The kinetic energy of an electron, α-particle and a proton are given as 4 K, 2 K and K respectively. The de-Broglie wavelength associated with electron (λₑ), α-particle (λₐ) and the proton (λₚ) are as follows:
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Question

The kinetic energy of an electron, α-particle and a proton are given as 4 K, 2 K and K respectively. The de-Broglie wavelength associated with electron (λₑ), α-particle (λₐ) and the proton (λₚ) are as follows:

Options

G1 Physics · Accepted Answer

For a particle, de Broglie wavelength is λ = h/p and, for non-relativistic motion, p = √(2mK). Therefore λ ∝ 1/√(mK). Electron: m = mₑ and K = 4K, so λₑ ∝ 1/√(4mₑK). Proton: m = mₚ and K = K, so λₚ ∝ 1/√(mₚK). Alpha particle: mₐ ≈ 4mₚ and K = 2K, so λₐ ∝ 1/√(8mₚK). Since mₚ ≫ mₑ, λₑ is the largest. Also λₐ < λₚ. Thus λₐ < λₚ < λₑ, so option D follows.