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(A) The radius $R$ of a circular path of a charged particle with mass $m$,charge $q$,and kinetic energy $K$ in a uniform magnetic field $B$ is given b

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(A) The radius $R$ of a circular path of a charged particle with mass $m$,charge $q$,and kinetic energy $K$ in a uniform magnetic field $B$ is given by:$R = \frac{\sqrt{2mK}}{Bq}$For a proton: $m_p = m$,$q_p = e$,$K_p = 1\, MeV$.$R_p = \frac{\sqrt{2m(1)}}{Be}$For an $\alpha$-particle: $m_{\alpha} = 4m$,$q_{\alpha} = 2e$.$R_{\alpha} = \frac{\sqrt{2(4m)K_{\alpha}}}{B(2e)} = \frac{\sqrt{8mK_{\alpha}}}{2Be} = \frac{\sqrt{2mK_{\alpha}}}{Be}$Since the radii are equal $(R_p = R_{\alpha})$:$\frac{\sqrt{2m(1)}}{Be} = \frac{\sqrt{2mK_{\alpha}}}{Be}$$1 = K_{\alpha}$Therefore,the kinetic energy of the $\alpha$-particle is $1\, MeV$.

$A$ proton carrying $1\, MeV$ kinetic energy is moving in a circular path of radius $R$ in a uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of the same radius in the same field? ........$MeV$

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