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An electron accelerated under a potential difference $\mathrm{V}$ volt has a certain wavelength $\lambda$. Mass of proton is some 2000 times of the mass of the electron. If the proton has to have the same wavelength $\lambda$, then it will have to be accelerated under a potential difference of
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The correct answer is:
$\frac{\mathrm{V}}{2000}$ volt
According to question $\lambda_{e}=\lambda_{P}$
$\begin{aligned}
& \text { or } \frac{h}{\sqrt{2 m_{e} Q_{e} V}}=\frac{h}{\sqrt{2 m_{p} Q_{p} V_{p}}} \\
\therefore \quad & m_{e} Q_{e} V=m_{p} Q_{p} V_{p} \\
\therefore \quad & V_{p}=\left(\frac{m_{e}}{m_{p}}\right)\left(\frac{Q_{e}}{Q_{p}}\right) V \\
&=\left(\frac{1}{2000}\right)(1) V \\
&=\frac{V}{2000} v o l t
\end{aligned}$
$\begin{aligned}
& \text { or } \frac{h}{\sqrt{2 m_{e} Q_{e} V}}=\frac{h}{\sqrt{2 m_{p} Q_{p} V_{p}}} \\
\therefore \quad & m_{e} Q_{e} V=m_{p} Q_{p} V_{p} \\
\therefore \quad & V_{p}=\left(\frac{m_{e}}{m_{p}}\right)\left(\frac{Q_{e}}{Q_{p}}\right) V \\
&=\left(\frac{1}{2000}\right)(1) V \\
&=\frac{V}{2000} v o l t
\end{aligned}$
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