Explanation

(b) For a moving charge in a perpendicular magnetic field,
$\frac{mv^{2}}{r} = Bqv$
$\Rightarrow \ \ r = \frac{mv}{Bq} = \frac{p}{Bq}$
or $\frac{r_{p}}{r_{d}} = \frac{p_{p}}{p_{d}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
(as $q$ is same for both)
Also, momentum $p = \sqrt{2mE}$
or $\frac{p_{p}}{p_{d}} = \sqrt{\frac{m_{p}}{m_{d}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs. (i) and (ii), we have,
$\frac{r_{p}}{r_{d}} = \sqrt{\frac{m_{p}}{m_{d}}} = \ \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$