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 :-

• A. $\frac{m_{p}}{m_{e}}$
• B. $1$
• C. $\sqrt{\frac{m_{p}}{m_{e}}}$
• D. $\frac{m_{e}}{m_{p}}$

Answer 1

Answer: (C)

$K E=e \Delta V$
$\lambda_{e}=\frac{h}{\sqrt{2 m_{e}(e \Delta V)}}$
$\lambda_{p}=\frac{h}{\sqrt{2 m_{p}(e \Delta V)}}$
$\Rightarrow \frac{\lambda_{e}}{\lambda_{p}}=\sqrt{\frac{m_{p}}{m_{e}}}$