A proton, a deuteron and an alpha particle having the same kinetic energy are allowed to pass through a uniform magnetic field perpendicular to their direction of motion. The ratio of the radii of their respective circular paths is

$1: \sqrt{2}:1$

The correct answer is Option (4) → $1: \sqrt{2}:1$

$ r = \frac{p}{qB} = \frac{\sqrt{2mK}}{qB} \Rightarrow r \propto \frac{\sqrt{m}}{q} $

$ \text{For proton: } m_p,\ q=e \Rightarrow r_p \propto \frac{\sqrt{m_p}}{e} = 1 $

$ \text{For deuteron: } m_d = 2m_p,\ q=e \Rightarrow r_d \propto \frac{\sqrt{2m_p}}{e} = \sqrt{2} $

$ \text{For alpha: } m_\alpha = 4m_p,\ q=2e \Rightarrow r_\alpha \propto \frac{\sqrt{4m_p}}{2e} = 1 $

$ therefore\; r_p : r_d : r_\alpha = 1 : \sqrt{2} : 1 $