Correct Option: (B) 1/2

Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

When a charged particle enters a uniform magnetic field $\vec{B}$ perpendicular to its velocity $\vec{v}$, it experiences a magnetic Lorentz force that provides the necessary centripetal force for circular motion. The radius $r$ of the path is given by:

$$r = \frac{mv}{qB}$$

Given: Velocity $v$ and magnetic field $B$ are constant. Therefore, $r \propto \frac{m}{q}$.

• For a proton ($p$): Mass = $m$, Charge = $e$
• For an $\alpha$-particle ($\alpha$): Mass = $4m$, Charge = $2e$

Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]

The ratio of the radii of the paths of the proton ($r_p$) and the $\alpha$-particle ($r_\alpha$) is:

$$\frac{r_p}{r_\alpha} = \frac{m_p / q_p}{m_\alpha / q_\alpha}$$

Substituting the values:

$$\frac{r_p}{r_\alpha} = \frac{m / e}{4m / 2e}$$ $$\frac{r_p}{r_\alpha} = \frac{m}{e} \times \frac{2e}{4m}$$ $$\frac{r_p}{r_\alpha} = \frac{2}{4} = \frac{1}{2}$$

Step 3 (Final Concluded Answer): [0.25 Mark]

The ratio of the radii of their paths is 1/2. Thus, option (B) is correct.

💡 EL10 Concept Insight: Think of the radius as a balance between "momentum" (which wants to keep the particle going straight) and "magnetic pull" (which tries to turn it). Since the alpha particle is four times heavier but only twice as strong in charge, it is harder to turn, resulting in a larger circle—making the proton's path exactly half the size of the alpha's.