To solve for the ratio of the radii of deutron and proton paths in a magnetic field, we use the formula for the radius $$r$$ of the circular path of a charged particle moving perpendicular to a uniform magnetic field:

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

where:

• $$m$$ is the mass of the particle,
• $$v$$ is the velocity of the particle,
• $$q$$ is the charge of the particle, and
• $$B$$ is the magnetic field strength.

The proton and the deutron are given to have the same kinetic energy. The kinetic energy $$K$$ of a particle is given by:

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

From the kinetic energy, we can express the velocity as:

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

Since both particles have the same kinetic energy and the charge of the deutron is the same as the charge of the proton $$q = e$$, but the mass of the deutron is twice that of the proton ($$m_d = 2m_p$$), substituting the expression for $$v$$ in the radius formula, we get:

For the deutron:

$$r_d = \frac{m_d\sqrt{2K/m_d}}{eB} = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_d}$$

For the proton ($$m_p = m$$):

$$r_p = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_p}$$

The ratio of the radius of the deutron path $$r_d$$ to the radius of the proton path $$r_p$$ is therefore:

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

So, the correct answer is:

Option C: $$\sqrt{2}: 1$$.