(2)

As, $r=\frac{mv}{qB}=\frac{p}{qB}$

For given $p$ and $B,$ $r\propto \frac{1}{q}$ $\Rightarrow \frac{r_H}{r_{\alpha }}=\frac{q_{\alpha }}{q_H}=\frac{2}{1}$

(2)

The kinetic energy acquired by a charged particle in a uniform magnetic field B is

                $K=\frac{q^2{B}^2{R}^2}{2m}$    $(\text{as }R=\frac{{\displaystyle mv}}{{\displaystyle qB}}=\frac{{\displaystyle \sqrt{2mK}}}{{\displaystyle qB}})$

where $q$ and $m$ are the charge and mass of the particle and $R$ is the radius of circular orbit.

$\therefore$   The kinetic energy acquired by proton is $K_p=\frac{q_p^2{B}^2{R}_p^2}{2m_p}$

and that by the alpha particle is $K_{\alpha }=\frac{q_{\alpha }^2{B}^2{R}_{\alpha }^2}{2m_{\alpha }}$

Thus, $\frac{K_{\alpha }}{K_p}={\left(\frac{{\displaystyle q_{\alpha }}}{{\displaystyle q_p}}\right)}^2\left(\frac{{\displaystyle m_p}}{{\displaystyle m_{\alpha }}}\right){\left(\frac{{\displaystyle R_{\alpha }}}{{\displaystyle R_p}}\right)}^2$

or   $K_{\alpha }=K_p{\left(\frac{{\displaystyle q_{\alpha }}}{{\displaystyle q_p}}\right)}^2\left(\frac{{\displaystyle m_p}}{{\displaystyle m_{\alpha }}}\right){\left(\frac{{\displaystyle R_{\alpha }}}{{\displaystyle R_p}}\right)}^2$

Here, $K_p=1$ MeV, $\frac{q_{\alpha }}{q_p}=2$, $\frac{m_p}{m_{\alpha }}=\frac{1}{4}$ and $\frac{R_{\alpha }}{R_p}=1$

$\therefore$ $K_{\alpha }=\left(1\text{ MeV}\right){\left(2\right)}^2\left(\frac{1}{4}\right){\left(1\right)}^2=1$ $\mathrm{MeV}$