Q 12-11-070JEE MainJEE Main 2024 (8 Apr, Shift 1)Easy
A proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume $h = 6.63\times10^{-34}\ \text{J s}$, $m_e = 9.0\times10^{-31}\ \text{kg}$ and $m_p = 1836$ times $m_e$)
Answer: (D) $1 : 1836$
Equal de Broglie wavelengths mean equal momenta $p$. With $K = \dfrac{p^2}{2m}$:
$$\frac{K_p}{K_e} = \frac{m_e}{m_p} = \frac{1}{1836}$$
So $K_p : K_e = 1 : 1836$.
Solution by Sreeraj P, M.Sc Physics