# Engineering Physics Questions and Answers – Mass Energy Relation

«
»

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Mass Energy Relation”.

1. The basic theorem/principle used to obtain mass-energy relation is _____________
a) Heisenberg’s Uncertainty Principle
b) Work-Energy Theorem
c) Momentum Conservation Theorem
d) Maxwell Theorem
View Answer

Answer: b
Explanation: To derive Einstein’s mass-energy relation, the basic principle used is Work-Energy Theorem. It states that the kinetic energy of a moving body is equal to the work done by the external force on the body from rest.
advertisement

2. Which of the following is Einstein’s mass energy relation?
a) Ek = (m – m0)c2
b) E = mc2
c) E2 – p2c2 = m02c4
d) Ek = mv2/c2
View Answer

Answer: b
Explanation: E = mc2 is the famous Einstein mass-energy relation. It states a universal equivalence between mass and energy.

3. For Pair Production phenomenon to occur to photon must have energy, greater than or equal to ____________
a) 0.51 MeV
b) 1.02 MeV
c) 0.32 MeV
d) 0.85 MeV
View Answer

Answer: b
Explanation: In Pair Production, a proton with energy, greater than or equal to 1.02 MeV is required to create a pair of electron and positron. 1.02 MeV is the minimum required energy for their creation.

4. Which of the following is the momentum-energy relation?
a) E2 – p2c2 = m02c2
b) E2 – p2c2 = m02c4
c) E2 – p2c2 = m02c3
d) E2 – p2c2 = m0c2
View Answer

Answer: b
Explanation: Einstein’s famous momentum-energy relation shows that a particle may have energy and momentum even if its rest mass is zero, i.e., if m0 = 0.

5. In Relativistic case, as the velocity of the particle approaches the speed of light, the Kinetic energy approaches ___________
a) Zero
b) Kinetic Energy as in Non-Relativistic case
c) Rest Energy
d) Infinite
View Answer

Answer: d
Explanation: In Relativistic case, the expression for Kinetic Energy is: Ek = $$(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} – m_0)c^2$$
Thus, as v -> c, v/c -> 1.
Thus, the Kinetic Energy approaches infinity, which would require an infinite amount of external work.
advertisement

6. If a star radiates energy at the rate of 5 x 1020 Js-1, what is the rate at which its mass is decreasing?
a) 5.54 x 109 kgs-1
b) 4.44 x 109 kgs-1
c) 3.44 x 109 kgs-1
d) 2.44 x 109 kgs-1
View Answer

Answer: a
Explanation: As we know from Einstein’s mass energy relation, E = mc2.
Therefore, ΔE = Δmc2
ΔE = 5 x 1020 Js-1, c = 3 x 108 ms-1.
Δm = 5 x 1020 Js-1/ 9 x 1016m2s-2
Δm = 5.54 x 103 kgs-1.

7. What will be the rest energy of an electron?
a) 0.41 MeV
b) 0.51 MeV
c) 0.61 MeV
d) 0.71 MeV
View Answer

Answer: b
Explanation: We Know, rest energy = mc2
Here, m = 3.1 X 10-31 kg, c = 3 X 108 m/s
Therefore, E = 9.109 X 10-31 kg X 9 X 1016 m2/s2
= 8.198 X 10-14 J
= 0.51 MeV.

8. The binding energy of an electron to a proton (i.e., hydrogen atom) is 13.6 eV. The loss of mass in the formation of one atom of hydrogen is _____________
a) 2.42 X 10-35 Kg
b) 3.34 X 10-35 Kg
c) 4.58 X 10-35 Kg
d) 5.19 X 10-35 Kg
View Answer

Answer: a
Explanation: E = 13.6 eV = 13.6 X 1.6 X 10-19 J
Using Einstein’s Mass-energy relation, the loss of mass Δm = E/c2
Δm = 13.6 X 1.6 X 10-19/9 X 1016
= 2.42 X 10-35 Kg.

9. The momentum of a photon having energy 10-17J is ____________
a) 1.11 X 10-26 Kg m/s
b) 2.22 X 10-26 Kg m/s
c) 3.33 X 10-26 Kg m/s
d) 4.44 X 10-26 Kg m/s
View Answer

Answer: c
Explanation: We know that the rest mass of a photon is zero.
Therefore, from the momentum-energy relation
P = E/c = 10-17/3 X 108
= 3.33 X 10-26 Kg m/s.

10. Which phenomenon is shown in the figure?

a) Pair Production
b) Photoelectric Effect
c) Compton effect
d) Pair annihilation
View Answer

Answer: d
Explanation: In the figure, an electron and a positron annihilate each other and release the equivalent amount of energy. This process is called Pair annihilation.
advertisement

Sanfoundry Global Education & Learning Series – Engineering Physics.

To practice all areas of Engineering Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn