Question

A proton (mass mp), a deuteron (m = 2mp, Q = e), and an alpha particle (m = 4mp, Q = 2e), are accelerated by the same potential difference V and then enter a uniform magnetic field B where they move in circular paths perpendicular to B. Determine the radius of the paths for the deuteron and alpha particle in terms of that for the proton.

R deuteron

R alpha

Answer 1

The radius comes from the centripetal force eq.

qvB = mv^2/R

R = mv/qB

The velocity comes from conservation of energy , with PE = qV (charge times potential difference);

(1/2)mv^2 = qV

v = SqRt[2qV/m]

So the radius of any one particle is, in terms of the potential diff.;

R = (m/qB)SqRt[2qV/m] = BSqRt(2V)SqRt[m/q]

Now since B and V are the same for all charges in the problem, the ratio of two radii is;

R1/R2 = SqRt[m1q2/m2q1]

Let R2 = R, m2 =m , q2= e, the proton;

R1 = SqRt[m1e/mq1]R

For Deuteron, m1 = 2m and q = e

R(D) = SqRt[2)R

For alpha, m1 =4m and q = 2e

R(alpha) = SqRt[2]R

So both the Deuteron and Alpha have the same radius, which is SqRt(2) times the Proton radius.