Question

An electron and a proton, both having the same kinetic energy, enter a region of uniform magnetic field, in a plane perpendicular to the field. If their masses are denoted by $m_{e}$ and $m_{p}$ respectively, then the ratio of the radii (electron to proton) of their circular orbits is

• A. $\sqrt{\frac{m_{p}}{m_{e}}}$
• B. $\sqrt{\frac{m_{e}}{m_{p}}}$
• C. $\frac{m_{e}}{m_{p}}$
• D. $ 1 $

Answer 1

Answer: (B)

Radius of circular path in a uniform magnetic field $r=\frac{m v}{q B}$ Kinetic energy of the charged particle
$K=\frac{1}{2} m v^{2} $
$v^{2}=\frac{2 K}{m}$
$\Rightarrow v=\sqrt{\frac{2 K}{m}} $
$\therefore r=\frac{m}{q B} \sqrt{\frac{2 K}{m}}$
$r=\frac{\sqrt{2 K m}}{q B}$
For the same value of $K$ and $B$
$r \propto \frac{\sqrt{m}}{q}$
$\therefore \frac{r_{e}}{r_{p}}=\frac{\sqrt{m_{e}}}{\sqrt{m_{p}}} \times \frac{q_{p}}{q_{e}} $
$\frac{r_{e}}{r_{p}}=\frac{\sqrt{m_{e}}}{\sqrt{m_{p}}} \times \frac{e}{e}=\sqrt{\frac{m_{e}}{m_{p}}}$