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NEET Dual Nature of Matter and Radiation MCQs with answers

236 questions on Dual Nature of Matter and Radiation for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easyNEET 2023

1. Photoelectric effect proves light has:

  1. Wave nature
  2. Particle nature
  3. Both
  4. Neither
Show answer and explanation

Answer: (B) Particle nature

Photons (E=hν) cannot be explained by wave theory → particle nature.

Quick tip: Photoelectric → particle. Diffraction/Interference → wave.

NCERT Class 12, Chapter 11, page 382

easy

2. The energy of a photon of wavelength 800 nm is (take hc = 1240 eV nm):

  1. 0.645 eV
  2. 1.55 eV
  3. 15.5 eV
  4. 0.775 eV
Show answer and explanation

Answer: (B) 1.55 eV

E = hc/λ = 1240 eV nm / 800 nm = 1.55 eV.

Quick tip: E (eV) = 1240 / λ (nm).

NCERT Class 12, Chapter 11

medium

3. In an experiment on photoelectric effect, the graph between the maximum kinetic energy (K_max) of emitted photoelectrons and the frequency (ν) of incident radiation is a straight line. The intercept on the negative energy axis represents:

  1. Planck's constant
  2. Work function of the metal
  3. Threshold frequency
  4. Stopping potential
Show answer and explanation

Answer: (B) Work function of the metal

Einstein's photoelectric equation is K_max = hν - Φ. When plotted as K_max versus ν, the equation corresponds to y = mx + c where m = h and the vertical intercept c = -Φ. Thus, the magnitude of the intercept on the negative energy axis equals the work function Φ.

hard

4. Two particles A and B have de Broglie wavelengths λA and λB respectively. They undergo a completely inelastic one-dimensional collision and stick together. If they were originally moving in opposite directions with equal speeds, the de Broglie wavelength of the combined mass is:

  1. |λA - λB|
  2. (λA λB) / |λA - λB|
  3. (λA λB) / (λA + λB)
  4. (λA + λB) / 2
Show answer and explanation

Answer: (B) (λA λB) / |λA - λB|

Let the particles have velocities v and -v. Their momenta are pA = mA v = h/λA and pB = -mB v, so magnitude pB = h/λB. By conservation of linear momentum, the total momentum of the combined body is p_total = pA + pB = h/λA - h/λB = h(λB - λA)/(λA λB). The magnitude of the wavelength of the combined mass is λ = h/|p_total| = (λA λB)/|λA - λB|.

medium

5. Starting from rest, an electron is accelerated through 700 V. Using λ = 1.227/√V nm, its de Broglie wavelength is:

  1. 0.00175 nm
  2. 0.325 nm
  3. 0.0464 nm
  4. 0.464 nm
Show answer and explanation

Answer: (C) 0.0464 nm

λ = h/√(2meV) = 1.227/√V nm = 1.227/√700 = 0.0464 nm.

Quick tip: For electrons, λ = 1.227/√V nm (or 12.27/√V Å). Watch the unit.

NCERT Class 12, Chapter 11

easy

6. The energy of a photon of wavelength 200 nm is (take hc = 1240 eV nm):

  1. 6.2 eV
  2. 0.161 eV
  3. 3.1 eV
  4. 62 eV
Show answer and explanation

Answer: (A) 6.2 eV

E = hc/λ = 1240 eV nm / 200 nm = 6.2 eV.

Quick tip: E (eV) = 1240 / λ (nm).

NCERT Class 12, Chapter 11

hard

7. Two identical photosensitive surfaces A and B are illuminated with light of wavelengths λ_A and λ_B respectively. If the maximum kinetic energies of the emitted photoelectrons are K_A and K_B such that K_A = 3 K_B, and the work function of the metal is Φ₀, then:

  1. λ_A < λ_B and hc/λ_A - 3hc/λ_B = 2Φ₀
  2. λ_A > λ_B and hc/λ_A - 3hc/λ_B = 2Φ₀
  3. λ_A < λ_B and hc/λ_A - 3hc/λ_B = -2Φ₀
  4. λ_A > λ_B and hc/λ_A - 3hc/λ_B = -2Φ₀
Show answer and explanation

Answer: (C) λ_A < λ_B and hc/λ_A - 3hc/λ_B = -2Φ₀

Since K_A > K_B, the photon energy for A must be higher, meaning λ_A < λ_B. Using Einstein's equation: K_A = hc/λ_A - Φ₀ and K_B = hc/λ_B - Φ₀. Given K_A = 3 K_B: hc/λ_A - Φ₀ = 3(hc/λ_B - Φ₀) = 3hc/λ_B - 3Φ₀. Rearranging terms gives hc/λ_A - 3hc/λ_B = -2Φ₀.

easy

8. Which of the following observations provides direct evidence for the particle nature of light rather than its wave nature?

  1. Existence of a threshold frequency below which no photoemission occurs regardless of intensity
  2. Light exhibits diffraction around sharp obstacles
  3. Interference fringes produced in Young's double-slit experiment
  4. Polarisation of light waves upon reflection from a dielectric surface
Show answer and explanation

Answer: (A) Existence of a threshold frequency below which no photoemission occurs regardless of intensity

Wave theory predicts that any light wave of sufficient intensity should eventually transfer enough energy to eject electrons, irrespective of frequency. The existence of an instantaneous threshold frequency strictly confirms that light interacts as discrete energy quanta (photons).

easy

9. Taking hc = 1240 eV nm, the energy of a photon of wavelength 620 nm is:

  1. 20 eV
  2. 0.5 eV
  3. 4 eV
  4. 2 eV
Show answer and explanation

Answer: (D) 2 eV

Photon energy is E = hc/λ; with hc = 1240 eV nm, E = 1240/620 = 2 eV.

easy

10. Which of the following graphs correctly represents the variation of the de Broglie wavelength λ of a particle with its momentum p?

  1. A straight line passing through the origin with positive slope
  2. A rectangular hyperbola asymptotic to both axes
  3. A parabola opening upwards along the λ-axis
  4. A horizontal straight line parallel to the p-axis
Show answer and explanation

Answer: (B) A rectangular hyperbola asymptotic to both axes

The relation is λ = h/p, which gives λp = h = constant. This is the equation of a rectangular hyperbola where λ decreases asymptotically to zero as p increases, and approaches infinity as p approaches zero.

easy

11. According to Einstein's photoelectric equation, the graph showing the variation of the maximum kinetic energy (K_max) of emitted photoelectrons with the frequency (ν) of the incident radiation has an intercept on the horizontal (frequency) axis equal to:

  1. The threshold wavelength
  2. The work function of the metal
  3. The stopping potential
  4. The work function divided by Planck's constant
Show answer and explanation

Answer: (D) The work function divided by Planck's constant

Einstein's photoelectric equation is K_max = hν - Φ₀. When K_max = 0, we find hν = Φ₀, which gives ν = Φ₀ / h = ν₀ (the threshold frequency). Thus, the intercept on the frequency axis is Φ₀ / h.

easy

12. Photons of energy 6 eV fall on a metal surface of work function 1 eV. The maximum kinetic energy of the emitted photoelectrons is:

  1. 7 eV
  2. 1 eV
  3. 5 eV
  4. 6 eV
Show answer and explanation

Answer: (C) 5 eV

Einstein’s photoelectric equation: K_max = hν − φ₀ = 6 − 1 = 5 eV.

Quick tip: K_max = hν − φ₀. The work function is subtracted, never added.

NCERT Class 12, Chapter 11

easy

13. Light of wavelength 500 nm consists of photons of energy (hc = 1240 eV nm):

  1. 1.24 eV
  2. 0.403 eV
  3. 2.48 eV
  4. 24.8 eV
Show answer and explanation

Answer: (C) 2.48 eV

E = hc/λ = 1240 eV nm / 500 nm = 2.48 eV.

Quick tip: E (eV) = 1240 / λ (nm).

NCERT Class 12, Chapter 11

medium

14. Light of wavelength 248 nm falls on a metal with work function 2.5 eV. Taking hc = 1240 eV nm, the maximum kinetic energy of the photoelectrons is:

  1. 5 eV
  2. 1.25 eV
  3. 2.5 eV
  4. 7.5 eV
Show answer and explanation

Answer: (C) 2.5 eV

Einstein's equation: K_max = hc/λ − φ = 1240/248 − 2.5 = 5 − 2.5 = 2.5 eV.

easy

15. Light whose photons carry 5 eV is incident on a metal with work function 1 eV. The maximum kinetic energy of the photoelectrons is:

  1. 6 eV
  2. 4 eV
  3. 1 eV
  4. 5 eV
Show answer and explanation

Answer: (B) 4 eV

Einstein’s photoelectric equation: K_max = hν − φ₀ = 5 − 1 = 4 eV.

Quick tip: K_max = hν − φ₀. The work function is subtracted, never added.

NCERT Class 12, Chapter 11

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