Q 12-11-110JEE MainJEE Main 2022 (25 Jul, Shift 2)Easy
The ratio of wavelengths of proton and deuteron accelerated by potentials $V_p$ and $V_d$ is $1 : \sqrt2$. Then the ratio of $V_p$ to $V_d$ will be
Answer: (D) $4 : 1$
$\lambda = \dfrac{h}{\sqrt{2mqV}}$ with equal charges and $m_d = 2m_p$:
$$\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{2m_pV_d}{m_pV_p}} = \frac{1}{\sqrt2}\ \Rightarrow\ \frac{2V_d}{V_p} = \frac12\ \Rightarrow\ \frac{V_p}{V_d} = 4$$
Solution by Sreeraj P, M.Sc Physics