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(A) The radius of a circular orbit for a charged particle in a uniform magnetic field $B$ is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, whe

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(A) The radius of a circular orbit for a charged particle in a uniform magnetic field $B$ is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, where $K$ is the kinetic energy.Rearranging for kinetic energy $K$, we get $K = \frac{q^2 B^2 R^2}{2m}$.For the proton $(p)$ and alpha particle $(\alpha)$, since $B$ and $R$ are equal, the ratio of their kinetic energies is:$\frac{K_{\alpha}}{K_{p}} = \left(\frac{q_{\alpha}}{q_{p}}ight)^2 \left(\frac{m_{p}}{m_{\alpha}}ight)$.Given $q_{\alpha} = 2q_{p}$ and $m_{\alpha} = 4m_{p}$, we have:$\frac{K_{\alpha}}{K_{p}} = (2)^2 	imes \left(\frac{1}{4}ight) = 4 	imes \frac{1}{4} = 1$.Therefore, $K_{\alpha} = K_{p} = 1 \, MeV$.

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