Question 39 of 200

An electron (mass $m$) with an initial velocity $\vec{v}={v}_{0}\hat{i}\left({v}_{0}\gt 0\right)$ is moving in an electric field $\vec{E}=-{E}_{0}\hat{i}\left({E}_{0}\gt 0\right)$ where ${E}_{0}$ is constant. If at $t=0$, de-Broglie wavelength is ${\lambda }_{0}=\dfrac{h}{m{v}_{0}}$, then its de-Broglie wavelength after time $t$ is given by

A${\lambda }_{0}$B${\lambda }_{0}\left(1+\dfrac{e{E}_{0}t}{m{v}_{0}}\right)$C${\lambda }_{0}t$D$\dfrac{{\lambda }_{0}}{\left(1+\dfrac{e{E}_{0}t}{m{v}_{0}}\right)}$

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Answer: (D) $\dfrac{{\lambda }_{0}}{\left(1+\dfrac{e{E}_{0}t}{m{v}_{0}}\right)}$

Initial de-Broglie wavelength of electron,
${\lambda }_{0}=\dfrac{h}{m{v}_{0}}$ ....(i)
Force on electron in electric field,
$F=-eE=-e\left[-{E}_{0}\hat{i}\right]=e{E}_{0}\hat{i}$
Acceleration of electron, $a=\dfrac{F}{m}=\dfrac{e{E}_{0}\hat{i}}{m}$
Velocity of electron after time $t$,
$v={v}_{0}\hat{i}+\left(\dfrac{e{E}_{0}\hat{i}}{m}\right)t=\left({v}_{0}+\dfrac{e{E}_{0}}{m}t\right)\hat{i}={v}_{0}\left(1+\dfrac{e{E}_{0}}{m{v}_{0}}t\right)\hat{i}$
de-Broglie wavelength associated with electron at time $t$ is
$\lambda =\dfrac{h}{mv}$
$\Rightarrow \lambda =\dfrac{h}{m\left[{v}_{0}\left(1+\dfrac{e{E}_{0}}{m{v}_{0}}t\right)\right]}=\dfrac{{\lambda }_{0}}{\left[1+\dfrac{e{E}_{0}}{m{v}_{0}}t\right]}$