A hydrogen atom has no single size. Two different radii are quoted for the ground state, they differ by half, and each is the correct answer to a different question.

§9.2 produced the energies and the eigenfunctions. This section asks the question a chemist would ask: where is the electron, and what shape does the atom have?

The radial distribution

The probability of finding the electron in a volume element $\mathrm d^3\mathbf r$ is $|\psi_{n_r,l,m_l}|^2\,\mathrm d^3\mathbf r$ . Writing $\mathrm d^3\mathbf r = r^2\,\mathrm dr\,\mathrm d\Omega$ and $\psi = (u_{n_r,l}/r)\,Y_{l,m_l}$ , the probability of being between $r$ and $r + \mathrm dr$ is

\left|\frac{u_{n_r,l}(r)}{r}\right|^2r^2\,\mathrm dr\int|Y_{l,m_l}|^2\,\mathrm d\Omega = |u_{n_r,l}(r)|^2\,\mathrm dr

because the spherical harmonic is normalized (Eq. 8.25). So the radial probability density is simply $|u(r)|^2$ — which is exactly why §9.1 defined $u = rR$ in the first place.

Why the r² and the 1/r² cancel, and why that is the point

There are two competing factors here and it is easy to lose track of which wins.

The wave function itself, $R = u/r$ , is largest at the origin for an s-state — it decays outward. But the volume of a shell at radius $r$ grows as $r^2$ , so there is more space to be found in further out.

The radial probability density multiplies them: $|R|^2 \cdot r^2 = |u|^2$ . The two factors of $r$ cancel exactly, which is the whole reason $u$ is the natural variable. $|u(r)|^2$ is what you should picture when you imagine “where the electron is” — not $|R|^2$ , and not $|\psi|^2$ at a point.

That cancellation is why the 1s density peaks at $a_0$ rather than at $r = 0$ , even though $|\psi|^2$ is largest at the nucleus. There is simply no volume at the origin.

Angular shapes: |Y_l,m(θ,φ)|² for l ≤ 3

n — principal3l — capped at n−11 (p)|m_l| — capped at l0

state 3p, n_r = 1

⟨r⟩ = 12.50 a₀, peak at 12.01 a₀

nodes: 1 angular, 1 radial

m = 0 gives the most nodal circles — 1 of them, the maximum for this l. With no angular momentum about z, the density is free to pile up at the poles. Note there is no dependence on φ, whatever l and m — because |e^imφ|² = 1. A definite L_z means the wave function is completely smeared around the z axis.

The mean radius can be evaluated in closed form:

\langle r\rangle_{n_r,l} = \tfrac12\left[3n_r^2 + 6n_r(l+1) + (l+1)(2l+3)\right]a_0\tag{9.25}

The mean radius is not the most probable radius

For the 1s state, the most probable radius is exactly $a_0$ — that is where $|u|^2$ peaks, and it is the Bohr radius’s original meaning. But Eq. (9.25) with $n_r = l = 0$ gives

\langle r\rangle = \tfrac12(1)(3)a_0 = 1.5\,a_0

Both are correct; they answer different questions. The radial density rises like $r^2$ and falls like $e^{-2r/a_0}$ , so it is skewed — a long tail to the right drags the mean out past the peak, exactly as a skewed distribution’s mean sits beyond its mode.

So “the size of a hydrogen atom is $a_0$ ” and “…is $1.5a_0$ ” are both defensible, and any page quoting one owes you which it means. Problem 3 asks for both precisely to force the distinction, and the widget above draws both lines.

Reproduce it in NumPy

The integration uses Table 9.1's own functions, so this checks the book's closed form against the book's own eigenfunctions rather than against theory. Note 4d is the one case where the peak lies BEYOND the mean — a state with a radial node has two lobes, and when the outer one dominates the skew reverses.

import numpy as np
s = np.sqrt

# Table 9.1 again, atomic units (a0 = 1, x = r/a0)
TABLE = [
 ("1s", 0, 0, lambda x: 2*x*np.exp(-x)),
 ("2s", 1, 0, lambda x: (1/s(2))*(1 - x/2)*x*np.exp(-x/2)),
 ("3s", 2, 0, lambda x: (2/(3*s(3)))*(1 - (2/3)*x + (2/27)*x**2)*x*np.exp(-x/3)),
 ("2p", 0, 1, lambda x: (1/(2*s(6)))*x**2*np.exp(-x/2)),
 ("3p", 1, 1, lambda x: (8/(27*s(6)))*(1 - x/6)*x**2*np.exp(-x/3)),
 ("3d", 0, 2, lambda x: (4/(81*s(30)))*x**3*np.exp(-x/3)),
 ("4d", 1, 2, lambda x: (1/(64*s(5)))*(1 - x/12)*x**3*np.exp(-x/4)),
]
x = np.linspace(1e-7, 300, 2_000_001)

print("Eq. (9.25) checked against direct integration, and the peak vs the mean")
print("=" * 78)
print(f"  {'state':>5} {'n':>3} {'<r> integrated':>15} {'Eq. (9.25)':>12} "
      f"{'(a0/2)[3n^2-l(l+1)]':>21} {'most probable':>14}")
ok = True
for name, nr, l, f in TABLE:
    u = f(x); p = u*u
    mean = np.trapezoid(x*p, x)
    book = 0.5*(3*nr**2 + 6*nr*(l+1) + (l+1)*(2*l+3))
    n = nr + l + 1
    std = 0.5*(3*n*n - l*(l+1))
    peak = x[int(np.argmax(p))]
    ok &= abs(mean-book) < 2e-4 and abs(book-std) < 1e-12
    print(f"  {name:>5} {n:>3} {mean:>15.6f} {book:>12.4f} {std:>21.4f} {peak:>14.4f}")
print(f"\n  all agree: {ok}")
print("\n  The two closed forms are algebraically identical -- the book writes it in")
print("  n_r and l, every other textbook writes it in n and l.\n")

print("  THE PEAK IS NOT THE MEAN, and 1s is where it bites:")
u = TABLE[0][3](x); p = u*u
print(f"    1s most probable radius = {x[int(np.argmax(p))]:.4f} a0   (exactly a0)")
print(f"    1s mean radius <r>      = {np.trapezoid(x*p, x):.4f} a0   (exactly 1.5 a0)")
print("    The radial density |u|^2 is skewed -- it rises like r^2 and falls like")
print("    e^{-2r}, so it has a long tail to the right that drags the mean out past")
print("    the peak.  Quote 'the size of a hydrogen atom' and you must say which.")

printsEq. (9.25) checked against direct integration, and the peak vs the mean ============================================================================== state n <r> integrated Eq. (9.25) (a0/2)[3n^2-l(l+1)] most probable 1s 1 1.500000 1.5000 1.5000 1.0001 2s 2 6.000000 6.0000 6.0000 5.2361 3s 3 13.500000 13.5000 13.5000 13.0740 2p 2 5.000000 5.0000 5.0000 4.0001 3p 3 12.500000 12.5000 12.5000 12.0000 3d 3 10.500000 10.5000 10.5000 9.0000 4d 4 21.000000 21.0000 21.0000 21.2111 all agree: True The two closed forms are algebraically identical -- the book writes it in n_r and l, every other textbook writes it in n and l. THE PEAK IS NOT THE MEAN, and 1s is where it bites: 1s most probable radius = 1.0001 a0 (exactly a0) 1s mean radius <r> = 1.5000 a0 (exactly 1.5 a0) The radial density |u|^2 is skewed -- it rises like r^2 and falls like e^{-2r}, so it has a long tail to the right that drags the mean out past the peak. Quote 'the size of a hydrogen atom' and you must say which.

How the atom grows with n and with l

Read Eq. (9.25) in the form $\langle r\rangle = \frac{a_0}{2}[3n^2 - l(l+1)]$ and two patterns fall out.

$n$ dominates, quadratically. The $3n^2$ term swamps everything else, so an atom’s size goes as $n^2$ :

state	$\langle r\rangle$
1s	$1.5\,a_0$	0.08 nm
3d	$10.5\,a_0$	0.56 nm
10s	$150\,a_0$	7.9 nm
100s	$15{,}000\,a_0$	0.79 μm

A Rydberg atom with $n = 100$ is about a micrometre across — bigger than many bacteria, and large enough to see the individual atom’s size in a microscope. Those states are real and routinely made in laboratories.

At fixed $n$ , higher $l$ is smaller. The $-l(l+1)$ term subtracts, so within $n = 3$ : 3s is $13.5a_0$ , 3p is $12.5a_0$ , 3d is $10.5a_0$ .

That looks backwards — the centrifugal barrier pushes higher- $l$ states outward, so why are they smaller? Because at fixed $n$ , raising $l$ means lowering $n_r$ : the s-state has two radial nodes and a far-reaching outer lobe, while 3d has none and sits in a single compact shell. The long tail costs more than the barrier.

The shape

The angular part contributes $|Y_{l,m_l}(\theta,\phi)|^2$ , which by Eq. (8.26) has no dependence on $\phi$ . So the probability density is unchanged by rotation about the $z$ axis, and everything about the state’s size and shape is visible on a single vertical plane through that axis — which is exactly what the book’s Figs. 9.6 and 9.7 draw.

Angular shapes: |Y_l,m(θ,φ)|² for l ≤ 3

n — principal3l — capped at n−11 (p)|m_l| — capped at l0

state 3p, n_r = 1

⟨r⟩ = 12.50 a₀, peak at 12.01 a₀

nodes: 1 angular, 1 radial

Reading an orbital picture

Two independent structures are superimposed, and separating them makes the pictures readable.

Radial nodes — spheres. There are $n_r = n - l - 1$ of them, and they show up as dark shells at fixed radius. The 3p state has one; the 3d state has none.

Angular nodes — cones and planes. There are $l - |m_l|$ nodal circles of latitude, from §8.3b. For $m_l = 0$ they are the familiar pinches along the axis; for $|m_l| = l$ there are none and the density crowds onto the equator.

Together they always total $n - 1$ . That is worth checking against the widget: dial any state and count. It is the three-dimensional version of the node-counting rule the site has used since chapter 4 — more nodes, more curvature, more energy — and it explains why every state with the same $n$ has the same energy despite looking completely different. They have the same total number of nodes, just distributed differently between radial and angular.

That is the clearest intuitive handle on the accidental degeneracy that §9.2 flagged, though it is a mnemonic rather than a derivation: node counting alone does not explain why radial and angular nodes should cost exactly the same energy.

Engineer's bridge — the book's figures were made by rejection sampling

Figs. 9.6 and 9.7 are dot-density plots, and the caption says how they were produced: “by selecting a point on the plane at random and deciding to plot or not to plot in accordance with the value of $|\psi|^2$ at that point.”

That is rejection sampling, and it is the same algorithm you would reach for to draw from any distribution you can evaluate but not invert. Propose a point uniformly, accept it with probability proportional to the density, repeat.

The widget above renders the density directly on a grid instead — deterministic, so it looks identical on every load and in server-side rendering, and it does not mislead the eye with sampling noise that could be mistaken for structure. The book’s dots have a texture that is pure Monte Carlo variance; the smooth version shows only what is actually there.

Worth knowing both. Rejection sampling is what you want when you need samples (a simulated measurement, as §3.4’s MeasureStrip does); direct evaluation is what you want when you need the distribution.

Check yourself

0 / 6 answered

1.Why is the radial probability density $∣ u (r) ∣^{2}$ rather than $∣ R (r) ∣^{2}$ ?
2.For the 1s state, the most probable radius is $a_{0}$ but $⟨ r ⟩ = 1.5 a_{0}$ . Why do they differ?
Within n = 3: 3s has ⟨r⟩ = 13.5a₀, 3p has 12.5a₀, 3d has 10.5a₀.
3.Higher $l$ feels a stronger centrifugal barrier, so why is 3d the *smallest*?
4.How many nodes does the $4 p$ state have, and of what kinds?
5.The book's Figs. 9.6 and 9.7 were made by "selecting a point at random and deciding to plot or not in accordance with $∣ ψ ∣^{2}$ ". What is that algorithm?
6.What does the shading in an orbital picture actually represent?