Answer
See answers.
Work Step by Step
a. In the 2s shell of an atom, $\ell=0$, so there are two states, both with $ m_{\ell}=0$, and $m_s=\pm \frac{1}{2}$.
When the 2s states of N atoms form the 2s band, each of the atoms provides two states. The total number of electron states in the 2s band is 2N.
b. In the 2p shell of an atom, $\ell=1$, so there are three values of $m_{\ell}$, which are -1, 0, and +1. As before, for each of these values, $m_s=\pm \frac{1}{2}$. The total is 6 states.
When the 2p states of N atoms form the 2p band, each of the atoms provides six states. The total number of electron states in the 2p band is 6N.
c. In the 3p shell of an atom, $\ell=1$, so there are three values of $m_{\ell}$, which are -1, 0, and +1. As before, for each of these values, $m_s=\pm \frac{1}{2}$. The total is 6 states.
When the 3p states of N atoms form the 3p band, each of the atoms provides six states. The total number of electron states in the 3p band is 6N.
d. Generally, for a value of $\ell$, there are $2\ell+1$ values of $m_{\ell}$. As before, for each of these value, $m_s=\pm \frac{1}{2}$. The total is $2(2\ell + 1)$ states. This is stated on page 828, problem 54.
When the $2(2\ell + 1)$ states of N atoms come together, they form a band. The total number of electron states in the band is $2N(2\ell + 1)$.
As a check, the general formula in part d matches the cases in parts a, b, and c.
