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According to de-Broglie hypothesis if an electron of mass ' $\mathrm{m}$ ' is accelerated by potential difference ' $\mathrm{V}$ ' then associated wavelength is ' $\lambda$ '. When a proton of mass ' $M$ ' is accelerated through potential difference ' $9 \mathrm{~V}$ ' then the wavelength associated with it is
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The correct answer is:
$\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$
Do Broglie relation
$$
\lambda=\frac{h}{\mathrm{P}} ...(1)
$$
Energy conservation,
$$
\mathrm{qV}=\mathrm{K} ...(2)
$$
Energy vs momentum relation
$$
K=\frac{p^2}{2 m} ...(3)
$$
Using equation (1), (2) and (3)
$$
\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{Km}}} \\
& \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{qVm}}}
\end{aligned}
$$
Now, for proton of mass $M$ that is accelerated through $9 \mathrm{~V}$, the de-broglie wavelength can be written as,
$$
\begin{aligned}
& \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{q}(9 \mathrm{~V}) \mathrm{M}}} \\
& \therefore \frac{\lambda}{\lambda_{\mathrm{p}}}=3 \sqrt{\frac{\mathrm{M}}{\mathrm{m}}} \\
& \Rightarrow \lambda_{\mathrm{p}}=\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}
\end{aligned}
$$
$$
\lambda=\frac{h}{\mathrm{P}} ...(1)
$$
Energy conservation,
$$
\mathrm{qV}=\mathrm{K} ...(2)
$$
Energy vs momentum relation
$$
K=\frac{p^2}{2 m} ...(3)
$$
Using equation (1), (2) and (3)
$$
\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{Km}}} \\
& \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{qVm}}}
\end{aligned}
$$
Now, for proton of mass $M$ that is accelerated through $9 \mathrm{~V}$, the de-broglie wavelength can be written as,
$$
\begin{aligned}
& \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{q}(9 \mathrm{~V}) \mathrm{M}}} \\
& \therefore \frac{\lambda}{\lambda_{\mathrm{p}}}=3 \sqrt{\frac{\mathrm{M}}{\mathrm{m}}} \\
& \Rightarrow \lambda_{\mathrm{p}}=\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}
\end{aligned}
$$
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