An electron of mass m is moving in an electric field vec E = -2E 0hat i ( E 0 = constant 0 ), with an initial velocity vec V = v ohat i ( v 0 = constant 0 ). If λ 0 = h/4mv 0 , its de Broglie wavelength at time t is . ( e = charge\; of\; electron )

Physics · Atomic Physics

JEE Main 2026 — 5 April, Morning Shift — Question 9

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ ( $E_0 = \text{constant} > 0$ ), with an initial velocity $\vec{V} = v_o\hat{i}$ ( $v_0 = \text{constant} > 0$ ). If $\lambda_0 = \frac{h}{4mv_0}$ , its de Broglie wavelength at time $t$ is __________. ( $e = \text{charge\; of\; electron}$ )

Option A:
$\frac{4\lambda_0}{\left[1 - \frac{E_0e}{2m}\frac{t}{v_0}\right]}$
Option B:
$\frac{4\lambda_0}{\left[1 + \frac{E_0e}{2m}\frac{t}{v_0}\right]}$
Option C:
$\frac{4\lambda_0}{\left[1 + \frac{2E_0e}{m}\frac{t}{v_0}\right]}$
Correct
Option D:
$\frac{4\lambda_0}{\left[1 - \frac{2E_0e}{m}\frac{t}{v_0}\right]}$

Answer: C

Step-by-step solution

$V = U + at$

$V = V_0 + \left(\frac{2Ee_0}{m}\right)t$

$\because a = \frac{(2E_0)e}{m}$

$\lambda = \frac{h}{mV}$

$\lambda = \frac{h}{m\left[V_0 + \frac{2eE_0t}{m}\right]}$

Put $\lambda_0 = \frac{h}{4mV_0} \Rightarrow h = 4\lambda_0 m V_0$

$\lambda = \frac{4m\lambda_0 V_0}{m\left[V_0 + \frac{2eE_0}{m}t\right]}$

$\lambda = \frac{4\lambda_0}{1 + \frac{2eE_0}{mV_0}t}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter

Answer: C

Step-by-step solution

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