객관식
Which statement is true when the principal quantum number n equals 1 for an electron in an atom?
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9. Quantum Mechanics
Quantum Numbers: Principal Quantum Number
객관식
Which of the following sets of quantum numbers is invalid for an electron in an atom?
A
n = 4, l = 0, m_l = 0, m_s = -1/2
B
n = 2, l = 1, m_l = 0, m_s = +1/2
C
n = 3, l = 2, m_l = -2, m_s = -1/2
D
n = 1, l = 1, m_l = 0, m_s = +1/2
검증된 단계별 안내1
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 \(-\frac{1}{2}\).
Check the given quantum numbers for each set against these rules. For example, for \(n=4\), \(l=0\) is valid because \(l\) can be from \(0\) to \(3\) when \(n=4\); similarly, \(m_l=0\) is valid since \(m_l\) ranges from \(-l\) to \(+l\).
Focus on the 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 violates the rule for \(l\).
Confirm that the other quantum numbers in this set (\(m_l=0\) and \(m_s=+\frac{1}{2}\)) are within their allowed ranges, but the invalidity arises solely from the \(l\) value relative to \(n\).
Conclude that the set with \(n=1\), \(l=1\), \(m_l=0\), \(m_s=+\frac{1}{2}\) is invalid because the azimuthal quantum number \(l\) cannot be equal to or greater than the principal quantum number \(n\).
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