Question

An electron of mass $m$ with an initial velocity $\vec{V}=V_0\hat{i}(V_0>0)$ enters an electric field $\vec{E}=-E_0\hat{i}(E_0=\text{constant }>0)$ at $t=0$. If ${\lambda }_o$ is its de-Broglie wavelength initially,. then its de-Broglie wavelength at time $t$ is

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

Answer 1

Answer: (B)

Initial de-Broglie wavelength
${\lambda }_0=\frac{h}{m{V}_0}$ ....(i)

Acceleration of electron
$a=\frac{e{E}_0}{m}$
Velocity after time ‘ $t$ ’
$V=\left(V_0+\frac{e{E}_0}{m}t\right)$
So, $\lambda =\frac{h}{mV}=\frac{h}{m\left(V_0+\frac{e{E}_0}{m}t\right)}$
$=\frac{h}{m{V}_0\left[1+\frac{e{E}_0}{m{V}_0}t\right]}=\frac{{\lambda }_0}{\left[1+\frac{e{E}_0}{m{V}_0}t\right]}$ ....(ii)
Divide (ii) by (i),
$\lambda =\frac{{\lambda }_0}{\left[1+\frac{e{E}_0}{m{V}_0}t\right]}$