A proton, a deutron and an$\alpha -$ particle whose kinetic energies are same enter at right angles to a uniform magnetic field. Then the ratio of the radii of their circular paths is

Charge of proton = q

Mass of Deuteron $=2{\text{m}}_{\text{p}}$  

Charge of Deuteron = q

Mass of Particle $=4{\text{m}}_{\text{p}}$

Charge of Particle = 2q

$\text{P}=\text{mv}=\text{RqB}=\sqrt{2\text{mK}}$

$\Rightarrow \text{R}=\frac{\sqrt{\text{2mK}}}{\text{qB}}\left(\text{R}\propto \frac{\sqrt{\text{m}}}{\text{q}}\right)$

$\Rightarrow {\text{R}}_{\text{p}}=\frac{\sqrt{\text{2}\times {\text{m}}_{\text{P}}\text{K}}}{\text{qB}},$

${\text{R}}_{\text{d}}=\frac{\sqrt{\text{2}\times 2{\text{m}}_{\text{P}}\text{K}}}{\text{qB}},$

${\text{R}}_{\alpha }=\frac{\sqrt{\text{2}\times 4{\text{m}}_{\text{P}}\text{K}}}{\text{2qB}}$

$\therefore {\text{R}}_{\text{p}}:{\text{R}}_{\text{d}}:{\text{R}}_{\alpha }=\sqrt{2}:2:\sqrt{2}$

$=1:\sqrt{2}:1$