## Solution to 2008 Problem 6

 There are $2$ $n = 1$ orbitals: \begin{enumerate}\item \text{$l = 0$, $m_l = 0$, $m_s = 1/2$}\item \text{$l = 0$, $m_l = 0$, $m_s = -1/2$}\end{enumerate}There are $8$ $n = 2$ orbitals:\begin{enumerate}\item \text{$l = 0$, $m_l = 0$, $m_s = 1/2$}\item \text{$l = 0$, $m_l = 0$, $m_s = -1/2$}\item \text{$l = 1$, $m_l = 1$, $m_s = 1/2$}\item \text{$l = 1$, $m_l = 1$, $m_s = -1/2$}\item \text{$l = 1$, $m_l = 0$, $m_s = 1/2$}\item \text{$l = 1$, $m_l = 0$, $m_s = -1/2$}\item \text{$l = 1$, $m_l = -1$, $m_s = 1/2$}\item \text{$l = 1$, $m_l = -1$, $m_s = -1/2$}\end{enumerate}This makes a total of $\boxed{10}$. Therefore, answer (E) is correct.