A proton and a deutron ($$q = +e, m = 2.0u$$) having same kinetic energies enter a region of uniform magnetic field $$\vec{B}$$, moving perpendicular to $$\vec{B}$$. The ratio of the radius of deutron path to the radius of the proton path is:

A proton and a deuteron with the same kinetic energy enter a uniform magnetic field perpendicular to $$\vec{B}$$. We need the ratio of their radii of circular paths.

When a charged particle moves perpendicular to a magnetic field, it follows a circular path with radius:

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

Since $$KE = \frac{1}{2}mv^2$$, we get $$mv = \sqrt{2m \cdot KE}$$. Therefore:

$$ r = \frac{\sqrt{2m \cdot KE}}{qB} $$

For the proton: mass = $$m_p$$, charge = $$e$$, so $$r_p = \frac{\sqrt{2m_p \cdot KE}}{eB}$$

For the deuteron: mass = $$2m_p$$ (approximately twice the proton mass), charge = $$e$$ (same charge as proton), so:

$$ r_d = \frac{\sqrt{2 \cdot 2m_p \cdot KE}}{eB} = \frac{\sqrt{2} \cdot \sqrt{2m_p \cdot KE}}{eB} = \sqrt{2} \cdot r_p $$

The ratio is:

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

The correct answer is Option (1): $$\sqrt{2} : 1$$.