Hydrogenic Wavefunctions and Radial Distributions

We take the machinery of quantum mechanics as given and ask a spectroscopist's question of it. The hydrogenic states $\psi_{n\ell m} = R_{n\ell}(r)\,Y_\ell^m(\theta,\phi)$ — the radial functions $R_{n\ell}$ , their node theorem, and the associated-Laguerre bookkeeping — are built and owned by aq-qm-hydrogenic-wavefunctions, and the angular factors $Y_\ell^m$ by aq-qm-spherical-harmonics; both are recalled here by name, not re-derived. What this article owns is the piece they hand forward: the radial distribution $P(r) = r^2|R_{n\ell}|^2$ , the honest "where is the electron" curve that atomic spectroscopy actually runs on. The wavefunction gives an amplitude at a point; $P(r)$ gives the probability of finding the electron in a shell — and the two peak in different places. That single distinction, carried by one geometric factor of $r^2$ , resolves the oldest confusion in atomic structure. The $1s$ density $|\psi|^2$ is largest at the nucleus, yet the electron is most likely found one Bohr radius out, exactly where Bohr drew his orbit. Push the same tool further and it explains why the sodium D lines glow at $589$ nm: low-angular-momentum electrons penetrate the inner shells, feel a less-screened nucleus, and drop in energy — the quantum defect. The radial distribution is where the abstract catalogue of states becomes the concrete language of spectra, screening, and the periodic table.

Key takeaways

The radial distribution $P(r) = r^2|R_{n\ell}(r)|^2$ — not $|R|^2$ — is the probability per unit radius of finding the electron in the shell $[r, r+\mathrm{d}r]$ ; the factor $r^2$ is the shell's volume, and $\int_0^\infty P\,\mathrm{d}r = 1$ .
For the ground state the density $|\psi_{1s}|^2$ is maximal at the nucleus, but $P(r)$ peaks at the Bohr radius $r = a_0 = 0.529$ Å — the most-probable radius, not the mean.
Peak $\ne$ mean: the most-probable radius of $1s$ is $a_0$ , but $\langle r\rangle_{1s} = \tfrac32 a_0$ , because $P(r)$ is right-skewed. Neither is a literal orbit.
Radial nodes number $n-\ell-1$ (recalled from the wavefunction article); the mean radius is $\langle r\rangle_{n\ell} = \tfrac{a_0}{2Z}[3n^2 - \ell(\ell+1)]$ .
Low- $\ell$ states penetrate the core: $2s$ reaches farther than $2p$ ( $\langle r\rangle = 6a_0$ vs $5a_0$ ) yet its inner lobe pokes deep inside $a_0$ — the origin of screening, the quantum defect, and alkali spectra.
Everything scales with nuclear charge: lengths as $1/Z$ , energies as $Z^2$ , so the He $^+$ $1s$ peaks at $a_0/2$ and binds at $-54.4$ eV.
Key idea: the radial distribution $P(r) = r^2|R_{n\ell}|^2$ turns the abstract wavefunction into the shell-by-shell probability that governs atomic size, penetration, screening, and every spectral line.

The core insight

The amplitude $|\psi|^2$ answers "how dense is the cloud at this point?"; the radial distribution $P(r) = r^2|R_{n\ell}(r)|^2$ answers the spectroscopist's question, "how likely is the electron to be found at this radius?" The two differ by the shell-volume factor $r^2$ , and that factor is why the $1s$ density peaks at the nucleus while the $1s$ shell probability peaks at $a_0$ — reconciling the quantum orbital with Bohr's radius, and setting up penetration, screening, and the quantum defect.

Where this lives today

The radial distribution is the quiet engine of computational chemistry: quantum-chemistry codes expand molecular orbitals in Slater-type functions carrying exactly the $r^\ell e^{-\zeta r}$ anatomy of the hydrogenic radial part, and effective-core-potential methods lean on the penetration-and-screening story to replace inner shells with a single potential. The quantum defect it explains is a precision tool — Rydberg-atom quantum simulators are engineered around the way high- $\ell$ states barely penetrate the core (small $\delta_\ell$ ) while low- $\ell$ states plunge in, and modern laser cooling of alkalis rides on transitions whose wavelengths are set, as in Worked Example 3, by quantum defects. Most strikingly, the curve is now directly observable: in 2013 a photoionization-microscopy experiment magnified the outgoing-electron pattern of hydrogen atoms onto a detector, resolving the nodal rings of individual states — a literal photograph of what a century of textbooks could only plot. The radial distribution function, once the reconciliation of Bohr with Schrödinger, is now a laboratory image.

The physical picture

Do not picture the electron on a track. Picture a cloud, densest at the nucleus for an $s$ -state and thinning outward, and then ask the honest question a detector asks: if I sweep a thin spherical shell of radius $r$ outward, in which shell am I most likely to catch the electron? A shell at radius $r$ has area $4\pi r^2$ , so even where the cloud is thin, a large shell can hold a lot of probability. The competition between a density that falls and a shell volume that grows is the whole story — and it has a definite winner at $r = a_0$ .

Figure

The r² lift: the 1s density is largest at the nucleus, but the shell volume ∝ r² carries the probability out to a peak at r = a₀ = 0.529 Å

The left curve is the density: fall, fall, fall from the origin. The right curve is $P(r)$ , the density multiplied by the shell volume — it must vanish at $r = 0$ (a shell of zero radius holds nothing), rise as $r^2$ opens up room, then collapse as the exponential density wins at large $r$ . The turnover lands exactly at $a_0$ . That is the reconciliation atomic physics keeps returning to: the electron's cloud is thickest at the nucleus, but the electron is most often found a Bohr radius away. Now stack the low states on one axis and the spectroscopy jumps out. Each shell sits farther out, each radial function crosses zero $n-\ell-1$ times, and — the subtle part — states of the same $n$ but different $\ell$ probe the core very differently.

Figure

The punchline: P(r) for 1s, 2s, 2p. The 1s peak is at a₀ = 0.529 Å but the mean ⟨r⟩ = 1.5 a₀ lies farther out; 2s reaches farther than 2p yet penetrates inside a₀, while 2p ∝ r⁴ is shut out of the core

Read the three curves. The $1s$ (blue) peaks at $a_0$ ; its mean, marked with the red tick at $1.5\,a_0$ , sits farther out because the long exponential tail drags the average past the peak — the distribution is right-skewed, and peak is not mean. The $2p$ (amber) is a clean single hump peaking at $4a_0$ ; because $R_{2p} \propto r$ , it is pushed out of the core by the centrifugal barrier and has almost no probability near the nucleus. The $2s$ (green) has a node at $r = 2a_0$ splitting it into a large outer lobe (peaking near $5.24\,a_0$ , farther than $2p$ ) and a small but crucial inner lobe that pokes deep inside $a_0$ . That inner lobe is penetration, and it is why sodium is yellow.

The core equations

Everything below rests on one recalled fact and one owned definition. The recalled fact is the separated form of the hydrogenic state, constructed in aq-qm-hydrogenic-wavefunctions (which supplies the radial table $R_{n\ell}$ and the node count) and aq-qm-spherical-harmonics (which supplies $Y_\ell^m$ ):

\psi_{n\ell m}(r,\theta,\phi) = R_{n\ell}(r)\,Y_\ell^m(\theta,\phi) \qquad (1)

where: $n$ — principal quantum number, $n = 1, 2, 3, \dots$ ; $\ell$ — orbital quantum number, $0 \le \ell \le n-1$ ; $m$ — magnetic quantum number, $-\ell \le m \le \ell$ ; $R_{n\ell}$ — radial function (units $a_0^{-3/2}$ ); $Y_\ell^m$ — spherical harmonic (normalized on the sphere). This article takes $R_{n\ell}$ as given and asks what it says about radii. Integrate $|\psi|^2$ over all angles and the spherical harmonic normalizes to one; what survives is a probability that depends on radius alone. The angular integration hands over a factor $r^2$ from the volume element $\mathrm{d}V = r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$ , and that factor is the entire subject of this article:

P(r) = r^2\,|R_{n\ell}(r)|^2, \qquad \int_0^\infty P(r)\,\mathrm{d}r = 1 \qquad (2)

where: $P(r)\,\mathrm{d}r$ — probability of finding the electron in the shell $[r, r+\mathrm{d}r]$ , independent of angle. The $r^2$ is geometry, not dynamics: the surface area of the shell, and what forces $P(0) = 0$ for every state, even the $s$ -states whose density is maximal at the origin. Forgetting it is the single most common error in the subject. The most-probable radius is the maximum of $P(r)$ , found by setting $\mathrm{d}P/\mathrm{d}r = 0$ . For the ground state this is a one-line calculus problem (Derivation A) with a famous answer:

r_{\text{peak}}^{(1s)} = \frac{a_0}{Z} \qquad (3)

where: $a_0 = 0.529177$ Å — the Bohr radius; $Z$ — nuclear charge. For hydrogen ( $Z = 1$ ) the $1s$ shell probability peaks exactly at $a_0$ , numerically identical to Bohr's orbit radius — but it is the peak of a broad smear, not a track. The mean radius is a different number entirely, obtained from the first moment of $P(r)$ :

\langle r\rangle_{n\ell} = \frac{a_0}{2Z}\big[\,3n^2 - \ell(\ell+1)\,\big] \qquad (4)

where: $\langle r\rangle$ — expectation value of the radius. It grows as $n^2$ and shrinks with $\ell$ at fixed $n$ . For $1s$ it gives $\tfrac32 a_0 > a_0 = r_{\text{peak}}$ : the right-skew made quantitative. For $2s$ and $2p$ it gives $6a_0$ and $5a_0$ — the $s$ -state's mean is larger. The inverse moment $\langle 1/r\rangle$ is fixed not by geometry but by energy, through the virial theorem (Derivation B), and it depends only on $n$ :

\left\langle\frac1r\right\rangle_{n\ell} = \frac{Z}{n^2 a_0}, \qquad \langle V\rangle = 2E_n = -\frac{Z^2 e^2}{4\pi\varepsilon_0\, a_0\, n^2} \qquad (5)

where: $\langle V\rangle$ — mean Coulomb potential energy; $E_n = -Z^2\,(13.606\ \mathrm{eV})/n^2$ — the level energy. Because $\langle 1/r\rangle$ is set by $E_n$ , which depends only on $n$ for a pure Coulomb field, this moment is blind to $\ell$ — the accidental degeneracy in a radial average. That degeneracy is the hydrogen story. In any other atom the radial distribution breaks it, and the break has a name. A penetrating low- $\ell$ electron spends part of its time inside the inner shells, sees a nuclear charge only partly screened, binds more tightly, and its energy is pushed down by an amount captured in a single empirical shift:

E_{n\ell} = -\frac{hc\,R_H}{(n - \delta_\ell)^2} \qquad (6)

where: $\delta_\ell$ — the quantum defect, large for penetrating low- $\ell$ orbitals and $\approx 0$ for high- $\ell$ orbitals that never reach the core; $R_H$ — the Rydberg constant. For sodium $\delta_s \approx 1.37$ , $\delta_p \approx 0.88$ , $\delta_d \approx 0.01$ — the $s$ -electron's penetration is worth more than a whole unit of $n$ . This is where the radial-distribution picture cashes out as real spectroscopy (Worked Example 3).

Derivation AThe most-probable radius of the ground state sits at the Bohr radius

The goal is to prove that the $1s$ radial distribution peaks at exactly $r = a_0/Z$ , reconciling the quantum probability cloud — whose density is maximal at the nucleus — with Bohr's semiclassical orbit at $a_0$ . The strategy is direct: build $P(r)$ from the recalled $1s$ radial function, differentiate, and solve for the stationary point at finite radius. The one thing to watch is that the maximum of $P(r)$ and the maximum of the density $|\psi|^2$ are genuinely different questions with different answers — the factor $r^2$ moves the peak from the origin all the way out to $a_0$ .

Step 1. Take the ground-state radial function from aq-qm-hydrogenic-wavefunctions, $R_{10} = 2(Z/a_0)^{3/2}e^{-Zr/a_0}$ , and form the distribution: $P(r) = r^2|R_{10}|^2 = 4(Z/a_0)^3\,r^2 e^{-2Zr/a_0}$ . Why: $P(r)$ , not $|R_{10}|^2$ , is the probability per unit radius — the $r^2$ is the shell area the electron could be found on.

Step 2. Differentiate, pulling the exponential out as a never-zero common factor: $\dfrac{\mathrm{d}P}{\mathrm{d}r} = 4(Z/a_0)^3\left(2r - \dfrac{2Z}{a_0}r^2\right)e^{-2Zr/a_0}$ . Why: only the polynomial bracket can vanish at finite $r$ , so the turning point is set by $2r - (2Z/a_0)r^2 = 0$ — the exponential merely guarantees the peak is a maximum, not a runaway.

Step 3. Solve the bracket: $2r = (2Z/a_0)r^2 \Rightarrow r = a_0/Z$ . For hydrogen this is $a_0$ exactly. Why: this is the owned result — the shell-volume factor has shifted the most-probable radius from $r = 0$ (where $|\psi|^2$ peaks) to $r = a_0$ (where Bohr placed his orbit), and the coincidence is exact, not approximate.

Step 4. Contrast with the mean. The peak is where $P$ is largest; the mean $\langle r\rangle = \int_0^\infty r\,P(r)\,\mathrm{d}r$ weights the long tail and lands at $\tfrac32 a_0$ (evaluated in Worked Example 1). Why: because $P(r)$ is right-skewed, the most-probable, mean, and Bohr-orbit radii are three distinct statements about one cloud — a distinction the rest of the article keeps sharp. *Verified symbolically ( $\mathrm{d}P/\mathrm{d}r = 0 \Rightarrow r = a_0$ for $Z=1$ ) and for He $^+$ ( $a_0/2$ ) in verify_aq-optics-hydrogenic-wavefunctions.py.*

Go deeper · Derivation B · 1 minThe inverse radius from the virial theorem, and why it ignores ℓ

Derivation BThe inverse radius from the virial theorem, and why it ignores ℓ

The goal is to obtain $\langle 1/r\rangle = Z/(n^2 a_0)$ without evaluating a single Laguerre integral, and to expose why this radial moment depends only on $n$ while $\langle r\rangle$ depends on both $n$ and $\ell$ . The strategy is the virial theorem for a $-1/r$ potential, which ties the average potential energy directly to the level energy — and the Coulomb level energy is $\ell$ -blind. The payoff is a clean bridge from the radial distribution to the energy spectrum.

Step 1. Invoke the virial theorem for a potential $V \propto r^{-1}$ : $\langle T\rangle = -E_n$ and $\langle V\rangle = 2E_n$ . Why: for a homogeneous potential of degree $-1$ the theorem fixes the split of energy between kinetic and potential parts once and for all, independent of the state's angular shape.

Step 2. Write the potential energy as a radial average: $\langle V\rangle = -\dfrac{Ze^2}{4\pi\varepsilon_0}\left\langle\dfrac1r\right\rangle$ . Setting this equal to $2E_n$ with $E_n = -Z^2 e^2/(8\pi\varepsilon_0 a_0 n^2)$ gives one equation for $\langle 1/r\rangle$ . Why: the only unknown is the inverse-radius moment, so the virial identity solves for it directly — no wavefunction integral required.

Step 3. Solve: $\left\langle\dfrac1r\right\rangle = \dfrac{Z}{n^2 a_0}$ , a function of $n$ alone. Why: it inherits the energy's $\ell$ -independence — the accidental Coulomb degeneracy, seen here as a radial average that cannot tell $2s$ from $2p$ , even though those two states have visibly different distributions.

Step 4. Note the contrast with $\langle r\rangle_{n\ell} = \tfrac{a_0}{2Z}[3n^2 - \ell(\ell+1)]$ , which does carry $\ell$ . Why: different moments weight different regions — $\langle 1/r\rangle$ is dominated by the small- $r$ core (tied to energy), $\langle r\rangle$ by the outer tail (tied to geometry) — so they need not, and do not, share the same symmetry. *Verified: $\langle 1/r\rangle = Z/(n^2 a_0)$ against direct integration for $(n,\ell) = (1,0),(2,0),(2,1),(3,0)$ , and the virial identity $\langle V\rangle = 2E_n$ numerically, in verify_aq-optics-hydrogenic-wavefunctions.py.*

Limiting cases

Case	Result	Meaning
Ground state $1s$	$r_{\text{peak}} = a_0$ , $\langle r\rangle = \tfrac32 a_0$	most-probable radius equals Bohr's orbit; the mean is larger — peak $\ne$ mean
Circular states $\ell = n-1$	single peak near $n^2 a_0/Z$ , no radial nodes	the most Bohr-like orbitals — one clean shell at the Bohr radius
Fixed $n$ , growing $\ell$	$\langle r\rangle$ falls, but core probability falls faster	high $\ell$ is pushed out of the center by the $r^\ell$ barrier yet has a smaller mean radius
Nuclear charge $Z$	$r \to r/Z$ , $E \to Z^2 E$	He $^+$ peaks at $a_0/2$ and binds at $-54.4$ eV; the whole cloud contracts
$s$ -state at the nucleus	$\lvert\psi_{ns}(0)\rvert^2 = Z^3/(\pi n^3 a_0^3) \ne 0$	only $\ell = 0$ reaches $r = 0$ — the seat of contact hyperfine and Lamb-shift physics
Non-Coulomb core (alkali)	$E_{n\ell} = -hcR_H/(n-\delta_\ell)^2$	penetration lifts the $\ell$ -degeneracy — the quantum defect, and the reason alkalis have rich optical spectra

Common trapThe 1s electron does not orbit at the Bohr radius, and |R|² is not the radial probability

Two errors, one root. The first: reading $|R_{n\ell}|^2$ (or $|\psi|^2$ ) as "the probability of finding the electron at radius $r$ ." It is not — the probability per unit radius is $P(r) = r^2|R|^2$ , and dropping the $r^2$ factor makes the $1s$ probability look maximal at the nucleus instead of at $a_0$ , misplacing every radial peak in the atom. The second: seeing the $1s$ peak land at $a_0$ and concluding the electron orbits there like a planet. It does not; $a_0$ is merely where a broad, static cloud is densest per shell. The same cloud has its most-probable radius at $a_0$ , its mean radius at $\tfrac32 a_0$ , and no definite radius at all — three numbers, one smear, zero orbits. Keep "density," "shell probability," "most-probable radius," and "mean radius" as four separate ideas and the subject stays honest.

Worked examples

1 · Probability inside one Bohr radius, and the mean radius (confidence builder)

For the hydrogen ground state, find the probability that the electron lies within one Bohr radius of the nucleus, and compute the mean radius $\langle r\rangle$ . Compare it to the most-probable radius $a_0$ .

Step 1. Build the distribution in units $x = r/a_0$ : $P(r)\,\mathrm{d}r = 4x^2 e^{-2x}\,\mathrm{d}x$ . The probability inside $a_0$ is $\displaystyle\int_0^1 4x^2 e^{-2x}\,\mathrm{d}x$ .

Step 2. Integrate by parts twice (or quote the standard result): $\displaystyle\int_0^1 4x^2 e^{-2x}\,\mathrm{d}x = 1 - 5e^{-2}$ . Numerically $5e^{-2} = 5(0.13534) = 0.6767$ , so the probability is $\mathbf{1 - 5e^{-2} = 0.323}$ — only about a third of the electron's probability lies inside the Bohr radius, even though that is where the shell probability peaks.

Step 3. Now the mean radius: $\langle r\rangle = \int_0^\infty r\,P(r)\,\mathrm{d}r = a_0\int_0^\infty 4x^3 e^{-2x}\,\mathrm{d}x = a_0 \cdot 4\cdot\dfrac{3!}{2^4} = a_0\cdot\dfrac{24}{16} = \mathbf{\tfrac32 a_0}$ , using $\int_0^\infty x^3 e^{-2x}\mathrm{d}x = 3!/2^4$ .

Step 4. Read the lesson off the numbers. The most-probable radius is $a_0$ , the mean is $1.5\,a_0$ , and two-thirds of the probability lies beyond $a_0$ : the long outer tail drags the average past the peak. P(inside $a_0$ ) = 0.323, $\langle r\rangle = 1.5\,a_0$ , versus $r_{\text{peak}} = a_0$ .

2 · The He⁺ ion by pure scaling (exam level)

The He $^+$ ion is hydrogenic with $Z = 2$ . Find its $1s$ most-probable radius and its ground-state binding energy, using only the scaling of the hydrogenic solutions.

Step 1. Length scaling: every radius carries the factor $a_0/Z$ , so the $1s$ peak moves in to $r_{\text{peak}} = a_0/Z = a_0/2 = \mathbf{0.265\ \text{Å}}$ . The cloud is half the size of hydrogen's — a stronger nucleus pulls it in.

Step 2. Energy scaling: $E_n = -Z^2\,(13.606\ \mathrm{eV})/n^2$ , so the ground state is $E_1 = -4 \times 13.606 = \mathbf{-54.4\ \text{eV}}$ . He $^+$ binds its electron four times more tightly than hydrogen — its ionization energy of $54.4$ eV is a standard atomic-physics benchmark.

Step 3. Consistency check via the mean radius: $\langle r\rangle_{1s} = \tfrac{a_0}{2Z}(3) = \tfrac{3a_0}{2Z} = \tfrac34 a_0$ for $Z = 2$ — again half of hydrogen's $\tfrac32 a_0$ , as the $1/Z$ scaling demands. Every length halves, every energy quadruples.

Step 4. The scaling is exact because the only length is $a_0/Z$ and the only energy is $Z^2\mathrm{Ry}$ . He $^+$ : $r_{\text{peak}} = a_0/2 = 0.265$ Å, $E_1 = -54.4$ eV.

3 · Sodium's yellow line from penetration and the quantum defect (exam level)

Sodium's valence electron sits in a $3s$ orbital outside a neon-like core. Using the measured quantum defects $\delta_s = 1.373$ and $\delta_p = 0.883$ , find sodium's ionization energy and the wavelength of the $3p \to 3s$ transition — the D line.

Step 1. The $3s$ electron penetrates the core (its inner lobe, exactly the $2s$ -style penetration seen in the punchline figure), feels a poorly screened nucleus, and binds more tightly than a pure-Coulomb $n = 3$ level. Equation (6) gives $E_{3s} = -13.606/(3 - 1.373)^2 = -13.606/(1.627)^2 = -13.606/2.647 = \mathbf{-5.14\ \text{eV}}$ .

Step 2. The binding energy is the ionization energy: sodium's first ionization energy is $5.14$ eV — measured value $5.139$ eV. The penetration of one $s$ -electron, worth $\delta_s = 1.373$ units of effective quantum number, reproduces it almost exactly.

Step 3. The $3p$ orbital penetrates less ( $\delta_p = 0.883$ ), so it binds less: $E_{3p} = -13.606/(3 - 0.883)^2 = -13.606/(2.117)^2 = \mathbf{-3.04\ \text{eV}}$ . The optical transition energy is $\Delta E = E_{3s} \to E_{3p} = 5.140 - 3.036 = 2.104$ eV.

Step 4. Convert to a wavelength: $\lambda = hc/\Delta E = 1239.8\ \mathrm{eV\,nm}/2.104\ \mathrm{eV} = \mathbf{589\ \text{nm}}$ — the sodium D line, the yellow of every streetlamp and flame test. The entire number traces back to a low- $\ell$ electron's radial distribution reaching inside the core. Na ionization $= 5.14$ eV, D line $\lambda = 589$ nm.

Physics of the portrait

The hero here is not a person but an object: the curve $P(r) = r^2|R_{n\ell}|^2$ itself, the radial distribution function. It is the quantity that made the quantum atom picturable — the bridge from an abstract complex amplitude to a plottable, measurable probability of catching the electron at a given distance.

Origins · 1 minradial distribution function and how this was discovered

radial distribution function

the radius-squared-weighted probability P(r) locating where an electron most likely sits · atomic spectroscopy

The radial distribution function was born the moment Schrödinger's 1926 wave equation produced the functions $R_{n\ell}(r)$ and Max Born, in the same year, reinterpreted $|\psi|^2$ as a probability density rather than a physical charge cloud. The decisive insight — that the honest radial curve is not $|R|^2$ but $r^2|R|^2$ , carrying the shell-volume Jacobian — is what reconciled the new quantum orbital, densest at the nucleus, with Bohr's old orbit at $a_0$ : the same electron, most probably found one Bohr radius out. It quickly became the working tool of atomic structure. Douglas Hartree's 1928 self-consistent-field method computed radial distributions for many-electron atoms, and John Slater's 1930 screening rules turned the penetration of low- $\ell$ orbitals into effective-charge bookkeeping — the quantum-defect picture that shapes the periodic table and sets the alkali optical lines. What began as a way to draw the invisible is now photographable: in 2013 Stodolna and collaborators used photoionization microscopy to image the nodal structure of hydrogen's electron distribution on a detector, turning the radial function from a calculated curve into a measured one.

History

1926

— Schrödinger's wave mechanics yields the hydrogenic radial functions R_nl(r)

1926

— Born interprets |ψ|² as a probability density, making P(r) = r²|R|² a probability curve

1928

— Hartree's self-consistent field computes radial distributions for many-electron atoms

1930

— Slater's screening rules recast low-ℓ penetration as effective nuclear charge and the quantum defect

2013

— Stodolna and coworkers directly image hydrogen's nodal structure by photoionization microscopy

Going further

Texts. Foot, Atomic Physics (2005), Ch. 2 and the hydrogen-wavefunction appendix, and Ch. 4 on the quantum defect and alkali spectra · Bransden and Joachain, Physics of Atoms and Molecules, Ch. 3 · Griffiths and Schroeter, Introduction to Quantum Mechanics, 3rd ed., Ch. 4 · Demtröder, Atoms, Molecules and Photons, atomic-structure chapters · Loudon, The Quantum Theory of Light, hydrogen matrix-element appendix.

Provenance. Radial functions $R_{10}, R_{20}, R_{21}$ and the spherical harmonics are standard (Griffiths Ch. 4; Foot Appendix). A verifier should confirm: (i) normalization $\int_0^\infty R_{n\ell}^2 r^2\mathrm{d}r = 1$ for the listed forms; (ii) the ground-state peak $r = a_0$ by differentiating $P(r)$ ; (iii) $\langle r\rangle_{1s} = \tfrac32 a_0$ and $\langle 1/r\rangle = Z/n^2 a_0$ from the moment formula and virial theorem; (iv) contact density $|\psi_{ns}(0)|^2 = Z^3/(\pi n^3 a_0^3)$ ; (v) node counts $n-\ell-1$ (radial) and $\ell$ (angular). Bohr radius $a_0 \approx 5.29\times10^{-11}\,\mathrm{m}$ (CODATA). Cross-check the moment $\langle 1/r^3\rangle$ formula against the Bransden and Joachain table of hydrogenic expectation values. All of (i)–(v), the He $^+$ scaling, and the sodium quantum-defect numbers are reproduced in verify_aq-optics-hydrogenic-wavefunctions.py.

Prerequisites. Solution of the Schrödinger equation for hydrogen · Quantum numbers and orbital angular momentum · The Bohr model of hydrogen. Leads to. Fine structure and spin-orbit coupling · Hyperfine structure · The Lamb shift · The Stark effect · Multi-electron atoms and the central-field approximation.

Why exactly r², and not r or r³?

Because a spherical shell of radius $r$ and thickness $\mathrm{d}r$ has volume $4\pi r^2\,\mathrm{d}r$ — its area grows as $r^2$ . The probability of finding the electron in that shell is the density $|\psi|^2$ times the shell volume, and after integrating $|\psi|^2 = |R_{n\ell}|^2|Y_\ell^m|^2$ over the angles (which normalizes the spherical harmonic to one) exactly one factor of $r^2$ survives: $P(r) = r^2|R_{n\ell}|^2$ . Not $r$ (that would be a ring), not $r^3$ (that would be a solid ball's volume, not a shell) — $r^2$ is the surface area of the sphere at radius $r$ , and it is pure geometry, identical for every central-potential problem.

Is the 1s peak at a₀ really the same as Bohr's orbit?

Numerically yes, conceptually no. Bohr's model puts the electron on a sharp circular orbit of radius $a_0$ ; the Schrödinger $1s$ state is a static, spherically symmetric probability cloud whose most-probable shell happens to sit at the same $a_0$ . The coincidence is exact and not accidental — both are set by the same balance of Coulomb attraction and quantum-mechanical spread — but the quantum object has no orbit, no fixed radius, and in fact a mean radius of $\tfrac32 a_0$ , larger than the peak. Bohr got the right length scale for the wrong reason: he imagined a track where there is only a smear that happens to be densest per shell at his radius.

Why can only s-electrons reach the nucleus, and why does it matter?

Near the origin the radial function behaves as $R_{n\ell} \propto r^\ell$ , so only $\ell = 0$ ( $s$ -states) has nonzero amplitude at $r = 0$ ; every $\ell > 0$ state is pushed out of the center by the centrifugal barrier $\ell(\ell+1)\hbar^2/2m_e r^2$ , which the $r^\ell$ prefactor encodes. The contact density $|\psi_{ns}(0)|^2 = Z^3/(\pi n^3 a_0^3)$ is therefore nonzero only for $s$ -states. This is not bookkeeping: the Fermi contact hyperfine interaction, the leading Lamb shift, and electron-capture rates all scale with $|\psi(0)|^2$ , so they act on $s$ -electrons alone. The $21$ -cm hydrogen line that maps the galaxy is a $1s$ contact-hyperfine transition — it exists because the ground state is an $s$ -state that touches the proton.

How can 2s reach farther than 2p but also penetrate deeper?

Because "farther" and "deeper" are different features of the same distribution. The $2s$ mean radius is $6a_0$ , larger than $2p$ 's $5a_0$ , because $2s$ has a large outer lobe beyond the $2p$ hump — that is "farther." But $2s$ also has a small inner lobe, inside its node at $2a_0$ , carrying probability into the core, whereas $2p \propto r^2$ is shut out near the origin — that is "deeper." No single number captures both; you must look at the whole curve. The inner lobe is small (a few percent inside $a_0$ ) but decisive in energy, since binding is dominated by the attractive core.

What is the quantum defect, and how does the radial distribution explain alkali spectra?

In hydrogen the energy depends only on $n$ ; in an alkali the single valence electron orbits a compact noble-gas core, and its energy depends on $\ell$ too, because low- $\ell$ orbitals penetrate that core. Inside it the electron sees the full, poorly screened nuclear charge; outside, only $+e$ . A penetrating orbital binds more tightly, a shift summarized as $E_{n\ell} = -hcR_H/(n-\delta_\ell)^2$ , replacing $n$ by $n-\delta_\ell$ . The defect $\delta_\ell$ is large for penetrating $s$ and $p$ orbitals and nearly zero for $d$ and $f$ orbitals that stay outside. This idea, rooted entirely in the radial distribution, reproduces sodium's ionization energy and D-line wavelength (Worked Example 3) and organizes every alkali's optical spectrum.

How does everything scale for hydrogenic ions like He⁺ or Li²⁺?

The hydrogenic solutions depend on $Z$ only through the combination $Zr/a_0$ , so a single substitution rescales the entire catalogue: replace $a_0 \to a_0/Z$ everywhere. Consequently every radius contracts as $1/Z$ (the He $^+$ $1s$ peaks at $a_0/2$ , Li $^{2+}$ at $a_0/3$ ), while every energy grows as $Z^2$ (He $^+$ binds at $-54.4$ eV, Li $^{2+}$ at $-122.5$ eV). The shapes of the distributions — node positions in units of $a_0/Z$ , the peak-to-mean ratio, the penetration pattern — are identical to hydrogen's; only the scale changes. This is why one solved atom serves an entire isoelectronic sequence.