Question

An electron having mass $m$ with an initial velocity $v=v_0\hat{i}$ in an electric field $E=E_0\hat{i}.$ If ${\lambda }_0=\frac{h}{m{v}_0}$, its de-Broglie wave length at time $t$ is given by-

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

Answer 1

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

Given,

$E=E_0\hat{i}$ ; initial velocity and de-Broglie wavelength as $v=v_0\hat{i}\&{\lambda }_0=\frac{h}{m{v}_0}$

When, the electron subjected to an electric field $E=E_0\hat{i}$,

$F_e=ma$

$q{E}_0\hat{i}=ma\Rightarrow a=\frac{e{E}_0}{m}\hat{i}$

At any time $t$, net velocity of the electron, $v=v_0+at$

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

$\therefore v=v_0(1+\frac{e{E}_0}{m{v}_0}t)\hat{i}$

Now, de-Broglie wavelength at time $t$ is,

$\lambda =\frac{h}{mv}=\frac{h}{m\left[v_0(1+\frac{e{E}_0}{m{v}_0}t)\right]}$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.