For the principal quantum number n = 3, how many possible combinations of the angular momentum quantum number l and the magnetic quantum number m_l are there?

• A. 15
• B. 6
• C. 9
• D. 12

Answer: (A)

Verified step by step guidance

1. Recall that the principal quantum number \(n\) determines the possible values of the angular momentum quantum number \(l\), where \(l\) can take integer values from \$0\( up to \)n-1\(. For \)n=3\(, list all possible \)l\( values: \)l = 0, 1, 2$.
2. For each value of \(l\), the magnetic quantum number \(m_l\) can take integer values from \(-l\) to \(+l\), including zero. This means the number of possible \(m_l\) values for each \(l\) is \$2l + 1$.
3. Calculate the number of \(m_l\) values for each \(l\): - For \(l=0\), number of \(m_l\) values = \$2(0) + 1 = 1\( - For \)l=1\(, number of \)m_l\( values = \)2(1) + 1 = 3\( - For \)l=2\(, number of \)m_l\( values = \)2(2) + 1 = 5$.
4. Add the number of \(m_l\) values for all \(l\) values to find the total number of possible \((l, m_l)\) combinations: \$1 + 3 + 5$.
5. The sum from the previous step gives the total number of possible combinations of \(l\) and \(m_l\) for \(n=3\). This total represents all the unique pairs of angular momentum and magnetic quantum numbers for that principal quantum number.