1. Speed gained after acceleration through potential difference $V$

When a charged particle is accelerated from rest through a potential difference $V$, the gained kinetic energy is:

$qV = \frac{1}{2}mv^2$

So,

$v = \sqrt{\frac{2qV}{m}}$

2. Radius of circular path in magnetic field

If the particle enters a uniform magnetic field $B$ perpendicular to its velocity, then the magnetic force provides centripetal force:

$qvB = \frac{mv^2}{r}$

Thus,

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

Substitute $v = \sqrt{\frac{2qV}{m}}$:

$r = \frac{m}{qB}\sqrt{\frac{2qV}{m}}$

Simplifying,

$r = \frac{1}{B}\sqrt{\frac{2mV}{q}}$

Hence,

$r \propto \sqrt{\frac{m}{q}}$

3. Apply to proton and $\alpha$-particle

Given:

• Mass ratio: $m_p : m_\alpha = 1:4$
• Charge ratio: $q_p : q_\alpha = 1:2$

Therefore,

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

Substitute the ratios:

$\frac{r_p}{r_\alpha} = \sqrt{\frac{1/1}{4/2}} = \sqrt{\frac{1}{2}}$

So,

$r_p : r_\alpha = 1 : \sqrt{2}$

4. Check options

• A: $1: \sqrt{3}$ ❌
• B: $1:3$ ❌
• C: $1: \sqrt{2}$ ✅
• D: $1:2$ ❌

Therefore, the correct option is C.