Atoms · Quantum Mechanics (Phillips)

Atoms are identical, stable, and of a definite size. Those are three separate puzzles for a solar-system model, and one standing-wave condition closes all three at once.

Everyone knows atoms have discrete energy levels. Phillips spends about half a page on that and then says something more interesting: quantized energy is not the most amazing property of atoms.

The amazing properties are the ones so familiar that nobody thinks to be surprised by them. Atoms are resilient — knock one about and it returns to exactly the condition it started in. Atoms of an element are identical, with no manufacturing tolerance whatsoever. And despite eighty-fold differences in electron count, they are all about the same size.

No model built from little orbiting planets can deliver any of those three. This section is about why waves can.

Quantized levels

Later chapters derive it; for now, take it as given that a bound electron and proton have energies

E_n = -\frac{13.6}{n^2}\ \mathrm{eV}\tag{1.9}

where $n$ is the principal quantum number , an integer $1, 2, 3, \dots$

What this really says — read the picture, not the formula

Three features of that ladder matter more than the algebra.

The ground state is isolated. From $n=1$ to $n=2$ is 10.2 eV — an enormous gap on the atomic scale, five times a visible photon. That is why hydrogen gas is transparent: no photon your eye can see carries enough energy to excite it, so the light goes straight through.

The levels converge. They pile up towards $E = 0$ rather than continuing upward, because the gaps go as $1/n^3$ . An electron in a high $n$ state is barely bound at all.

Zero is not the bottom, it is the exit. Negative energy means bound; $E = 0$ means just barely free; above zero the electron has left — the atom is ionized — and its energy is no longer quantized at all — the continuum. Quantization is a property of being trapped, not a property of electrons. That is the single most transferable idea in the chapter, and §0.3 already showed you the mechanism: boundary conditions admit solutions only at particular eigenvalues.

The evidence: spectral lines

You cannot see an energy level. What you can see is the light emitted when an atom drops from one to another, carrying off the difference as a single photon :

\frac{hc}{\lambda} = |E_{n_i} - E_{n_f}|

Because the levels are discrete, the possible differences are discrete, so the emitted wavelengths are discrete. An atom does not glow in a smear of colour — it glows in lines, and the pattern of those lines is a fingerprint of the ladder that produced them.

Click a gold level for its energy and degeneracy.

Click a coloured arrow to see the photon it emits, and watch its line light up in the spectrum.

What this really says — which series a line belongs to is set by where it lands, not where it starts

Switch between the two series and watch where they land. The Lyman series ends on $n = 1$ , so each transition dumps at least 10.2 eV — deep ultraviolet, 91–122 nm, invisible. The Balmer series ends on $n = 2$ , starting 10.2 eV up the ladder, so the leftover energies are only 1.9–3.4 eV, which is exactly the visible band.

That is why Balmer was found first, in 1885, by a Swiss schoolteacher fitting a formula to four numbers — they were the only hydrogen lines anyone could see. The far larger Lyman series stayed hidden until people could work in the ultraviolet. The ladder was always the same; only part of it was visible.

Put a number on it — the red line in every hydrogen tube

The 3 → 2 transition:

\Delta E = 13.6\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \times \frac{5}{36} = 1.889\ \mathrm{eV}

\lambda = \frac{hc}{\Delta E} = \frac{1239.84\ \mathrm{eV\,nm}}{1.889\ \mathrm{eV}} = 656.4\ \mathrm{nm}

Deep red — H-α, the line that makes hydrogen discharge tubes and much of the night sky glow pink. It is measured at 656.28 nm. Our two-significant-figure $13.6$ eV reproduces it to four digits.

Worth keeping the shortcut $hc = 1240\ \mathrm{eV\,nm}$ : divide it by an energy in eV and you get a wavelength in nm, directly.

The book's mercury example does not add up — and it is worth checking

Phillips illustrates the same idea with mercury (p. 8): an electron passing through mercury vapour loses energy once it exceeds 4.2 eV, the gap between the ground and first excited state, and the excited atoms then emit photons of ” $\epsilon = 4.2\ \mathrm{eV}$ and wavelength $\lambda = hc/\epsilon = 254$ nm.”

Those two numbers contradict each other. Run it:

\frac{1239.84\ \mathrm{eV\,nm}}{4.2\ \mathrm{eV}} = 295\ \mathrm{nm}, \qquad \frac{1239.84\ \mathrm{eV\,nm}}{254\ \mathrm{nm}} = 4.88\ \mathrm{eV}

254 nm is right — it is the famous mercury resonance line, the one in every germicidal UV lamp, measured at 253.7 nm. The excitation energy that goes with it is about 4.9 eV, not 4.2, and 4.9 V is precisely the value Franck and Hertz measured in 1914 in the experiment this passage is describing.

So the wavelength is correct and the energy is a slip. Nothing else in the book depends on it. The point of flagging it rather than quietly printing 4.9 is that you should be able to open Phillips alongside this site and find the same numbers — and that you now have the habit that caught it: when a source gives you two numbers linked by a formula, spend the five seconds to check they agree.

The properties that really need explaining

Now the part Phillips actually cares about. Set the spectra aside and ask why atoms behave like this at all:

They are resilient. Atoms collide constantly and are mostly unaffected; when they are excited they return to their original pristine condition.
They are identical. Every atom with the same number of electrons has the same properties. Not similar — identical, with no spread.
They barely vary in size. Mercury has 80 electrons to hydrogen’s one, and is only about three times bigger.

Physics you need first — why a classical atom cannot exist

By 1911 Rutherford had established the picture: a tiny dense positive nucleus with electrons somewhere around it. The obvious model is a solar system, with the Coulomb attraction playing the role of gravity. It fails, and it fails catastrophically, for a reason from classical electromagnetism:

An accelerating charge radiates.

This is the Larmor result, and there is no way around it. An electron in a circular orbit is accelerating — centripetally — every moment it orbits. So it must continuously radiate away energy. Radiating means losing energy, losing energy means falling to a smaller orbit, and a smaller orbit means accelerating harder and radiating faster. It runs away.

You can put a number on the disaster: an electron starting at the Bohr radius spirals into the nucleus in about 16 picoseconds. Not “eventually” — essentially immediately. Classical physics does not predict that atoms are fragile; it predicts that atoms cannot exist for even one nanosecond.

And even if you waved that away, the model would still fail the other two tests. Orbits come in a continuum, so no two atoms would be alike; and every collision would leave an atom on a different orbit, so nothing would be resilient or reproducible.

The same electron, two descriptions. Dashed slate is always the classical prediction on this site; the classical one here is not merely inaccurate, it is catastrophic.

Reproduce it in NumPy

The whole section in one script: the two constants of Eqs. (1.10) and (1.11), the classical disaster, and the spectrum from Eq. (1.9).

import numpy as np

hbar, e, me, c   = 1.055e-34, 1.602e-19, 9.109e-31, 2.998e8
eps0, eV         = 8.854e-12, 1.602e-19

# Eq (1.10): the ONLY length you can build from these four constants
a0 = (4*np.pi*eps0 / e**2) * (hbar**2 / me)
print(f"a0  = {a0:.4e} m      (book: 0.529e-10 m)")

# Eq (1.11): and then the only energy
E_R = e**2 / (8*np.pi*eps0*a0)
print(f"E_R = {E_R/eV:.3f} eV        (book: 13.6 eV)")

# the classical disaster: Larmor radiation collapses the orbit
r_e = e**2 / (4*np.pi*eps0*me*c**2)          # classical electron radius
t   = a0**3 / (4 * r_e**2 * c)
print(f"classical collapse time = {t:.2e} s   ({t*1e12:.1f} picoseconds)")

# the spectrum, from Eq (1.9)
hc = 1239.84
for ni, nf, name in [(2,1,"Lyman a"), (3,1,"Lyman b"), (3,2,"H-alpha"), (4,2,"H-beta")]:
    dE = 13.6*(1/nf**2 - 1/ni**2)
    print(f"{name:9s} {ni}->{nf}:  dE = {dE:6.3f} eV,  lambda = {hc/dE:6.1f} nm")

printsa0 = 5.2973e-11 m (book: 0.529e-10 m) E_R = 13.590 eV (book: 13.6 eV) classical collapse time = 1.56e-11 s (15.6 picoseconds) Lyman a 2->1: dE = 10.200 eV, lambda = 121.6 nm Lyman b 3->1: dE = 12.089 eV, lambda = 102.6 nm H-alpha 3->2: dE = 1.889 eV, lambda = 656.4 nm H-beta 4->2: dE = 2.550 eV, lambda = 486.2 nm

Waves fix all three at once

Phillips’ resolution is one sentence long, and it is worth reading slowly:

To some extent atoms behave like musical instruments. When a violin string vibrates with definite frequency, it forms a standing wave pattern of specific shape. When wave-like electrons, with definite energy, are confined inside an atom, they form a wave pattern of specific shape.

Take the three puzzles in turn.

Resilience. A standing wave pattern has a lowest mode. Disturb a violin string and it rings; leave it alone and it settles back to the shape it prefers. There is nothing below the fundamental to decay into. An atom left alone assumes its lowest-energy wave pattern, and in that state there is no tendency to radiate and fall inward — because there is nowhere to fall.

Identity. Modes are not adjustable. A string of a given length and tension has exactly one fundamental, determined by the string, not by its history. Two atoms with the same nucleus and the same number of electrons support exactly the same wave patterns, so they are identical in the strong sense — not similar, the same.

Size. The wave has to fit. Its wavelength is $\lambda = h/p$ , so the size of the pattern is fixed by $h$ , the electron mass, and how hard the nucleus pulls. Adding electrons does not change any of those much, which is why an 80-electron atom is only a few times bigger than a one-electron atom.

Engineer's bridge — an atom is a resonator, and this is an eigenvalue problem

Every word in that violin analogy is literally true, and you already own the mathematics.

A string clamped at both ends supports only frequencies $f_n = nv/2L$ . Nothing forbids you driving it at another frequency, but only those survive as free oscillations, because only they satisfy the boundary conditions at both ends. The allowed set is discrete because the constraint is a boundary-value problem, and a boundary-value problem has non-trivial solutions only at particular eigenvalues.

String	Atom
clamped at both ends	wave must decay to zero far from the nucleus
allowed frequencies $f_n$	allowed energies $E_n$
the fundamental $f_1$	the ground state $E_1$
harmonics	excited states
mode shape	orbital
a mode of a given string is unique	atoms of an element are identical
you cannot get below the fundamental	atoms do not collapse

This is the same statement as §0.3: put the problem on a grid, build the matrix, and the discrete spectrum falls out of eigh. Chapter 4 §4.2 makes the string comparison exact, and chapter 9 solves the atom this way.

Where the analogy breaks — and it matters.

A string’s displacement is a real, observable thing you can photograph. The electron’s wave is not. $\psi$ is complex and unobservable; only $|\psi|^2$ has meaning, as a probability density. Nothing is physically waving.
A string vibrates in time at frequency $f_n$ ; the atom’s ground state does nothing at all. A stationary state is stationary — its $|\psi|^2$ is frozen. The “frequency” $E_n/h$ is a phase rotation you cannot observe.
A string is a one-dimensional line; the atom is three-dimensional, so its modes need three quantum numbers, not one. $n$ alone is the whole story only for hydrogen’s energy, and even that is a special coincidence of the Coulomb potential (chapter 9).

Building the atom out of constants

The last argument in §1.3 is a piece of pure dimensional analysis, and it is the same trick as the ⇄ box in §0.2. If a wave-like electron is bound by the Coulomb attraction, then only three quantities can possibly be involved: $\hbar$ (it is a wave), $m_e$ (it is that particle), and $e^2/4\pi\epsilon_0$ (that is the force). Not $c$ — the footnote on p. 10 notes the electron is non-relativistic, which §0.2 confirmed with $v/c = \alpha = 1/137$ .

step 1 of 5

Three ingredients, one target. Nothing here solves the Schrödinger equation; we are only asking what a length COULD be made of.

1List the ingredients and their dimensions. This is the whole input to the argument.
$[ℏ] = M L^{2} T^{- 1}, [m_{e}] = M, [\frac{e ^{2}}{4 π ϵ _{0}}] = M L^{3} T^{- 2}$
The last one is an energy times a length, because e²/4πε₀r is a potential energy.
2Ask for the combination that is a pure length. Write it with unknown powers and solve for them.
$a_{0} = ℏ^{α} m_{e}^{β} (\frac{e ^{2}}{4 π ϵ _{0}})^{γ} =! L$
3Match the exponent of each dimension separately — three equations, three unknowns. This is the only solution.
$α = 2, β = - 1, γ = - 1$
4Substitute back. The answer was forced; there was never any freedom in it.
$a_{0} = [\frac{4 π ϵ _{0}}{e ^{2}}] \frac{ℏ ^{2}}{m _{e}} = 0.529 \times 1 0^{- 10} m$
Equation (1.10) — the Bohr radius. And it is right: atoms really are about an ångström across.
5Now the energy comes free, because a length and a charge already make one.
$E_{R} = \frac{e ^{2}}{8 π ϵ _{0} a _{0}} = 13.6 eV$
Equation (1.11) — the Rydberg energy, exactly the 13.6 eV asserted back in Eq. (1.9). The binding energy of state n is E_R/n².

What this really says — the argument is stronger than it looks

No physics was solved here. We wrote down three constants, demanded a length, and only one combination existed. That the Bohr radius it produces is the actual size of an atom is the evidence.

The logic is worth stating plainly, because it recurs throughout the book: if the answer must be built from these ingredients, and only one combination has the right dimensions, then the answer is that combination times a pure number. Dimensional analysis cannot give you the pure number — it might have been $4\pi$ or $1/2$ — but it fixes the scale, and the scale is what you had no feel for. The energy that comes with it is the Rydberg energy , and it is exactly the 13.6 eV Eq. (1.9) opened with.

Problem 4 at the end of the chapter drives the point home from the other side: try it with a relativistic classical electron, using $m_e$ , $e^2/4\pi\epsilon_0$ and $c$ instead, and the only length available comes out four orders of magnitude too small. Choosing $\hbar$ over $c$ — treating the electron as a wave rather than as a fast classical particle — is what makes the number right.

Bohr saw this in 1913, before there was any wave mechanics to justify it:

Whatever alteration in the laws of motion of electrons may be, it seems necessary to introduce in the laws in question a quantity foreign to the classical electrodynamics; i.e., Planck’s constant… as this constant is of such dimensions and magnitude that it, together with the mass and the charge of the particles, can determine a length of the order of the magnitude required.

He had the right constant and the right conclusion a decade before anyone knew why. The reason, as Phillips puts it, is that Planck’s constant “links the particle-like and wave-like properties of atomic electrons” — which is what §1.2 established and what chapter 2 will turn into an equation.

Check yourself

0 / 6 answered

Look at the gap between $n = 1$ and $n = 2$ on the ladder.
1.Why is hydrogen gas transparent to visible light?
2.The Balmer series is visible and the Lyman series is not. What causes the difference?
3.A classical electron orbiting a proton at the Bohr radius spirals in within about 16 picoseconds. What does that argue?
4.The violin-string analogy explains three things at once. Which is NOT one of them?
5.The book gives the mercury excitation as 4.2 eV and the emitted wavelength as 254 nm. What should you conclude?
6.Dimensional analysis gives $a_{0} = (4 π ϵ_{0} / e^{2}) (ℏ^{2} / m_{e})$ . What has it actually established?