An electron of mass ' m ' with an initial velocity v =v_0 i (v_0 0) enters an electric field E =-E_ o k . If the initial de Broglie wavelength is λ_0 , the value after time t would be

Correct answer: A. λ_0 1+ e^2 E_0 ^2 t^2 ~m^2 v_0 ^2 aligned & v =v_0 i - E_0 e m t k & | v |= v_0^2+ E_0^2 e^2 t^2 ~m^2 & λ_0= h mv_0 & λ^ = h mv_0 1+ E_0^2 e^2 t^2 v_0^2 ~m^2 & λ^ = λ_0 1+ E_0^2 e^2 t^2 v_0^2 ~m^2 aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2025 · 24 January · Morning Shift · Physics · Dual Nature of Matter

An electron of mass ' m ' with an initial velocity v =v_0 i (v_0 0)...

An electron of mass ' m ' with an initial velocity

\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{i}\left(\mathrm{v}_0\gt0\right)

enters an electric field

\overrightarrow{\mathrm{E}}=-\mathrm{E}_{\mathrm{o}} \hat{\mathrm{k}}

. If the initial de Broglie wavelength is

\lambda_0

, the value after time t would be

Options

A $\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0{ }^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_0{ }^2}}}$
B $\lambda_{\mathrm{o}} \sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_{\mathrm{o}}^2}}$
C $\frac{\lambda_0}{\sqrt{1-\frac{\mathrm{e}^2 \mathrm{E}_{\mathrm{o}}{ }^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_{\mathrm{o}}^2}}}$
D $\lambda_0$

Correct answer

A. $\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0{ }^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_0{ }^2}}}$

Step-by-step solution

\begin{aligned} & \overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\mathrm{i}}-\frac{\mathrm{E}_0 \mathrm{e}}{\mathrm{m}} t \hat{\mathrm{k}} \\ & |\overrightarrow{\mathrm{v}}|=\sqrt{\mathrm{v}_0^2+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{~m}^2}} \\ & \lambda_0=\frac{\mathrm{h}}{\mathrm{mv}_0} \\ & \lambda^{\prime}=\frac{\mathrm{h}}{\mathrm{mv}_0 \sqrt{1+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{v}_0^2 \mathrm{~m}^2}}} \\ & \lambda^{\prime}=\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{v}_0^2 \mathrm{~m}^2}}}\end{aligned}

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