Engineering Physics Multiple Choice Questions on “Einstein Coefficients”.
1. Which of the following is the correct expression for the relation between Einstein’s coefficients A and B?
a) (frac{8pi v^3 h}{c^3})
b) (frac{8pi v^2 h}{c^3})
c) (frac{8pi v^2 h}{c^2})
d) (frac{8pi hv}{c^3})
Answer: a
Clarification: The expression (frac{8pi v^3 h}{c^3}) is the correct expression for the relation between the two Einstein’s coefficients. This expression is known as the Einstein’s relation.
2. What is the relationship between B21 and B12?
a) B12 > B21
b) B12 < B21
c) B12 = B21
d) No specific relation
Answer: c
Clarification: B21 is the coefficient for the stimulated emission while B12 is the coefficient for stimulated absorption. Both the processes are mutually reverse processes and their probabilities are equal. Therefore, B12 = B21.
3. Which of the following Einstein’s coefficient represents spontaneous emission?
a) A12
b) A21
c) B12
d) B21
Answer: b
Clarification: A21 represents the spontaneous emission of photons. A12 signifies spontaneous absorption.B12 is for stimulated absorption while B21 is for stimulated emission.
4. The correct expression for the rate of stimulated emission is _______________
a) Rse = A21N2
b) Rse = A21uN2
c) Rse = B21N2
d) Rse = B21uN2
Answer: d
Clarification: The stimulates emission is directly proportional to the energy density u, of the external radiation field. Also, stimulated emission rate increases with the increase in number N2 of exited atoms.
5. Which law is used for achieving the relation between the Einstein’s coefficients?
a) Heisenberg’s Uncertainty Principle
b) Planck’s radiation law
c) Einstein’s equation
d) Quantum law
Answer: b
Clarification: Planck’s radiation law, which gives the energy density u = (frac{8πhv^3}{c^3} frac{1}{e^α-1}), is used as the formula resembles the one for the energy density of the external radiation field in stimulated emission, u = (frac{A_{21}}{B_{21}(frac{B_{12}}{B_{21}}e^α-1)}).
6. The probability of spontaneous emission increases rapidly with the energy difference between the two states.
a) True
b) False
Answer: a
Clarification: From Einstein’s relation we know that the ratio of Einstein’s coefficients is (frac{8pi v^3 h}{c^3}). Thus, the ration of Einstein’s coefficients is proportional to the cube of the frequency. Hence, the probability of spontaneous emission increases rapidly with the energy difference between the two states.
7. If the frequency of emitted photon is 10 Hz, the ratio of Einstein’s coefficient is _____________
a) 2.177 X 10-51
b) 3.177 X 10-51
c) 5.177 X 10-51
d) 6.177 X 10-51
Answer: d
Clarification: We know Einstein’s relation = (frac{8pi v^3 h}{c^3})
v = 10 Hz, h = 6.63 X 10-34 Js, c = 3 X 108m/s
Therefore, the ratio of Einstein’s coefficients is: 6.177 X 10-51.
8. What is the unit of the coefficient of spontaneous emission?
a) s-1
b) s
c) J-1
d) J
Answer: a
Clarification: For spontaneous emission, the expression for the rate is = A21N2, where N2 is the number of particles in exited state. As the unit of rate is Number of particles per second, the unit of A21 is s-1.
9. What is the unit for the coefficient of stimulated emission?
a) s-2
b) m3 s-2
c) J−1 m3
d) J−1 m3 s-2
Answer: d
Clarification: For stimulated emission, the expression for the rate is B21uN2 where u stands for the energy density and N is the number of exited atoms. Therefore, the unit of B turns out to be J−1 m3 s-2.
10. Which Einstein’s coefficient should be used in this case?
a) A12
b) A21
c) B12
d) B21
Answer: d
Clarification: The given figure shows stimulated emission. Hence, the Einstein coefficient for stimulated emission is B21. If it had been spontaneous emission, then A21 would have been used.