Question

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

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

Answer 1

The correct option is A $\frac{{\lambda }_0}{(1+\frac{e{E}_0}{m{v}_0}t)}$

Given,
$v=v_0\hat{i}$, $\overrightarrow{E}=E_0\hat{i}$

Now, $\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{q\overrightarrow{E}}{m}=\frac{(-e)(-E_0)\hat{i}}{m}$

$\therefore \overrightarrow{a}=\frac{e{E}_0}{m}\hat{i}$

Also, $v=u+at=v_0+\frac{e{E}_0}{m}t$

And, $\lambda =\frac{h}{mv}$

$=\frac{h}{m[v_0+\frac{e{E}_{0}t}{m}]}$

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

$=\frac{{\lambda }_0}{1+\frac{e{E}_{0}t}{m{v}_0}}(\because {\lambda }_0=\frac{h}{m{v}_0})$

Hence, $\left(A\right)$ is the correct answer.