Explanation

(b) Kinetic energy of proton = Kinetic energy of electron
$\frac{1}{2}\ m_{p}\ v_{p}^{2} = \frac{1}{2}m_{e}v_{e}^{2}$
$\Rightarrow \ \frac{m_{p}}{m_{e}} = \left( \frac{v_{e}}{v_{p}} \right)^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
If $B$ is the strength of the magnetic field and $m,\ v\ \ and\ q,\ $the mass, velocity and charge of the positive ion, then
$Bqv = \frac{mv^{2}}{r}$
As $q_{p} = q_{e}\ $ and $B$ is same for both electron and proton
$r = mv$
$\therefore\frac{r_{e}}{r_{p}} = \frac{m_{e}v_{e}}{m_{p}v_{p}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs (i) and (ii)
$\frac{r_{e}^{2}}{r_{p}^{2}} = \frac{m_{e}}{m_{p}}$
$\therefore\ \ r^{2} \propto m$
Mass of a proton is more than that of electron. Therefore, radius of proton will be more. Hence, the path of proton will be less curved.