An electron of mass m is moving in an electric field E = -2Eoi (Eo = constant > 0), with an initial velocity V = voi (vo = constant > 0). If λo = (h)/(4mvo), its de Broglie wavelength at time t is ______. ( e = charge of electron) (a) (4λo)/([1-(Eoe)/(2m)t/vo]) (b) (4λo)/([1+(Eoe)/(2m)t/vo]) (c) (4λo)/([1+(2Eoe)/(m)t/vo]) (d) (4λo)/([1-(2Eoe)/(m)t/vo])

Answer: (c). Force on electron = -e E = 2eEo i, along v: v = vo + (2eEo)/(m)t. λ = h/mv = (h)/(mvo(1+(2Eoe)/(m)t/vo)) h/mvo = 4λo, so λ = (4λo)/(1+(2Eoe)/(m)t/vo) So the answer is (c).

An electron of mass m is moving in an electric field E = -2Eoi (Eo =… | JEE Main 2026 (5 Apr, Shift 1)

Q27JEE Main 2026 · 5 APR · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_o\hat{i}\ (E_o =$ constant $> 0)$ , with an initial velocity $\vec{V} = v_o\hat{i}\ (v_o =$ constant $> 0)$ . If $\lambda_o = \frac{h}{4mv_o}$ , its de Broglie wavelength at time $t$ is ______.

( $e =$ charge of electron)

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Answer: (c)

Force on electron $= -e\vec E = 2eE_o\hat i$ , along $\vec v$ : $v = v_o + \dfrac{2eE_o}{m}t$ .
$\lambda = \dfrac{h}{mv} = \dfrac{h}{mv_o\left(1+\frac{2E_oe}{m}\frac{t}{v_o}\right)}$
$\dfrac{h}{mv_o} = 4\lambda_o$ , so $\lambda = \dfrac{4\lambda_o}{1+\frac{2E_oe}{m}\frac{t}{v_o}}$
So the answer is (c).

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Matter Waves: de Broglie Wavelength question from JEE Main 2026 (5 Apr, Shift 1)

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