Question 42.7: Excited States of Hydrogen Suppose a hydrogen atom’s energy ......

Interpret and Anticipate
This example is much like Example 42.3, where we used the energy-level diagram in Figure 42.7 to find that the principal quantum number in this case is n = 3. Now we must find the possible values for \ell, m, and  m_s..

SOLVE

From Table 42.4, use \ell=0,1,2, \ldots(n-1) to find the values for the orbital quantum number when n = 3.

Consulting Table 42.4 again, use m=0, \pm 1, \pm 2, \ldots \pm \ell to find the values for the orbital magnetic quantum number for each possible value of \ell. Count the number of states for each unique value of \ell and m.

For each state, the electron can be either spin up (m_s = 1/2) or spin down (m_s = -1/2). Add up the number of unique states specified by \ell and m, and then multiply by 2.

Number of states = 2(1 + 3 + 5) = 18

CHECK AND THINK
In all 18 of these states, hydrogen has the same energy E (-1.51 eV). So the 18 states form a single shell. This shell is broken into three subshells for the three values of l. Each subshell has the same energy and the same orbital angular momentum L, but within the subshells for \ell=1 \text { and } \ell=2 there is variation in the orbital magnetic moment (\mu_{orbit})_z. In addition, in each state the spin magnetic moment (\mu_{spin})_z is either up or down.