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Class XII · Physics

Dual Nature of Matter and Radiation: NEET MCQ practice

127 practice questions on Dual Nature of Matter and Radiation, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

127

Practice questions

63

Easy

61

Medium

3

Hard

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What Dual Nature of Matter and Radiation covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Dual nature of radiation, photoelectric effect, Hertz and Lenard's observations
  • Einstein's photoelectric equation, particle nature of light
  • Matter waves- wave nature of particles, de Broglie relation

Sample questions on Dual Nature of Matter and Radiation

Try each one, then open the answer and explanation.

Question 1 · Practice question

Read the assertion and reason carefully to mark the correct option out of the options given below: Assertion : For higher temperature, the peak emission wavelength of a black body shifts to lower wavelengths. Reason : Peak emission wavelength of a blackbody is proportional to the fourth power of temperature.

  1. A.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.
  2. B.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.
  3. C.C. If assertion is true but reason is false.
  4. D.D. If the assertion and reason both are false.
Show answer

Answer: C. C. If assertion is true but reason is false.

The assertion is correct by Wien's displacement law (λₘₐₓ ∝ 1/T), so as temperature rises the peak wavelength decreases. The reason is false because peak wavelength is inversely proportional to temperature, not proportional to T⁴. The T⁴ relation is for total radiated power (Stefan-Boltzmann law), which is often confused with Wien's law by students.

Question 2 · Practice question

Q. The energy of electrons of atoms of a substance will be same if:

  1. A.A. atoms are isolated.
  2. B.B. atoms are closely spaced.
  3. C.C. atoms are excited.
  4. D.D. atoms are charged.
Show answer

Answer: A. A. atoms are isolated.

In isolated atoms, electrons occupy well-defined discrete energy levels without external influence, so their energies remain unchanged. When atoms are closely spaced (as in solids), interactions cause energy levels to broaden into bands, changing electron energies.

Question 3 · Practice question

To reduce the de Broglie wavelength of an electron from 100 𝑝𝑚 to 50 𝑝𝑚, the required increase in energy will be:

  1. A.600 𝑒𝑉
  2. B.450 𝑒𝑉
  3. C.300 𝑒𝑉
  4. D.150 𝑒𝑉
Show answer

Answer: B. 450 𝑒𝑉

The de Broglie wavelength λ is inversely proportional to the square root of kinetic energy for a non-relativistic electron: λ ∝ 1/√E. Halving the wavelength (from 100 pm to 50 pm) means λ₂ = λ₁/2, so √(E₂/E₁) = λ₁/λ₂ = 2, thus E₂/E₁ = 4. If initial energy E₁ corresponds to λ₁ = 100 pm and equals 150 eV, then E₂ = 4 × 150 = 600 eV; the energy increase is E₂ - E₁ = 600 - 150 = 450 eV. A common mistake is to think λ ∝ 1/E and incorrectly get a factor of 2 increase, leading to 150 eV.

Question 4 · Practice question

The work function of a metal is 1 eV. When the surface of this metal is illuminated with light of wavelength 64000 Å, then the maximum kinetic energy of emitted photoelectron will be

  1. A.31 × 19 –19 J
  2. B.3.1 × 10 –19 J
  3. C.0.15 × 10 –19 J
  4. D.1.5 × 10 –19 J
Show answer

Answer: D. 1.5 × 10 –19 J

Using Einstein's photoelectric equation Kₘₐₓ = hc/λ – work function: Work function φ = 1 eV = 1.6×10⁻¹⁹ J, λ likely intended as 6400 Å (640 nm). Photon energy ≈ 1.94 eV, so Kₘₐₓ ≈ 0.94 eV ≈ 1.5×10⁻¹⁹ J. Option (d) correct. The other options arise from miscalculating λ or forgetting unit conversions.

Question 5 · Practice question

If K 1 and K 2 are maximum kinetic energies of photoelectrons emitted when lights of wavelength λ 1 and λ 2 respectively incident on a metallic surface and λ 1 = 3λ 2 then

  1. A.K 1 > (K 2 / 3)
  2. B.K 1 < (K 2 / 3)
  3. C.K 1 = 2K 2
  4. D.K 2 = 2K 1
Show answer

Answer: B. K 1 < (K 2 / 3)

Use Einstein's photoelectric equation: Kₘₐₓ = hc/λ − φ. With λ₁ = 3λ₂, substitute to find K₁ = hc/(3λ₂) − φ and K₂/3 = (hc/λ₂ − φ)/3 = hc/(3λ₂) − φ/3. Since φ > 0, φ > φ/3, so K₁ is less than K₂/3. Trap: If you forget the work function φ, you might think K₁ = K₂/3. But φ reduces K₁ more than it reduces K₂/3, making K₁ smaller.

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