An $$\alpha$$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

We have an $$\alpha$$ particle (mass $$m_\alpha = 4m_p$$, charge $$q_\alpha = 2e$$) and a proton (mass $$m_p$$, charge $$q_p = e$$), both accelerated from rest through the same potential difference $$V$$.

When a charged particle is accelerated through a potential difference $$V$$, the kinetic energy gained is $$K = qV$$. Since $$K = \frac{p^2}{2m}$$, the momentum is $$p = \sqrt{2mK} = \sqrt{2mqV}$$.

For the $$\alpha$$ particle: $$p_\alpha = \sqrt{2 \cdot 4m_p \cdot 2eV} = \sqrt{16\,m_p\,eV}$$.

For the proton: $$p_p = \sqrt{2 \cdot m_p \cdot eV} = \sqrt{2\,m_p\,eV}$$.

The ratio is: $$\frac{p_\alpha}{p_p} = \frac{\sqrt{16\,m_p\,eV}}{\sqrt{2\,m_p\,eV}} = \sqrt{\frac{16}{2}} = \sqrt{8} = 2\sqrt{2}$$

Hence the ratio of momenta is $$2\sqrt{2} : 1$$.

Hence, the correct answer is Option 2.