Michael S. answered • 14d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
Part 1: L = √6 ℏ ≈ 2.45 ℏ | Part 2: ⟨Lz⟩ = 0
Part 1 — magnitude for a 3d electron
L = √(ℓ(ℓ+1)) ℏ
The letter gives ℓ, and the number in front does not enter this formula at all: s → 0, p → 1, d → 2, f → 3. So for 3d, ℓ = 2.
L = √(2 × 3) ℏ = √6 ℏ ≈ 2.45 ℏ
The "3" is the principal quantum number, which sets energy and size — not angular momentum. A 4d electron has exactly the same L. Substituting n = 3 here is the standard error.
Also worth noticing: L is never a whole multiple of ℏ. The naive guess ℓℏ = 2ℏ is wrong; the true value √6 ℏ is always larger than ℓℏ, which is the mathematical root of the uncertainty principle for angular momentum — the vector can never lie perfectly along any axis.
Part 2 — average Lz with no external field
For a 3p electron, ℓ = 1, so mℓ can be −1, 0, or +1, giving Lz = −ℏ, 0, or +ℏ.
Here is the key: with no external magnetic field, those three states are degenerate — identical in energy. Nothing favours one over another, so across a large sample the electrons populate all three equally.
⟨Lz⟩ = [(−1) + (0) + (+1)] / 3 × ℏ = 0
The deeper reason
Without a field there is no physically meaningful z-axis. Space is isotropic — every direction is equivalent — so a "preferred" component of angular momentum cannot exist. You could rotate your coordinate system arbitrarily and nothing measurable would change. The average has to be zero by symmetry, before you compute anything.
Applying a magnetic field breaks that symmetry: it defines a z-direction, splits the three mℓ levels in energy (the Zeeman effect), and then the populations are no longer equal — so ⟨Lz⟩ becomes nonzero. That contrast is exactly what the phrase "in the absence of an external magnetic field" is testing.
Careful: zero average does not mean zero angular momentum. Every individual electron still has L = √2 ℏ and an Lz of −ℏ, 0, or +ℏ. It is the ensemble average that vanishes.
