An electron of mass ' m ' with an initial velocity v = v₀i (v₀ > 0) enters an electric field E = -Eok. If the initial de Broglie wavelength is λ₀, the value after time t would be (a) (λo)/(√(1-(e²Eo²t²)/(m²vo²))) (b) λ₀ (c) (λo)/(√(1+(e²Eo²t²)/(m²vo²))) (d) λo√(1 + (e²Eo²t²)/(m²vo²))

Answer: (c). Force on electron = -e E = eE₀ k, perpendicular to v₀. vx = v₀, vz = eE₀/mt, so v = v₀√(1+(e²E₀²t²)/(m²v₀²)). λ = h/mv = (λ₀)/(√(1+(e²E₀²t²)/(m²v₀²))) So the answer is (c).

An electron of mass ' m ' with an initial velocity v = v₀i (v₀ > 0)… | JEE Main 2025 (24 Jan, Shift 1)

Q39JEE Main 2025 · 24 JAN · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

An electron of mass ' m ' with an initial velocity $\vec{v} = v_0\hat{i}\ (v_0 > 0)$ enters an electric field $\vec{E} = -E_o\hat{k}$ . If the initial de Broglie wavelength is $\lambda_0$ , the value after time t would be

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Answer: (c)

Force on electron $= -e\vec E = eE_0\hat k$ , perpendicular to $\vec v_0$ .
$v_x = v_0$ , $v_z = \dfrac{eE_0}{m}t$ , so $v = v_0\sqrt{1+\dfrac{e^2E_0^2t^2}{m^2v_0^2}}$ .
$\lambda = \dfrac{h}{mv} = \dfrac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}}$
So the answer is (c).

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