Let the momentum of $\mathrm{e}^{-}$at any time $t$ is $p$ and its de-broglie wavelength is $Z$. Then, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $$ \begin{aligned} & \frac{d p}{d t}=\frac{-h}{\lambda^2} \frac{d \lambda}{d t} \\ & m a=F=-\frac{h}{\lambda} \frac{d \lambda}{d t} \quad\left[m=\text { mass } d e^{-}\right] \end{aligned} $$ Where, se sign represents decrease in $\lambda$ with time $$ \begin{aligned} & \mathrm{ma}=\frac{-\mathrm{h}}{(\mathrm{~h} / \mathrm{p})^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\ & \mathrm{a}=-\frac{\mathrm{p}^2}{\mathrm{mh}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\ & \mathrm{a}=-\frac{\mathrm{mv}}{\mathrm{~h}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\ & \frac{\mathrm{~d} \lambda}{\mathrm{dt}}=-\frac{\mathrm{ah}}{\mathrm{mv} \mathrm{v}^2} \end{aligned} $$ here, $\mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}=\frac{\mathrm{e}}{\mathrm{m}} \frac{\sigma}{2 \varepsilon_0}$ $$ \mathrm{a}=\frac{\sigma \mathrm{e}}{2 \mathrm{~m} \varepsilon_0} $$ and $\mathrm{v}=\mathrm{u}+$ at $$ v=a t $$ Substituting values of a $\& \&$ in equation (1) $$ \begin{aligned} & \frac{\mathrm{d} \lambda}{\mathrm{dt}}=-\frac{2 \mathrm{~h} \varepsilon_0}{\sigma \mathrm{et}^2} \\ & \Rightarrow \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \propto \frac{1}{\mathrm{t}^2} \\ & \Rightarrow \mathrm{n}=2 \end{aligned} $$