For an electron in the n = 4 shell with a magnetic quantum number ml = -2, how many distinct sets of all four quantum numbers (n, l, ml, ms) are possible?

• A. 4
• B. 8
• C. 2
• D. 6

Answer: (C)

Verified step by step guidance

1. Identify the principal quantum number \(n\), which is given as 4. This means the electron is in the fourth energy level.
2. Determine the possible values of the azimuthal quantum number \(l\) for \(n=4\). Recall that \(l\) can take integer values from 0 up to \(n-1\), so \(l = 0, 1, 2, 3\).
3. Since the magnetic quantum number \(m_l\) is given as \(-2\), find which values of \(l\) allow \(m_l = -2\). Remember that \(m_l\) ranges from \(-l\) to \(+l\) in integer steps. Therefore, \(m_l = -2\) is only possible if \(l \\geq 2\).
4. From the previous step, the possible \(l\) values are 2 and 3. For each valid \(l\), \(m_l = -2\) is allowed, so we have two possible \((n, l, m_l)\) sets: (4, 2, -2) and (4, 3, -2).
5. Finally, consider the spin quantum number \(m_s\), which can be either \(+\\frac{1}{2}\) or \(-\\frac{1}{2}\). For each \((n, l, m_l)\) set, there are 2 possible \(m_s\) values, so multiply the number of \((n, l, m_l)\) sets by 2 to get the total number of distinct sets of all four quantum numbers.