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

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Question

(C) The initial velocity is $\overrightarrow{v} = v_{0} \hat{i} + v_{0} \hat{j}$. The initial speed is $v = \sqrt{v_{0}^{2} + v_{0}^{2}} = v_{0} \sqrt

Vedclass · Accepted Answer

$\frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{2 m^{2} v_{0}^{2}}}}$

Vedclass · Answer

(C) The initial velocity is $\overrightarrow{v} = v_{0} \hat{i} + v_{0} \hat{j}$. The initial speed is $v = \sqrt{v_{0}^{2} + v_{0}^{2}} = v_{0} \sqrt{2}$.The initial de-Broglie wavelength is $\lambda_{0} = \frac{h}{m v_{0} \sqrt{2}} \implies h = \lambda_{0} m v_{0} \sqrt{2}$.The electric field is $\overrightarrow{E} = -E_{0} \hat{k}$. The force on the electron is $\overrightarrow{F} = -e \overrightarrow{E} = e E_{0} \hat{k}$.The acceleration is $\overrightarrow{a} = \frac{e E_{0}}{m} \hat{k}$.At time $t$,the velocity is $\overrightarrow{v}(t) = v_{0} \hat{i} + v_{0} \hat{j} + \frac{e E_{0} t}{m} \hat{k}$.The speed at time $t$ is $v(t) = \sqrt{v_{0}^{2} + v_{0}^{2} + \left(\frac{e E_{0} t}{m}\right)^{2}} = \sqrt{2 v_{0}^{2} + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2}}}$.The de-Broglie wavelength at time $t$ is $\lambda = \frac{h}{m v(t)} = \frac{\lambda_{0} m v_{0} \sqrt{2}}{m \sqrt{2 v_{0}^{2} + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2}}}} = \frac{\lambda_{0} v_{0} \sqrt{2}}{\sqrt{2 v_{0}^{2} + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2}}}} = \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{2 m^{2} v_{0}^{2}}}}$.

An electron (mass $m$) with initial velocity $\overrightarrow{v} = v_{0} \hat{i} + v_{0} \hat{j}$ is in an electric field $\overrightarrow{E} = -E_{0} \hat{k}$. If $\lambda_{0}$ is the initial de-Broglie wavelength of the electron,its de-Broglie wavelength at time $t$ is given by:

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