De Broglie Wavelength of Electron Acceleration in Electric Field

To find the de Broglie wavelength of the electron at time $t$, we analyze its motion in the given electric field step-by-step:

1. Force on the Electron: The force $\vec{F}$ acting on an electron of charge $-e$ moving in an electric field $\vec{E} = -2E_0\hat{i}$ is given by: $\vec{F} = q\vec{E} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}$

2. Acceleration of the Electron: Using Newton's second law, the acceleration $\vec{a}$ of the electron is: $\vec{a} = \frac{\vec{F}}{m} = \frac{2eE_0}{m}\hat{i}$

3. Velocity at Time $t$: Since the acceleration is constant and parallel to the initial velocity $\vec{V} = v_0\hat{i}$, the velocity $\vec{v}(t)$ at time $t$ is: $\vec{v}(t) = \vec{V} + \vec{a}t = v_0\hat{i} + \left(\frac{2eE_0}{m}t\right)\hat{i} = \left(v_0 + \frac{2eE_0 t}{m}\right)\hat{i}$

   The magnitude of the velocity at time $t$ is: $v(t) = v_0 + \frac{2eE_0 t}{m} = v_0 \left(1 + \frac{2eE_0 t}{m v_0}\right)$

4. De Broglie Wavelength at Time $t$: The de Broglie wavelength $\lambda(t)$ at time $t$ is defined as: $\lambda(t) = \frac{h}{m v(t)} = \frac{h}{m v_0 \left(1 + \frac{2eE_0 t}{m v_0}\right)}$

5. Expressing in terms of $\lambda_0$: We are given that: $\lambda_0 = \frac{h}{4mv_0} \implies \frac{h}{mv_0} = 4\lambda_0$

   Substituting $\frac{h}{mv_0} = 4\lambda_0$ into the expression for $\lambda(t)$, we get: $\lambda(t) = \frac{4\lambda_0}{1 + \frac{2E_0 e t}{m v_0}}$

This corresponds to Option C.