For a principal quantum number n = 3, what is the full range of possible values for the angular momentum quantum number l?

• A. 0, 1, 2, 3
• B. 0, 1, 2
• C. 1, 2
• D. 1, 2, 3

Answer: (B)

Verified step by step guidance

1. Recall that the principal quantum number \(n\) determines the energy level or shell of an electron in an atom, and it must be a positive integer (\(n = 1, 2, 3, \ldots\)).
2. Understand that the angular momentum quantum number \(l\) defines the shape of the orbital and can take on integer values from \$0\( up to \)n-1\( for a given principal quantum number \)n$.
3. For \(n = 3\), list all possible values of \(l\) by applying the rule \(l = 0, 1, 2, \ldots, (n-1)\), which means \(l\) can be \$0\(, \)1\(, or \)2$.
4. Recognize that \(l = 3\) is not possible for \(n = 3\) because \(l\) cannot be equal to or greater than \(n\).
5. Therefore, the full range of possible values for \(l\) when \(n = 3\) is \$0\(, \)1\(, and \)2$.