A deuteron and a proton moving with equal kinetic energy enter into a uniform magnetic field at right angle to the field. If $$r_d$$ and $$r_p$$ are the radii of their circular paths respectively, then the ratio $$\frac{r_d}{r_p}$$ will be $$\sqrt{x} : 1$$ where $$x$$ is ______

Recall the formula for radius of circular motion in a magnetic field.

A charged particle moving with kinetic energy $$K$$ in a magnetic field $$B$$ follows a circular path of radius:

$$r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$$

(since $$K = \frac{1}{2}mv^2 \Rightarrow mv = \sqrt{2mK}$$)

Write the radii for deuteron and proton.

For the proton: mass $$m_p$$, charge $$q_p = e$$

$$r_p = \frac{\sqrt{2m_pK}}{eB}$$

For the deuteron: mass $$m_d = 2m_p$$, charge $$q_d = e$$

$$r_d = \frac{\sqrt{2 \times 2m_p \times K}}{eB} = \frac{\sqrt{2} \cdot \sqrt{2m_pK}}{eB}$$

Find the ratio.

$$\frac{r_d}{r_p} = \frac{\sqrt{2} \cdot \sqrt{2m_pK}}{eB} \times \frac{eB}{\sqrt{2m_pK}} = \sqrt{2}$$

So $$\frac{r_d}{r_p} = \sqrt{2} : 1$$.

Comparing with $$\sqrt{x} : 1$$, we get $$x = 2$$.

The answer is $$\boxed{2}$$.