Question

An electron of mass $m$ with an initial velocity $\overrightarrow{v}=v_0\hat{i}$ $(v_0>0)$ enters an electric field $\overrightarrow{E}=E_0\hat{i}$ ($E_0$ = constant) at $t=0$. If ${\lambda }_0$ is its De-Broglie wavelength initially, then its De-Broglie wavelength at time $t$ is

• A. ${\lambda }_{0}t$
• B. $\frac{{\lambda }_0}{1+\frac{e{E}_0}{v_{0}m}t}$
• C. ${\lambda }_0(1+\frac{e{E}_0}{m{v}_0}t)$
• D. $\frac{{\lambda }_0}{t}$

Answer 1

The correct option is B $\frac{{\lambda }_0}{1+\frac{e{E}_0}{v_{0}m}t}$

$\overrightarrow{v}=v_0\hat{i}$
$\overrightarrow{E}=-E_0\hat{i}$
$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{q\overrightarrow{E}}{m}=\frac{(-e)(-E_0)\hat{i}}{m}$
$\Rightarrow \overrightarrow{a}=\frac{e{E}_0}{m}\hat{i}$

Using, equation of motion,
$v=u+at$
$v=v_0+\frac{e{E}_0}{m}t$

Now, De-Broglie wavelength is given by
$\lambda =\frac{h}{mv}=\frac{h}{m[v_0+\frac{e{E}_{0}t}{m}]}$

$=\frac{h}{m{v}_0[1+\frac{e{E}_{0}t}{m{v}_0}]}=\frac{{\lambda }_0}{1+\frac{e{E}_{0}t}{m{v}_0}}$