Constants, Units and Orders of Magnitude · Quantum Mechanics (Phillips)

Almost every mistake in this subject is a factor of a thousand rather than a sign error. Four constants and a length ladder are enough to catch them.

Quantum mechanics is a subject you can get badly lost in without ever making an algebra mistake, simply by having no feel for how big things are. An answer of $10^{-19}$ joules is either exactly right or off by twenty orders of magnitude, and nothing in the formula tells you which.

This page is the antidote. It is the book’s own table of constants, the units it uses, and — most usefully — a ladder of landmarks so you can tell at a glance whether a number is plausible before you check it.

Everything here is also one click away from every other page: the ħ constants button in the top-right corner opens the same table, converter and calculators wherever you are.

Four numbers, and everything else

The whole book is built from a handful of constants. Four of them do most of the work:


$\hbar = 1.055\times10^{-34}\ \mathrm{J\,s}$	the quantum of action	how big quantum effects are
$e = 1.602\times10^{-19}\ \mathrm{C}$	the elementary charge	how hard the nucleus pulls
$m_e = 9.109\times10^{-31}\ \mathrm{kg}$	the electron mass	how reluctantly it moves
$c = 2.998\times10^{8}\ \mathrm{m\,s^{-1}}$	the speed of light	when relativity starts to matter

Nearly every other quantity in the book — the size of an atom, the energy to ionize it, the strength of its magnetism — is these four combined in some way. That is not a coincidence and it is not numerology: if a hydrogen atom is built only out of an electron, a proton and the Coulomb force, then the only length its size could be made of is the one you can assemble from $\hbar$ , $e$ and $m_e$ . Chapter 1 §1.3 makes exactly this argument, and chapter 9 does the algebra properly.

Physics you need first — what an electronvolt is

You will see almost every energy in this book quoted in eV, so it is worth thirty seconds on where the unit comes from.

Push a charge $q$ through a potential difference $V$ and it gains energy $qV$ . That is the definition of voltage, and it is the same relation as in any circuit. Now take the smallest charge there is — one electron, $e = 1.602 \times 10^{-19}\ \mathrm{C}$ — and push it through one volt:

1\ \mathrm{eV} = (1.602\times10^{-19}\ \mathrm{C})(1\ \mathrm{V}) = 1.602\times10^{-19}\ \mathrm{J}

So the electronvolt is not a new physical idea. It is a convenient size. Atomic energies are around $10^{-19}$ J, which is a miserable number to carry around; in eV the same energies are 1 to 100. When Phillips writes that hydrogen’s ground state is $-13.6\ \mathrm{eV}$ , he means “it takes 13.6 volts’ worth of push to tear the electron off” — and that is genuinely how you would measure it.

The unit scales the way you would expect: $\mathrm{keV} = 10^3$ , $\mathrm{MeV} = 10^6$ , $\mathrm{GeV} = 10^9\ \mathrm{eV}$ .

Where things live

Click any landmark for why it is worth remembering. The rows are decades, so one row apart means ten times bigger.

What this really says — five decades of nothing separate the nucleus from the atom

Look at the length ladder and notice how narrow the interesting band is. From the nucleus at $10^{-15}\ \mathrm{m}$ to the atom at $10^{-10}\ \mathrm{m}$ is five decades of nothing — the atom is some fifty thousand times wider than the thing at its centre, and essentially all of it is empty. Then from the atom up to visible light is another three and a half decades. This book lives almost entirely in the two decades around $10^{-10}\ \mathrm{m}$ and the two decades around $1\ \mathrm{eV}$ .

That is why quantum mechanics went unnoticed until 1900. Nothing you can see or touch is anywhere near this scale, and $\hbar$ is small enough that its effects average away completely before you get to anything a nineteenth-century laboratory could measure.

Engineer's bridge — units are a type system

Dimensional analysis is type checking, and you should use it exactly the way you use a compiler.

Every physical quantity carries a type: a length is L, a time is T, a mass is M, an energy is M·L²·T⁻². The rules are the ones you would guess:

you may only add quantities of the same type — energy + length is a type error, and if your algebra produces one, the algebra is wrong;
multiplying multiplies the types, so force × distance really does have the type of energy;
the argument of $\exp$ , $\sin$ , $\ln$ must be dimensionless. This one catches more mistakes than any other. If you derive $e^{-x/a}$ then $a$ is forced to be a length, and you have learnt something about your answer for free.

Two habits worth stealing:

Check the type of every result before you check its value. In chapter 5 you will meet $\psi \sim e^{-\beta x}$ with $\beta = \sqrt{2m(V-E)}/\hbar$ . Is that right? $m(V-E)$ has type M · M L² T⁻² = M² L² T⁻², whose square root is M L T⁻¹ — a momentum. Divided by $\hbar$ (type M L² T⁻¹) that leaves L⁻¹. So $\beta x$ is dimensionless and the expression type-checks. You now believe the formula more than you did, without having followed the derivation.

Build the answer out of the constants before you solve anything. If a problem only involves $\hbar$ , $m$ and a length $L$ , then the only energy you can construct is $\hbar^2/mL^2$ times some pure number. Every energy in chapters 4, 5 and 6 has this form. You will often get the answer to within a factor of $\pi^2$ before doing any calculus at all.

Where it breaks: type-checking proves an expression is possible, never that it is right. $\hbar^2/mL^2$ and $17\hbar^2/mL^2$ type-check equally well, and dimensionless groups like $\alpha$ or $Z$ are invisible to it — you can multiply by them freely without the types complaining. It is a linter, not a proof.

The atom, built from scratch

Here is the argument that shows the four constants really are enough. From $\hbar$ , $e$ , $m_e$ and $\epsilon_0$ there is exactly one length you can build — the Bohr radius — and it turns out to be the size of a hydrogen atom:

a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} = 0.5292\times10^{-10}\ \mathrm{m}

What this really says — the atom's size is a tug of war, and both sides are calculable

Read it as a tug of war. Quantum mechanics pushes the electron out: confining it to a small box costs kinetic energy, and that cost goes as $\hbar^2/m_e r^2$ — fiercely expensive as $r$ shrinks. The Coulomb attraction pulls it in, at a cost of $-e^2/4\pi\epsilon_0 r$ . One term blows up faster than the other as $r \to 0$ , so there is a stand-off, and $a_0$ is where it settles.

This is why atoms have a size at all — and it is the answer to the question that broke classical physics. Classically, nothing stops the electron spiralling into the nucleus. Quantum mechanically, squeezing it into a smaller space raises its kinetic energy faster than the Coulomb attraction can pay for, so the collapse stops. Chapter 1 §1.3 makes this argument in words; chapter 9 makes it exactly.

Pair it with an energy, and the same four constants give the ionization energy of hydrogen — the Rydberg energy :

E_R = \frac{\hbar^2}{2m_e a_0^2} = \tfrac{1}{2}\alpha^2 m_e c^2 = 13.61\ \mathrm{eV}

Put a number on it — is the electron relativistic?

The second form above is worth pausing on, because it answers a question you should be asking: do we need relativity for atoms?

The fine structure constant is the dimensionless combination

\alpha = \frac{e^2}{4\pi\epsilon_0\hbar c} = \frac{1}{137.0}

and it turns out to be the electron’s orbital speed as a fraction of $c$ :

\frac{v}{c} = \alpha \approx \frac{1}{137} \quad\Longrightarrow\quad v \approx 2.19\times10^{6}\ \mathrm{m\,s^{-1}}

So the electron in hydrogen moves at about 0.7% of the speed of light. The relativistic correction to its energy goes as $(v/c)^2 = \alpha^2 \approx 5\times10^{-5}$ — about one part in twenty thousand. Small enough to ignore for the whole book, which is exactly why Phillips can write a non-relativistic textbook and still get atomic physics right.

It is not zero, though, and it is measurable: that one-part-in-20 000 splitting of spectral lines is called fine structure, which is where $\alpha$ gets its name. Chapter 9 §9.6 works out the correction.

And $E_R = \tfrac{1}{2}\alpha^2 m_ec^2$ now reads as a sentence: the binding energy of hydrogen is $\alpha^2/2$ of the electron’s rest-mass energy — $\tfrac{1}{2}(1/137)^2 \times 511\,000\ \mathrm{eV} = 13.6\ \mathrm{eV}$ .

Reproduce it in NumPy

Building the atom from nothing but the four constants. The disagreements in the third and fourth digits come from the book's own rounded inputs, not from the physics — see the caution below.

import numpy as np

hbar = 1.055e-34      # J s
e    = 1.602e-19      # C
me   = 9.109e-31      # kg
c    = 2.998e8        # m/s
eps0 = 8.854e-12      # C^2 N^-1 m^-2
eV   = 1.602e-19      # J

a0    = 4*np.pi*eps0*hbar**2 / (me*e**2)
E_R   = me*e**4 / (2*(4*np.pi*eps0)**2 * hbar**2)
alpha = e**2 / (4*np.pi*eps0*hbar*c)

print(f"a0  = {a0:.4e} m   (book: 0.5292e-10 m)")
print(f"E_R = {E_R/eV:.4f} eV    (book: 13.61 eV)")
print(f"alpha = 1/{1/alpha:.2f}")
print(f"v/c   = {alpha:.7f}")

printsa0 = 5.2973e-11 m (book: 0.5292e-10 m) E_R = 13.5902 eV (book: 13.61 eV) alpha = 1/137.12 v/c = 0.0072928

Common misconception — the book's numbers are not CODATA's

Every constant on this site is quoted exactly as Phillips quotes it on the inside back cover: $\hbar = 1.055\times10^{-34}$ , $\alpha = 1/137.0$ , $E_R = 13.61\ \mathrm{eV}$ . Modern reference values differ in the last digit or two ( $\alpha^{-1} = 137.036$ , $E_R = 13.6057\ \mathrm{eV}$ ).

That is deliberate, and the site will not “correct” it. The point is that a worked example here reproduces the book’s arithmetic, so when your answer disagrees with the page you know it is your algebra and not a silently updated constant. At this book’s precision — three or four significant figures — the difference never matters.

There is a second, subtler effect worth noticing. The code block above computes $\alpha$ from the book’s own $\hbar$ , $e$ , $\epsilon_0$ and $c$ — and gets $1/137.12$ , not the $1/137.0$ the same table quotes. Nothing is wrong: the table rounds each input to four significant figures, and $\alpha$ is built from four of them at once, so the rounding errors compound. A quantity derived from several rounded constants is less precise than any of them. That is worth internalising before you start checking your own answers against the page to five digits.

Time scales

The time ladder is the one people find most surprising. An electron goes round a hydrogen atom in $1.5\times10^{-16}\ \mathrm{s}$ , but the atom then sits in an excited state for something like a nanosecond before emitting — about seven million orbits. On the atom’s own clock, a radiative transition is a geologically slow event, which is why treating an excited state as a well-defined state with a definite energy works as well as it does.

The full table

The book’s own table, from the inside back cover, plus a converter and the three calculations this book asks for over and over.

Electromagnetism

$c$	velocity of light in vacuum	$299792458$ $m s^{- 1}$ exact
$μ_{0}$	permeability of vacuum	$4 π \times 1 0^{- 7}$ $N A^{- 2}$ exact
$ϵ_{0}$	permittivity of vacuum $c = 1/ ϵ_{0} μ_{0}$	$8.854 \times 1 0^{- 12}$ $C^{2} N^{- 1} m^{- 2}$

Quantum

$h$	Planck constant	$6.626 \times 1 0^{- 34}$ $J s$
$ℏ$	reduced Planck constant $ℏ = h /2 π$	$1.055 \times 1 0^{- 34}$ $J s$

Gravitation

G

gravitational constant

6.674 \times 1 0^{- 11}

m^{3} k g^{- 1} s^{- 2}

Charge and energy

$e$	elementary charge	$1.602 \times 1 0^{- 19}$ $C$
$eV$	electronvolt	$1.602 \times 1 0^{- 19}$ $J$
$α$	fine structure constant $α = e^{2} /4 π ϵ_{0} ℏ c$	$1/137.0$

The electron

$m_{e}$	electron mass	$9.109 \times 1 0^{- 31}$ $kg$
$m_{e} c^{2}$	electron rest-mass energy	$0.511 MeV$ $J$
$μ_{B}$	Bohr magneton $μ_{B} = e ℏ/2 m_{e}$	$9.274 \times 1 0^{- 24}$ $J T^{- 1}$

Atomic scale

$E_{R}$	Rydberg energy $E_{R} = α^{2} m_{e} c^{2} /2$	$13.61 eV$ $J$
$a_{0}$	Bohr radius $a_{0} = (1/ α) (ℏ/ m_{e} c)$	$0.5292 \times 1 0^{- 10}$ $m$
$\overset{˚}{A}$	angstrom	$1 0^{- 10}$ $m$ exact

Nucleons

$m_{p}$	proton mass	$1.673 \times 1 0^{- 27}$ $kg$
$m_{p} c^{2}$	proton rest-mass energy	$938.272 MeV$ $J$
$m_{n} c^{2}$	neutron rest-mass energy	$939.566 MeV$ $J$
$μ_{N}$	nuclear magneton $μ_{N} = e ℏ/2 m_{p}$	$5.051 \times 1 0^{- 27}$ $J T^{- 1}$
$fm$	femtometre or fermi	$1 0^{- 15}$ $m$ exact
$b$	barn	$1 0^{- 28}$ $m^{2}$ exact

Chemistry

$u$	atomic mass unit $u = \frac{1}{12} m (^{12} C atom)$	$1.661 \times 1 0^{- 27}$ $kg$
$N_{A}$	Avogadro constant, atoms in gram mol	$6.022 \times 1 0^{23}$ $mo l^{- 1}$

Thermal

$T_{t}$	triple-point temperature	$273.16$ $K$
$k$	Boltzmann constant	$1.381 \times 1 0^{- 23}$ $J K^{- 1}$
$R$	molar gas constant $R = N_{A} k$	$8.315$ $J mo l^{- 1} K^{- 1}$
$σ$	Stefan–Boltzmann constant $σ = (π^{2} /60) (k^{4} / ℏ^{3} c^{2})$	$5.670 \times 1 0^{- 8}$ $W m^{- 2} K^{- 4}$

Transcribed from the inside back cover of Phillips. These are the book's own values — a worked example on any page reproduces this arithmetic, not the latest CODATA digits.

Engineer's bridge — ħ over an energy is a time

One trick pays for itself repeatedly. $\hbar$ has units of energy × time, so:

\frac{\hbar}{E} = \text{a time}, \qquad \frac{\hbar}{p} = \text{a length}, \qquad \hbar\omega = \text{an energy}

$\hbar$ is a unit converter between the wave description and the particle description. Given any energy, $\hbar/E$ is immediately the timescale on which that energy’s physics happens; given any momentum, $\hbar/p$ is the length scale over which the wave oscillates.

If you have done signal processing this should feel familiar, because it is the same statement as bandwidth × duration ≈ 1. A signal confined to a short time must be spread over a wide band; a state confined to a small energy spread must persist for a long time. In chapter 3 this stops being an analogy and becomes the same theorem: $\Delta x\,\Delta p \ge \hbar/2$ is the time–bandwidth product with $\hbar$ setting the scale.

Where it breaks: $\hbar/E$ gives you an order of magnitude, not an exact period. The factors of $2\pi$ are real and they matter once you are computing rather than estimating — note that the electron’s orbital period on the ladder above is about $2\pi$ times $\hbar/E_R$ , not equal to it.

Check yourself

0 / 6 answered

1.You derive a result $ψ \sim e^{- β x}$ and want to check it before trusting the algebra. What must $β$ be?
2.A problem involves only $ℏ$ , a mass $m$ and a length $L$ . Without doing any calculus, what must the energy scale be?
3.Room-temperature $k T$ is about 0.026 eV. A CO molecule's vibrational quantum is $ℏ ω = 0.27$ eV. What follows?
4.The fine structure constant is $α = 1/137$ , and it equals the electron's orbital speed in hydrogen divided by $c$ . What does that let you conclude?
5.The code on this page computes $α$ from the book's own $ℏ$ , $e$ , $ϵ_{0}$ and $c$ , and gets $1/137.12$ — while the same table quotes $1/137.0$ . What is going on?
Compare the two on the time ladder above.
6.An electron orbits hydrogen in about $1.5 \times 1 0^{- 16}$ s, but an excited atom takes roughly a nanosecond to emit a photon. Why does that matter?