An electron of mass $$m$$ and magnitude of charge $$e$$ at rest, gets accelerated by a constant electric field $$E$$. The rate of change of de-Broglie wavelength of this electron at time $$t$$ is (ignore relativistic effects):

We begin with an electron that is initially at rest. Because it is placed in a constant, uniform electric field of magnitude $$E$$, it experiences a constant force. According to Newton’s second law,

$$F = m a.$$ But the electric force on a charge $$e$$ in an electric field $$E$$ is

$$F = eE.$$

Equating the two forces gives

$$m a = eE \quad\Longrightarrow\quad a = \frac{eE}{m}.$$

Since the acceleration $$a$$ is constant and the electron starts from rest, the kinematic relation for velocity after time $$t$$ is

$$v = a t.$$

Substituting the value of $$a$$ we just found, we have

$$v = \left(\frac{eE}{m}\right) t.$$

Linear momentum is defined as $$p = m v$$. Substituting for $$v$$, we obtain the momentum of the electron at time $$t$$:

$$p = m \left(\frac{eE}{m}\right) t = eE t.$$

The de-Broglie wavelength formula is stated as

$$\lambda = \frac{h}{p},$$

where $$h$$ is Planck’s constant. Inserting the expression for $$p$$ gives

$$\lambda = \frac{h}{eE t}.$$

We are asked for the rate of change of this wavelength with respect to time. Therefore, we differentiate $$\lambda$$ with respect to $$t$$. Writing $$\lambda(t) = \dfrac{h}{eE}\, t^{-1}$$, we differentiate using the power rule $$\dfrac{d}{dt}\bigl(t^{-1}\bigr) = -t^{-2}$$:

$$\frac{d\lambda}{dt} = \frac{h}{eE}\left(-t^{-2}\right) = -\frac{h}{eE t^{2}}.$$

The negative sign simply indicates that the wavelength decreases as the time increases, which is expected because the momentum is increasing. All algebraic steps are complete and no relativistic corrections were necessary (as instructed).

Hence, the correct answer is Option D.