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 $>0$) at $t=0$. If ${\lambda }_0$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is?

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

Answer 1

Answer: (C)

$\frac{{\lambda }_0}{\left(1+\frac{e{E}_0}{m{V}_0}t\right)}$

R.E.F image

The diagram will be :-

Initial velocity $=V_0$

$\because$ Acceleration is constant :-

Final velocity $=v=v_0+\frac{e{E}_0}{m}t$

(at time t)

Now, at $t=0\Rightarrow$ de-Broglie wavelength $={\lambda }_0$

$\Rightarrow {\lambda }_0=\frac{h}{m{v}_0}$

Let at time t, de-Broglie wavelength $=\lambda$

Then, $\lambda =\frac{h}{mV}$

$=\frac{h}{m(V_0+\frac{e{E}_{0}t}{m})}$

$=\frac{h}{m{V}_0+e{E}_{0}t}$

$=\frac{1}{1/{\lambda }_0+e{E}_{0}t/h}$

$=\frac{\lambda .h}{h+e{E}_0{\lambda }_{0}t}$

$=\frac{{\lambda }_0}{1+\frac{e{E}_o{\lambda }_{0}t}{h}}$

$=\lambda =\frac{{\lambda }_0}{1+\left(\frac{e{E}_0}{m{V}_0}\right)t}$