According to de-Broglie hypothesis if an electron of mass ' $\mathrm{m}$ ' is accelerated by potential…
- $\frac{\lambda}{3} \sqrt{\frac{M}{m}}$
- $\frac{\lambda}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$
- $\frac{\lambda}{6} \sqrt{\frac{m}{M}}$
- $\frac{\lambda}{6} \sqrt{\frac{M}{m}}$
Solution
Energy conservation,
Energy vs momentum relation
Using equation (1), (2) and (3)
$\begin{aligned}
& \lambda=\frac{h}{p}=\frac{h}{\sqrt{2 K m}} \\
& \lambda=\frac{h}{\sqrt{2 q V m}}
\end{aligned}$
Now, for proton of mass $M$ that is accelerated through $9 \mathrm{~V}$, the de-broglie wavelength can be written as,
$\begin{aligned}
& \lambda_p=\frac{h}{\sqrt{2 q(9 V) M}} \\
& \therefore \frac{\lambda}{\lambda_p}=3 \sqrt{\frac{M}{m}} \\
& \Rightarrow \lambda_p=\frac{\lambda}{3} \sqrt{\frac{m}{M}}
\end{aligned}$Asked in: MHT CET 2022 (08 Aug Shift 2)