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

When a charged particle is accelerated by a potential difference \(V\) volts, then the radius of the trajectory of the path of the particle is given as,
\(r =\frac{\sqrt{2 K m}}{B q}=\frac{\sqrt{2 q V m}}{B q} \)
\( \Rightarrow K =\frac{1}{2} \frac{B^{2} q^{2} r^{2}}{m} \text { or } K \propto \frac{q^{2}}{m} ...(i)\)
For protons \(\left({ }_{1}^{1} \mathrm{H}\right)\), deuterons \(\left({ }_{1}^{2} \mathrm{H}\right)\) and \(\alpha\)-particles \(\left({ }_{2}^{4} \mathrm{He}\right)\),
\(q_{p}: q_{d}: q_{\alpha}=1: 1: 2 ...(ii)\)
\( m_{p}: m_{d}: m_{\alpha} =1: 2: 4 ...(iii)\)
From Eqs. (i), (ii) and (iii), we can write
\(\Rightarrow K_{p}: K_{d}: K_{\alpha}=1: \frac{1}{4}: 1\)
Thus, minimum kinetic energy is gained by deuterons.