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.

Solution to 2008 Problem 6