An electron (mass $m$) with initial velocity $\vec{v} = v_0\hat{i} + v_0\hat{j}$ is in an electric field $\vec{E} = -E_0\hat{k}$. If $\lambda_0$ is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time $t$ is given by:
Answer: (C) $\dfrac{\lambda_0}{\sqrt{1 + \dfrac{e^2E_0^2t^2}{2m^2v_0^2}}}$
Force on the electron: $\vec{F} = -e\vec{E} = eE_0\hat{k}$, so it gains a velocity component $\dfrac{eE_0t}{m}$ along $\hat{k}$, while the $x$ and $y$ components stay $v_0$.
$$v^2 = 2v_0^2 + \frac{e^2E_0^2t^2}{m^2}$$
Initially $\lambda_0 = \dfrac{h}{m\sqrt{2}\,v_0}$. At time $t$:
$$\lambda = \frac{h}{mv} = \frac{\lambda_0\sqrt{2}\,v_0}{\sqrt{2v_0^2 + \dfrac{e^2E_0^2t^2}{m^2}}} = \frac{\lambda_0}{\sqrt{1 + \dfrac{e^2E_0^2t^2}{2m^2v_0^2}}}$$
Solution by Sreeraj P, M.Sc Physics