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)

The force experienced by the electron in the electric field is given by \(\vec{F} = q\vec{E}\).

Since the charge of an electron is \(-e\) and the electric field is \(\vec{E} = -2E_0\hat{i}\), the force is:

\(\vec{F} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}\)

The acceleration of the electron is:

\(\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{2eE_0}{m}\hat{i}\)

Using the first equation of motion, the velocity of the electron at time \(t\) is:

\(\vec{v}(t) = \vec{V} + \vec{a}t = \left(v_0 + \dfrac{2eE_0}{m}t\right)\hat{i}\)

The de Broglie wavelength of the electron at time \(t\) is:

\(\lambda = \dfrac{h}{m|\vec{v}(t)|} = \dfrac{h}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}\)

We are given that \(\lambda_0 = \dfrac{h}{4mv_0}\), which implies \(h = 4mv_0\lambda_0\). Substituting this into the expression for \(\lambda\):

\(\lambda = \dfrac{4mv_0\lambda_0}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}\)

Dividing the numerator and the denominator by \(mv_0\), we get:

\(\lambda = \dfrac{4\lambda_0}{1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}}\)

Answer: \(\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}\)