Question

An electron $\left(\text{mass}m\right)$ 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 initial de-Broglie wavelength of electron, its de-Broglie wave length at time $t$ is given by

• A. $\frac{{\lambda }_0\sqrt{2}}{\sqrt{1+\frac{e^2{E}_0^2{t}^2}{m^2{v}_0^2}}}$
• B. $\frac{{\lambda }_0}{\sqrt{1+\frac{e^2{E}_0^2{t}^2}{m^2{v}_0^2}}}$
• C. $\frac{{\lambda }_0}{\sqrt{1+\frac{e^2{E}_0^2{t}^2}{2m^2{v}_0^2}}}$
• D. $\frac{{\lambda }_0}{\sqrt{2+\frac{e^2{E}_0^2{t}^2}{m^2{v}_0^2}}}$

Answer 1

The correct option is C $\frac{{\lambda }_0}{\sqrt{1+\frac{e^2{E}_0^2{t}^2}{2m^2{v}_0^2}}}$

$\text{Given, Inital velocity}u=v_0\hat{i}+v_0\hat{j}$

$\text{Acceleration a}=\frac{q{E}_0}{m}=\frac{e{E}_0}{m}$

$\text{Using v = u + at}$

$v=v_0\hat{i}+v_0\hat{j}+\frac{e{E}_0}{m}t\hat{k}$

$\therefore |\overrightarrow{v}|=\sqrt{2v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}$

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

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

$\text{Initial wavelength,}{\lambda }_0=\frac{h}{m{v}_0\sqrt{2}}$

Final wavelength,

$\lambda =\frac{h}{m\sqrt{2v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}}$

$\frac{\lambda }{{\lambda }_0}=\frac{1}{\sqrt{1+{\left(\frac{e{E}_{0}t}{\sqrt{2}m{v}_0}\right)}^2}}$

$\Rightarrow \lambda =\frac{{\lambda }_0}{\sqrt{1+\frac{e^2{E}_0^2{t}^2}{2m^2{v}_0^2}}}$