Radius of circular path of a charged particle in uniform magnetic field, when it enters perpendicular direction of magnetic field.
$$
r=\frac{m v}{B q} \quad \text{...(i)}
$$
where, $m=$ mass, $v=$ velocity and $q=$ charge.
We know that, kinetic energy,
$$
K=\frac{1}{2} m v^{2}
$$
$\Rightarrow \quad v=\sqrt{\frac{2 K}{m}} \quad \text{...(ii)}$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
&r=\frac{m}{B q} \cdot \sqrt{\frac{2 K}{m}} \\
&r=\frac{\sqrt{2 K m}}{B q}
\end{aligned}
$$
For the same value of kinetic energy,
$$
\begin{aligned}
r & \propto \frac{\sqrt{m}}{q} \\
\Rightarrow \quad \frac{r_{p}}{r_{\alpha}} &=\sqrt{\frac{m_{p}}{m_{\alpha}}} \cdot \frac{q_{\alpha}}{q_{p}}=\sqrt{\frac{m_{p}}{4 m_{p}}} \cdot\left(\frac{2 q_{p}}{q_{p}}\right)=\frac{1}{1} \text { or } 1: 1
\end{aligned}
$$