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 \(2\).