An electron of mass $\mathrm{m}$ with an initial velocity $\overrightarrow{\mathbf{V}}=\mathbf{V}_{0} \hat{\hat{i}}\left(\mathrm{~V}_{0}>0\right)$ enters an electric field $\vec{E}=-E_{0} \hat{i}\left(E_{0}=\right.$ constant $>0$ ) at $t=0$. If $\lambda_{0}$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time $t$ is
(1) $\lambda_{0} t$
(2) $\lambda_{0}\left(1+\frac{\mathrm{e} \mathrm{E}_{0}}{\mathrm{mV}_{0}} \mathrm{t}\right)$
(3) $\frac{\lambda_{0}}{\left(1+\frac{\mathrm{eE}_{0}}{\mathrm{mV}_{0}} \mathrm{t}\right)}$
(4) $\lambda_{0}$

Answer is (3)

Initial de-broglie wavelength will be,

$\lambda_0=\frac{h}{mV_0}$ …(i)

Acceleration of electron will be,

$a=\frac{eE_0}{m}$

Velocity after time t,

$\mathbf{V}=\left(\mathbf{V}_{0}+\frac{\mathrm{eE}_{0}}{\mathrm{~m}} \mathrm{t}\right)$

So, $\lambda=\frac{h}{m V}=\frac{h}{m\left(v_{0}+\frac{e E_{0}}{m} t\right)}$

$=\frac{h}{m V_{0}\left[1+\frac{e E_{0}}{m V_{0}} t\right]}=\frac{\lambda_{0}}{\left[1+\frac{e E_{0}}{m V_{0}} t\right]}$…(ii)

Divide (ii) by (i) we get,

$\lambda=\frac{\lambda_{0}}{\left[1+\frac{\mathrm{e} \mathrm{E}_{0}}{\mathrm{~m} \mathrm{~V}_{0}} \mathrm{t}\right.}$