An electron of mass m with an initial velocity \(\overrightarrow{ V }= V_0 \hat i \,(V_0 > 0)\) enters an electric field \(\vec E = -\vec E_0 \hat i \,(E_0 =\) constant \(> 0)\) at \(t = 0.\) If \(\lambda_0\) is its de-Broglie wavelength initially,then its de-Broglie wavelenght at time \(t\) is

\(\bar{F}=(-e)\left(-E_{0} \hat{i}\right)=e E_{0} \hat{i}\)

Acceleration produced in the electron, \(\vec{a}=\frac{\vec{F}}{m}=\frac{e E_{0}}{m} \hat{i}\)

Now, velocity of electron after time \(t\)

\(\vec{v}_{t}=\vec{v}+\vec{a} t=\left(v_{0}+\frac{e E_{0} t}{m}\right) \hat{i}\)

or \(\left|\vec{v}_{t}\right|=v_{0}+\frac{e E_{0} t}{m}\)

Now, \(\lambda_{t}=\frac{h}{m v_{t}}=\frac{h}{m\left(v_{0}+\frac{e E_{0} t}{m}\right)}=\frac{h}{m v_{0}\left(1+\frac{e E_{0} t}{m v_{0}}\right)}\)

\(=\frac{\lambda_{0}}{\left(1+\frac{e E_{0} t}{m v_{0}}\right)} \quad \quad \quad\left(\because \lambda_{0}=\frac{h}{m v_{0}}\right)\)