Dual Nature of Matter and Radiation

12 NEET questions on Dual Nature of Matter and Radiation. A sample is shown below with full explanations; the rest are available free after signing up.

Dual Nature of Matter and Radiation2025medium

Q1.De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)

  • A0.067 nm
  • B0.67 nmAnswer
  • C1.67 nm
  • D2.67 nm

Explanation

Bohr quantisation makes the orbit circumference an exact whole number of de Broglie wavelengths: 2πrn = nλ. With r₂ = 0.052 x 2² = 0.208 nm, this gives λ = 2πr₂/2 = π x 0.208 ≈ 0.65 nm.

Source: NEET 2025 (4 May), Code 45, Q34

Dual Nature of Matter and Radiation2025medium

Q2.A photon and an electron (mass m) have the same energy E. The ratio (λphoton/λelectron) of their de Broglie wavelengths is: (c is the speed of light)

  • A√(E/2m)
  • Bc√(2mE)
  • Cc√(2m/E)Answer
  • D(1/c)√(E/2m)

Explanation

A photon obeys E = pc, so λphoton = hc/E, whereas a non-relativistic electron obeys E = p²/2m, so λelectron = h/√(2mE). Dividing one by the other gives (hc/E) x (√(2mE)/h) = c√(2m/E), which is the only dimensionless option.

Source: NEET 2025 (4 May), Code 45, Q37

Dual Nature of Matter and Radiation2026medium

Q3.A ray of light with wavelength λ is incident on three different photoelectric cells namely 1, 2 and 3. The threshold wavelength of these photoelectric cells are λ₁, λ₂ and λ₃, respectively, and the magnitudes of stopping potentials of these cells are V₁, V₂ and V₃, respectively. The relation between λ and the threshold wavelengths are λ₁ < λ, λ₂ > λ and λ₃ >> λ. The correct option is:

  • AV₁ = 0, V₂ < V₃Answer
  • BV₁ = 0, V₂ > V₃
  • CV₁ > V₂, V₃ = 0
  • DV₁ < V₂, V₃ = 0

Explanation

Emission needs the photon energy above the work function, which means the incident wavelength must be BELOW the threshold. Cell 1 has λ₁ < λ, so the light is too red for it and nothing is emitted: V₁ = 0. For the other two a longer threshold wavelength means a smaller work function and so a larger stopping potential, and since λ₃ >> λ₂ it follows that V₃ > V₂.

Source: NEET 2026 re-exam (21 June), Code 50, Q18

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