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. $\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2}E^{2}_{0}t^{2}}{m^{2}v^{2}_{0}}}}$
• B. $\frac{\lambda_{0}}{\sqrt{2+\frac{e^{2}E^{2}_{0}t^{2}}{m^{2}v^{2}_{0}}}}$
• C. $\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2}E^{2}_{0}t^{2}}{2m^{2}v^{2}_{0}}}}$
• D. $\frac{\lambda_{0}\sqrt{2}}{\sqrt{1+\frac{e^{2}E^{2}_{0}t^{2}}{m^{2}v^{2}_{0}}}}$

Answer 1

Answer: (C)

By de-Broglie hypothesis
$\lambda=\frac{h}{mv}$
$\lambda_{0}=\frac{h}{m\sqrt{2}v_{0}}\,........\left(1\right)$
$\lambda '=\frac{h}{\sqrt{v^{2}_{0}+v^{2}_{0}+\left(\frac{eE_{0}t}{m}\right)^{2}}}$
$=\frac{h}{\sqrt{2v^{2}_{0}+\frac{e^{2}E^{2}_{0}t^{2}}{m^{2}}}}\,........\left(2\right)$
By $\left(1\right)$ and $\left(2\right)$
$\lambda '=\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2}E^{2}_{0}t^{2}}{2m^{2}v^{2}_{0}}}}$