A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is

A proton and an alpha particle with the same velocity enter a uniform magnetic field perpendicular to their direction of motion.

The radius of the circular motion in a magnetic field is given by $$r = \frac{mv}{qB}$$.

For the proton, the mass is $$m_p$$ and the charge is $$e$$, while the alpha particle has mass $$4m_p$$ and charge $$2e$$.

Substituting into the formula gives for the proton $$r_p = \frac{m_p v}{eB}$$ and for the alpha particle $$r_\alpha = \frac{4m_p v}{2eB} = \frac{2m_p v}{eB}$$.

Therefore, the ratio of their radii is $$\frac{r_\alpha}{r_p} = \frac{\frac{2m_p v}{eB}}{\frac{m_p v}{eB}} = 2$$, which implies $$r_\alpha : r_p = 2 : 1$$.

Hence, the correct answer is Option B.