An electron of mass $m$ is moving in an electric field $\overrightarrow{\mathrm{E}}=-2 \mathrm{E}_{0} \hat{\mathrm{i}}$ ($\mathrm{E}_{0}=$ constant $>0$), with an initial velocity $\overrightarrow{\mathrm{V}}=\mathrm{v}_{\mathrm{o}} \hat{\mathrm{i}}$ ($\mathrm{v}_{0}=$ constant $>0$). If $\lambda_{0}=\frac{\mathrm{h}}{4 \mathrm{mv}_{0}}$, its de Broglie wavelength at time t is ______. ($\mathrm{e}=$ charge of electron)

(C) $\frac{4 \lambda_{0}}{\left[1+\frac{2 E_{0} e}{m} \frac{t}{v_{0}}\right]}$

(A) $\frac{4 \lambda_{0}}{\left[1-\frac{E_{0} e}{2 m} \frac{t}{v_{0}}\right]}$

(B) $\frac{4 \lambda_{0}}{\left[1+\frac{E_{0} e}{2 m} \frac{t}{v_{0}}\right]}$

(D) $\frac{4 \lambda_{0}}{\left[1-\frac{2 \mathrm{E}_{0} \mathrm{e}}{\mathrm{m}} \frac{\mathrm{t}}{\mathrm{v}_{0}}\right]}$

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Modern Physics

An electron of mass

m

is moving in an electric field

E = - 2 E_{0} \hat{i}

(

E_{0} =

constant

> 0

), with an initial velocity

V = v_{o} \hat{i}

(

v_{0} =

constant

> 0

). If

λ_{0} = \frac{h}{4 mv _{0}}

, its de Broglie wavelength at time t is ______. (

e =

charge of electron)

Answer: (C)

With

E = - 2 E_{0} \hat{i}

and electron charge

- e

, the force is

F = (- e) (- 2 E_{0} \hat{i}) = 2 e E_{0} \hat{i}

, so the acceleration

a = \frac{2 e E _{0}}{m}

is along

+ \hat{i}

, the direction of the initial velocity. The speed grows as

v = v_{0} + \frac{2 e E _{0}}{m} t

. The de Broglie wavelength is

λ = \frac{h}{m v}

; writing

λ_{0} = \frac{h}{4 m v _{0}}

so that

h = 4 λ_{0} m v_{0}

λ = \frac{4 λ _{0} m v _{0}}{m ( v _{0} + \frac{2 e E _{0}}{m} t )} = \frac{4 λ _{0}}{1 + \frac{2 E _{0} e}{m} \frac{t}{v _{0}}}

. This is the de Broglie wavelength at time

t

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option