Q 12-11-118JEE MainJEE Main 2021 (31 Aug, Shift 1)Easy
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:
Answer: (B) $K_p < K_e$ and $P_p = P_e$
$\lambda = \dfrac hP$, so equal wavelengths mean equal momenta: $P_p = P_e$.
$K = \dfrac{P^2}{2m}$ and $m_p > m_e$, so $K_p < K_e$.
Solution by Sreeraj P, M.Sc Physics