The ratio of wavelengths of proton and deuteron accelerated by potential V p and V d is 1 : 2 . Then, the ratio of V p to V d will be

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TestHub · Accepted Answer

Kinetic energy gained by a charged particle accelerated by a potential $V$ is $q V$
$K E = q V$
Using relation between momentum and kinetic energy
$\Rightarrow \frac{p^{2}}{2 m} = q V \Rightarrow p = \sqrt{2 mqV}$
We know that momentum of charged particle is $p = \frac{h}{λ}$ .
Thus, $λ = \frac{h}{\sqrt{2 m q V}}$
Now, $\frac{λ_{p}}{λ_{d}} = \sqrt{\frac{m_{d} V_{d}}{m_{p} V_{p}}}$
$\Rightarrow \frac{1}{\sqrt{2}} = \sqrt{\frac{2 V_{d}}{V_{p}}} \Rightarrow \frac{V_{p}}{V_{d}} = 4$

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