Question

An electron (mass m) with an initial velocity $\overline{v}=v_0\hat{i}(v_0>0)$ is in an electric field $\overline{E}=-E_0\hat{i}$ ($E_0=constant>0$). Its de Broglie wavelength at time t is given by

Assume:$({\lambda }_0=\frac{h}{m{v}_0})$

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

Answer 1

The correct option is A $\frac{{\lambda }_0}{(1+\frac{e{E}_{0}t}{m{v}_0})}$

Here, $\overline{E}=-E_0\hat{i}$ ; initial velocity $\overline{v}=v_0\hat{i}$ Force action on electron due to electric field $\overline{F}=(-e)(-E_0\hat{i})=e{E}_0\hat{i}$
Acceleration produced in the electron, $\overline{a}=\frac{\overline{F}}{m}=\frac{e{E}_0}{m}\hat{i}$
Now, velocity of electron after time t, ${\overline{v}}_t=\overline{v}+\overline{a}t=(v_0+\frac{e{E}_0}{m})\hat{i}or\left|{\overline{v}}_t\right|=v_0+\frac{e{E}_{0}t}{m}$

Now,

${\lambda }_t=\frac{h}{m{v}_t}=\frac{h}{m(v_0+\frac{e{E}_{0}t}{m})}=\frac{h}{m{v}_0(1+\frac{e{E}_{0}t}{m{v}_0})}$

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