An electron is released from rest near an infinite non-conducting sheet of uniform charge density ' -σ '. The rate of change of de-Broglie wave length associated with the electron varies inversely as n^ th power of

Correct answer: 0 Let the momentum of e^ - at any time t is p and its de-broglie wavelength is λ . Then, p= h λ aligned & dp dt = -h λ^2 ~d λ dt & ma=F=- h λ ~d λ dt [~m= mass of e] aligned Where, -ve sign represents decrease in λ with time ma= -h (~h / p)^2 ~d λ dt aligned & a=- p^2 mh ~d λ dt & a=- mv^2 ~h ~d λ dt & ~d λ dt =- ah mv^2 ...(1) aligned here, a= qE m = e m (σ)/(2 _0) a= σ e 2 ~m _0 and v=u+at v=at Substituting values of a \& v in equation (1) aligned & d λ dt =- 2 ~h _0 σ et^2 & ~d λ dt 1 t^2 & n=2 aligned

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Electrostatics

JEE Main 2025 · 8 April · Evening Shift · Physics · Electrostatics

An electron is released from rest near an infinite non-conducting...

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 ________.

Correct answer

Step-by-step 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}<br/>& \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}]<br/>\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}<br/>& \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)<br/>\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}<br/>& \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<br/>\end{aligned}

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