Q 12-11-113JEE MainJEE Main 2022 (27 Jun, Shift 1)Easy
An $\alpha$ particle and a carbon $12$ atom has same kinetic energy $K$. The ratio of their de-Broglie wavelengths $(\lambda_\alpha : \lambda_{C12})$ is
Answer: (A) $\sqrt3 : 1$
$\lambda = \dfrac{h}{\sqrt{2mK}}$, so at equal $K$, $\lambda \propto \dfrac{1}{\sqrt m}$.
$$\frac{\lambda_\alpha}{\lambda_C} = \sqrt{\frac{m_C}{m_\alpha}} = \sqrt{\frac{12}{4}} = \sqrt3$$
Solution by Sreeraj P, M.Sc Physics