1. Radius of circular motion in a magnetic field

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

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

Since all particles have the same kinetic energy $K$,

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

Substitute into the radius formula:

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

Thus,

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

when $K$ and $B$ are the same.

2. Now compare the three particles

(i) Proton

• Mass $= m_p$
• Charge $= e$

So,

$r_p \propto \frac{\sqrt{m_p}}{e}$

(ii) Deuteron

A deuteron has:

• Mass $\approx 2m_p$
• Charge $= e$

So,

$r_d \propto \frac{\sqrt{2m_p}}{e} = \sqrt{2}\,\frac{\sqrt{m_p}}{e}$

Hence,

$r_d = \sqrt{2}\,r_p$

(iii) $\alpha$-particle

An $\alpha$-particle has:

• Mass $\approx 4m_p$
• Charge $= 2e$

So,

$r_\alpha \propto \frac{\sqrt{4m_p}}{2e} = \frac{2\sqrt{m_p}}{2e} = \frac{\sqrt{m_p}}{e}$

Hence,

$r_\alpha = r_p$

3. Required ratio

Therefore,

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

4. Option check

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

So the correct option is D.