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JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Physics question on Dual Nature of Radiation

An α\alpha-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:
  1. A.
    λα>λp>λe\lambda_{\alpha} > \lambda_{p} > \lambda_{e}
  2. B.
    λα=λp=λe\lambda_{\alpha} = \lambda_{p} = \lambda_{e}
  3. C.
    λα<λp<λe\lambda_{\alpha} < \lambda_{p} < \lambda_{e}
  4. D.
    λα>λp<λe\lambda_{\alpha} > \lambda_{p} < \lambda_{e}
Show answer and solution

Correct option: C

Solution

For a particle with kinetic energy K, the de-Broglie wavelength λ = h/p = h/√(2mK). Since K is the same for all three, λ ∝ 1/√m. The masses satisfy me << mp << m_α, so the wavelengths satisfy λ_α < λ_p < λ_e.

[JEE Main 2023 · 24 Jan · Shift 2]

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