Question:
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is :-
Correct Option: , 3
Solution:
$\mathrm{KE}=\mathrm{e} \Delta \mathrm{V}$
$\lambda_{e}=\frac{\mathrm{h}}{\sqrt{2 m_{e}(e \Delta V)}}$
$\lambda_{p}=\frac{h}{\sqrt{2 m_{p}(e \Delta V)}}$
$\Rightarrow \quad \frac{\lambda_{e}}{\lambda_{\mathrm{P}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{e}}}$
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