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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Initially,the velocity is $v_0 \hat{i}$,so the initial de-Broglie wav

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Initially,the velocity is $v_0 \hat{i}$,so the initial de-Broglie wavelength is $\lambda_0 = \frac{h}{mv_0}$.The force on the electron due to the electric field is $F = eE_0 \hat{j}$.The acceleration is $a = \frac{F}{m} = \frac{eE_0}{m} \hat{j}$.At time $t$,the velocity of the electron is $v = v_0 \hat{i} + \frac{eE_0 t}{m} \hat{j}$.The magnitude of the velocity is $|v| = \sqrt{v_0^2 + \left(\frac{eE_0 t}{m}\right)^2} = v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}$.The de-Broglie wavelength at time $t$ is $\lambda = \frac{h}{m|v|} = \frac{h}{m v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$.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 of charge $e$ and mass $m$ moving with an initial velocity $v_0 \hat{i}$ is subjected to an electric field $E_0 \hat{j}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron $= \lambda_0$)

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