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(C) The radius of a charged particle moving in a circular path in a uniform magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$,where $m$ is the ma

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$8$

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(C) The radius of a charged particle moving in a circular path in a uniform magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$,where $m$ is the mass,$K$ is the kinetic energy,and $q$ is the charge of the particle.Since the radius $r$ and the magnetic field $B$ are the same for both particles,we have $q \propto \sqrt{mK}$,which implies $K \propto \frac{q^2}{m}$.For a proton,$q_p = e$ and $m_p = m$. For an alpha particle,$q_{\alpha} = 2e$ and $m_{\alpha} = 4m$.Taking the ratio of kinetic energies: $\frac{K_{\alpha}}{K_p} = \left( \frac{q_{\alpha}}{q_p} ight)^2 	imes \frac{m_p}{m_{\alpha}}$.Substituting the values: $\frac{K_{\alpha}}{8} = \left( \frac{2e}{e} ight)^2 	imes \frac{m}{4m} = 4 	imes \frac{1}{4} = 1$.Therefore,$K_{\alpha} = 8\, eV$.

$A$ proton of energy $8\, eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be.....$eV$

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