Multiple Choice
Which type of orbital has the lowest energy for a given principal quantum number n?
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9. Quantum Mechanics
Quantum Numbers: Principal Quantum Number
Multiple Choice
Which of the following sets of quantum numbers is correct and consistent with n = 4?
A
n = 4, l = 1, m_l = -2, m_s = -1/2
B
n = 4, l = 2, m_l = -2, m_s = +1/2
C
n = 4, l = 4, m_l = 0, m_s = -1/2
D
n = 4, l = 3, m_l = +4, m_s = +1/2
Verified step by step guidance1
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\(, so for \)n=4\(, \)l\( can be \)0, 1, 2,\( or \)3$.
The magnetic quantum number \(m_l\) depends on \(l\) and can take integer values from \(-l\) to \(+l\), including zero. For example, if \(l=2\), then \(m_l\) can be \(-2, -1, 0, 1,\) or \$2$.
The spin quantum number \(m_s\) can only be \(+\frac{1}{2}\) or \(-\frac{1}{2}\), representing the two possible spin states of an electron.
Check each set of quantum numbers against these rules: verify that \(l\) is less than \(n\), \(m_l\) is within the range \(-l\) to \(+l\), and \(m_s\) is either \(+\frac{1}{2}\) or \(-\frac{1}{2}\).
Identify the set where all these conditions are met. For example, if \(n=4\), \(l=2\), \(m_l=-2\), and \(m_s=+\frac{1}{2}\), this set is valid because \(l=2\) is less than \(n=4\), \(m_l=-2\) is within \(-2\) to \(+2\), and \(m_s=+\frac{1}{2}\) is an allowed spin value.
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