A proton and an $$\alpha$$-particle, having kinetic energies $$K_p$$ and $$K_\alpha$$, respectively, enter into a magnetic field at right angles. The ratio of the radii of the trajectory of proton to that of $$\alpha$$-particle is 2 : 1. The ratio of $$K_p : K_\alpha$$ is:

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

For a proton: $$r_p = \frac{\sqrt{2m_p K_p}}{eB}$$. For an $$\alpha$$-particle (mass $$4m_p$$, charge $$2e$$): $$r_\alpha = \frac{\sqrt{2 \cdot 4m_p \cdot K_\alpha}}{2eB} = \frac{\sqrt{8m_p K_\alpha}}{2eB}$$.

The ratio is: $$\frac{r_p}{r_\alpha} = \frac{\sqrt{2m_p K_p}}{eB} \cdot \frac{2eB}{\sqrt{8m_p K_\alpha}} = \frac{2\sqrt{K_p}}{2\sqrt{K_\alpha}} = \sqrt{\frac{K_p}{K_\alpha}}$$.

Given $$\frac{r_p}{r_\alpha} = 2$$, we get $$\frac{K_p}{K_\alpha} = 4$$, so $$K_p : K_\alpha = 4 : 1$$.