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

\(K=\frac{q^{2} B^{2} R^{2}}{2 m}\left(\text { as } R=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B}\right)\)

where \(q\) and \(m\) are the charge and mass of the partic and \(R\) is the radius of circular orbit

\(\therefore \) The kinetic energy acquired by proton is

\(K_{p}=\frac{q_{p}^{2} B^{2} R_{p}^{2}}{2 m_{p}}\)

and that by the alpha particle is

\(K_{\alpha}=\frac{q_{\alpha}^{2} B^{2} R_{\alpha}^{2}}{2 m_{\alpha}}\)

Thus, \(\frac{K_{\alpha}}{K_{p}}=\left(\frac{q_{\alpha}}{q_{p}}\right)^{2}\left(\frac{m_{p}}{m_{\alpha}}\right)\left(\frac{R_{\alpha}}{R_{p}}\right)^{2}\)

or \(\quad K_{\alpha}=K_{p}\left(\frac{q_{\alpha}}{q_{p}}\right)^{2}\left(\frac{m_{p}}{m_{\alpha}}\right)\left(\frac{R_{\alpha}}{R_{p}}\right)^{2}\)

Here, \(K_{p}=1\) \(MeV\) \(, \frac{q_{\alpha}}{q_{p}}=2, \frac{m_{p}}{m_{\alpha}}=\frac{1}{4}\)

and \(\frac{R_{\alpha}}{R_{p}}=1\)

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