Multiple Choice
Which of the following correctly describes the allowed values for the angular momentum quantum number l in an atom?
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9. Quantum Mechanics
Quantum Numbers: Angular Momentum Quantum Number
Multiple Choice
For the principal quantum number n = 5, how many possible combinations of the angular momentum quantum number l and the magnetic quantum number m_l are there?
A
25
B
35
C
5
D
10
Verified step by step guidance1
Recall that the principal quantum number \(n\) determines the possible values of the angular momentum quantum number \(l\), which range from \$0\( to \)n-1\(. For \)n=5\(, \)l\( can be \)0, 1, 2, 3,\( or \)4$.
For each value of \(l\), the magnetic quantum number \(m_l\) can take integer values from \(-l\) to \(+l\), inclusive. This means for each \(l\), the number of possible \(m_l\) values is \$2l + 1$.
Calculate the number of \(m_l\) values for each \(l\):
- For \(l=0\), number of \(m_l\) values = \$2(0) + 1 = 1$
- For \(l=1\), number of \(m_l\) values = \$2(1) + 1 = 3$
- For \(l=2\), number of \(m_l\) values = \$2(2) + 1 = 5$
- For \(l=3\), number of \(m_l\) values = \$2(3) + 1 = 7$
- For \(l=4\), number of \(m_l\) values = \$2(4) + 1 = 9$.
Sum all the \(m_l\) values for \(l=0\) to \$4\( to find the total number of possible \)(l, m_l)\( combinations:
\)total = 1 + 3 + 5 + 7 + 9$.
This sum gives the total number of possible combinations of \(l\) and \(m_l\) for \(n=5\). This is the answer to the problem.
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