Q 12-11-037JEE MainJEE Main 2026 (5 Apr, Shift 1)Hard
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) $\dfrac{4\lambda_0}{\left[1 + \frac{2E_0e}{m}\frac{t}{v_0}\right]}$
The force on the electron is $-e\vec{E} = 2eE_0\hat{i}$, along its motion, so it speeds up:
$$v = v_0 + \frac{2eE_0}{m}t$$
$$\lambda = \frac{h}{mv} = \frac{h}{mv_0\left(1 + \frac{2eE_0t}{mv_0}\right)} = \frac{4\lambda_0}{1 + \frac{2E_0e}{m}\frac{t}{v_0}}$$
Solution by Sreeraj P, M.Sc Physics