JEE Main 2026 · 5 Apr · Shift 1 · Physics Q35

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

\dfrac{4\lambda_0}{\left[1-\dfrac{E_0e\,t}{2m\,v_0}\right]}

\dfrac{4\lambda_0}{\left[1+\dfrac{E_0e\,t}{2m\,v_0}\right]}

\dfrac{4\lambda_0}{\left[1+\dfrac{2E_0e\,t}{m\,v_0}\right]}

\dfrac{4\lambda_0}{\left[1-\dfrac{2E_0e\,t}{m\,v_0}\right]}

Answer and worked solution

Official answer: C

Motion. The force on the electron is $-e\vec E=2eE_0\hat i$ , along its velocity. It speeds up:
$v=v_0+\frac{2eE_0}m t.$

Wavelength.
$\lambda=\frac h{mv}=\frac{h/(mv_0)}{1+\frac{2eE_0t}{mv_0}}=\frac{4\lambda_0}{1+\frac{2E_0et}{mv_0}}.$

(C)

Topics:de Broglie waves Electric field

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JEE Main 2026 · 5 Apr · Shift 1, Physics question 35