# Physics Questions and Answers – Einstein’s Photoelectric Equation : Energy Quantum of Radiation

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Einstein’s Photoelectric Equation : Energy Quantum of Radiation”.

1. Two beams, one of red light and the other of blue light, of the same intensity are incident on a metallic surface to emit photoelectrons. Which emits electrons of greater frequency?
a) Both
b) Red light
c) Blue light
d) Neither

Explanation: From Einstein’s photoelectric equation, we have
$$\frac {1}{2}$$ mv2 = h(v – v0).
Since the frequency of blue light is greater than that of the red light, blue light emits electrons of greater kinetic energy.

2. If the intensity of incident radiation in a photo-cell is increased, how does the stopping potential vary?
a) Increases
b) Remains the same
c) Decreases
d) Infinite

Explanation: There is no effect on stopping potential. The intensity of incident radiation is independent of stopping potential. Therefore, even if the incident radiation in a photo-cell is increased, the stopping potential remains unchanged.

3. If the frequency of the incident radiation is equal to the threshold frequency, what will be the value of the stopping potential?
a) 0
b) Infinite
c) 180 V
d) 1220 V

Explanation: From Einstein’s photoelectric equation, we have
$$\frac {1}{2}$$ mv2 = h(v – v0).
When v = v0,
hv0 = hv0 + eV0
Therefore, V0 = 0.
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4. How does retarding potential vary with the frequency of light causing photoelectric effect?
a) Infinite
b) Zero
c) Decreases
d) Increases

Explanation: The stopping potential is directly proportional to the frequency of light.
Hence, the stopping potential increases with an increase in the frequency of the incident light.

5. If the intensity of the radiation incident on a photo-sensitive plate is doubled, how does the stopping potential change?
a) Increases
b) Decreases
c) No effect
d) Infinite

Explanation: No effect. This is because the intensity of radiation incident on a photo-sensitive plate is independent on stopping potential. So, the stopping potential remains the same, even if the intensity is doubled.

6. The stopping potential in photoelectric emission does not depend upon the frequency of the incident radiation.
a) True
b) False

Explanation: The stopping potential depends on the frequency of incident radiation. Above the threshold frequency, the stopping potential is directly proportional to the frequency of incident radiation.

7. The maximum kinetic energy of a photoelectron is 3 eV. What is the stopping potential?
a) 0 V
b) 3 V
c) 9 V
d) 12 V

Explanation: Stopping potential is given as:
V0 = $$\frac {K_{max}}{e}$$
V0 = $$\frac {3eV}{e}$$
V0 = 3 V
Therefore, the stopping potential is 3 V.

8. The stopping potential in an experiment on the photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?
a) 1.5 eV
b) 3 eV
c) 4.5 eV
d) 6 eV

Explanation: Kmax = eV0
Kmax = 1.6 × 10-19 C × 1.5 V
Kmax = 24 × 10-19 J
It can also be converted into units of electron volts that will be:
Kmax = 1.5 eV

9. For a photosensitive surface, the work function is 3.3 × 10-19 J. Calculate the threshold frequency.
a) 15 × 1014 Hz
b) 25 × 1014 Hz
c) 5 × 1014 Hz
d) 55 × 1014 Hz

Explanation: Threshold frequency is given as:
v0 = $$\frac {W_0}{h}$$
v0 = $$\frac {W_0}{h}$$
v0 = $$\frac {(3.3 \times 10^{-19})}{(6.6 \times 10^{-34})}$$
v0 = 5 × 1014 Hz.

10. Calculate the kinetic energy of a photoelectron (in eV) emitted on shining light of wavelength 6.2 × 10-6 m on a metal surface. The work function of the metal is 0.1 eV.
a) 10 eV
b) 0.1 eV
c) 0.01 eV
d) 1000 eV

Explanation: Kmax = $$( \frac {hc}{\lambda })$$ – W0
Kmax = $$[ \frac {(6.6 \times 10^{-34} \times 3 \times 10^8 )}{(6.2 \times 10^{-6} \times 1.6 \times 10^{-19}) }]$$ – 0.1.
Kmax = 0.2 – 0.1
Kmax = 0.1 eV

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