Which of the following sets of quantum numbers is valid for an electron in an atom?

• A. n = 1, l = 1, m_l = 0, m_s = +1/2
• B. n = 2, l = 0, m_l = 2, m_s = +1/2
• C. n = 3, l = 3, m_l = 2, m_s = -1/2
• D. n = 2, l = 1, m_l = 0, m_s = +1/2

Answer: (D)

Verified step by step guidance

1. Recall the allowed ranges for each quantum number: the principal quantum number \(n\) must be a positive integer (\(n = 1, 2, 3, \ldots\)), the azimuthal quantum number \(l\) can take integer values from \$0\( up to \)n-1\(, the magnetic quantum number \)m_l\( ranges from \)-l\( to \)+l\( in integer steps, and the spin quantum number \)m_s\( can be either \)+\frac{1}{2}\( or \)-\frac{1}{2}$.
2. Check the first set: \(n = 1\), \(l = 1\), \(m_l = 0\), \(m_s = +\frac{1}{2}\). Since \(l\) must be less than \(n\), and here \(l = 1\) is not less than \(n = 1\), this set is invalid.
3. Check the second set: \(n = 2\), \(l = 0\), \(m_l = 2\), \(m_s = +\frac{1}{2}\). For \(l = 0\), \(m_l\) must be between \(-0\) and \$0\(, so \)m_l\( can only be \)0\(. Here, \)m_l = 2$ is outside this range, so this set is invalid.
4. Check the third set: \(n = 3\), \(l = 3\), \(m_l = 2\), \(m_s = -\frac{1}{2}\). Since \(l\) must be less than \(n\), and here \(l = 3\) is not less than \(n = 3\), this set is invalid.
5. Therefore, the only valid set must have \(l\) less than \(n\), \(m_l\) within \(-l\) to \(+l\), and \(m_s\) equal to \(+\frac{1}{2}\) or \(-\frac{1}{2}\). The correct valid set is \(n = 2\), \(l = 1\), \(m_l = 0\), \(m_s = +\frac{1}{2}\).