A proton and an alpha particle, initially at rest, are accelerated...

The kinetic energy of a particle accelerated through a potential difference $V$ is $K = qV$.

The magnetic moment of a charged particle moving in a circular orbit is given by $M = IA$.

The time period of revolution is $T = \dfrac{2\pi m}{qB}$, so the equivalent current is $I = \dfrac{q}{T} = \dfrac{q^2 B}{2\pi m}$.

The radius of the circular path is $r = \dfrac{\sqrt{2mK}}{qB}$, so the area of the orbit is $A = \pi r^2 = \pi \left(\dfrac{\sqrt{2mK}}{qB}\right)^2 = \dfrac{2\pi m K}{q^2 B^2}$.

Therefore, the magnetic moment is $M = I A = \left(\dfrac{q^2 B}{2\pi m}\right) \left(\dfrac{2\pi m K}{q^2 B^2}\right) = \dfrac{K}{B} = \dfrac{qV}{B}$.

For a proton, the charge is $q$, so its magnetic moment is $M_p = \dfrac{qV}{B}$.

For an alpha particle, the charge is $2q$, so its magnetic moment is $M_{\alpha} = \dfrac{2qV}{B}$.

The ratio of the magnetic moment of the alpha particle to that of the proton is $\dfrac{M_{\alpha}}{M_p} = \dfrac{2qV/B}{qV/B} = 2$.

Thus, the ratio is $2 : 1$.

Answer: $2 : 1$