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Q.
The ratio of wavelengths of proton and deuteron accelerated by potential $V_p$ and $V_d$ is $1: \sqrt{2}$. Then, the ratio of $V_p$ to $V_d$ will be
JEE MainJEE Main 2022Dual Nature of Radiation and Matter
Solution:
Kinetic energy gained by a charged particle accelerated by a potential $V$ is $qV$
$KE = qV $
$ \Rightarrow \frac{ p ^2}{2 m }= qV \Rightarrow p =\sqrt{2 mqV }$
$ p =\frac{ h }{\lambda}, \text { thus } \lambda=\frac{ h }{\sqrt{2 mqV }} $
now $ \frac{\lambda_{ p }}{\lambda_{ d }}=\sqrt{\frac{ m _{ d } V _{ d }}{ m _{ p } V _{ p }}}$
$ \Rightarrow \frac{1}{\sqrt{2}}=\sqrt{\frac{2 V _{ d }}{ V _{ p }}} \Rightarrow \frac{ V _{ p }}{ V _{ d }}=4$