Answer
(a) The diameter of the orbit of the 14C+ ions is $29~cm$
(b) The ratio of the frequency of revolution of the 12C+ ions to the frequency of revolution of the 14C+ ions is $\frac{7}{6}$
Work Step by Step
(a) $F = \frac{mv^2}{r}$
$F = qvB$
We can equate the two expressions for $F$ to find an expression for the speed $v$:
$\frac{mv^2}{r} = qvB$
$v = \frac{q~B~r}{m}$
We can write an expression for the velocity of the 12C+ ions:
$v_{12} = \frac{q~B~r_{12}}{m_{12}}$
We can write an expression for the velocity of the 14C+ ions:
$v_{14} = \frac{q~B~r_{14}}{m_{14}}$
Since the velocities are equal, we can equate the two expressions for velocity:
$\frac{q~B~r_{12}}{m_{12}} = \frac{q~B~r_{14}}{m_{14}}$
$r_{14} = (\frac{m_{14}}{m_{12}})~(r_{12})$
$r_{14} = (\frac{14~u}{12~u})~(r_{12})$
$r_{14} = (\frac{7}{6})~(r_{12})$
The radius of the 14C+ ions' orbit is $\frac{7}{6}$ times the radius of the 12C+ ions' orbit. We can find the diameter of the 14C+ ions' orbit:
$d = \frac{7}{6}~(25~cm) = 29~cm$
(b) We can find the ratio of the frequencies:
$\frac{f_{12}}{f_{14}} = \frac{\frac{v_{12}}{2\pi~r_{12}}}{\frac{v_{14}}{2\pi~r_{14}}}$
$\frac{f_{12}}{f_{14}} = \frac{r_{14}}{r_{12}}$
$\frac{f_{12}}{f_{14}} = \frac{7}{6}$
The ratio of the frequency of revolution of the 12C+ ions to the frequency of revolution of the 14C+ ions is $\frac{7}{6}$.
