Question

Electrons $(e,m)$ emitted with negligible speed from an electron gun are accelerated through a potential difference $V_0$ along the $\text{x-}$axis. These electrons emerge from a narrow hole into a uniform magnetic field of strength $B$ directed along $\text{x-}$axis. Some electrons emerging at slightly divergent angles as shown. These electrons are refocused on the $\text{x -}$axis at a distance ___.

• A. $\frac{8{\pi }^{2}m{V}_0}{3eB}$
• B. $\sqrt{\frac{8{\pi }^{2}m{V}_0}{e{B}^2}}$
• C. $\sqrt{\frac{2{\pi }^{2}m{V}_0}{e{B}^2}}$
• D. $\sqrt{\frac{4{\pi }^{2}m{V}_0}{e{B}^2}}$

Answer 1

The correct option is B $\sqrt{\frac{8{\pi }^{2}m{V}_0}{e{B}^2}}$

From the figure, the electrons will be refocussed after one time period of revolution for their helical path.

Thus, they will cover a distance equal to pitch, till they are refocussed.

Let, $d=$ pitch,
$v=$ speed of electron along $x-$ axis
$K=$ Kinetic energy of electron

$\therefore d=v_x\times T$

$d=v\left(\frac{2\pi m}{eB}\right)........\left(1\right)$

Here,

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2e{V}_0}{m}}.....\left(2\right)$

From equations $\left(1\right)$ & $\left(2\right)$;

$d=\sqrt{\frac{2e{V}_0}{m}}\times \frac{2\pi m}{eB}$

$d=\sqrt{{\left(\frac{2\pi m}{eB}\right)}^2\frac{2e{V}_0}{m}}$

$\therefore d=\sqrt{\frac{8{\pi }^{2}m{V}_0}{e{B}^2}}$

Hence, option (b) is the correct answer.

Why this question ? Tip: When electrons are accelerated through a potential difference of $V_0\text{Volt}$, they will gain K.E equal to loss in P.E, i.e., $\frac{1}{2}m{v}^2=e{V}_0$