An electron of mass m is moving in an electric field E = -2E_0 i (...

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}$, the force is:<br><br>$\vec{F} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}$<br><br>The acceleration of the electron is:<br><br>$\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{2eE_0}{m}\hat{i}$<br><br>Using the first equation of motion, the velocity of the electron at time$t$is:<br><br>$\vec{v}(t) = \vec{V} + \vec{a}t = \left(v_0 + \dfrac{2eE_0}{m}t\right)\hat{i}$<br><br>The de Broglie wavelength of the electron at time$t$is:<br><br>$\lambda = \dfrac{h}{m|\vec{v}(t)|} = \dfrac{h}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}$<br><br>We are given that$\lambda_0 = \dfrac{h}{4mv_0}$, which implies$h = 4mv_0\lambda_0$. Substituting this into the expression for$\lambda$:<br><br>$\lambda = \dfrac{4mv_0\lambda_0}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}$<br><br>Dividing the numerator and the denominator by$mv_0$, we get:<br><br>$\lambda = \dfrac{4\lambda_0}{1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}}$<br><br>Answer:$\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}