A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $$r_d$$ and $$r_\alpha$$ be their respective radii of circular path. The value of $$\frac{r_d}{r_\alpha}$$ is equal to:

For a charged particle moving in a magnetic field, the radius of the circular path is $$r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$$, where $$K$$ is the kinetic energy.

For the deuteron: mass $$m_d = 2u$$, charge $$q_d = e$$. For the alpha particle: mass $$m_\alpha = 4u$$, charge $$q_\alpha = 2e$$.

Both particles have equal kinetic energy $$K$$ and enter the same magnetic field $$B$$. Taking the ratio:

$$\frac{r_d}{r_\alpha} = \frac{\sqrt{2m_d K}/(q_d B)}{\sqrt{2m_\alpha K}/(q_\alpha B)} = \frac{q_\alpha}{q_d} \cdot \sqrt{\frac{m_d}{m_\alpha}} = \frac{2e}{e} \cdot \sqrt{\frac{2u}{4u}} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$$