\(K=\frac{1}{2} m v^{2} \text { or } v=\sqrt{\frac{2 K}{m}}\)

Radius of the circular path of a charged particle in uniform magnetic field is given by

\(R=\frac{m v}{B q}=\frac{m}{B q} \sqrt{\frac{2 K}{m}}=\frac{\sqrt{2 m K}}{B q}\)

Mass of a proton, \(m_{p}=m\) 

Mass of an \(\alpha\) -particle, \(m_{\alpha}=4 m\) 

Charge of a proton, \(q_{p}=e\) 

Charge of an \(\alpha\) -particle, \(q_{\alpha}=2 e\)

\(\therefore \quad R_{p}=\frac{\sqrt{2 m_{p} K_{p}}}{B q_{p}}=\frac{\sqrt{2 m K_{p}}}{B e}\)

and \(R_{\alpha}=\frac{\sqrt{2 m_{\alpha} K_{\alpha}}}{B q_{\alpha}}=\frac{\sqrt{2(4 m) K_{\alpha}}}{B(2 e)}=\frac{\sqrt{2 m K_{\alpha}}}{B e}\)

\(\therefore \quad \frac{R_{p}}{R_{\alpha}}=\sqrt{\frac{K_{p}}{K_{\alpha}}}\)

As \(R_{p}=R_{\alpha}\) (given) \(\therefore K_{\alpha}=K_{p}=1\, \mathrm{MeV}\)