An electron of mass ' m ' with an initial velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_{0} \hat{\mathrm{i}}\left(\mathrm{v}_{0}>0\right)$ enters an electric field $\overrightarrow{\mathrm{E}}=-\mathrm{E}_{0} \hat{\mathrm{k}}$. If the initial de Broglie wavelength is $\lambda_{0}$, the value after time $t$ would be :-

(A) $\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}}$

(B) $\frac{\lambda_{0}}{\sqrt{1-\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}}$

(D) $\lambda_{0} \sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}$

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Modern Physics

An electron of mass ' m ' with an initial velocity

v = v_{0} \hat{i} (v_{0} > 0)

enters an electric field

E = - E_{0} \hat{k}

. If the initial de Broglie wavelength is

λ_{0}

, the value after time

t

would be :-

(A) $\frac{λ _{0}}{1 + \frac{e ^{2} E _{0}^{2} t ^{2}}{m ^{2} v _{0}^{2}}}$ ✓ Correct
(B) $\frac{λ _{0}}{1 - \frac{e ^{2} E _{0}^{2} t ^{2}}{m ^{2} v _{0}^{2}}}$
(C) $λ_{0}$
(D) $λ_{0} 1 + \frac{e ^{2} E _{0}^{2} t ^{2}}{m ^{2} v _{0}^{2}}$

Answer: (A)

The electron starts with velocity

v_{0} = v_{0} \hat{i}

and experiences a force

F = - e E = - e (- E_{0} \hat{k}) = e E_{0} \hat{k}

(its charge is

- e

), giving acceleration

a = \frac{e E _{0}}{m} \hat{k}

.

After time

t

, the velocity is

v = v_{0} \hat{i} + \frac{e E _{0} t}{m} \hat{k}

Since these components are perpendicular, the speed is

∣ v ∣ = v_{0}^{2} + (\frac{e E _{0} t}{m})^{2} = v_{0} 1 + \frac{e ^{2} E _{0}^{2} t ^{2}}{m ^{2} v _{0}^{2}}

The de Broglie wavelength is

λ = \frac{h}{m v}

, inversely proportional to speed. With

λ_{0} = \frac{h}{m v _{0}}

λ (t) = \frac{h}{m ∣ v ∣} = \frac{λ _{0}}{1 + \frac{e ^{2} E _{0}^{2} t ^{2}}{m ^{2} v _{0}^{2}}}

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option