Physics Multiple Choice Questions on “Particle Nature of Light : The Photon”.
1. What is the frequency of a photon whose energy is 66.3 eV?
a) 12.6 × 1016 Hz
b) 91.6 × 1016 Hz
c) 1.6 × 1016 Hz
d) 81.6 × 1016 Hz
Answer: c
Clarification: Frequency can be written as, v = (frac {E}{h})
v = (frac {E}{h})
v = (frac {(66.3 times 1.6 times 10^{-19})}{(6.63 times 10^{-34})} )
v = 1.6 × 1016 Hz.
2. Calculate the energy of a photon of wavelength 6600 angstroms.
a) 0.3 × 10-19 J
b) 3 × 10-19 J
c) 30 × 10-19 J
d) 300 × 10-19 J
Answer: b
Clarification: λ = 6600 angstroms = 6600 × 10-10m.
Energy of 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.
3. Among the following four spectral regions, in which of them, the photon has the highest energy in?
a) Infrared
b) Violet
c) Red
d) Blue
Answer: b
Clarification: According to the equation:
E = (frac {hc}{lambda })
Energy is inversely proportional to wavelength. Since a photon in the violet region has the least wavelength, it implies, that photon has the highest energy.
4. What will be the photon energy for a wavelength of 5000 angstroms, if the energy of a photon corresponding to a wavelength of 7000 angstroms is 4.23 × 10-19 J?
a) 0.456 eV
b) 5.879 eV
c) 3.701 eV
d) 1.6 × 10-19 eV
Answer: c
Clarification: (frac {E_2}{E_1} = frac {lambda_1}{lambda_2})
(frac {lambda_1}{lambda_2} = frac {1.4 times 4.23 times 10^{-19}}{1.6 times 10^{-19}}) eV
(frac {lambda_1}{lambda_2}) = 3.701 eV
5. Photons of energy 10.25 eV fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 5.0 eV. What is the stopping voltage required for these electrons?
a) 10 V
b) 4 V
c) 8 V
d) 5 V
Answer: d
Clarification: The required equation is given as:
Stopping voltage = (frac {K_{max}}{e}).
(frac {K_{max}}{e} = frac {5.0 eV}{e})
(frac {K_{max}}{e}) = 5.0 V
Therefore, the stopping potential = 5 V.
6. When a proton is accelerated through 1 V, then its kinetic energy will be 1 V.
a) True
b) False
Answer: b
Clarification: Kinetic energy = qV.
K = qV
K = e × (1V)
K = 1 eV.
7. What is the energy of a photon of wavelength λ?
a) hcλ
b) (frac {hc}{lambda })
c) (frac {lambda }{hc})
d) (frac {lambda h}{c})
Answer: b
Clarification: Energy of a photon, E = hv
V = (frac {c}{lambda }).
Therefore, E = (frac {hc}{lambda })
8. What is the momentum of a photon of wavelength λ?
a) (frac {hv}{c})
b) Zero
c) (frac {hlambda }{c^2})
d) (frac {hlambda }{c})
Answer: a
Clarification: Momentum is given as:
p = mc
p = mc
p = (frac {mc^2}{c})
p = (frac {hv}{c}).
9. Which among the following shows the particle nature of light?
a) Photoelectric effect
b) Interference
c) Refraction
d) Polarization
Answer: a
Clarification: Photoelectric effect can only be explained based on the particle nature of light. This effect is caused due to the ejection of electrons from a metal plate when light falls on it. The others do not show the particle nature of light.
10. What will be the number of photons emitted per second, if the power of the radio transmitter is 15 kW and it operates at a frequency of 700 kHz?
a) 3.24 × 1031
b) 3.87 × 1025
c) 2.77 × 1037
d) 3.24 × 1045
Answer: a
Clarification: Number of photons emitted per second is given as:
N = ( frac {Power}{Energy , of , a , photon} = frac {P}{hv}).
N = ( frac {15 times 10^3}{6.6 times 10^{-34} times 700 times 10^3})
N = 3.24 × 1031