An electron mass (m) with initial velocity

Question:

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

1. (1) $\frac{\lambda_{0} \sqrt{2}}{\sqrt{1+\frac{e^{2} E^{2} t^{2}}{m^{2} v_{0}^{2}}}}$

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

3. (3) $\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2} E^{2} t^{2}}{2 m^{2} v_{0}^{2}}}}$

4. (4) $\frac{\lambda_{0}}{\sqrt{2+\frac{e^{2} E^{2} t^{2}}{m^{2} v_{0}^{2}}}}$

Correct Option: , 3

Solution:

(3) Given, Initial velocity, $u=v_{0} \hat{i}+v_{0} \hat{j}$

Acceleration, $a=\frac{q E_{0}}{m}=\frac{e E_{0}}{m}$

Using $v=u+a t$

$v=v_{0} \hat{i}+v_{0} \hat{j}+\frac{e E_{0}}{m} t \hat{k}$

$\therefore|\vec{v}|=\sqrt{2 v_{0}^{2}+\left(\frac{e E_{0} t}{m}\right)^{2}}$

de-Broglie wavelength, $\lambda=\frac{h}{p}$

$\Rightarrow \lambda=\frac{h}{m v}(\because p=m v)$

Initial wavelength, $\lambda_{0}=\frac{h}{m v_{0} \sqrt{2}}$

Final wavelength,

$\lambda=\frac{h}{\sqrt[m]{2 v_{0}^{2}+\left(\frac{e E_{0} t}{m}\right)^{2}}}$

$\frac{\lambda}{\lambda_{0}}=\frac{1}{\sqrt{1+\left(\frac{e E_{0} t}{\sqrt{2} m v_{0}}\right)^{2}}}$

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