Counting electronic states in a principal shell For a given principal quantum number n, how many one-electron states (including all allowed orbital and spin states in that shell) are possible?

Introduction / Context:
Electronic states in atoms are specified by quantum numbers. For each principal quantum number n, permitted orbital angular momentum values l range from 0 to n − 1, and for each l there are 2l + 1 magnetic sublevels m_l. Each spatial orbital can be occupied by two spin states (m_s = ±1/2). Counting these systematically yields the total number of one-electron states in the shell.

Given Data / Assumptions:

• Principal quantum number n ∈ {1, 2, 3, …}.
• l = 0, 1, …, n − 1; m_l = −l, …, +l.
• Each orbital carries two spin states.
• Nonrelativistic, hydrogenic counting; fine structure ignored.

Concept / Approach:
First count the spatial orbitals in shell n. The number of orbitals is the sum over l of (2l + 1). This sum equals n^2. Then include spin multiplicity (×2) to obtain the total number of one-electron states as 2 n^2.

Step-by-Step Solution:
Compute spatial orbitals: Σ_{l=0}^{n−1} (2l + 1) = n^2.Include spin degeneracy: two spin states per spatial orbital.Total one-electron states in shell n = 2 * n^2.Therefore, choose 2 n^2.

Verification / Alternative check:
Check small n: for n = 1, l = 0 only ⇒ 1 orbital × 2 spins = 2 = 2 n^2; for n = 2, orbitals = 1 (s) + 3 (p) = 4 ⇒ ×2 = 8 = 2 n^2, confirming the formula.

Why Other Options Are Wrong:
n and n(n + 1) undercount; n^2 omits spin multiplicity; 2 n^2 captures both orbital and spin states.

Common Pitfalls:
Forgetting spin; confusing number of orbitals (n^2) with number of states (2 n^2).

Final Answer:
2 n^2