What is the maximum number of electrons that can have principal quantum number n = 3 and spin quantum number ms = +1/2?

• A. 9
• B. 12
• C. 6
• D. 18

Answer: (B)

Verified step by step guidance

1. Recall that the principal quantum number \(n\) defines the energy level or shell of an electron. For \(n = 3\), the possible subshells are \(l = 0\) (3s), \(l = 1\) (3p), and \(l = 2\) (3d).
2. Determine the number of orbitals in each subshell using the formula: number of orbitals = \$2l + 1\(. For \)l = 0\(, orbitals = 1; for \)l = 1\(, orbitals = 3; for \)l = 2$, orbitals = 5.
3. Calculate the total number of orbitals for \(n = 3\) by summing the orbitals from all subshells: \$1 + 3 + 5 = 9$ orbitals.
4. Each orbital can hold a maximum of 2 electrons with opposite spins (\(m_s = +\frac{1}{2}\) and \(m_s = -\frac{1}{2}\)). Since we are only considering electrons with spin quantum number \(m_s = +\frac{1}{2}\), each orbital can hold only 1 such electron.
5. Multiply the total number of orbitals (9) by 1 electron per orbital with \(m_s = +\frac{1}{2}\) to find the maximum number of electrons with \(n = 3\) and \(m_s = +\frac{1}{2}\): \$9 \times 1 = 9$ electrons.