JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Modern Physics

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A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K .E. and momentum respectively. Then choose the correct option :

(A) $K_{p} = K_{e}$ and $P_{p} = P_{e}$
(B) $K_{p} < K_{e}$ and $P_{p} < P_{e}$
(C) $K_{p} < K_{e}$ and $P_{p} = P_{e}$ ✓ Correct
(D) $K_{p} > K_{e}$ and $P_{p} = P_{e}$

Answer: (C)

Solution

Equal de-Broglie wavelengths mean equal momenta, since

λ = \frac{h}{p}

; thus

P_{p} = P_{e}

. The kinetic energy is

K = \frac{p ^{2}}{2 m}

, so for the same momentum the heavier proton has the smaller kinetic energy,

K_{p} < K_{e}

. Hence

K_{p} < K_{e}

while

P_{p} = P_{e}

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option

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