This document contains 12 numerical problems related to quantum mechanics. The problems calculate things like the velocity, wavelength, and kinetic energy of electrons and protons using concepts like the de Broglie wavelength equation and Bragg's law. The document provides the equations and step-by-step workings to arrive at the solutions for each problem.

1. QUANTUM MECHANICS
1
MODULE - I
Dr. C. R. Kesavulu, Associate Professor, Dept. of Physics

2. Numerical Problems
1. Calculate the velocity and kinetic energy of an
electron of wavelength 0.21nm.
2. Calculate the de Broglie wavelength associated
with a proton moving with a velocity of 1/10 of
velocity of light. (Mass of proton = 1.674 x 10-27
kg).
3. Calculate the wavelength of an electron raised to
a potential 15kV.
4. If the kinetic energy of the neutron is 0.025eV
calculate its de-Broglie wavelength (mass of
neutron =1.674 x 10-27 Kg)
5. Calculate the velocity and kinetic energy of an
electron of wavelength 1.66 x 10-10 m.
Numerical Problems

3. Numerical Problems
1. Calculate the velocity and kinetic energy of an electron of
wavelength 0.21nm.
Solution:
𝑑e- Broglie wavelength 𝜆 =
ℎ
𝑚𝜐
𝜐 =
ℎ
𝑚𝜆
𝜐 =
6.626×10−34
9.1 × 10−31 ×2.1 × 10−10
= 34.67 x 105 m/s
Kinetic energy of electron
E =
1
2
𝑚𝜐2
=
1
2
× 9.1 × 10−31 × 34.67 × 34.67 × 10 10
= 0.5469 × 10-17 J
=
0.5469×10−17
1.6 × 10−19 𝑒𝑉
= 34.182eV

4. Numerical Problems
2. Calculate the de Broglie wavelength associated with a proton
moving with a velocity
of 1/10 of velocity of light. (Mass of proton = 1.674 x 10-27 kg).
Solution:
𝑑e- Broglie wavelength 𝜆 =
ℎ
𝑚𝜐
𝜆 =
6.626×10−34
1.674× 10−27 ×
1
10
×3 × 108
= 1.31 x10-14 m

5. Numerical Problems
3. Calculate the wavelength of an electron raised to a potential
15kV.
Solution: de-Broglie wavelength
𝜆 =
12.26
𝑉
A0
=
12.26
15000
=
12.26
122.47
= 0.1Ao

6. Numerical Problems
4. If the kinetic energy of the neutron is 0.025eV calculate its
de-Broglie wavelength (mass of neutron =1.674 x 10-27 Kg)
Solution: Kinetic energy of neutron
E =
1
2
𝑚𝜐2
= 0.025eV
= 0.025 × 1.6x10-19 J
𝜐=
2𝑋0.025×1.6×10−19
1.674×10−27
1
2
= 0.04779 × 108
1
2
= 0.2186x104m/s
∴ 𝑑e- Broglie wavelength 𝜆 =
ℎ
𝑚𝜐
𝜆 =
6.626×10−34
1.67 × 410−27×0.2186×104
=0.181nm

7. Numerical Problems
5. Calculate the velocity and kinetic energy of an electron of
wavelength 1.66 x 10-10 m.
Solution:
𝑑e- Broglie wavelength 𝜆 =
ℎ
𝑚𝜐
𝜐 =
ℎ
𝑚𝜆
𝜐 =
6.626×10−34
9.1 × 10−31 ×1.66 × 10−10
= 43.86 x 105 m/s
Kinetic energy of electron
E =
1
2
𝑚𝜐2
=
1
2
× 9.1 × 10−31
× 43.86 × 43.86 × 10 10
= 8752.83 × 10-21 J = 0.875× 10-17 J
=
0.875×10−17
1.6 × 10−19 𝑒𝑉
= 54.68 eV

8. Numerical Problems
6. Calculate the wavelength of an electron raised to a potential of 1600
V.
7. Calculate the energies that can be possessed by a particle of mass
8.50 x10-31kg which is placed in an infinite potential box of width 10-
9cm.
8. Find the lowest energy of an electron confined in a 3-D box of side 0.1
nm
9. An electron is bound in 1-dimensional infinite well of width 1 x 10-10 m.
Find the energy values of ground state and first two excited states.
10.An electron is bound in one-dimensional box of size 4x 10-10 m. What
will be its minimum energy?
11.Calculate the wavelength of matter wave associated with a neutron
whose kinetic energy is 1.5 times the rest mass of electron.[Given that
mass of neutron=1.676 × 10-27kg, mass of electron 9.1 × 10-31Kg,
Planck’s constant = 6.62 × 10-34J-Sec,velocity of light is 3×108m/s].
12.Electrons are accelerated by 344 volts and are reflected from a
crystal. The first reflection maximum occurs when the glancing angle
is 60o . Determine the spacing of the crystal

9. Numerical Problems
6. Calculate the wavelength of an electron raised to a potential
1600V.
Solution: de-Broglie wavelength
𝜆 =
12.26
𝑉
A0
=
12.26
1600
=
12.26
40
= 0.3065A0

10. Numerical Problems
7. Calculate the energies that can be possessed by a particle
of mass 8.50 x10-31kg which is placed in an infinite potential
box of width 10-9cm.
Solution: The possible energies of a particle in an infinite potential
box of width L is given by En =
𝑛2ℎ2
8𝑚𝐿2
m =8.50 x 10-31Kg
L =1 x 10-11m
h=6.626× 10-34J-s
For ground state n=1
E1 =
6.626 × 10−34 2
8 8.50 × 10−31 1 × 10−11 2
=6.456 × 10-16 joule
For first excited state, E2 = 4 x 6.4456 x 10-16
= 25.8268 x10-16Joule

11. Numerical Problems
8. Find the lowest energy of an electron confined in a
square box of side 0.1nm.
Solution: The possible energies of a particle in an infinite
potential box of width L is given by En =
𝑛2ℎ2
8𝑚𝐿2
m =9.1 x 10-31Kg
L =0.1 x 10-9m
h=6.626× 10-34J-s
For lowest energy n=1
E1 =
6.626 × 10−34 2
8 9.1 × 10−31 0.1 × 10−9 2
=60.307 × 10-19 joule

12. Numerical Problems
9. An electron is bound in 1-dimensional infinite well of width
1 x 10-10 m. Find the energy values of ground state and first
two excited states.
Solution: The possible energies of a particle in an infinite potential
box of width L is given by En =
𝑛2ℎ2
8𝑚𝐿2
m =9.1 x 10-31Kg
L =1 x 10-10m
h=6.626× 10-34J-s
For ground state n=1
E1 =
6.626 × 10−34 2
8 9.1 × 10−31 10−10 2
= 0.6031 × 10-17 joule
For first excited state, E2 = 4 x 0.6031 x 10-17
= 2.412 x10-17Joule
For second excited state, E3 = 9 x 0.6031 x 10-17
= 5.428 x10-17Joule

13. Numerical Problems
10. An electron is bound in one-dimensional box of size
4x 10-10 m. What will be its minimum energy?
Solution: The possible energies of a particle in an infinite
potential box of width L is given by En =
𝑛2ℎ2
8𝑚𝐿2
m =9.1 x 10-31Kg
L =4 x 10-10m
h=6.626× 10-34J-s
For minimum energy n=1
E1 =
6.626 × 10−34 2
8 9.1 × 10−31 4 × 10−10 2
=0.346 × 10-18 joule

14. 11. Calculate the wavelength of matter wave associated with a neutron whose kinetic
energy is 1.5times the rest mass of electron.[Given that mass of neutron=1.676 ×
10-27
kg, mass of electron 9.1 × 10-31
Kg, Planck’s constant = 6.62 × 10-34
J-Sec,velocity
of light is 3×108
m/s].
Solution: For neutron
1
2
m𝜐2
=1.5×9.1×10-31
Joules
Or 𝑣2
=
2 1.5 × 9.1×10−31
1.676 × 10−27
= 16.288x10-4
𝜐 = 16.288 × 10−4
= 4.046 ×10-2
m/s
The de-Broglie wavelength expression is
𝜆 =
ℎ
𝑚𝜐
=
6.62×10−34
1.676 𝑋 10−2× 4.046𝑋10−2
= 9.76X10-6
m
Numerical Problems

15. Numerical Problems
12. Electrons are accelerated by 344 volts and are reflected
from a crystal. The first reflection maximum occurs when the
glancing angle is 60o . Determine the spacing of the crystal
Solution:
Given V= 344 V; ϴ=600
We have, 𝝀 =
𝟏𝟐.𝟐𝟕
𝟑𝟒𝟒
=
𝟏𝟐.𝟐𝟕
𝟏𝟖.𝟓𝟒𝟕
= 0.6615 A0.
From Bragg’s law we have 𝟐𝒅 𝐬𝐢𝐧 𝜽 = 𝒏𝝀
First reflection maximum i.e., n=1. Spacing of crystal d=?
So, 𝑑 =
𝑛𝜆
2 sin 𝜃
=
1×0.6615
2×sin 60
=
1×0.6615
2×0.8660
= 0.3819 A0

16. 16