An electron (mass m) with an initial velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_{0} \hat{\mathrm{i}}$ ($\mathrm{v}_{0}>0$) is moving in an electric field $\overrightarrow{\mathrm{E}}=-\mathrm{E}_{0} \hat{\mathrm{i}}$ ($\mathrm{E}_{0}>0$) where $\mathrm{E}_{0}$ is constant. If at $\mathrm{t}=0$ de Broglie wavelength is $\lambda_{0}=\frac{\mathrm{h}}{\mathrm{mv}_{0}}$, then its de Broglie wavelength after time $t$ is given by

(D) $\frac{\lambda_{0}}{\left(1+\frac{e E_{0} t}{m v_{0}}\right)}$

(B) $\lambda_{0}\left(1+\frac{\mathrm{eE}_{0} \mathrm{t}}{\mathrm{mv}_{0}}\right)$

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Modern Physics

An electron (mass m) with an initial velocity

v = v_{0} \hat{i}

(

v_{0} > 0

) is moving in an electric field

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

(

E_{0} > 0

) where

E_{0}

is constant. If at

t = 0

de Broglie wavelength is

λ_{0} = \frac{h}{mv _{0}}

, then its de Broglie wavelength after time

t

is given by

Answer: (D)

The electric field

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

exerts a force

F = - e E_{0}

on the electron (opposite to initial velocity), causing deceleration. The velocity after time

t

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

(taking the direction into account, the magnitude increases as the electron is negatively charged). Actually, the force on electron is

F = - e (- E_{0}) = e E_{0}

in the positive direction, so

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

. The de Broglie wavelength is

λ = h / (m v) = h / (m (v_{0} + e E_{0} t / m)) = λ_{0} / (1 + e E_{0} t / (m v_{0}))

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option