Delete one rule and rebuild the periodic table. Everything else in the theory is left untouched, so whatever changes can only have been the exclusion principle holding it up.

§11.2 built the periodic table from two ideas. One of them — that electrons occupy orbitals with definite energies — is ordinary quantum mechanics. The other is the Pauli exclusion principle , which has no classical analogue at all: it is a statement about the symmetry of a wave function under relabelling.

So the book ends by asking the obvious question. What would atoms be like without it?

The answer is not a hand-wave. It is a calculation, built on nothing more than the uncertainty principle and Coulomb’s law, and it produces a specific prediction that can be laid beside the measured properties of real atoms.

The model: two terms and a minimum

Confine an electron to a region of size $R$ around a nucleus of charge $e$ . Its potential energy is about

\bar V \approx -\frac{e^2}{4\pi\epsilon_0 R}

and its kinetic energy cannot be less than the localization cost. The uncertainty principle gives $\Delta p \gtrsim \hbar/R$ , and since the average momentum is comparable to its own uncertainty,

\bar T \approx \frac{\hbar^2}{2m_eR^2}

So the total is

E \approx \frac{\hbar^2}{2m_eR^2} - \frac{e^2}{4\pi\epsilon_0 R}\tag{11.9}

which the book abbreviates

E = \frac{A_1}{R^2} - \frac{B_1}{R}, \qquad A_1 = \frac{\hbar^2}{2m_e}, \quad B_1 = \frac{e^2}{4\pi\epsilon_0}\tag{11.10–11.11}

step 1 of 4

Two competing terms: localization is expensive and attraction is cheap, and they scale differently with R. One derivative finds the balance.

1Write the energy as a function of one variable. The 1/R-squared term is the cost of squeezing the electron; the 1/R term is the reward for keeping it near the nucleus.
$E (R) = \frac{A _{1}}{R ^{2}} - \frac{B _{1}}{R}$
At large R the attraction dominates and E rises toward zero; at small R the kinetic cost blows up. So there is a minimum in between, and that is the atom.
2Set the derivative to zero — the standard condition for a minimum, and the only calculus in the whole section.
$\frac{d E}{d R} = - \frac{2 A _{1}}{R ^{3}} + \frac{B _{1}}{R ^{2}} = 0$
3Multiply through by R-cubed and solve. This is the size the atom settles at.
$R_{1} = \frac{2 A _{1}}{B _{1}}$
Substituting the constants gives 2 x (hbar-squared / 2 m_e) / (e-squared / 4 pi epsilon-nought), which is the Bohr radius exactly.
4Put that R back into E to get the binding energy.
$E_{1} = - \frac{B _{1}^{2}}{4 A _{1}}$
And that is minus the Rydberg energy, exactly. No fitting, no adjustable parameter — the hydrogen atom from one inequality and one derivative.

E_1 = -E_R = -13.6\ \mathrm{eV} \qquad\text{and}\qquad R_1 = a_0\tag{11.12–11.13}

Why this works, and why it is not luck

Getting hydrogen exactly from a one-line estimate looks too good. It is not luck, and it is not quite a coincidence either.

The uncertainty principle supplies the only scale in the problem. Once you have decided that kinetic energy goes as $\hbar^2/2m_eR^2$ and potential as $-e^2/4\pi\epsilon_0R$ , there is only one length you can build from $\hbar$ , $m_e$ , $e$ and $\epsilon_0$ — and it is $a_0$ . The minimisation cannot produce anything else; the most it can get wrong is a numerical factor.

What is genuinely fortunate is that the factor comes out to be 1. Had we written $\Delta p \approx \hbar/2R$ the answer would have been off by four. The book’s choice is the one that works, and it works because $-B^2/4A$ happens to reproduce the exact Coulomb ground state.

The right attitude is the one §9.7 earned: this estimate gets the scale right for a reason, and the exact answer required solving a differential equation. That the two agree to the last digit here is a gift, not a method.

Helium, and the one fitted number

Two electrons around a nucleus of charge $2e$ , both in the same single-particle state — which is allowed, since two electrons of opposite spin may share an orbital:

E_2 \approx 2\frac{\hbar^2}{2m_eR^2} - 4\frac{e^2}{4\pi\epsilon_0 R} + \frac{e^2}{4\pi\epsilon_0 R_{ee}}\tag{11.14}

Two kinetic terms; attraction $2 \times 2e^2$ ; and one new term, the repulsion between the electrons themselves at separation $R_{ee}$ . That separation is not known, so the book parametrizes it:

R_{ee} = fR\tag{11.15}

where $f$ is the electron–electron avoidance parameter — expected to be around 1, and larger than 1 if the electrons are good at keeping apart. With that,

E = \frac{A_2}{R^2} - \frac{B_2}{R}, \qquad A_2 = 2A_1, \quad B_2 = \left(4 - \frac{1}{f}\right)B_1\tag{11.16–11.17}

and the same minimisation gives

E_2 = -\frac{\left(4 - \tfrac1f\right)^2}{2}E_R \qquad R_2 = \frac{2}{4 - \tfrac1f}\,a_0\tag{11.18}

Now remove the Pauli principle

Here is the move the whole section exists for. Put all $Z$ electrons in the same single-particle state. Nothing in electrostatics forbids it; only Pauli does, and Pauli is what we are removing.

There are now $Z$ kinetic terms, an attraction $Z \times Ze^2$ , and $\tfrac12 Z(Z-1)$ pairs each repelling:

E_Z \approx \frac{\hbar^2}{2m_eR^2}Z - \frac{e^2}{4\pi\epsilon_0R}\left[Z^2 - \frac{Z(Z-1)}{2f}\right]\tag{11.19–11.21}

Minimising exactly as before:

E_Z = -\frac{\left[Z^2 - \frac{Z(Z-1)}{2f}\right]^2}{Z}E_R \qquad R_Z = \frac{Z}{Z^2 - \frac{Z(Z-1)}{2f}}\,a_0\tag{11.22}

Removing one electron gives the corresponding ion (problem 8),

E_{Z^-} = -\frac{\left[Z(Z-1) - \frac{(Z-1)(Z-2)}{2f}\right]^2}{Z-1}E_R\tag{11.23}

and the ionization energy is the difference:

E_I = E_{Z^-} - E_Z\tag{11.24}

which the book fits with

E_I \approx (0.07Z^2 + 0.57Z + 0.36)E_R \qquad R_Z \approx \frac{1}{0.7Z + 0.3}\,a_0\tag{11.25}

Look at the form of those. The ionization energy grows like $Z^2$ — smoothly, monotonically, forever. The radius shrinks like $1/Z$ — smoothly, monotonically, forever. There is no structure in either, because there is nothing in the model that could produce structure.

Fig. 11.5 — atoms with the Pauli principle, and atoms without it

electron–electron avoidance parameter ff = 1.67show the no-Pauli atoms

real atoms — Pauli in forcehypothetical atoms — no Pauli principle

helium, the atom that fixes f: E = -5.78 E_R, R = 0.588 a₀ vs measured 5.81 and 0.586

lithium, where they part: 2.71 vs 0.40 E_R — a factor of 6.8

Move f and watch what does not change. The hypothetical curve shifts up and down, but it never develops a kink — no value of f produces a peak at helium or a collapse at lithium, because a model in which every electron occupies the same state has no shell to close. The measured points oscillate; the model cannot. That gap is the Pauli principle, and it is the whole of chemistry. Note also that hydrogen and helium sit on top of each other in both panels: with one electron, or two in the same state, Pauli forbids nothing and the hypothetical atom is the real one. They separate at lithium — the first element whose third electron has nowhere legal to go.

Reproduce it in NumPy

Radii are Clementi's calculated atomic radii, which is the convention the book is using — its own helium check quotes 0.6 a0 and Clementi gives 0.586. Ionization energies are NIST's. Note that step 1 involves no fitting whatever: hydrogen comes out exactly, and f is introduced only when a second electron appears.

import numpy as np
ER=13.6057
# Clementi's calculated atomic radii (1963), Angstrom -> a0. The book's own check
# is helium at 0.6 a0, which this reproduces, so it is the intended convention.
A2a0=0.529177
RAD_A={1:0.53,2:0.31,3:1.67,4:1.12,5:0.87,6:0.67,7:0.56,8:0.48,9:0.42,10:0.38,11:1.90}
RAD={z:r/A2a0 for z,r in RAD_A.items()}
EI={1:13.598,2:24.587,3:5.392,4:9.323,5:8.298,6:11.260,7:14.534,8:13.618,9:17.423,10:21.565,11:5.139}

f=1.67
def EZ(Z,f):   return -(Z**2 - Z*(Z-1)/(2*f))**2 / Z            # Eq. (11.22), in E_R
def RZ(Z,f):   return Z / (Z**2 - Z*(Z-1)/(2*f))                # Eq. (11.22), in a0
def EZm(Z,f):  return -(Z*(Z-1) - (Z-1)*(Z-2)/(2*f))**2/(Z-1)   # Eq. (11.23)
def EI_model(Z,f): return (EZm(Z,f) - EZ(Z,f)) if Z>1 else 1.0

print("Step 1 — the model reproduces hydrogen with NO fitting at all")
print("=" * 74)
print(f"  E = A/R^2 - B/R  is minimised at R = 2A/B with E = -B^2/4A")
print(f"  A1 = hbar^2/2m_e, B1 = e^2/4*pi*eps0  ->  E1 = {EZ(1,f):.4f} E_R, R1 = {RZ(1,f):.4f} a0")
print(f"  measured: E_I(H) = {EI[1]/ER:.4f} E_R, radius = {RAD[1]:.4f} a0")
print("  Exact. One line of calculus, two constants, and the hydrogen atom.")

print("\nStep 2 — helium needs ONE fitted number, the avoidance parameter f")
print("=" * 74)
for ff in (1.0, 1.5, 1.67, 2.0):
    print(f"  f = {ff:<5} E2 = {EZ(2,ff):>7.3f} E_R   R2 = {RZ(2,ff):.3f} a0")
print(f"  measured: total binding 5.81 E_R (79.0 eV), radius {RAD[2]:.2f} a0")
print(f"  -> f = 1.67 is the choice that fits, and the book keeps it for every Z.")

print("\nStep 3 — atoms WITHOUT the Pauli principle, Eq. (11.25)")
print("=" * 74)
print(f"  {'Z':>3}{'no-Pauli E_I':>14}{'real E_I':>11}{'no-Pauli R':>13}{'real R':>10}")
for Z in range(1,12):
    print(f"  {Z:>3}{EI_model(Z,f):>14.2f}{EI[Z]/ER:>11.2f}{RZ(Z,f):>13.3f}{RAD[Z]:>10.3f}")
print(f"""
  Read the columns, not the rows. Without Pauli the ionization energy climbs
  monotonically -- {EI_model(1,f):.1f}, {EI_model(2,f):.1f}, {EI_model(3,f):.1f}, ... {EI_model(11,f):.1f} E_R -- and the radius
  falls monotonically, {RZ(1,f):.2f} a0 down to {RZ(11,f):.2f} a0. Every atom is smaller and
  more tightly bound than the one before it.

  The real columns oscillate. E_I peaks at helium ({EI[2]/ER:.2f}) and neon ({EI[10]/ER:.2f}) and
  collapses at lithium ({EI[3]/ER:.2f}) and sodium ({EI[11]/ER:.2f}); the radius does the reverse.
  THAT oscillation is the periodic table, and it is entirely absent from the
  model. Same electrostatics, same quantum mechanics, one principle removed.""")

print("\n  The polynomial fit the book quotes, Eq. (11.25):")
print(f"    {'Z':>3}{'exact E_I':>12}{'0.07Z^2+0.57Z+0.36':>21}{'error':>9}")
for Z in (1,2,5,8,11):
    ex=EI_model(Z,f); fit=0.07*Z*Z+0.57*Z+0.36
    print(f"    {Z:>3}{ex:>12.3f}{fit:>21.3f}{fit-ex:>9.3f}")
print("    Good to a few per cent -- fine for a figure, but the model itself is exact.")

print("\nStep 4 — the two atoms Pauli does NOT touch")
print("=" * 74)
for Z in (1,2):
    print(f"  Z = {Z}: model E_I = {EI_model(Z,f):.3f} E_R, real {EI[Z]/ER:.3f} E_R"
          f"   |   model R = {RZ(Z,f):.3f} a0, real {RAD[Z]:.3f} a0")
print("""
  Hydrogen has one electron and helium has two in the SAME state, which is
  exactly what the Pauli principle permits. So for Z = 1 and 2 the hypothetical
  atom IS the real atom, and the curves start together. They separate at
  lithium, the first element where a third electron has nowhere legal to go.""")

printsStep 1 — the model reproduces hydrogen with NO fitting at all ========================================================================== E = A/R^2 - B/R is minimised at R = 2A/B with E = -B^2/4A A1 = hbar^2/2m_e, B1 = e^2/4*pi*eps0 -> E1 = -1.0000 E_R, R1 = 1.0000 a0 measured: E_I(H) = 0.9994 E_R, radius = 1.0016 a0 Exact. One line of calculus, two constants, and the hydrogen atom. Step 2 — helium needs ONE fitted number, the avoidance parameter f ========================================================================== f = 1.0 E2 = -4.500 E_R R2 = 0.667 a0 f = 1.5 E2 = -5.556 E_R R2 = 0.600 a0 f = 1.67 E2 = -5.784 E_R R2 = 0.588 a0 f = 2.0 E2 = -6.125 E_R R2 = 0.571 a0 measured: total binding 5.81 E_R (79.0 eV), radius 0.59 a0 -> f = 1.67 is the choice that fits, and the book keeps it for every Z. Step 3 — atoms WITHOUT the Pauli principle, Eq. (11.25) ========================================================================== Z no-Pauli E_I real E_I no-Pauli R real R 1 1.00 1.00 1.000 1.002 2 1.78 1.81 0.588 0.586 3 2.71 0.40 0.416 3.156 4 3.78 0.69 0.322 2.116 5 4.99 0.61 0.263 1.644 6 6.35 0.83 0.222 1.266 7 7.84 1.07 0.192 1.058 8 9.48 1.00 0.169 0.907 9 11.27 1.28 0.151 0.794 10 13.19 1.58 0.137 0.718 11 15.26 0.38 0.125 3.590 Read the columns, not the rows. Without Pauli the ionization energy climbs monotonically -- 1.0, 1.8, 2.7, ... 15.3 E_R -- and the radius falls monotonically, 1.00 a0 down to 0.12 a0. Every atom is smaller and more tightly bound than the one before it. The real columns oscillate. E_I peaks at helium (1.81) and neon (1.58) and collapses at lithium (0.40) and sodium (0.38); the radius does the reverse. THAT oscillation is the periodic table, and it is entirely absent from the model. Same electrostatics, same quantum mechanics, one principle removed. The polynomial fit the book quotes, Eq. (11.25): Z exact E_I 0.07Z^2+0.57Z+0.36 error 1 1.000 1.000 -0.000 2 1.784 1.780 -0.004 5 4.992 4.960 -0.032 8 9.484 9.400 -0.084 11 15.259 15.100 -0.159 Good to a few per cent -- fine for a figure, but the model itself is exact. Step 4 — the two atoms Pauli does NOT touch ========================================================================== Z = 1: model E_I = 1.000 E_R, real 0.999 E_R | model R = 1.000 a0, real 1.002 a0 Z = 2: model E_I = 1.784 E_R, real 1.807 E_R | model R = 0.588 a0, real 0.586 a0 Hydrogen has one electron and helium has two in the SAME state, which is exactly what the Pauli principle permits. So for Z = 1 and 2 the hypothetical atom IS the real atom, and the curves start together. They separate at lithium, the first element where a third electron has nowhere legal to go.

Hydrogen and helium sit on top of each other, and that is the control

In both panels the first two points coincide. That is not a fitting artefact — it is the check that the comparison is fair.

Hydrogen has one electron, so there is no pair to exclude. Helium has two, and two electrons of opposite spin are permitted to share a single-particle state. So for $Z = 1$ and $Z = 2$ the hypothetical atom and the real atom are the same atom, and any model worth trusting must give the same answer for both. It does.

The curves separate at lithium — the first element where a third electron arrives and, in the real world, has nowhere legal to go but the $n = 2$ shell, far out and well screened. The model puts it in the 1s state with the others, and gets an ionization energy 6.8 times too large and a radius 7.6 times too small.

What the world would be like

Read the two panels as a chemist would.

Without the Pauli principle, ionization energy rises steadily with $Z$ and radius falls steadily. There are no families. Nothing is especially inert and nothing is especially reactive; each element is simply a slightly smaller, slightly more tightly bound version of the one before. Every atom would be less reactive than helium, and getting steadily more so.

There would be no valence, because valence is about what sits outside a closed shell and nothing ever closes. There would be no periodic table, because there is no period. There would be no ionic bonding, because that needs one atom eager to give and another eager to take.

The book’s closing line is worth quoting exactly:

A world without the Pauli exclusion principle would be very different. One thing is for certain: it would be a world with no chemists.

Where the site began, and where it ends

§0.3 opened with a promise: one recipe — discretize space, build a tridiagonal Hamiltonian, ask for its eigenvalues — would generate every result in this book. Eleven chapters later that recipe has produced the square well, the oscillator, tunnelling, the hydrogen atom and, in §11.1a, the screened potential that gives an atom its shells.

But the recipe alone was never enough. It produces levels. What turns a list of levels into an atom is ch10’s antisymmetry — a constraint that comes not from any equation about energy but from the observation that identical particles cannot be labelled.

This page is the measurement of that second ingredient. Take it away, keep every equation, and the levels are all still there — but everything gets to pile into the lowest one, and the world flattens into a smooth monotonic function of $Z$ .

Quantum mechanics gives you the energies. Indistinguishability gives you chemistry.

The problems extend the model to ions — problem 9 applies it to seven two-electron ions and reproduces every measured ionization energy to within about half a per cent, which is a striking return for a model with one fitted constant.

Check yourself

0 / 6 answered

1.The model of §11.3 reproduces hydrogen's binding energy and radius exactly, with no fitted parameter. Why should you not be too impressed?
The avoidance parameter $f$ is defined by $R_{ee} = f R$ and fitted to helium at $f = 1.67$ .
2.What is $f$ standing in for?
3.In Fig. 11.5 the hypothetical and real curves coincide at $Z = 1$ and $Z = 2$ . Why is that reassuring rather than suspicious?
4.Moving the $f$ slider changes the hypothetical curve's height but never its shape. Why does that matter more than the numerical disagreement?
5.Without the Pauli principle, what would chemistry look like?
6.This section is the site's closing argument. What is it really measuring?