An electron is released from rest near an infinite non-conducting sheet of uniform charge density '…

An electron is released from rest near an infinite non-conducting sheet of uniform charge density ' $-\sigma$ '. The rate of change of de-Broglie wave length associated with the electron varies inversely as $\mathrm{n}^{\text {th }}$ power of time. The numerical value of $n$ is ________.

Solution

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

Asked in: JEE Main 2025 (08 Apr Shift 2)

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