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Q.
The ratio of de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{ m }}$, the value of $m$ is
JEE MainJEE Main 2023Dual Nature of Radiation and Matter
Solution:
$\frac{\lambda_{\alpha}}{\lambda_{p}}=\frac{h}{\sqrt{\frac{2m_{\alpha}q_{\alpha}V}{\frac{h}{\sqrt{2m_{p}q_{p}V}}}}}$
$\frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{1}{8}} \,\,\,\, m =8$