Answer
The atomic mass of the rarer isotope is $13.0~u$
Work Step by Step
$F = \frac{mv^2}{r}$
$F = qvB$
We can equate the two expressions for $F$ to find an expression for the speed $v$:
$\frac{mv^2}{r} = qvB$
$v = \frac{q~B~r}{m}$
Since both ions have the same charge and they are accelerated through the same potential difference, then they have the same kinetic energy.
Each ion has a kinetic energy of $q~\Delta V$. We can find a general expression for the mass $m$:
$K = q~\Delta V$
$\frac{1}{2}mv^2 = q~\Delta V$
$\frac{1}{2}(m)(\frac{q~B~r}{m})^2 = q~\Delta V$
$m = \frac{q~B^2~r^2}{2~\Delta V}$
We can write an expression for $m_1$, the mass of the ion with an atomic mass of $12.00~u$:
$m_1 = \frac{q~B^2~r_1^2}{2~\Delta V}$
We can write an expression for $m_2$, the mass of the other ion:
$m_2 = \frac{q~B^2~r_2^2}{2~\Delta V}$
We can divide $m_2$ by $m_1$:
$\frac{m_2}{m_1} = \frac{\frac{q~B^2~r_2^2}{2~\Delta V}}{\frac{q~B^2~r_1^2}{2~\Delta V}}$
$m_2 = (\frac{r_2}{r_1})^2~m_1$
$m_2 = (\frac{15.6~cm}{15.0~cm})^2~(12.00~u)$
$m_2 = 13.0~u$
The atomic mass of the rarer isotope is $13.0~u$
