Question

An electron (mass 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:

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

Answer 1

Answer: (C)

By de-Broglie hypothesis
$\lambda =\frac{h}{mv}$
${\lambda }_0=\frac{h}{m\sqrt{2}{v}_0}........\left(1\right)$
${\lambda }^{\prime }=\frac{h}{\sqrt{v_0^2+v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}}$
$=\frac{h}{\sqrt{2v_0^2+\frac{e^2{E}_0^2{t}^2}{m^2}}}........\left(2\right)$
By $\left(1\right)$ and $\left(2\right)$
${\lambda }^{\prime }=\frac{{\lambda }_0}{\sqrt{1+\frac{e^2{E}_0^2{t}^2}{2m^2{v}_0^2}}}$