An electron of mass m with an initial velocity $\overrightarrow{v}=v_0\hat{i}\left(v_0>0\right)$ enters an electric field $\overrightarrow{E}=-{\overrightarrow{E}}_0\hat{i} \left(E_0=\right.\text{ constant }>0\text{ ) at }t=0.$
 If ${\lambda }_0$ is its de-Broglie wavelength initially, then its de- Broglie wavelength at time t is

Here, = $\frac{{\lambda }_0}{\left(1+\frac{e{E}_0}{m{v}_0}t\right)}$initial; velocity $\overrightarrow{v}=v_0\hat{i}$

Force action on electron due to electric field

$\overrightarrow{F}=(-e)\left(-E_0\hat{i}\right)=e{E}_0\hat{i}$

$\text{ Acceleration produced in the electron, }\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{e{E}_0}{m}\hat{i}$

Now, velocity of electron after time t

${\overrightarrow{v}}_t=\overrightarrow{v}+\overrightarrow{a}t=\left(v_0+\frac{e{E}_{0}t}{m}\right)\hat{i} \text{ or } \left|{\overrightarrow{v}}_t\right|=v_0+\frac{e{E}_{0}t}{m}$

$\text{ Now, }{\lambda }_t=\frac{h}{m{v}_t}=\frac{h}{m\left(v_0+\frac{e{E}_{0}t}{m}\right)}=\frac{h}{m{v}_0\left(1+\frac{e{E}_{0}t}{m{v}_0}\right)}$

$=\frac{{\lambda }_0}{\left(1+\frac{e{E}_{0}t}{m{v}_0}\right)} \left(\because {\lambda }_0=\frac{h}{m{v}_0}\right)$