Q 12-11-123JEE MainJEE Main 2021 (25 Feb, Shift 1)Easy
An $\alpha$ particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\dfrac{\lambda_p}{\lambda_\alpha}$ is:
Answer: (C) 2.8
$\lambda = \dfrac{h}{\sqrt{2mqV}}$, so
$$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_pq_p}} = \sqrt{4\times2} = 2\sqrt2 \approx 2.8$$
Solution by Sreeraj P, M.Sc Physics