Question

An electron in an electron microscope with initial velocity $v_{0}i$ enters a region of a stray transverse electric field $E_{0}j$. The time taken for the change in its de Broglie wavelength from the initial value of $\lambda$ to $\frac{\lambda }{3}$ is proportional to.

• A. $E_0$
• B. $\frac{1}{E_0}$
• C. $\frac{1}{\sqrt{E_0}}$
• D. $\sqrt{E_0}$

Answer 1

The correct option is B $\frac{1}{E_0}$

Path traced by the electron is shown in the figure such that electron reaches the point P when its de Broglie wavelength becomes $\frac{\lambda }{3}$.
Acceleration of electron is zero in x direction.
Acceleration of electron in y direction $a^{\prime }=\frac{e{E}_o}{m}$
Velocity of electron initially $v_o=\frac{h}{\lambda }$
At P, $v_p=\frac{h}{\left(\lambda /3\right)}=3v_o$
So, velocity is y direction $v^{\prime }=\sqrt{(v^{\prime }{)}^2-v_o^2}=\sqrt{9v_o^2-v_o^2}=2\sqrt{2}{v}_o$
Using $v^{\prime }=a^{\prime }t$
$\therefore$ $2\sqrt{2}{v}_o=\frac{e{E}_o}{m}t$
We get $t=\frac{2\sqrt{2}{v}_{o}m}{e{E}_o}$
$⟹t\propto \frac{1}{E_o}$