JEE Main 2025 — Physics Electromagnetism

Let the momentum of $\mathrm{e}^{-}$e−at any time t is p and its de-broglie wavelength is $\lambda$λ.

Then, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$p=λh​

$\begin{aligned}
& \frac{\mathrm{dp}}{\mathrm{dt}}=\frac{-\mathrm{h}}{\lambda^2} \frac{\mathrm{d} \lambda}{\mathrm{dt}} \
& \mathrm{ma}=\mathrm{F}=-\frac{\mathrm{h}}{\lambda} \frac{\mathrm{d} \lambda}{\mathrm{dt}} \quad[\mathrm{~m}=\text { mass of } \mathrm{e}]
\end{aligned}$

Where, -ve sign represents decrease in $\lambda$λ with time

$$\mathrm{ma}=\frac{-\mathrm{h}}{(\mathrm{~h} / \mathrm{p})^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}}$$ma=( h/p)2−h​dt dλ​

$\begin{aligned}
& \mathrm{a}=-\frac{\mathrm{p}^2}{\mathrm{mh}} \frac{\mathrm{d} \lambda}{\mathrm{dt}} \
& \mathrm{a}=-\frac{\mathrm{mv}^2}{\mathrm{h}} \frac{\mathrm{d} \lambda}{\mathrm{dt}} \
& \frac{\mathrm{d} \lambda}{\mathrm{dt}}=-\frac{\mathrm{ah}}{\mathrm{mv}^2}...(1)
\end{aligned}$

here, $\mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}=\frac{\mathrm{e}}{\mathrm{m}} \frac{\sigma}{2 \varepsilon_0}$a=mqE​=me​2ε0​σ​

$$\mathrm{a}=\frac{\sigma \mathrm{e}}{2 \mathrm{~m} \varepsilon_0}$$a=2 mε0​σe​

and $\mathrm{v}=\mathrm{u}+\mathrm{at}$v=u+at

$$\mathrm{v}=\mathrm{at}$$v=at

Substituting values of a & v in equation (1)

$\begin{aligned}
& \frac{\mathrm{d} \lambda}{\mathrm{dt}}=-\frac{2 \mathrm{h} \varepsilon_0}{\sigma \mathrm{et}^2} \
& \Rightarrow \frac{\mathrm{d} \lambda}{\mathrm{dt}} \propto \frac{1}{\mathrm{t}^2} \
& \Rightarrow \mathrm{n}=2
\end{aligned}$