1. Radius of circular path in a magnetic field

For a charged particle entering perpendicular to a uniform magnetic field,

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

Since kinetic energy is given same for both particles,

$K = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2K}{m}}$

Substitute into the radius formula:

$r = \frac{m}{qB}\sqrt{\frac{2K}{m}} = \frac{\sqrt{2Km}}{qB}$

So,

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

2. Apply to proton and deuteron

Both have the same charge:

$q_p = q_d = e$

Masses are:

• Proton: $m_p = 1\,u$
• Deuteron: $m_d = 2\,u$

Therefore,

$\frac{r_d}{r_p} = \sqrt{\frac{m_d}{m_p}} = \sqrt{\frac{2u}{1u}} = \sqrt{2}$

Thus,

$r_d : r_p = \sqrt{2} : 1$

3. Check options

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

So the correct option is C.