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Question

(A) The initial velocity of the electron is $\overrightarrow{v}_i = v_0 \hat{i}$.The force on the electron in the electric field $\overrightarrow{E} =

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The initial velocity of the electron is $\overrightarrow{v}_i = v_0 \hat{i}$.The force on the electron in the electric field $\overrightarrow{E} = -E_0 \hat{k}$ is $\overrightarrow{F} = q\overrightarrow{E} = -e(-E_0 \hat{k}) = e E_0 \hat{k}$.The acceleration is $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{e E_0}{m} \hat{k}$.The velocity at time $t$ is $\overrightarrow{v}(t) = \overrightarrow{v}_i + \overrightarrow{a}t = v_0 \hat{i} + \frac{e E_0 t}{m} \hat{k}$.The magnitude of the velocity is $|\overrightarrow{v}(t)| = \sqrt{v_0^2 + \left(\frac{e E_0 t}{m}\right)^2} = v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}$.The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m|\overrightarrow{v}|}$.The initial wavelength is $\lambda_0 = \frac{h}{m v_0}$.The wavelength at time $t$ is $\lambda' = \frac{h}{m v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} = \frac{\lambda_0}{\sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$.

An electron of mass $m$ with an initial velocity $\overrightarrow{v}=v_0 \hat{i}$ $(v_0>0)$ enters an electric field $\overrightarrow{E}=-E_0 \hat{k}$. If the initial de Broglie wavelength is $\lambda_0$,the value after time $t$ would be $:-$

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