Q 12-11-168JEE MainJEE Main 2019 (12 Jan, Shift 1)Easy
A particle A of mass $m$ and charge $q$ is accelerated by a potential difference of $50$ V. Another particle B of mass $4m$ and charge $q$ is accelerated by a potential difference of $2500$ V. The ratio of de Broglie wavelengths $\dfrac{\lambda_A}{\lambda_B}$ is close to:
Answer: (D) $14.14$
$\lambda = \dfrac{h}{\sqrt{2mqV}}$, so
$$\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{m_BV_B}{m_AV_A}} = \sqrt{\frac{4m\times2500}{m\times50}} = \sqrt{200} \approx 14.14$$
Solution by Sreeraj P, M.Sc Physics