A proton accelerated from rest through a potential difference V enters a uniform magnetic field perpendicular to its velocity and moves on a circle…

A proton accelerated from rest through a potential difference $V$ enters a uniform magnetic field perpendicular to its velocity and moves on a circle of radius $r_0$. In List-I a different particle (alpha particle: mass $4m_p$, charge $2e$; deuteron: mass $2m_p$, charge $e$) enters the same field under the stated condition. Match the ratio of its orbit radius to $r_0$ with List-II.

List-I: (P) alpha particle accelerated through the same $V$; (Q) deuteron with the same speed as the proton; (R) alpha particle with the same momentum as the proton; (S) alpha particle with the same kinetic energy as the proton.

List-II: (1) $1/2$; (2) $1$; (3) $\sqrt2$; (4) $2$; (5) $2\sqrt2$.