A proton and an alpha particle are projected with equal velocities into a magnetic field perpendicular to their velocity. What is the ratio of the radius of the path of the proton to that of the alpha particle?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 : 2
A charged particle moving perpendicular to a uniform magnetic field goes round a circle. The magnetic force \( qvB \) supplies the centripetal force \( mv^2/r \), so the radius is \( r=mv/qB \). For the same \( v \) and \( B \), the radius is proportional to \( m/q \).
Taking the proton as mass \( m \), charge \( e \), the alpha particle (two protons and two neutrons) has mass \( 4m \) and charge \( 2e \):
\( \begin{aligned} \frac{r_p}{r_\alpha} &= \frac{m/e}{4m/2e} \\ &= \frac{m}{e}\times\frac{2e}{4m} \\ &= \frac{1}{2} \end{aligned} \)
Therefore, \( r_p:r_\alpha=1:2 \).