Class 12 Physics MCQ – Photoelectric Effect

This set of Class 12 Physics Chapter 11 Multiple Choice Questions & Answers (MCQs) focuses on “Photoelectric Effect”.

1. Which photon is more energetic: A red one or a violet one?
a) Both
b) Red
c) Violet
d) Neither
View Answer

Answer: c
Explanation: Violet photon has more energy because the energy of a photon is given as:
E = hv
Since energy is directly proportional to wavelength, i.e. vvoilet > vred, the violet photon is more energetic than the red photon.

2. If the wavelength of electromagnetic radiation is doubled, what will happen to the energy of photons?
a) Remains the same
b) Doubled
c) Halved
d) Infinite
View Answer

Answer: c
Explanation: Energy of a photon is given as:
E = hv = \(\frac {hc}{\lambda}\)
The energy of the photon reduces to one-half when the wavelength of radiation is doubled.

3. What happens to the wavelength of a photon after it collides with an electron?
a) Increases
b) Decreases
c) Remains the same
d) Infinite
View Answer

Answer: a
Explanation: A photon transfers a part of its energy to the colliding electron, so its energy decreases, and consequently wavelength increases. This is because the energy of a photon is inversely proportional to the wavelength of a photon.

4. Why are alkali metals most suited as photo-sensitive metals?
a) High frequency
b) Zero rest mass
c) High work function
d) Low work function
View Answer

Answer: d
Explanation: Alkali metals have low work functions. Even visible radiation can eject out electrons from them. So alkali metals are the most suitable photo-sensitive metals.

5. Which radiations will be most effective for the emission of electrons from a metallic surface?
a) Microwaves
b) X rays
c) Ultraviolet
d) Infrared
View Answer

Answer: c
Explanation: Ultraviolet rays are most effective for photoelectric emission because they have the highest frequency when compared to the other electromagnetic waves, and hence the most energetic among all of them.

6. Photoelectric emission is possible at all frequencies.
a) True
b) False
View Answer

Answer: b
Explanation: No. Photoelectric emission is possible only if the energy of the incident photon is greater than the work function (W0 = hv0) of the metal. Hence, the frequency v of the incident radiation must be greater than the threshold frequency v0.

7. Two metals A and B have work functions 4 eV and 10 eV respectively. Which metal has a higher threshold wavelength?
a) Metal A
b) Metal B
c) Both
d) Neither
View Answer

Answer: a
Explanation: According to the equation, W0 = hv0 = \(\frac {hc}{\lambda_0}\), the work function is inversely proportional to the wavelength.
So metal A with lower work function has a higher threshold wavelength.

8. Give the unit of work function.
a) Electron volt
b) Joule
c) Hertz
d) Watt
View Answer

Answer: a
Explanation: One electron volt is the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt.
Therefore, 1 eV = 1.602 × 10-19 J.

9. What is the frequency of a photon whose energy is 66.3 eV?
a) 196 × 1016 Hz
b) 1336 × 1016 Hz
c) 1.6 × 1016 Hz
d) 16 × 1016 Hz
View Answer

Answer: c
Explanation: Frequency can be calculated as follows:
V = \(\frac {E}{h}\)
V = \(\frac {(66.3 \times 1.6 \times 10^{-19})}{(6.63 \times 10^{-34})}\)
V = 1.6 × 1016Hz

10. Calculate the energy of a photon of wavelength 6600 angstroms.
a) 30 × 10-19 J
b) 3 × 10-19 J
c) 300 × 10-19 J
d) 3000 × 10-19 J
View Answer

Answer: b
Explanation: Given: λ = 6600 angstrom = 6600 × 10-10 m.
Energy of a photon = \(\frac {hc}{\lambda }\)
E = \(\frac {(6.6 \times 10^{-34} \times 3 \times 10^8)}{(6600 \times 10^{-10})}\)
E = 3 × 10-19 J

Sanfoundry Global Education & Learning Series – Physics – Class 12.

To practice all chapters and topics of class 12 Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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