객관식
Which of the following combinations of principal quantum number n and angular momentum quantum number l represent possible atomic orbitals?
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9. Quantum Mechanics
Quantum Numbers: Principal Quantum Number
객관식
Which one of the following sets of quantum numbers is not possible for an electron in an atom?
A
n = 2, l = 1, m_l = 0, m_s = +1/2
B
n = 3, l = 2, m_l = -2, m_s = -1/2
C
n = 1, l = 1, m_l = 0, m_s = +1/2
D
n = 4, l = 0, 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 set \(n = 2, l = 1, m_l = 0, m_s = +\frac{1}{2}\): since \(l\) must be less than \(n\), and \(1 < 2\), this set is possible. Also, \(m_l = 0\) is within \(-1\) to \(+1\), and \(m_s\) is valid.
Check the set \(n = 3, l = 2, m_l = -2, m_s = -\frac{1}{2}\): here, \(l = 2\) is less than \(n = 3\), \(m_l = -2\) is within \(-2\) to \(+2\), and \(m_s\) is valid, so this set is possible.
Check the set \(n = 1, l = 1, m_l = 0, m_s = +\frac{1}{2}\): since \(l\) must be less than \(n\), but here \(l = 1\) is not less than \(n = 1\), this set violates the rule and is not possible.
Check the set \(n = 4, l = 0, m_l = 0, m_s = -\frac{1}{2}\): \(l = 0\) is less than \(n = 4\), \(m_l = 0\) is valid for \(l = 0\), and \(m_s\) is valid, so this set is possible.
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