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 :

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K_p < K_e and P_p < P_e

K_p = K_e and P_p = P_e

K_p < K_e and P_p = P_e

K_p > K_e and P_p = P_e

answer is A.

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Detailed Solution

$\begin{array}{l} λ_{P} = \frac{h}{P_{P}} λ_{e} = \frac{h}{P_{e}} \\ ∵ λ_{P} = λ_{e} \\ \Rightarrow P_{P} = P_{e} \\ (K)_{P} = \frac{P_{P}^{2}}{2 m_{P}} \\ (K)_{e} = \frac{P_{e}^{2}}{2 m_{e}} \end{array}$
K_P < K_e as m_P > m_e
Option (1)

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