## College Physics (4th Edition)

The atomic mass of the rarer isotope is $13.0~u$
$F = \frac{mv^2}{r}$ $F = qvB$ 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 $13.0~u$