Verified Solution • 2025–26 Session moderate

QJEEM22Ph3075: JEE Main 2022 Physics - Class 12 JEEM22Ph3075 Solution: An electron (mass ) with an initial velocity is moving in an…

An electron (mass m\mathrm{m} ) with an initial velocity \vec{v}=v_{0} \hat{i}\left(v_{0}>0\right) is moving in an electric field \vec{E}=-E_{0} \hat{i}\left(E_{0}>0\right) where E0E_{0} is constant. If at t=0\mathrm{t}=0 de Broglie wavelength is λ0=hmv0\lambda_{0}=\frac{h}{m v_{0}} , then its de Broglie wavelength after time t is given by

Multiple Choice Options

  • A λ0\lambda_{0}
  • B λ0(1+eE0tmv0)\lambda_{0}\left(1+\frac{e E_{0} t}{m v_{0}}\right)
  • C λ0t\lambda_{0} t
  • D λ0(1+eE0tmv0)\frac{\lambda_{0}}{\left(1+\frac{e E_{0} t}{m v_{0}}\right)}

Step-by-Step Verified Solution

 At t=0,λ0=hmv0\text { At } t=0, \lambda_0=\frac{h}{m v_0}

Since v⃗=v0i^\vec{v}=v_0 \hat{i} and E→=E0i^\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \hat{i}

its velocity vv at any time tt is given by

v=vo+εEomtv=v_o+\frac{\varepsilon \mathrm{E}_{\mathrm{o}}}{m} t

De Broglie wavelength λ\lambda at any time tt is given by

λamp;=hmv=hm(v0+eE0mt)amp;=hmv0(1+eE0mv0t)amp;=λ01+eE0mv0t\begin{aligned}\lambda & =\frac{h}{m v}=\frac{h}{m\left(v_0+\frac{e \mathrm{E}_0}{m} t\right)} \\\\& =\frac{h}{m v_0\left(1+\frac{e \mathrm{E}_0}{m v_0} t\right)} \\\\& =\frac{\lambda_0}{1+\frac{e \mathrm{E}_0}{m v_0} t}\end{aligned}

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