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(C) When a charged particle is accelerated in a cyclotron,the radius $r$ of its path is given by $r = \frac{\sqrt{2Km}}{Bq}$.Squaring both sides,we ge

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deuterons

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(C) When a charged particle is accelerated in a cyclotron,the radius $r$ of its path is given by $r = \frac{\sqrt{2Km}}{Bq}$.Squaring both sides,we get $r^2 = \frac{2Km}{B^2q^2}$,which implies $K = \frac{B^2q^2r^2}{2m}$.Since $B$ and $r$ are the same for all particles at the exit,$K \propto \frac{q^2}{m}$.For protons $\left({ }_{1}^{1} H\right)$,deuterons $\left({ }_{1}^{2} H\right)$,and $\alpha$-particles $\left({ }_{2}^{4} He\right)$:Charge ratio: $q_p : q_d : q_{\alpha} = 1 : 1 : 2$.Mass ratio: $m_p : m_d : m_{\alpha} = 1 : 2 : 4$.Calculating the ratio of kinetic energies $K \propto \frac{q^2}{m}$:$K_p \propto \frac{1^2}{1} = 1$.$K_d \propto \frac{1^2}{2} = 0.5$.$K_{\alpha} \propto \frac{2^2}{4} = 1$.Comparing the values,the minimum kinetic energy is gained by deuterons.

$A$ cyclotron is used to accelerate protons $\left({ }_{1}^{1} H\right)$,deuterons $\left({ }_{1}^{2} H\right)$,and $\alpha$-particles $\left({ }_{2}^{4} He\right)$. While exiting under similar conditions,the minimum kinetic energy $(KE)$ is gained by:

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