Particle Nature of Light – The Photon: NEET MCQs with Answers

15 free questions on Particle Nature of Light – The Photon for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: The Photon: particle nature of light, Dual nature of radiation.

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Question 1

Previous-year question
The energy of a photon of wavelength λ is [AIPMT 1988]
  1. hcλ
  2. hc/λ
  3. λ/hc
  4. λh/c
Show answer
Answer: (B) hc/λ
Explanation: The energy of a photon is E = hν, and since ν = c/λ, this becomes E = hc/λ. The energy is thus inversely proportional to the wavelength.

Question 2

Previous-year question
A radio transmitter operates at a frequency 880 kHz and a power of 10 kW. The number of photons emitted per second is [AIPMT 1990]
  1. 1.72 × 10³¹
  2. 1.327 × 10²⁵
  3. 1.327 × 10³⁷
  4. 1.327 × 10⁴⁵
Show answer
Answer: (A) 1.72 × 10³¹
Explanation: n = P/hν = 10⁴/(6.63 × 10⁻³⁴ × 880 × 10³) = 10⁴/(5.83 × 10⁻²⁸) ≈ 1.72 × 10³¹ photons per second. The enormous number explains why radio waves show no observable quantum granularity.

Question 3

Previous-year question
The momentum of a photon of electromagnetic radiation is 3.3 × 10⁻²⁹ kg m s⁻¹. What is the frequency of the associated waves? (h = 6.6 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹) [AIPMT 1990]
  1. 1.5 × 10¹³ Hz
  2. 7.5 × 10¹² Hz
  3. 6.0 × 10¹³ Hz
  4. 3.0 × 10³ Hz
Show answer
Answer: (A) 1.5 × 10¹³ Hz
Explanation: For a photon p = hν/c, so ν = pc/h = (3.3 × 10⁻²⁹ × 3 × 10⁸)/(6.6 × 10⁻³⁴) = 9.9 × 10⁻²¹/6.6 × 10⁻³⁴ = 1.5 × 10¹³ Hz.

Question 4

Previous-year question
The wavelength of a 1 keV photon is 1.24 × 10⁻⁹ m. What is the frequency of a 1 MeV photon? [AIPMT 1991]
  1. 1.24 × 10¹⁵ Hz
  2. 2.4 × 10²⁰ Hz
  3. 1.24 × 10¹⁸ Hz
  4. 2.4 × 10²³ Hz
Show answer
Answer: (B) 2.4 × 10²⁰ Hz
Explanation: Since E = hc/λ, the wavelength is inversely proportional to energy. Raising the energy from 1 keV to 1 MeV — a factor of 10³ — reduces the wavelength to 1.24 × 10⁻¹² m. Hence ν = c/λ = 3 × 10⁸/1.24 × 10⁻¹² ≈ 2.4 × 10²⁰ Hz.

Question 5

Previous-year question
The momentum of a photon of wavelength λ is [AIPMT 1993]
  1. h/λ
  2. zero
  3. hλ/c²
  4. hλ/c
Show answer
Answer: (A) h/λ
Explanation: A photon travels at c, so its momentum is p = E/c = hν/c. Since ν/c = 1/λ, this gives p = h/λ — the de Broglie relation applied to the photon.

Question 6

Previous-year question
The nature of ions knocked out from hot surfaces is [AIPMT 1995]
  1. protons
  2. electrons
  3. neutrons
  4. nuclei
Show answer
Answer: (B) electrons
Explanation: In thermionic emission the particles liberated from a heated metal surface are electrons, which are far lighter and more mobile than any positive ion in the lattice. They are commonly called thermions.

Question 7

Previous-year question
The 21 cm radiowave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly [AIPMT 1998]
  1. 10⁻¹⁷ J
  2. 1 J
  3. 7 × 10⁻⁶ J
  4. 10⁻²⁴ J
Show answer
Answer: (D) 10⁻²⁴ J
Explanation: E = hc/λ = (6.6 × 10⁻³⁴ × 3 × 10⁸)/0.21 = 1.98 × 10⁻²⁵/0.21 ≈ 9.4 × 10⁻²⁵ J, which is of the order of 10⁻²⁴ J.
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Question 8

Previous-year question
Which one among the following shows the particle nature of light? [AIPMT 2001]
  1. Photoelectric effect
  2. Interference
  3. Refraction
  4. Polarization
Show answer
Answer: (A) Photoelectric effect
Explanation: In the photoelectric effect a single photon transfers all its energy to a single electron, and emission begins the instant the light arrives — behaviour that only a particle picture explains. Interference, refraction and polarisation are all wave phenomena.

Question 9

Previous-year question
The momentum of a photon of energy 1 MeV in kg m/s will be [AIPMT 2006]
  1. 0.33 × 10⁶
  2. 7 × 10⁻²⁴
  3. 10⁻²²
  4. 5 × 10⁻²²
Show answer
Answer: (D) 5 × 10⁻²²
Explanation: For a photon, p = E/c. With E = 1 MeV = 10⁶ × 1.6 × 10⁻¹⁹ = 1.6 × 10⁻¹³ J, p = 1.6 × 10⁻¹³/3 × 10⁸ ≈ 5 × 10⁻²² kg m/s.

Question 10

Previous-year question
Monochromatic light of frequency 6.0 × 10¹⁴ Hz is produced by a laser. The power emitted is 2 × 10⁻³ W. The number of photons emitted, on the average, by the source per second is [AIPMT 2007]
  1. 5 × 10¹⁵
  2. 5 × 10¹⁶
  3. 5 × 10¹⁷
  4. 5 × 10¹⁴
Show answer
Answer: (A) 5 × 10¹⁵
Explanation: The energy of each photon is E = hν = 6.6 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.96 × 10⁻¹⁹ J. Hence n = P/E = 2 × 10⁻³/3.96 × 10⁻¹⁹ ≈ 5 × 10¹⁵ photons per second.

Question 11

Previous-year question
Monochromatic light of wavelength 667 nm is produced by a helium–neon laser. The power emitted is 9 mW. The number of photons arriving per second on the average at a target irradiated by this beam is [AIPMT 2009]
  1. 9 × 10¹⁷
  2. 3 × 10¹⁶
  3. 9 × 10¹⁵
  4. 3 × 10¹⁹
Show answer
Answer: (B) 3 × 10¹⁶
Explanation: N = Pλ/hc = (9 × 10⁻³ × 667 × 10⁻⁹)/(6.6 × 10⁻³⁴ × 3 × 10⁸) = 6.0 × 10⁻⁹/1.98 × 10⁻²⁵ ≈ 3 × 10¹⁶ photons per second.

Question 12

Previous-year question
A source S₁ is producing 10¹⁵ photons/s of wavelength 5000 Å. Another source S₂ is producing 1.02 × 10¹⁵ photons per second of wavelength 5100 Å. Then (power of S₂)/(power of S₁) is equal to [AIPMT 2010]
  1. 1.00
  2. 1.02
  3. 1.04
  4. 0.98
Show answer
Answer: (A) 1.00
Explanation: Since P = nhc/λ, the ratio is P₂/P₁ = (n₂/n₁)(λ₁/λ₂) = (1.02 × 10¹⁵/10¹⁵) × (5000/5100) = 1.02 × 0.9804 = 1.00.

Question 13

Previous-year question
A 200 W sodium street lamp emits yellow light of wavelength 0.6 μm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is [AIPMT 2012]
  1. 15 × 10²⁰
  2. 6 × 10¹⁸
  3. 62 × 10²⁰
  4. 3 × 10¹⁹
Show answer
Answer: (A) 15 × 10²⁰
Explanation: The useful light power is 25% of 200 W, i.e. 50 W. Then n = Pλ/hc = (50 × 0.6 × 10⁻⁶)/(6.6 × 10⁻³⁴ × 3 × 10⁸) = 3 × 10⁻⁵/1.98 × 10⁻²⁵ ≈ 15 × 10²⁰ photons per second.

Question 14

Previous-year question
The number of photons per second on average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3 × 10⁻³ W will be (h = 6.6 × 10⁻³⁴ J s) [NEET 2021]
  1. 10¹⁸
  2. 10¹⁷
  3. 10¹⁶
  4. 10¹⁵
Show answer
Answer: (C) 10¹⁶
Explanation: The power is P = nhc/λ, so n = Pλ/hc = (3.3 × 10⁻³ × 600 × 10⁻⁹)/(6.6 × 10⁻³⁴ × 3 × 10⁸) = 1.98 × 10⁻⁹/1.98 × 10⁻²⁵ = 10¹⁶ photons per second.

Question 15

Seeing an object by the eye involves two stages: the gathering and focussing of light by the eye-lens, and the absorption of light by the rods and cones of the retina. Which picture of light best describes each stage?
  1. Photon picture for both stages
  2. Wave picture for focussing, photon picture for absorption
  3. Wave picture for both stages
  4. Photon picture for focussing, wave picture for absorption
Show answer
Answer: (B) Wave picture for focussing, photon picture for absorption
Explanation: Which description suits an experiment depends on the nature of that experiment. Propagation and focussing of light through a lens are well described by waves. Absorption involves transfer of energy to the receptor cells, which needs the photon picture, so seeing uses both descriptions.

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