Physics - Modern Physics Question with Solution | TestHub

Let the momentum of ${\mathrm{e}}^{-}$at any time t is p and its de-broglie wavelength is $\lambda$.
Then, $\mathrm{p}=\frac{\mathrm{h}}{\lambda }$
$\begin{array}{rl}& \frac{\mathrm{dp}}{\mathrm{dt}}=\frac{-\mathrm{h}}{{\lambda }^2}\frac{\mathrm{d}\lambda }{\mathrm{dt}}\\ & \mathrm{ma}=\mathrm{F}=-\frac{\mathrm{h}}{\lambda }\frac{\mathrm{d}\lambda }{\mathrm{dt}}[\mathrm{m}=\text{ mass of }\mathrm{e}]\end{array}$
Where, -ve sign represents decrease in $\lambda$ with time
$\mathrm{ma}=\frac{-\mathrm{h}}{(\mathrm{h}/\mathrm{p}{)}^2}\frac{\mathrm{d}\lambda }{\mathrm{dt}}$
$\begin{array}{rl}& \mathrm{a}=-\frac{{\mathrm{p}}^2}{\mathrm{mh}}\frac{\mathrm{d}\lambda }{\mathrm{dt}}\\ & \mathrm{a}=-\frac{{\mathrm{mv}}^2}{\mathrm{h}}\frac{\mathrm{d}\lambda }{\mathrm{dt}}\\ & \frac{\mathrm{d}\lambda }{\mathrm{dt}}=-\frac{\mathrm{ah}}{{\mathrm{mv}}^2}...(1)\end{array}$
here, $\mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}=\frac{\mathrm{e}}{\mathrm{m}}\frac{\sigma }{2{\epsilon }_0}$
$\mathrm{a}=\frac{\sigma \mathrm{e}}{2\mathrm{m}{\epsilon }_0}$
and $\mathrm{v}=\mathrm{u}+\mathrm{at}$
$\mathrm{v}=\mathrm{at}$
Substituting values of a \& v in equation (1)
$\begin{array}{rl}& \frac{\mathrm{d}\lambda }{\mathrm{dt}}=-\frac{2\mathrm{h}{\epsilon }_0}{\sigma {\mathrm{et}}^2}\\ & \Rightarrow \frac{\mathrm{d}\lambda }{\mathrm{dt}}\propto \frac{1}{{\mathrm{t}}^2}\\ & \Rightarrow \mathrm{n}=2\end{array}$