Q 12-11-114JEE MainJEE Main 2022 (27 Jul, Shift 1)Medium
An electron (mass $m$) with an initial velocity $\vec v = v_0\hat i$ $(v_0 > 0)$ is moving in an electric field $\vec E = -E_0\hat i$ $(E_0 > 0)$ where $E_0$ is constant. If at $t = 0$, de-Broglie wavelength is $\lambda_0 = \dfrac{h}{mv_0}$, then its de-Broglie wavelength after time $t$ is given by
Answer: (D) $\dfrac{\lambda_0}{\left(1 + \dfrac{eE_0t}{mv_0}\right)}$
Force on the electron: $\vec F = (-e)(-E_0\hat i) = eE_0\hat i$, along its motion, so it speeds up:
$$v = v_0 + \frac{eE_0}{m}t$$
$$\lambda = \frac{h}{mv} = \frac{h}{mv_0\left(1 + \dfrac{eE_0t}{mv_0}\right)} = \frac{\lambda_0}{1 + \dfrac{eE_0t}{mv_0}}$$
Solution by Sreeraj P, M.Sc Physics