A proton, a deuteron and an $$\alpha$$ particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is ______ and their speed is ______ in the ratio.

All three particles — proton, deuteron, and $$\alpha$$ particle — move with the same momentum $$p$$ in a uniform magnetic field $$B$$.

The magnetic force on a charged particle is $$F = qvB$$. Since $$p = mv$$, we have $$v = \frac{p}{m}$$, so $$F = \frac{qpB}{m}$$. The force ratio depends on $$\frac{q}{m}$$ for each particle.

For a proton: $$q_p = e$$, $$m_p = m$$, so $$\frac{q_p}{m_p} = \frac{e}{m}$$.

For a deuteron: $$q_d = e$$, $$m_d = 2m$$, so $$\frac{q_d}{m_d} = \frac{e}{2m}$$.

For an $$\alpha$$ particle: $$q_\alpha = 2e$$, $$m_\alpha = 4m$$, so $$\frac{q_\alpha}{m_\alpha} = \frac{2e}{4m} = \frac{e}{2m}$$.

The ratio of magnetic forces is $$F_p : F_d : F_\alpha = \frac{e}{m} : \frac{e}{2m} : \frac{e}{2m} = 2 : 1 : 1$$.

The speed of each particle is $$v = \frac{p}{m}$$, so the speed ratio is $$v_p : v_d : v_\alpha = \frac{1}{m} : \frac{1}{2m} : \frac{1}{4m} = 4 : 2 : 1$$.

Therefore the ratio of forces is $$2 : 1 : 1$$ and the speeds are in the ratio $$4 : 2 : 1$$, which corresponds to Option (1).