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

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,
\(
\begin{aligned}
& \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{q}(9 \mathrm{~V}) \mathrm{M}}} \\
& \therefore \frac{\lambda}{\lambda_{\mathrm{p}}}=3 \sqrt{\frac{\mathrm{M}}{\mathrm{m}}} \\
& \Rightarrow \lambda_{\mathrm{p}}=\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}
\end{aligned}
\)