Pauli exclusion principle: Worked examples — Physics deep dive

This page is a workout. The Pauli exclusion principle parent note built the idea (no two identical fermions share a full quantum-number set) and the reason (antisymmetry makes the shared-state wavefunction vanish). Here we run that rule through every kind of situation it can face — so that when an exam or a real problem hands you a scenario, you've already seen its shape.

Before anything else: a quantum "address" for an atomic electron is the four numbers $(n, ℓ, m_{ℓ}, m_{s})$ . Read them as street–number–floor–door plus a "which-way-it-spins" tag. Pauli's whole job is: every electron in one atom gets a different one of these four-part addresses. Keep that picture; we lean on it constantly.

The scenario matrix

Here is the full space of cases this topic throws at you. Every example below is tagged with the cell it covers, so you can see the grid fill up.

Cell	Scenario class	The "edge" it tests
A	Fill a full shell, count capacity	the $2 n^{2}$ machinery, no leftovers
B	Same orbital, spins save the day	two electrons legal because $m_{s}$ differs
C	The forbidden move	a repeated address → what breaks
D	Partial subshell (Hund territory)	many valid arrangements, pick the ground one
E	Degenerate / zero input	$ℓ = 0$ (single orbital), $n = 1$ (smallest system)
F	Limiting / large behaviour	big $n$ : how fast capacity grows
G	Non-electron fermions	protons & neutrons obey it too (nucleus, neutron star)
H	Boson contrast (the sign flip)	same math, $+$ sign → no exclusion
I	Real-world word problem	degeneracy pressure holding up a star
J	Exam-style twist	"how many electrons have $m_{s} = + \frac{1}{2}$ ?" counting subsets

We hit all ten cells across the examples.

Example 1 — Fill the $n = 3$ shell completely · Cell A, E

Worked example Count every electron the third shell can legally hold.

Forecast: guess the number before reading. Shell 1 held 2, shell 2 held 8… what's your bet for shell 3?

Step 1 — List the allowed $ℓ$ . For $n = 3$ , the rule is $ℓ = 0, 1, \dots, n - 1 = 0, 1, 2$ . Why this step? $ℓ$ is capped by $n$ ; you can't have more orbital "wiggliness" than the shell allows. This is where the degenerate low case $ℓ = 0$ (Cell E) already appears — it's a perfectly valid single orbital.

Step 2 — Count orbitals in each $ℓ$ . Each $ℓ$ carries $m_{ℓ} = - ℓ, \dots, + ℓ$ , which is $(2 ℓ + 1)$ values.

$ℓ = 0$ : $m_{ℓ} = 0$ → $1$ orbital.
$ℓ = 1$ : $m_{ℓ} = - 1, 0, + 1$ → $3$ orbitals.
$ℓ = 2$ : $m_{ℓ} = - 2, - 1, 0, + 1, + 2$ → $5$ orbitals. Why this step? Each distinct $(n, ℓ, m_{ℓ})$ is one "parking spot". We count spots first, then double for spin.

Step 3 — Double for spin, then add. Each orbital holds $2$ electrons ( $m_{s} = \pm \frac{1}{2}$ ). $N = 2 (1 + 3 + 5) = 2 \cdot 9 = 18.$ Why this step? The factor $2$ is exactly the freedom Pauli does allow: same orbital, opposite spin.

Verify: the formula $N = 2 n^{2}$ gives $2 \cdot 3^{2} = 18$ . ✓ Also notice $1 + 3 + 5 = 9$ is the sum of the first $3$ odd numbers $= 3^{2}$ — the identity the parent note used. Units-check: these are pure counts (dimensionless). ✓

The next figure shows the whole grid of addresses filling up.

Worked example Ground-state helium (

$Z = 2$ ). Both electrons want the lowest, smallest state $n = 1$ . Is that legal? Forecast: does Pauli block the second electron, or let it in?

Step 1 — Write the first electron's full address. Lowest energy: $n = 1 \Rightarrow ℓ = 0 \Rightarrow m_{ℓ} = 0$ . Pick a spin: $m_{s} = + \frac{1}{2}$ . Address $= (1, 0, 0, + \frac{1}{2})$ . Why this step? Pauli forbids identical addresses, so we must write the full four numbers, spin included — not just the orbital.

Step 2 — Try the second electron in the same orbital. Same $n, ℓ, m_{ℓ}$ . To stay legal the last slot must differ, so $m_{s} = - \frac{1}{2}$ . Address $= (1, 0, 0, - \frac{1}{2})$ . Why this step? The only free coordinate left is spin; it's a two-value switch, so exactly one alternative exists.

Step 3 — Compare the two sets. $(1, 0, 0, + \frac{1}{2}) \neq = (1, 0, 0, - \frac{1}{2})$ — they differ in the last entry. Legal. Why this step? "Same orbital" is not "same state." This is the classic Cell-B rescue.

Verify: shell $n = 1$ capacity is $2 n^{2} = 2 \cdot 1^{2} = 2$ . Helium places exactly $2$ electrons → shell full, no violation. ✓

Example 3 — Lithium's third electron: the forbidden move · Cell C

Worked example Lithium (

$Z = 3$ ). Can all three electrons live in $n = 1$ ? Forecast: where does electron #3 have to go?

Step 1 — Enumerate the $n = 1$ addresses. Only $(1, 0, 0, + \frac{1}{2})$ and $(1, 0, 0, - \frac{1}{2})$ exist. That's two addresses total. Why this step? With $ℓ$ forced to $0$ and $m_{ℓ}$ forced to $0$ , spin is the only variable — and it has just two values.

Step 2 — Try to seat electron #3 in $n = 1$ . Every $n = 1$ address is already used. Any choice repeats one → antisymmetric wavefunction collapses to $Ψ = 0$ . Why this step? This is Cell C: forcing a duplicate. Recall the parent's result $a = b \Rightarrow Ψ = 0$ — a zero-probability, i.e. impossible, state.

Step 3 — Promote it to the next shell. Electron #3 goes to $n = 2$ , giving Li the configuration $1 s^{2} 2 s^{1}$ . Why this step? Because $n = 1$ is genuinely full, the periodic table's new row begins — Pauli creates shell structure.

Verify: $n = 1$ capacity $= 2$ ; Li has $3$ electrons; $3 > 2$ so one must spill to $n = 2$ . ✓ Lithium indeed starts period 2. ✓

Example 4 — Carbon's $2 p^{2}$ : which arrangement wins? · Cell D

Worked example Carbon (

$Z = 6$ ): $1 s^{2} 2 s^{2} 2 p^{2}$ . The two $2 p$ electrons — same orbital or different? Forecast: paired in one $p$ -orbital, or spread across two?

Step 1 — List the $2 p$ orbitals. $ℓ = 1 \Rightarrow m_{ℓ} = - 1, 0, + 1$ : three orbitals, six slots. Why this step? We need the menu of legal seats before deciding which two to occupy.

Step 2 — Count Pauli-legal ways to place 2 electrons. Any two distinct addresses among the six are Pauli-allowed. So there are $(2 6) = 15$ legal placements. Why this step? Pauli alone permits all fifteen — it does not single one out. That's the subtle point of Cell D.

Step 3 — Let Hund's rule pick the ground state. Among the 15, the lowest energy puts the electrons in different $m_{ℓ}$ orbitals with parallel spins, e.g. $(2, 1, - 1, + \frac{1}{2})$ and $(2, 1, 0, + \frac{1}{2})$ . Why this step? Pauli sets the boundary of the legal; Hund selects the cheapest inside it. Note Pauli forbids two parallel-spin electrons in the same $m_{ℓ}$ (that would duplicate the address), so "parallel spins" forces "different orbitals" — Pauli and Hund cooperate.

Verify: the two chosen addresses differ (in $m_{ℓ}$ ), so legal. ✓ Count check: $(2 6) = 15$ . ✓

Example 5 — How fast does capacity grow? · Cell F

Worked example Compare the electron capacity of shells

$n = 1$ through $n = 5$ , and describe the growth. Forecast: linear? Doubling? Something faster?

Step 1 — Tabulate $2 n^{2}$ . $2, 8, 18, 32, 50 for n = 1, 2, 3, 4, 5.$ Why this step? Direct use of the derived capacity — the cleanest way to see the trend.

Step 2 — Look at the differences. Consecutive gaps: $6, 10, 14, 18$ — an arithmetic sequence rising by $4$ . Why this step? Capacity is quadratic in $n$ , so first differences are linear. This tells you far shells swallow electrons ever faster.

Step 3 — State the limiting behaviour. As $n \to \infty$ , capacity $\sim 2 n^{2} \to \infty$ : no ceiling on how many electrons a (hypothetical) shell holds. Why this step? Cell F is about limits — Pauli never runs out of addresses; it just demands each new one differ.

Verify: $2 \cdot 5^{2} = 50$ ; and $2 + 8 + 18 + 32 + 50 = 110$ , matching $\sum_{n = 1}^{5} 2 n^{2}$ . ✓ Differences $8 - 2, 18 - 8, 32 - 18, 50 - 32 = 6, 10, 14, 18$ . ✓

Example 6 — The alpha particle vs. the deuteron: fermion or boson? · Cell G, H

Worked example An

$α$ -particle (2 protons + 2 neutrons) and a lone neutron. Which obeys Pauli exclusion, and can two of each pile into one state? Forecast: guess "which is antisocial" before reading.

Step 1 — Neutron: half-integer spin. A neutron has spin $\frac{1}{2}$ . Half-integer ⇒ fermion ⇒ antisymmetric ⇒ obeys Pauli. Two identical neutrons cannot share a state. Why this step? Spin-statistics theorem ties spin- $\frac{1}{2}$ to the minus sign; the parent's $a = b \Rightarrow Ψ = 0$ applies (Cell G).

Step 2 — $α$ -particle: add up the spins. Four spin- $\frac{1}{2}$ constituents can combine to total spin $0$ (an integer). Integer total spin ⇒ boson. Why this step? Statistics follow the total spin of the composite. Even integer means symmetric, $+$ sign.

Step 3 — Test sharing for the boson. For bosons $Ψ (1, 2) = + Ψ (2, 1)$ ; set $a = b$ : $Ψ = \frac{1}{2} [ψ_{a} ψ_{a} + ψ_{a} ψ_{a}] = 2 ψ_{a} ψ_{a} \neq = 0$ . Sharing allowed. Why this step? Cell H: same algebra, opposite sign — the wavefunction survives, so no exclusion. This is why $α$ -particles (and cold atoms) can crowd into one state — see Bose-Einstein condensate.

Verify: neutron spin $\frac{1}{2}$ (half-integer) ⇒ fermion ✓; $α$ spin $0$ (integer) ⇒ boson ✓. Boson shared amplitude $2 \neq = 0$ . ✓

Example 7 — Why a white dwarf doesn't collapse · Cell I (word problem)

Worked example A white dwarf star is a ball of nuclei bathed in electrons, no longer burning fuel. Gravity crushes inward. What holds it up, in terms of Pauli?

Forecast: is it electric repulsion, or something Pauli-flavoured?

Step 1 — Identify the fermions. Electrons, spin $\frac{1}{2}$ → fermions. Each demands a distinct address. Why this step? Only fermions give degeneracy pressure; we must confirm the players first.

Step 2 — Gravity tries to squeeze them into the same low state. But the low-energy addresses are already all taken. New electrons must go into higher-momentum (higher-energy) states. Why this step? This is the parent's key correction to the "Pauli is a force" mistake: it's the energy cost of climbing to empty states, not an electromagnetic push. Compressing the star forces electrons up the ladder — that costs energy — and that resistance is the pressure.

Step 3 — Balance. The star settles where outward degeneracy pressure equals inward gravity. No fuel needed — Pauli alone provides the support. Why this step? Cell I links the microscopic rule to a macroscopic object; see White dwarf and neutron star. (In neutron stars the same story runs with neutrons — Cell G at stellar scale.)

Verify (qualitative, dimensional): degeneracy pressure arises from filled fermion states → requires exclusion → requires fermions. Bosons give no such pressure (they'd share the ground state and collapse). Consistent. ✓

Example 8 — Exam twist: count the spin-up electrons in argon · Cell J

Worked example Argon (

$Z = 18$ ) is a filled-shell atom: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$ . How many of its electrons have $m_{s} = + \frac{1}{2}$ ? Forecast: most students blurt "9" instinctively — is that right, and why?

Step 1 — Understand full subshells. Every orbital that is doubly occupied holds exactly one $+ \frac{1}{2}$ and one $- \frac{1}{2}$ electron (Pauli forces the pair to differ in spin). Why this step? Cell J tests whether you can slice the electron set by a single quantum number.

Step 2 — Are all argon's orbitals full? Yes: $2 + 2 + 6 + 2 + 6 = 18$ , and every listed subshell is completely filled (no half-filled Hund case). Why this step? Only when orbitals are fully paired is the spin split exactly even.

Step 3 — Split evenly. With all $9$ orbitals doubly occupied, exactly half the $18$ electrons are spin-up: $N_{+ 1/2} = \frac{18}{2} = 9.$ Why this step? Each full orbital contributes one $+ \frac{1}{2}$ ; there are $9$ full orbitals.

Verify: orbitals $= 1 (1 s) + 1 (2 s) + 3 (2 p) + 1 (3 s) + 3 (3 p) = 9$ ; spin-up count $= 9$ ; spin-down count $= 9$ ; total $= 18 = Z$ . ✓ The instinct "9" is correct here, but only because every orbital is paired — for a Hund-partial atom like carbon it would fail.

Active Recall

Recall Full-shell capacity for

$n = 3$ How many electrons? ::: $2 n^{2} = 18$ .

Recall Why Lithium's 3rd electron leaves shell 1

Reason? ::: Both $n = 1$ addresses $(1, 0, 0, \pm \frac{1}{2})$ are taken; a third would duplicate one, giving $Ψ = 0$ .

Recall Legal placements of 2 electrons in a

$2 p$ subshell Count (Pauli only)? ::: $(2 6) = 15$ ; Hund then selects the ground state.

Recall Is an

$α$ -particle a boson or fermion? Answer? ::: Boson — total spin $0$ (integer); it does not obey Pauli exclusion.

Recall Spin-up electrons in filled argon

Count? ::: $9$ (all $9$ orbitals doubly occupied, split evenly).

Connections

Pauli exclusion principle — the parent rule these examples exercise.
Quantum numbers — the four-part addresses we enumerate.
Spin and intrinsic angular momentum — the $m_{s} = \pm \frac{1}{2}$ doubling.
Hund's rule — picks the ground state inside Pauli-legal options (Example 4, 8).
Aufbau principle and electron configuration — filling order behind Li, C, Ar.
Periodic table structure — why full shells start new rows (Example 3).
Spin-statistics theorem — fermion vs boson sign (Example 6).
Bose-Einstein condensate — the boson world where sharing is allowed.
White dwarf and neutron star — degeneracy pressure (Example 7).