1. Radius of circular motion in a magnetic field

For a charged particle entering perpendicular to a magnetic field, $r = \frac{mv}{qB}$

2. Use kinetic energy condition

Given both particles have equal kinetic energy: $K = \frac{1}{2}mv^2$ So, $v = \sqrt{\frac{2K}{m}}$

Substitute into the radius formula: $r = \frac{m}{qB}\sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$

Since $K$ and $B$ are same for both particles, $r \propto \frac{\sqrt{m}}{q}$

3. Properties of deuteron and alpha particle

• Deuteron: mass $m_d \approx 2u$, charge $q_d = e$
• Alpha particle: mass $m_\alpha \approx 4u$, charge $q_\alpha = 2e$

Thus, $\frac{r_d}{r_\alpha} = \frac{\sqrt{m_d}/q_d}{\sqrt{m_\alpha}/q_\alpha}$

Substitute values: $\frac{r_d}{r_\alpha} = \frac{\sqrt{2u}/e}{\sqrt{4u}/(2e)}$

Now simplify: $\frac{r_d}{r_\alpha} = \frac{\sqrt{2u}}{e} \cdot \frac{2e}{2\sqrt{u}}$

Since $\sqrt{4u} = 2\sqrt{u}$, $\frac{r_d}{r_\alpha} = \frac{\sqrt{2u}}{e} \cdot \frac{2e}{2\sqrt{u}} = \frac{\sqrt{2u}}{\sqrt{u}} = \sqrt{2}$

4. Check options

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

Therefore, the correct answer is: $\boxed{\sqrt{2}}$