Quantum numbers n, l, mₗ, mₛ: Worked examples

For the full derivation, see the parent note.

The rules and the one symbol you need

Everything on this page rests on four rules and one physical constant. We state them here so this page stands on its own.

Definition The core quantum-number rules

Symbol	Meaning	Allowed values
$n$	principal (size/energy)	$1, 2, 3, \dots$
$l$	azimuthal (shape / angular momentum)	$0, 1, 2, \dots, (n - 1)$ — stops one short of $n$
$m_{l}$	magnetic (orientation)	$- l, - l + 1, \dots, 0, \dots, + l$ — that is $2 l + 1$ values, including negatives
$m_{s}$	spin projection	$+ \frac{1}{2}$ or $- \frac{1}{2}$ — only these two

Rule chain to memorise: $l < n$ , then $∣ m_{l} ∣ \leq l$ , then $m_{s} = \pm \frac{1}{2}$ .

The scenario matrix

Every question this topic can throw is one of these case classes. Each worked example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky
A	Smallest shell, $n = 1$	degenerate: only $l = 0$ exists, no orbital angular momentum
B	List all states for a shell	must sum $2 (2 l + 1)$ over all $l$ , land on $2 n^{2}$
C	Angular momentum magnitude	$l (l + 1) ℏ$ , never $l ℏ$ (Bohr trap)
D	All orientations $L_{z} = m_{l} ℏ$	negative projections; tilt angle of $L$
E	Legality check of a 4-tuple	violate one rule ( $l \geq n$ , $
F	Reverse problem: given a fact, find $(n, l)$	e.g. "this shell holds 18 electrons"
G	Real-world / spectroscopy word problem	subshell notation ↔ quantum numbers
H	Limiting / degenerate spin & energy	$l = 0$ vs large $l$ ; energy scaling with $n$

We now cover A–H with eight examples.

Worked example Example 1 — Cell A: the smallest, most degenerate shell (

$n = 1$ ) List every quantum state in the $n = 1$ shell and give the electron's orbital angular momentum.

Forecast: guess first — how many states? What is $∣ L ∣$ for a 1s electron?

Step 1 — Find allowed $l$ . $l$ runs $0, 1, \dots, (n - 1)$ . With $n = 1$ , that is just $l = 0$ . Why this step? The rule $l \leq n - 1$ is the master gate; $n = 1$ leaves only one value, so there is no 1p, 1d, …

Step 2 — Find allowed $m_{l}$ . For $l = 0$ : $m_{l}$ runs $- l, \dots, + l = 0$ . Only one value. Why this step? $m_{l}$ counts orientations; a round ( $l = 0$ ) cloud has nothing to orient, so exactly one.

Step 3 — Attach spin. Each $(n, l, m_{l})$ orbital gets $m_{s} = \pm \frac{1}{2}$ : 2 states. Why this step? Spin doubles every spatial orbital regardless of $l$ .

Step 4 — Orbital angular momentum. $∣ L ∣ = l (l + 1) ℏ = 0 \cdot 1 ℏ = 0.$ Why this step? An s-electron ( $l = 0$ ) is a spherical, non-rotating cloud — the degenerate case.

Verify: $2 n^{2} = 2 (1)^{2} = 2$ states ✓. Two states = the two electrons of helium's full 1s shell ✓.

The figure below turns this arithmetic into a picture. Look at the two bars for each subshell: the blue bar is the orbital count $2 l + 1$ (1, 3, 5) and the yellow bar is the electron count $2 (2 l + 1)$ (2, 6, 10). Notice how the yellow bars are always exactly twice the blue — that "×2" is the spin doubling from Step 3. Their green total, 18, is the $2 n^{2}$ we predicted.

Worked example Example 3 — Cell C: angular-momentum magnitude, dodging the Bohr trap

Find $∣ L ∣$ for a 4f electron. Compare it to Bohr's naive guess.

Forecast: the letter f corresponds to which $l$ ? Guess $∣ L ∣$ in units of $ℏ$ .

Step 1 — Read off $l$ from the letter. s,p,d,f $\to l = 0, 1, 2, 3$ . So 4f means $l = 3$ (and $n = 4$ ). Why this step? Subshell letters are a fixed dictionary for $l$ .

Step 2 — Apply the correct formula. $∣ L ∣ = l (l + 1) ℏ = 3 \cdot 4 ℏ = 12 ℏ = 23 ℏ \approx 3.464 ℏ.$ Why this step? Schrödinger's result is $l (l + 1)$ , not $l ℏ$ . This is the classic exam trap.

Step 3 — Show the Bohr guess fails. Naive Bohr-style $l ℏ = 3ℏ$ . But the true value is $\approx 3.464ℏ$ , larger. Why this step? The gap $l (l + 1) > l$ is exactly what makes full alignment with $z$ impossible.

Verify: $(23)^{2} = 12 = l (l + 1) = 3 \cdot 4$ ✓. And $3.464 > 3$ , so the quantum magnitude exceeds the largest projection $l ℏ = 3ℏ$ ✓.

Worked example Example 4 — Cell D: all orientations and the tilt angle

For a 3d electron, list every allowed $L_{z}$ , and find the smallest angle $L$ can make with the $z$ -axis.

Forecast: how many orientations? Can the angle ever be $0^{\circ}$ ?

Step 1 — Identify $l$ and $∣ L ∣$ . 3d $\Rightarrow n = 3, l = 2$ , so $∣ L ∣ = 2 \cdot 3 ℏ = 6 ℏ \approx 2.449 ℏ.$ Why this step? Magnitude is fixed by $l$ before we talk orientation.

Step 2 — List $m_{l}$ and thus $L_{z} = m_{l} ℏ$ . $m_{l} = - 2, - 1, 0, + 1, + 2$ , giving $L_{z} \in {- 2ℏ, - ℏ, 0, + ℏ, + 2ℏ} .$ Why this step? $2 l + 1 = 5$ orientations; note the negatives — projections point down too.

Step 3 — Smallest tilt angle. The vector tilts least when $L_{z}$ is biggest, $L_{z} = + 2ℏ$ . The angle $α$ from the $z$ -axis satisfies $cos α = \frac{L _{z}}{∣ L ∣} = \frac{2ℏ}{6 ℏ} = \frac{2}{6} \approx 0.8165.$ $α = arccos (0.8165) \approx 35. 3^{\circ} .$ Why this step? $cos α =$ adjacent/hypotenuse on the right triangle formed by $L_{z}$ (along $z$ ) and $∣ L ∣$ (the hypotenuse). Since $2/ 6 < 1$ , $α$ can never be $0$ — the vector is always tilted, the geometric face of "space quantization".

Verify: $cos^{2} α = 4/6 = 2/3$ , so $sin^{2} α = 1/3$ , giving $L_{x}^{2} + L_{y}^{2} = ∣ L ∣^{2} sin^{2} α = 6 ℏ^{2} \cdot \frac{1}{3} = 2 ℏ^{2} > 0$ ✓ — the perpendicular part never vanishes, consistent with the uncertainty argument.

The figure makes this "always tilted" fact visible. Each coloured arrow is one allowed orientation of $L$ for the 3d electron; its tip sits at the height $L_{z} = m_{l} ℏ$ marked on the white $z$ -axis. All five arrows have the same length $6 ℏ$ (the magnitude never changes) — only their tilt differs. Look at the top red arrow ( $m_{l} = + 2$ ): even this "most upright" one leans over by $\approx 35. 3^{\circ}$ and never lies flat along $z$ , because its dashed horizontal projection can never shrink to zero.

Worked example Example 5 — Cell E: legality checks (four traps in one)

Which of these $(n, l, m_{l}, m_{s})$ sets are physically allowed? (a) $(2, 1, - 1, + \frac{1}{2})$ (b) $(3, 3, 0, - \frac{1}{2})$ (c) $(4, 2, + 3, + \frac{1}{2})$ (d) $(1, 0, 0, + 1)$

Forecast: guess how many of the four survive.

Step 1 — Check the $l < n$ gate for each. (a) $l = 1 < n = 2$ ✓. (b) $l = 3$ but $n = 3$ → need $l \leq 2$ : fails. (c) $l = 2 < 4$ ✓. (d) $l = 0 < 1$ ✓. Why this step? This is the first, strictest rule; kills (b) immediately.

Step 2 — Check $∣ m_{l} ∣ \leq l$ . (a) $∣ - 1∣ = 1 \leq 1$ ✓. (c) $∣ + 3∣ = 3$ but $l = 2$ → need $∣ m_{l} ∣ \leq 2$ : fails. (d) $∣0∣ \leq 0$ ✓. Why this step? $m_{l}$ is a projection; it can't exceed the magnitude cap $l$ .

Step 3 — Check $m_{s} = \pm \frac{1}{2}$ only. (d) has $m_{s} = + 1$ → fails (spin can only be $\pm \frac{1}{2}$ ). (a) $+ \frac{1}{2}$ ✓. Why this step? Spin has exactly two values; anything else is unphysical.

Step 4 — Verdict. Only (a) is allowed. (b) breaks $l < n$ , (c) breaks $∣ m_{l} ∣ \leq l$ , (d) breaks $m_{s}$ .

Verify: (a) satisfies all three chained rules $l < n$ , $∣ m_{l} ∣ \leq l$ , $m_{s} \in {\pm \frac{1}{2}}$ simultaneously ✓ — exactly one of four survives.

Worked example Example 7 — Cell G: spectroscopy word problem

A hydrogen electron drops from the $n = 3$ shell to the $n = 1$ shell. Find the emitted photon energy, and state how many distinct upper quantum states ( $n = 3$ ) it could have started from.

Forecast: guess the energy in eV, and the state count from Example 2.

Step 1 — Energy of each level. $E_{n} = - \frac{13.6 eV}{n ^{2}}$ . $E_{3} = - \frac{13.6}{9} = - 1.511 eV, E_{1} = - \frac{13.6}{1} = - 13.6 eV .$ Why this step? Photon energy is the difference between levels (energy conservation).

Step 2 — Photon energy. $Δ E = E_{3} - E_{1} = - 1.511 - (- 13.6) = 12.09 eV .$ Why this step? The electron loses this much; a photon of exactly this energy is emitted.

Step 3 — Count possible starting states. From Example 2, the $n = 3$ shell has 18 distinct $(n, l, m_{l}, m_{s})$ states. Why this step? Energy in hydrogen depends only on $n$ , so all 18 states share $E_{3}$ — any of them could be the origin.

Verify: $Δ E = 13.6 (1 - \frac{1}{9}) = 13.6 \cdot \frac{8}{9} = 12.09 eV$ ✓; units: eV (energy) ✓; state count $2 n^{2} = 18$ ✓.

Worked example Example 8 — Cell H: limiting behaviour of energy and spin

(i) As $n \to \infty$ , what happens to $E_{n}$ ? (ii) Does the electron's spin magnitude $∣ S ∣$ depend on $n$ , $l$ , or $m_{l}$ ?

Forecast: does the energy go to $0$ , $- \infty$ , or a finite non-zero value? Does spin magnitude ever change?

Step 1 — Limit of the energy. $E_{n} = - \frac{13.6 eV}{n ^{2}} n \to \infty 0^{-} .$ Why this step? Larger $n$ = looser binding; the electron approaches the free (ionised) state at $E = 0$ . Levels crowd together toward $0$ .

Step 2 — Ionisation reading. From the ground state $E_{1} = - 13.6$ eV, reaching $E_{\infty} = 0$ costs $+ 13.6$ eV — the ionisation energy of hydrogen. Why this step? The limit is not abstract; it is the ionisation threshold.

Step 3 — Spin magnitude. $∣ S ∣ = s (s + 1) ℏ, s = \frac{1}{2} \Rightarrow ∣ S ∣ = \frac{1}{2} \cdot \frac{3}{2} ℏ = \frac{3}{2} ℏ \approx 0.866 ℏ.$ Why this step? $s = \frac{1}{2}$ is a fixed intrinsic constant of the electron — it never depends on $n, l, m_{l}$ . Only the projection $m_{s} = \pm \frac{1}{2}$ flips.

Verify: $(\frac{3}{2})^{2} = \frac{3}{4} = s (s + 1) = \frac{1}{2} \cdot \frac{3}{2}$ ✓; and $lim_{n \to \infty} (- 13.6/ n^{2}) = 0$ ✓.

Recall Active recall — cover the answers

How many states in the $n = 3$ shell? ::: $2 n^{2} = 18$ . $∣ L ∣$ for a 4f electron? ::: $12 ℏ = 23 ℏ \approx 3.46ℏ$ . Smallest tilt angle of $L$ for a 3d electron? ::: $arccos (2/ 6) \approx 35. 3^{\circ}$ . Photon energy for $n = 3 \to n = 1$ in hydrogen? ::: $12.09$ eV. Spin magnitude $∣ S ∣$ of any electron? ::: $\frac{3}{2} ℏ \approx 0.866ℏ$ , independent of $n, l, m_{l}$ . What is $ℏ$ ? ::: The reduced Planck constant $h /2 π \approx 1.055 \times 1 0^{- 34}$ J·s — the natural unit of angular momentum.

See also: Hydrogen Atom Energy Levels · Angular Momentum in Quantum Mechanics · Pauli Exclusion Principle · Electron Configuration & Periodic Table · Stern-Gerlach Experiment · Bohr Model · Schrödinger Equation.

How many distinct quantum states does the

n = 3

shell hold?

18

2 \cdot 3^{2}

For a 4f electron, what is

∣ L ∣

l (l + 1) ℏ = 12 ℏ = 23 ℏ \approx 3.46ℏ

Why can

L

never lie exactly along the

z

-axis?

Because

max L_{z} = l ℏ < l (l + 1) ℏ = ∣ L ∣

, so the smallest tilt angle is nonzero.

Photon energy emitted for hydrogen

n = 3 \to n = 1

E_{3} - E_{1} = 13.6 (1 - 1/9) = 12.09

eV.

Does spin magnitude depend on

n, l, m_{l}

No —

∣ S ∣ = \frac{3}{2} ℏ

always; only

m_{s} = \pm \frac{1}{2}

changes.

What does

ℏ

mean?

Reduced Planck constant

= h /2 π

, the fundamental packet size of angular momentum.