.39 ×10−34 J·s
6.55 ×10−38 kg ·n2
En= 6.71 ×10−3J·m−1·n2
3. The allowed energy levels are given by plugging in different values of ninto the formula.
For n= 1,2,3, ...:
n= 1 : E1= 6.71×10−3J·m−1·(1)2= 6.71×10−3Jn= 2 : E2= 6.71×10−3J·m−1·(2)2=
26.84 ×10−3Jn= 3 : E3= 6.71 ×10−3J·m−1·(3)2= 60.57 ×10−3J and so on.
Therefore, the allowed energy levels for the electron in the one-dimensional box with L= 3
nm are 6.71 ×10−3J, 26.84 ×10−3J, 60.57 ×10−3J, and so on.
12. Question 12: A particle is in a one-dimensional potential well given by V(x) = 0 for
0<x<Land V(x) = ∞otherwise. Calculate the first three energy levels, En, of the particle
in this potential well.
Ans. To find the energy levels of the particle in the potential well, we need to solve the
time-independent Schrödinger equation:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
Where ψis the wave function, Eis the energy, and mis the mass of the particle. Since the
potential energy is zero in the well, the equation becomes:
1. Inside the well:
−ˉh2
2m
d2ψ
dx2=Eψ
This is the same as the time-independent Schrödinger equation for a free particle.
2. The general solution to the above equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
3. Applying the boundary conditions, ψ(0) = ψ(L) = 0, we get:
ψ(0) = Bcos(0) = B= 0
ψ(L) = Asin(kL) = 0
This implies that kL =nπ where n= 1,2,3, ....
4. Substituting back k=nπ
Linto the expression for energy, we find the energy levels:
En=ˉh2π2n2
2mL2
Therefore, the first three energy levels are:
E1=ˉh2π2
2mL2, E2=4ˉh2π2
2mL2, E3=9ˉh2π2
2mL2
13. Let’s consider a particle in a one-dimensional potential well defined by the following potential
energy function:
V(x) = {0if 0≤x≤a
∞otherwise
The particle is described by the time-independent Schrödinger equation
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
We want to find the energy eigenvalues for this system.
Ans. The general solution to the time-independent Schrödinger equation in this case can be
written as:
ψ(x) = {Asin(kx) + Bcos(kx)if 0≤x≤a
0otherwise
where k=√2mE
ˉh.
To find the energy eigenvalues, we need to apply the boundary conditions. The wavefunction
should be continuous and its derivative should be discontinuous at the boundaries of the potential
well. Thus, we have:
1. At x= 0: The wavefunction continuity condition gives us:
B= 0
2. At x=a: The wavefunction continuity condition gives us:
Asin(ka) = 0
This condition gives us two possibilities:
a) A= 0: This corresponds to the trivial solution where the particle doesn’t exist in the well.
b) sin(ka) = 0: This condition gives us the quantization condition for the energy levels:
ka =nπ
where nis a positive integer. Solving this equation gives us the allowed values of E.
Therefore, the energy eigenvalues for this system are given by:
En=n2π2ˉh2
2ma2
where n= 1,2,3, ....
14. Suppose an electron is confined to a one-dimensional box of length L. Calculate the energy
levels of the electron in this system.
Ans. To calculate the energy levels of the electron in a one-dimensional box of length L, we
can use the formula for the energy levels of a particle in a box:
En=n2π2ˉh2
2mL2,
where nis the quantum number, ˉhis the reduced Planck constant, and mis the mass of the
electron.
1. The energy levels Enof the electron in the system can be calculated using the formula
above. The lowest energy level (n= 1) corresponds to the ground state energy:
E1=π2ˉh2
2mL2.
2. The first excited state (n= 2) energy level can be calculated as:
E2=4π2ˉh2
2mL2.
3. Continuing this pattern, the energy level for the n-th state is given by:
En=n2π2ˉh2
2mL2.
So, the energy levels of the electron in the one-dimensional box of length Lare quantized
and given by En=n2π2ˉh2
2mL2for n= 1,2,3, ...
15. Question 15:
Consider a particle of mass min a one-dimensional infinite square well potential defined by
V(x) = 0 for 0< x < a and V(x) = ∞otherwise.
Determine the first three energy levels of the particle in this potential well.
Ans. To find the energy levels of the particle in the infinite square well potential, we need to
solve the time-independent Schrödinger equation for the given potential.
1. Formulation of the Schrödinger equation: The time-independent Schrödinger equa-
tion for the particle in the potential well is given by:
−ˉh2
2m
d2ψ(x)
dx2+V(x)ψ(x) = Eψ(x)
For the infinite square well potential, V(x) = 0 for 0< x < a and ∞otherwise, which means
ψ(x) = 0 for x≤0and x≥a.
2. General solution in the well: Since V(x) = 0 in the region 0< x < a, we have:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
The general solution to this differential equation is:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
Applying the boundary conditions ψ(0) = 0 and ψ(a) = 0, we find that B= 0 and k=nπ
a,
where nis a positive integer.
3. Determining energy levels: The energy levels are quantized and given by:
En=n2ˉh2π2
2ma2
where nis a positive integer.
For the first three energy levels (n= 1,2,3), we have: 1. E1=ˉh2π2
2ma22. E2=4ˉh2π2
2ma2= 4E1
3. E3=9ˉh2π2
2ma2= 9E1
Therefore, the first three energy levels of the particle in the infinite square well potential are
ˉh2π2
2ma2,4ˉh2π2
2ma2, and 9ˉh2π2
2ma2respectively.
16. Question: Consider an electron confined in a one-dimensional infinite potential well of width
L. Find the energy levels of the electron in terms of the fundamental constants and the integer
n.
Ans. Let’s denote the energy levels of the electron as En, where nis a positive integer
representing the quantum number. The energy levels are quantized and can be found by solving
the time-independent Schrödinger equation for a particle in a one-dimensional infinite potential
well.
1. Setting 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(x)
dx2=Eψ(x)
where ˉhis the reduced Planck’s constant, mis the mass of the electron, ψ(x)is the wave function,
Eis the energy, and xis the position within the well.
2. Applying boundary conditions: Inside the well, the potential energy is zero, so the
Schrödinger equation becomes:
−ˉ
h2
2m
d2ψ(x)
dx2=Eψ(x)
with the boundary conditions that ψ(0) = ψ(L) = 0.
3. Solving the Schrödinger equation: The solution to the Schrödinger equation can be
separated into three regions: 0< x < L,x= 0, and x=L. By solving the differential equation
within the well and applying the boundary conditions, the solutions for ψ(x)are found to be:
ψn(x) = √2
Lsin (nπx
L)
where nis a positive integer.
4. Determining the energy levels: The energy levels can be found by substituting the
wave function solutions back into the Schrödinger equation:
En=n2π2ˉh2
2mL2
where nis a positive integer representing the quantized energy levels. Hence, the energy levels
of the electron in the one-dimensional infinite potential well are given by En=n2π2ˉh2
2mL2.
17. Question 17:
Consider a particle in a one-dimensional potential well defined by the following piecewise
function:
V(x) = {0if 0≤x≤a
V0if x > a
where V0is a positive constant. Determine the expression for the energy levels of this particle
in the potential well.
Ans. To find the energy levels of the particle in the potential well described, we need to solve
the time-independent Schrödinger equation for this system and apply the appropriate boundary
conditions.
1. Set up the Schrödinger equation:
The time-independent Schrödinger equation for one-dimensional motion is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
Substitute the potential function into the equation:
−ˉh2
2m
d2ψ
dx2= (E−V(x))ψ
2. Solve the Schrödinger equation in different regions:
In region 0≤x≤a, the potential is zero (V(x) = 0), so the Schrödinger equation becomes:
−ˉh2
2m
d2ψ
dx2=Eψ
The solutions for this region are general solutions of the form ψ(x) = Aeikx +Be−ikx, where
k=√2mE
ˉ
h.
In region x > a, the potential is V0, so the Schrödinger equation becomes:
−ˉh2
2m
d2ψ
dx2+V0ψ=Eψ
The solutions for this region are general solutions of the form ψ(x) = Ceqx +De−qx, where
q=√2m(V0−E)
ˉh.
3. Apply boundary conditions:
The wavefunction ψ(x)and its derivative must be continuous at x=a. This gives us two
conditions:
1) ψ(a−) = ψ(a+)2) dψ
dx (a−) = dψ
dx (a+)
4. Solve for the energy levels:
By applying the boundary conditions and solving the resulting equations, we can find the
expression for the energy levels of the particle in the potential well.
This process involves working through the matching conditions at the boundary, solving the
transcendental equation that arises, and finding the allowed energy levels.
The final expression for the energy levels will involve constants a,V0,ˉh, and m.
Therefore, the expression for the energy levels of the particle in the potential well described
by the given piecewise function can be obtained by solving the Schrödinger equation and applying
the appropriate boundary conditions.
18. Question: Consider a one-dimensional particle in a box with a length of 2 nm. Calculate
the energy levels of the particle in electron volts (eV) compared to its ground state energy.
Ans. Step-by-step solution: Let’s first determine the ground state energy of the particle in the
box. 1. The energy of a particle in a box is given by the equation:
En=n2π2ˉh2
2mL2
2. Substituting the given values: n= 1,ˉh= 1.054572 ×10−34 Js, m= 9.10938356 ×10−31
kg, and L= 2 ×10−9m, into the equation, we get:
E1=(1)2π2(1.054572 ×10−34)2
2(9.10938356 ×10−31)(2 ×10−9)2
3. Calculating the ground state energy E1, we find:
E1≈2.448 ×10−18 J
4. To convert the ground state energy from joules to electron volts, we use the conversion
factor:
1eV = 1.602 ×10−19 J
5. Therefore, to obtain the ground state energy in electron volts, we divide 2.448 ×10−18 J
by 1.602 ×10−19 J/eV:
E1≈15.29 eV
6. Now, let’s calculate the first excited state energy E2by plugging in n= 2 into the energy
equation:
E2=(2)2π2(1.054572 ×10−34)2
2(9.10938356 ×10−31)(2 ×10−9)2
7. Calculating E2, we get:
E2≈9.794 ×10−18 J
8. Converting E2from joules to electron volts, we find:
E2≈61.17 eV
Therefore, the energy levels of the particle in electron volts compared to its ground state
energy are approximately 15.29 eV for the ground state and 61.17 eV for the first excited state.
19. Question: Consider an electron in a one-dimensional infinite square well potential of width
L. Calculate the energy levels in terms of the well width L.
Ans. Let’s denote the energy levels of the electron in the infinite square well potential as En,
where nis a positive integer representing the quantum number.
1. Setting up the problem: Inside the square well potential, the electron obeys the time-
independent Schrödinger equation:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where ψ(x)is the wave function for the electron at position x,mis the mass of the electron, E
is the total energy of the electron, and ˉhis the reduced Planck constant.
2. General solution of the Schrödinger equation: Since the electron is in a potential
well, the wave function ψ(x)must satisfy boundary conditions. Therefore, the general solution
for ψ(x)is:
ψn(x) = Asin(knx) + Bcos(knx)
where kn=nπ
L, and Aand Bare constants to be determined.
3. Applying boundary conditions: The boundary conditions for the infinite square well
potential are that the wave function must be zero at x= 0 and x=L. Therefore, ψ(0) =
ψ(L) = 0.
Applying the boundary condition ψ(0) = 0, we get:
B= 0
Applying the boundary condition ψ(L) = 0, we get:
Asin(knL) = 0
This implies that knL=nπ. Therefore, the allowed values of knare:
kn=nπ
L
4. Determining the energy levels: The total energy Enof the electron is related to knby
the equation:
En=ˉh2k2
n
2m
Substitute the value of kninto the expression for En, we have:
En=n2π2ˉh2
2mL2
Therefore, the energy levels of the electron in the one-dimensional infinite square well potential
are given by:
En=n2π2ˉh2
2mL2
where nis a positive integer.
20. Question: Determine the energy levels of a particle in a one-dimensional box of length Lif
the particle is subject to a potential energy function given by V(x) = V0cos2(πx
L).
Ans. To find the energy levels of the particle in the box, we need to solve the time-independent
Schrödinger equation Hψ =Eψ, where H=−ˉh2
2m
d2
dx2+V(x)and ψis the wave function.
1. First, we need to write the Schrödinger equation for the given potential:
Hψ =Eψ
(−ˉh2
2m
d2
dx2+V0cos2(πx
L))ψ=Eψ
2. Next, let’s separate the Schrödinger equation into two parts: the kinetic energy operator
and the potential operator:
−ˉh2
2m
d2ψ
dx2+V0cos2(πx
L)ψ=Eψ
3. Now, we can simplify the equation by substituting k=π
L:
−ˉh2
2m
d2ψ
dx2+V0cos2(kx)ψ=Eψ
4. The potential energy function V(x) = V0cos2(kx)can be rewritten as:
V(x) = V0
2(1 + cos(2kx))
5. We can now substitute the potential energy function back into the Schrödinger equation:
−ˉh2
2m
d2ψ
dx2+V0
2(1 + cos(2kx)) ψ=Eψ
6. Solve the Schrödinger equation to find the energy levels Eand the corresponding wave
functions ψ(x).
21. Find the first three energy levels for a particle in a one-dimensional box of length L.
Ans. Let’s denote the energy levels as Enwhere nis a positive integer. The energy levels for
a particle in a one-dimensional box are given by the formula:
En=n2π2ˉh2
2mL2
1. To find the first energy level E1, substitute n= 1 into the formula:
E1=(1)2π2ˉh2
2mL2=π2ˉh2
2mL2
2. For the second energy level E2, substitute n= 2:
E2=(2)2π2ˉh2
2mL2=4π2ˉh2
2mL2=2π2ˉh2
mL2
3. Finally, for the third energy level E3, substitute n= 3:
E3=(3)2π2ˉh2
2mL2=9π2ˉh2
2mL2=9π2ˉh2
2mL2
22. Question: Consider a potential energy function given by V(x) = 1
2kx2−1
3gx3, where k
and gare positive constants. Calculate the energy levels for a particle in this potential.
Ans. Let Ebe the total energy of the particle. The energy levels are given by the solutions to
the equation V(x) = E.
Step 1. Substitute the given potential function into the equation V(x) = E:
1
2kx2−1
3gx3=E
Step 2. Rearrange the equation to isolate x:
1
2kx2−1
3gx3−E= 0
Step 3. Multiply through by 6 to clear the fractions:
3kx2−2gx3−6E= 0
Step 4. This is a cubic equation in the form ax3+bx2+cx +d= 0, where a=−2g,
b= 0,c= 3k, and d=−6E. We are interested in finding the values of Ethat give physically
meaningful solutions for x.
Step 5. The energy levels correspond to the values of Efor which the particle is bound
within the potential. This means that there will be a finite number of real solutions for x.
Step 6. The number of energy levels can be determined by analyzing the shape of the
potential (V(x)) and considering the number of turning points in the graph.
Step 7. Once you have found the energy levels (E), you can find the corresponding wave-
functions and probabilities of finding the particle at different positions using the Schrödinger
equation and normalization conditions.
23. Suppose an electron is confined to a one-dimensional box of length L= 1 nm. Calculate
the energy levels of the system in eV.
Ans. To calculate the energy levels of the system, we can use the formula for the energy levels
of a particle in a one-dimensional box:
En=n2π2ˉh2
2mL2
where nis the quantum number (positive integer), 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. Let’s first convert the given quantities to SI units:
L= 1 nm = 1 ×10−9m
2. Now, we can substitute the values into the energy formula:
En=n2π2ˉh2
2mL2=n2π2(1.0545718 ×10−34)2
2(9.11 ×10−31)(1 ×10−9)2
3. Simplifying the expression, we get:
En=n2π2(1.0545718 ×10−34)2
2(9.11 ×10−31)(1 ×10−9)2
En=n2×9.8696 ×10−68
1.65 ×10−39
En=n2×6×10−29
4. Therefore, the energy levels of the electron in the one-dimensional box are given by
En=n2×6×10−29 J. To convert this to eV, we use the conversion factor 1eV = 1.602 ×10−19
J:
En=n2×6×10−29 ×1.602 ×10−19
En=n2×9.61 ×10−10 eV
Hence, the energy levels in eV are En=n2×9.61 ×10−10.
24. Question: Determine the energy levels of an electron in a one-dimensional harmonic
oscillator with a restoring force constant of k= 8.0N/m.
Ans. Let’s denote the energy levels as En, where nis the quantum number.
1. The energy levels of a one-dimensional harmonic oscillator are given by the formula:
En=(n+1
2)ˉhω
where ωis the angular frequency of the oscillator and is related to the force constant kas
ω=√k
m.
2. We need to find the mass mof the electron. The mass of an electron is 9.11 ×10−31 kg.
3. Calculate the angular frequency ω:
ω=√k
m=√8.0N/m
9.11 ×10−31 kg
4. Substitute the angular frequency back into the formula for energy levels:
En=(n+1
2)ˉh√k
m
5. Calculate the energy levels with n= 0,1,2.
E0=(0 + 1
2)ˉhω
E1=(1 + 1
2)ˉhω
E2=(2 + 1
2)ˉhω
25. Question 25:
An electron in a one-dimensional infinite square well of width Lis in the ground state.
Calculate the uncertainty in its kinetic energy.
Ans. Let’s denote the uncertainty in kinetic energy as ∆K. The ground state wavefunction for
an infinite square well of width Lis given by:
ψ(x) = √2
Lsin (πx
L)
The expectation value ⟨K⟩for kinetic energy can be calculated as:
⟨K⟩=∫L
0
ψ∗(−ˉh2
2m
∂2
∂x2)ψ dx
Since the electron is in the ground state, the kinetic energy is:
K=π2ˉh2
2mL2
The uncertainty in kinetic energy ∆Kcan be calculated using the uncertainty principle:
∆K∆x≥ˉh
2
Given that the electron is in the ground state, the uncertainty in position can be considered
as the width of the well L. Therefore,
∆x=L
Hence,
∆K≥ˉh
2L
Therefore, the uncertainty in kinetic energy of the electron is ∆K≥ˉh
2L.
26. What is the energy (in eV) of an electron in the third energy level of a hydrogen atom?
(Given: energy of the first energy level is -13.6 eV)
Ans. Let’s denote the energy of the electron in the third energy level as E3. We know that the
energy of an electron in the nth energy level of a hydrogen atom is given by the formula:
En=−13.6
n2eV
where nis the principal quantum number.
1. Substitute n= 3 into the formula:
E3=−13.6
32eV
E3=−13.6
9eV
E3=−1.5111 eV
Therefore, the energy of an electron in the third energy level of a hydrogen atom is −1.5111 eV.
27. Question 27:
Consider a particle in a one-dimensional potential well of width a. The potential energy is
given by
V(x) = {0 0 < x < a
∞otherwise
Determine the energy levels for this system using the time-independent Schrödinger equation.
Ans. Let’s solve for the energy levels step by step:
1. The time-independent Schrödinger equation for this system is given by:
−ˉh2
2m
d2ψ
dx2+V(x)ψ=Eψ
2. Since the potential energy is zero within the well, the equation simplifies to:
−ˉh2
2m
d2ψ
dx2=Eψ
3. Let’s denote k=√2mE
ˉh2. The general solution to the differential equation in the well
(0< x < a) is:
ψ(x) = Asin(kx) + Bcos(kx)
4. Applying the boundary conditions at x= 0 and x=a, we have:
ψ(0) = 0 ⇒B= 0
5. This gives us:
ψ(x) = Asin(kx)
6. Applying the boundary condition at x=a, we have:
ψ(a) = 0 ⇒Asin(ka) = 0
Since sin(ka)= 0, we must have A= 0. This implies that there are no acceptable solutions
for the wavefunction, making the energy levels in this potential well undefined.
Thus, no quantized energy levels exist for the given potential well.
28. Question:
An electron in a one-dimensional infinite potential well of length Lhas an energy level of E1.
If the electron is in the ground state of another one-dimensional infinite potential well of length
2L, what is the energy level it occupies in this second well?
Ans. Step-by-step solution:
Let’s denote the energy level in the second potential well as E2.
1. The energy of the electron in the first potential well is given by the formula:
E1=n2
1h2
8mL2
where n1is the quantum number corresponding to the energy level E1,his the Planck constant,
mis the mass of the electron, and Lis the length of the first potential well.
2. The energy of the electron in the second potential well can also be expressed in terms of
the quantum number n2corresponding to the energy level E2:
E2=n2
2h2
8m(2L)2
3. Since the electron is in the ground state of the second potential well, n2= 1. Substituting
n2= 1 into the equation for E2gives:
E2=h2
32mL2
4. To relate E1and E2, we know that the energy levels are inversely proportional to the
square of the lengths of the potential wells. Thus, we can write the following relationship:
E1
E2
=(L
2L)2
=1
4
5. Substituting the expression for E2into the equation above and rearranging, we find:
E1= 4E2
6. Therefore, the energy level of the electron in the second potential well is four times the
energy level E1. Thus, the electron in the second well occupies the energy level equivalent to
4E1.
29. Find the energy levels of a particle in a one-dimensional infinite square well of width a.
Ans. Let’s first recall the formula for the energy levels of a particle in a one-dimensional infinite
square well:
En=n2π2ˉh2
2ma2
where nis a positive integer representing the quantum number associated with the energy
level, ˉhis the reduced Planck’s constant, and mis the mass of the particle.
Solution:
1. Given the formula for the energy levels, we can now plug in the values of the constants:
n= 1,2,3, ...;mis the mass of the particle; ais the width of the well.
2. The energy level for n= 1 is:
E1=(1)2π2ˉh2
2ma2
3. The energy level for n= 2 is:
E2=(2)2π2ˉh2
2ma2
4. The energy level for n= 3 is:
E3=(3)2π2ˉh2
2ma2
5. The energy level for a general nis:
En=n2π2ˉh2
2ma2
Therefore, the energy levels of a particle in a one-dimensional infinite square well are quantized
and given by the above formula for all positive integers n.
30. Question 30: Determine the energy levels of a particle confined to a one-dimensional box of
length L. Let’s denote the energy levels as En, where nis a positive integer. Find an expression
for Enin terms of nand fundamental constants.
Ans. Let’s start by using the Schrödinger equation for a particle in a one-dimensional box:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
Where ˉhis the reduced Planck constant, mis the mass of the particle, Eis the energy of the
particle, and ψ(x)is the wave function.
1. The general solution to the above differential equation is given by:
ψ(x) = Asin(kx) + Bcos(kx)
where k=√2mE
ˉh2.
2. To satisfy the boundary conditions for a particle in a box of length L, we impose that
ψ(0) = ψ(L) = 0. This leads to the quantization condition:
kL =nπ
where nis a positive integer.
3. Solving for the energy levels En, we get:
k=nπ
L
En=ˉh2π2n2
2mL2
Therefore, the expression for the energy levels Enin terms of nis:
En=ˉh2π2n2
2mL2
31. Question 31: Consider a particle of mass mmoving in one dimension under the influence of
a potential energy function given by V(x) = 1
2kx2+1
4rx4, where kand rare positive constants.
Determine the energy levels of this system.
Ans. To determine the energy levels of the system, we need to solve the time-independent
Schrödinger equation, Hψ =Eψ, where His the Hamiltonian operator, ψis the wave function,
Eis the energy eigenvalue, and we have H=−ˉh2
2m
d2
dx2+V(x).
1. Writing the Hamiltonian operator: The Hamiltonian operator for this system is given
by
H=−ˉh2
2m
d2
dx2+1
2kx2+1
4rx4.
2. Writing the time-independent Schrödinger equation: The time-independent Schrödinger
equation for this system is
−ˉh2
2m
d2ψ
dx2+(1
2kx2+1
4rx4)ψ=Eψ.
3. Solving the Schrödinger equation: Let ψ(x) = u(x)e−αx2with α=√mk
ˉh2as the trial
solution.
Substitute ψ(x)into the Schrödinger equation and simplify to get
d2u
dx2= 2α(2αx2−k)u+ 4α2x2u−2rα2x4u−2mE
ˉh2u.
4. Simplifying the equation: The above differential equation simplifies to
d2u
dx2= (4α2x2−2ku + 4α(2α−rx2)u−2mE
ˉh2u.
5. Determining the energy levels: By inspection, we see that the potential term is
a harmonic oscillator (1
2kx2) and an anharmonic oscillator (1
4rx4). The energy levels for this
system will involve solving the above differential equation, which may not have a simple analytical
solution. Numerical methods or approximation techniques may be required to find the energy
levels in this case.
32. Question 32:
Consider a particle of mass mmoving in a one-dimensional potential well defined by V(x) =
V0x2, where V0is a positive constant. Determine the allowed energy levels of this system.
Ans. To find the allowed energy levels of the system, we will solve the time-independent
Schrödinger equation for this potential and apply boundary conditions to ensure the wavefunction
is well-behaved within the potential well.
1. Write down the time-independent Schrödinger equation:
The time-independent Schrödinger equation for a one-dimensional system is given by:
−ˉh2
2m
d2
dx2ψ(x) + V(x)ψ(x) = Eψ(x)
Substituting the given potential V(x) = V0x2, the equation becomes:
−ˉh2
2m
d2
dx2ψ(x) + V0x2ψ(x) = Eψ(x)
2. Solve the Schrödinger equation in different regions:
We will consider two regions: x < 0and x > 0. In these regions, the potential term simplifies
to −V0x2.
The general solution in region x < 0can be written as:
ψ(x) = Aekx +Be−kx
where k=√2mE
ˉ
h2.
The general solution in region x > 0can be written as:
ψ(x) = Ceik′x+De−ik′x
where k′=√2m(E+V0x2)
ˉh2.
3. Apply boundary conditions:
At the boundaries x= 0 and x=L, the wavefunction ψ(x)must be continuous. This means:
ψ(0−) = ψ(0+)
and
ψ(L−) = ψ(L+)
4. Find the allowed energy levels:
By solving the Schrödinger equation in each region and applying the boundary conditions, we
can find the allowed energy levels of the system for different values of V0and the length of the
potential well.
33. Question 33:
A particle of mass mmoves in a one-dimensional box of length L. What are the energy levels
for this system?
Ans. To find the energy levels for a particle in a one-dimensional box, we can use the time-
independent Schrödinger equation.
1. Set up the Schrödinger equation: The time-independent Schrödinger equation for a
particle in a one-dimensional box is given by:
−h2
8m
d2ψ
dx2=Eψ
where ψis the wavefunction of the particle, his the reduced Planck constant, mis the mass of
the particle, xis the position, and Eis the energy.
2. Solve the differential equation: The general solution to the differential equation is:
ψ(x) = Asin (nπx
L)
where Ais a normalization constant and nis a positive integer corresponding to the energy level.
3. Apply boundary conditions: At x= 0 and x=L, the wavefunction must go to zero
to satisfy the boundary conditions. Therefore, we have:
ψ(0) = Asin(0) = 0 =⇒A= 0
ψ(L) = Asin(nπ) = 0 =⇒nπ =mπ =⇒n= 1,2,3, ...
4. Determine the energy levels: The energy levels are quantized and given by:
En=n2π2ˉh2
2mL2
where nis the quantum number.
Therefore, the energy levels for the particle in a one-dimensional box are quantized and depend
on the quantum number n.
34. Question: Determine the energy levels of a particle in a one-dimensional infinite potential
well of width L.
Ans. Step-by-step solution: 1. In a one-dimensional infinite potential well, the Schrödinger
equation for a particle of mass mis given by:
−ˉh2
2m
d2ψ(x)
dx2=Eψ(x)
where Eis the energy eigenvalue and ψ(x)is the wavefunction.
2. The general solution to the Schrödinger equation inside the well can be written as:
ψn(x) = Asin (nπx
L)
where Ais a normalization constant and nis a positive integer representing the quantum
number associated with the energy level.
3. To determine the energy levels, we substitute the general solution back into the Schrödinger
equation:
−ˉh2
2m
d2
dx2[Asin (nπx
L)]=EA sin (nπx
L)
4. By taking the second derivative of ψ(x)and equating it to Eψ(x), we find:
n2π2ˉh2
2mL2=E
5. The energy levels Enfor the particle in the infinite potential well are quantized and given
by:
En=n2π2ˉh2
2mL2
where n= 1,2,3, ... are the quantum numbers corresponding to the allowed energy levels.
6. Therefore, the energy levels of a particle in a one-dimensional infinite potential well of
width Lare quantized and determined by the equation En=n2π2ˉh2
2mL2.
35. Question: Consider a particle of mass mconfined to a one-dimensional box of length L.
Determine the first three energy levels of the particle in terms of the constants m,L, and h
(Planck’s constant).
Ans. Let’s denote the energy levels of the particle in the box by En, where nis the quantum
number indicating the energy level. The energy levels are given by the formula:
En=n2π2ˉh2
2mL2,
where n= 1,2,3, ....
1. To find the first energy level E1, substitute n= 1 into the formula:
E1=(1)2π2ˉh2
2mL2=π2ˉh2
2mL2.
2. Next, let’s calculate the second energy level E2:
E2=(2)2π2ˉh2
2mL2=4π2ˉh2
2mL2= 2 (π2ˉh2
2mL2)= 2E1.
3. Finally, we find the third energy level E3by substituting n= 3 into the formula:
E3=(3)2π2ˉh2
2mL2=9π2ˉh2
2mL2= 3 (π2ˉh2
2mL2)= 3E1.
Therefore, the first three energy levels of the particle in the box are:
E1=π2ˉh2
2mL2,
E2= 2 (π2ˉh2
2mL2)= 2E1,
E3= 3 (π2ˉh2
2mL2)= 3E1.
