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(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.For the same r

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(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.For the same radius $r$ and same magnetic field $B$,we have $r \propto \frac{\sqrt{mK}}{q}$.Squaring both sides,$r^2 \propto \frac{mK}{q^2}$,which implies $K \propto \frac{q^2}{m}$.Therefore,$\frac{K_p}{K_\alpha} = \left( \frac{q_p}{q_\alpha} ight)^2 	imes \frac{m_\alpha}{m_p}$.Given $q_p = e$,$q_\alpha = 2e$,$m_p = m$,and $m_\alpha = 4m$.Substituting these values: $\frac{1\, MeV}{K_\alpha} = \left( \frac{e}{2e} ight)^2 	imes \frac{4m}{m} = \left( \frac{1}{4} ight) 	imes 4 = 1$.Thus,$K_\alpha = 1\, MeV$.

$A$ proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha$-particle (mass = $4m$ and charge = $+2e$) so that it can revolve in the path of the same radius?

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