An electron (mass m) with an initial velocity | vedclass

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Question

(C) Initial de Broglie wavelength of the electron is $\lambda_0 = \frac{h}{mv_0}$.Force on the electron in the electric field is $\vec{F} = -e\vec{E}

Vedclass · Accepted Answer

$\frac{\lambda_0}{\sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$

Vedclass · Answer

(C) Initial de Broglie wavelength of the electron is $\lambda_0 = \frac{h}{mv_0}$.Force on the electron in the electric field is $\vec{F} = -e\vec{E} = -eE_0 \hat{j}$.Acceleration of the electron is $\vec{a} = \frac{\vec{F}}{m} = -\frac{eE_0}{m} \hat{j}$.The initial velocity is $\vec{v}_0 = v_0 \hat{i}$.Velocity at time $t$ is $\vec{v} = \vec{v}_0 + \vec{a}t = v_0 \hat{i} - \frac{eE_0 t}{m} \hat{j}$.The magnitude of velocity at time $t$ is $v = \sqrt{v_0^2 + (\frac{eE_0 t}{m})^2} = v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}$.The de Broglie wavelength at time $t$ is $\lambda = \frac{h}{mv} = \frac{h}{m v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$.Substituting $\lambda_0 = \frac{h}{mv_0}$,we get $\lambda = \frac{\lambda_0}{\sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$.

An electron (mass $m$) with an initial velocity $v = v_0 \hat{i}$ is in an electric field $E = E_0 \hat{j}$. If $\lambda_0 = h/mv_0$,its de Broglie wavelength at time $t$ is given by

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