For the principal quantum number n = 5, how many possible combinations of the angular momentum quantum number l and the magnetic quantum number m_l are there?

• A. 25
• B. 35
• C. 5
• D. 10

Answer: (B)

Verified step by step guidance

1. Recall that the principal quantum number \(n\) determines the possible values of the angular momentum quantum number \(l\), which range from \$0\( to \)n-1\(. For \)n=5\(, \)l\( can be \)0, 1, 2, 3,\( or \)4$.
2. For each value of \(l\), the magnetic quantum number \(m_l\) can take integer values from \(-l\) to \(+l\), inclusive. This means for each \(l\), the number of possible \(m_l\) values 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$ - For \(l=3\), number of \(m_l\) values = \$2(3) + 1 = 7$ - For \(l=4\), number of \(m_l\) values = \$2(4) + 1 = 9$.
4. Sum all the \(m_l\) values for \(l=0\) to \$4\( to find the total number of possible \)(l, m_l)\( combinations: \)total = 1 + 3 + 5 + 7 + 9$.
5. This sum gives the total number of possible combinations of \(l\) and \(m_l\) for \(n=5\). This is the answer to the problem.