According to the sources, a charged particle moving perpendicular to a uniform magnetic field follows a circular path with a radius $r$ given by $r = \frac{mv}{qB}$ . Kinetic energy ($K$) is defined as $K = \frac{1}{2}mv^2$, which implies $mv = \sqrt{2mK}$ . Substituting this into the radius formula, we get $r = \frac{\sqrt{2mK}}{qB}$.

For a proton ($p$) and an \alpha particle ($\alpha$), the masses and charges are related as $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$ . Since the radii are equal ($r_p = r_\alpha$) and they move in the same field $B$, we have: $\frac{\sqrt{2m_p K_p}}{q_p B} = \frac{\sqrt{2m_\alpha K_\alpha}}{q_\alpha B}$ Squaring both sides and simplifying: $\frac{m_p K_p}{q_p^2} = \frac{m_\alpha K_\alpha}{q_\alpha^2}$ Substituting the values for the \alpha particle: $\frac{m_p \cdot 1\text{ MeV}}{q_p^2} = \frac{4m_p \cdot K_\alpha}{(2q_p)^2}$ $\frac{m_p}{q_p^2} = \frac{4m_p \cdot K_\alpha}{4q_p^2}$ Solving this yields $K_\alpha = 1\text{ MeV}$.