According to the sources, a charged particle moving perpendicular to a uniform magnetic field follows a circular path with a radius $r$ defined by the formula $r = \frac{mv}{qB}$ . Kinetic energy ($K$) is related to the mass ($m$) and velocity ($v$) of a particle such that its momentum $mv$ can be expressed as $\sqrt{2mK}$ . Substituting this into the radius equation yields $r = \frac{\sqrt{2mK}}{qB}$ . An $\alpha$-particle possesses a mass approximately four times that of a proton ($m_\alpha \approx 4m_p$) and a charge twice that of a proton ($q_\alpha = 2q_p$) . Given that the radius ($R$) and magnetic field ($B$) are identical for both particles, the relationship $\frac{m_p K_p}{q_p^2} = \frac{m_\alpha K_\alpha}{q_\alpha^2}$ must be satisfied . Substituting the known ratios and the proton's kinetic energy ($K_p = 1 \text{ MeV}$): $\frac{m_p \cdot 1}{q_p^2} = \frac{4m_p \cdot K_\alpha}{(2q_p)^2}$, which simplifies to $1 = \frac{4 K_\alpha}{4}$. Thus, the energy of the $\alpha$-particle is $1 \text{ MeV}$.