An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ ($E_0 =$ constant $> 0$), with an initial velocity $\vec{V} = v_0\hat{i}$ ($v_0 =$ constant $> 0$). If $\lambda_0 = \dfrac{h}{4mv_0}$, its de Broglie wavelength at time $t$ is ______. ($e =$ charge of electron)
Answer: (C) $\dfrac{4\lambda_0}{\left[1 + \frac{2E_0e}{m}\frac{t}{v_0}\right]}$
Concept. The force $-e\vec E = 2eE_0\hat i$ is along the motion, so the electron speeds up; $\lambda = \dfrac{h}{mv}$.
Step 1. Speed. $v = v_0 + \dfrac{2eE_0}{m}t$.
Step 2. Wavelength. With $\dfrac{h}{mv_0} = 4\lambda_0$:
$$\lambda = \frac{4\lambda_0}{1 + \frac{2E_0e}{m}\frac{t}{v_0}}$$
Answer. $\dfrac{4\lambda_0}{\left[1 + \frac{2E_0e}{m}\frac{t}{v_0}\right]}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
