A proton, a deuteron and an $$\alpha$$-particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is :

Given: A proton, a deuteron, and an $$\alpha$$-particle enter a uniform magnetic field with the same kinetic energy, perpendicular to the field.

Recall the formula for the radius of circular motion in a magnetic field: When a charged particle moves in a magnetic field, the radius of its circular path is:

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

Express radius in terms of kinetic energy: Since $$KE = \frac{1}{2}mv^2$$, we get $$v = \sqrt{\frac{2 \cdot KE}{m}}$$, so $$mv = \sqrt{2m \cdot KE}$$.

Therefore:

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

Since $$KE$$ and $$B$$ are the same for all three particles:

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

Identify mass and charge of each particle: Let $$m_p$$ be the proton mass and $$e$$ be the elementary charge.

Proton: $$m = m_p$$, $$q = e$$

Deuteron: $$m = 2m_p$$, $$q = e$$

$$\alpha$$-particle: $$m = 4m_p$$, $$q = 2e$$

Calculate the ratio of radii: $$r_p : r_d : r_\alpha = \frac{\sqrt{m_p}}{e} : \frac{\sqrt{2m_p}}{e} : \frac{\sqrt{4m_p}}{2e}$$

$$= \frac{\sqrt{m_p}}{e} : \frac{\sqrt{2m_p}}{e} : \frac{2\sqrt{m_p}}{2e}$$

$$= \frac{\sqrt{m_p}}{e} : \frac{\sqrt{2m_p}}{e} : \frac{\sqrt{m_p}}{e}$$

Cancelling the common factor $$\frac{\sqrt{m_p}}{e}$$:

$$= 1 : \sqrt{2} : 1$$

The ratio of the radii of their circular paths is $$1 : \sqrt{2} : 1$$.

The correct answer is Option B.