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(C) The maximum kinetic energy $KE_{\max}$ of a particle in a cyclotron is given by the formula $KE_{\max} = \frac{q^2 B^2 R^2}{2m}$, where $q$ is the

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Deuteron

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(C) The maximum kinetic energy $KE_{\max}$ of a particle in a cyclotron is given by the formula $KE_{\max} = \frac{q^2 B^2 R^2}{2m}$, where $q$ is the charge, $B$ is the magnetic field, $R$ is the radius of the dees, and $m$ is the mass of the particle.Since $B$ and $R$ are constant for a given cyclotron, $KE_{\max} \propto \frac{q^2}{m}$.For a proton $(p)$: $q = e, m = m_p \implies \frac{q^2}{m} = \frac{e^2}{m_p}$.For a deuteron $(d)$: $q = e, m = 2m_p \implies \frac{q^2}{m} = \frac{e^2}{2m_p}$.For an alpha particle $(\alpha)$: $q = 2e, m = 4m_p \implies \frac{q^2}{m} = \frac{4e^2}{4m_p} = \frac{e^2}{m_p}$.Comparing the values, the ratio $\frac{q^2}{m}$ is minimum for the deuteron. Therefore, the minimum kinetic energy gained is for the deuteron.

In a cyclotron experiment, if we assume that the maximum radius attained by any charged particle is equal to the radius of the dees, then the minimum kinetic energy gained by the particles will be for

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