A proton, a deuteron and an alpha particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $r_p, r_d$ and $r_{\alpha }$ denote respectively the radii of trajectories of these particles, then:

The radius of the circular path is given by

$r=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB},\text{ where }K=\frac{1}{2}m{v}^2$

$\text{ Thus }r\propto \frac{\sqrt{m}}{q}\text{. }$Since K and B are the same for the three particles.

If m_p is the mass of a proton and $q_p$ its charge, then $m_d=2m_p\text{ and }{q}_d=q_p\text{ and }{m}_{\alpha }=4m_p$ and $q_{\alpha }=2q_p$

From these it follows that   $r_{\alpha }=r_p<r_d$.