An electron (mass $$m$$) with an initial velocity $$\vec{v} = v_0\hat{i}$$ ($$v_0 > 0$$) is moving in an electric field $$\vec{E} = -E_0\hat{i}$$ ($$E_0$$ 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

An electron (mass $$m$$) with initial velocity $$\vec{v} = v_0\hat{i}$$ moves in an electric field $$\vec{E} = -E_0\hat{i}$$. We need to find the de-Broglie wavelength after time $$t$$.

The electron has charge $$-e$$. The force on it is:

$$\vec{F} = (-e)(-E_0\hat{i}) = eE_0\hat{i}$$

The force is in the $$+x$$ direction, same as the initial velocity.

Acceleration: $$a = \frac{eE_0}{m}$$

Velocity at time $$t$$:

$$v(t) = v_0 + at = v_0 + \frac{eE_0}{m}t$$

$$\lambda = \frac{h}{mv(t)} = \frac{h}{m\left(v_0 + \frac{eE_0 t}{m}\right)} = \frac{h}{mv_0 + eE_0 t}$$

$$= \frac{h}{mv_0\left(1 + \frac{eE_0 t}{mv_0}\right)} = \frac{\lambda_0}{1 + \frac{eE_0 t}{mv_0}}$$

where $$\lambda_0 = \frac{h}{mv_0}$$.

The correct answer is Option D: $$\dfrac{\lambda_0}{\left(1 + \dfrac{eE_0 t}{mv_0}\right)}$$.