Hydrogenic Ions and Scaling with Nuclear Charge

In 1896 the astronomer Edward Pickering found something troubling in the spectrum of the hot star ζ Puppis: lines at almost — but not exactly — hydrogen's Balmer positions, with extra lines sitting between them. Hydrogen with half-integer quantum numbers? A new element? The answer turned out to be stranger and simpler: it was helium with one electron stripped away, impersonating hydrogen so well that only a 0.04% wavelength shift gave it away. The impersonation works because a hydrogenic ion — any single electron bound to a nucleus of charge $+Ze$ — is not a new problem. It is the hydrogen problem with one knob turned. Replace $e^2$ by $Ze^2$ in the Coulomb potential and the entire hydrogen solution follows along: same equation, same quantum numbers, same wavefunctions in rescaled coordinates. Energies multiply by $Z^2$ ; lengths shrink by $1/Z$ . Solve one atom and you have solved $\mathrm{He}^+$ , $\mathrm{Li}^{2+}$ , the hydrogen-like iron blazing in X-ray telescopes, and one-electron uranium. This article is where that substitution earns its keep: the scaling laws from one change of variable, Pickering's impostor resolved line by line, Moseley's X-ray law that re-sorted the periodic table — and the fine print. The electron's speed grows as $Z\alpha c$ , so relativity, ignorable in hydrogen, becomes structural by uranium. Scaling is the cheapest powerful idea in atomic physics; here is how far it goes, and exactly where it stops.

Key takeaways

A hydrogenic ion is any single-electron system bound to a nucleus of charge $+Ze$ : $\mathrm{He}^+$ , $\mathrm{Li}^{2+}$ , $\mathrm{C}^{5+}$ , $\mathrm{U}^{91+}$ — hydrogen is the case $Z = 1$ .
One substitution $e^2 \to Ze^2$ maps the solved hydrogen problem onto every one of them; no new differential equation is solved.
Energies scale as $Z^2$ : $E_n = -13.6\,Z^2/n^2\ \mathrm{eV}$ — the ionization energy of $\mathrm{He}^+$ is $54.4\ \mathrm{eV}$ , of $\mathrm{Li}^{2+}$ is $122.5\ \mathrm{eV}$ .
Lengths contract as $1/Z$ : the effective Bohr radius is $a_Z = a_0/Z$ , and wavefunctions rescale rigidly, $\psi^{(Z)}(\mathbf r) = Z^{3/2}\psi^{(1)}(Z\mathbf r)$ .
Spectral lines blue-shift by $Z^2$ ; even- $n$ transitions of $\mathrm{He}^+$ mimic hydrogen's Balmer lines exactly until the reduced mass breaks the tie — the Pickering-series story.
The electron's speed is of order $Z\alpha c$ : relativistic corrections scale as $(Z\alpha)^2$ and grow from $10^{-4}$ in hydrogen to order one in uranium.
Key idea: every hydrogen result becomes a hydrogenic result via $e^2 \to Ze^2$ — energies multiply by $Z^2$ , lengths by $1/Z$ .

The core insight

Every hydrogen result becomes a hydrogenic result by the substitution $e^2 \to Ze^2$ ; equivalently, energies multiply by $Z^2$ and lengths by $1/Z$ — one solved problem covers the whole one-electron family.

Where this lives today

Hydrogenic scaling is a daily working tool in X-ray astronomy. Hot plasmas strip iron down to $\mathrm{Fe}^{25+}$ — a one-electron ion with $E_1 = 9.20\ \mathrm{keV}$ — whose Lyman-α line the scaling laws place at $6.90\ \mathrm{keV}$ , within $1\%$ of the observed $6.97\ \mathrm{keV}$ (the missing percent is the $(Z\alpha)^2$ relativistic correction, $Z\alpha = 0.19$ ). That line complex is a standard thermometer for black-hole accretion disks and tokamak plasmas alike. At the violent end, storage rings prepare one-electron uranium, $\mathrm{U}^{91+}$ , where $Z\alpha = 0.671$ makes the electron genuinely relativistic — a laboratory for bound-state quantum electrodynamics in fields no neutral atom can supply. And at the precision end, the mass knob complements the charge knob: replace the electron with a muon ( $207\times$ heavier) and the orbit contracts to $285\ \mathrm{fm}$ in muonic hydrogen — so close that the proton's own size shifts the levels measurably. The 2010 muonic Lamb-shift measurement returned a proton radius $4\%$ off the accepted value — the decade-long "proton-radius puzzle." In muonic lead the scaling becomes self-defeating in the most instructive way: $a_0/(207 \times 82) = 3.1\ \mathrm{fm}$ lies inside the $7.1\ \mathrm{fm}$ nucleus, and the point-charge premise of equation (1) dissolves — which is precisely why muonic X-rays are used to map nuclear charge distributions.

The physical picture

Think of the nuclear charge as a knob on the Coulomb well. Turning $Z$ up makes the well deeper and steeper at once — and because a $1/r$ potential has no built-in length scale, the problem does not deform as you turn the knob; it rescales rigidly. Measure distance in units of $a_0/Z$ and energy in units of $Z^2\,\mathrm{Ry}$ , and every hydrogenic ion collapses onto the same universal picture. The wavefunctions of $\mathrm{Li}^{2+}$ are hydrogen's wavefunctions viewed through a $3\times$ magnifier turned backwards.

Figure

One well, one knob — deeper and tighter as Z grows

Figure

The impostor comb — He II lines shadow and interleave the Balmer series

A virial-style estimate shows why the two scalings are one fact. Kinetic energy costs $\sim \hbar^2/2mr^2$ by confinement; potential energy pays $-Ze^2/4\pi\varepsilon_0 r$ . Minimizing the sum — the balancing act of The Bohr Radius and Characteristic Atomic Scales — lands at $r \propto 1/Z$ with depth $\propto Z^2$ . Deeper because tighter: the electron sits $Z$ times closer to a charge $Z$ times bigger. The speed is the fine print. Momentum scales as $\hbar/r \propto Z$ , so the characteristic velocity is $v_1 \sim Z\alpha c$ : hydrogen's electron ambles at $0.7\%$ of light speed, one-electron uranium's moves at $67\%$ of $c$ . The clean $Z^2$ law is the non-relativistic leading term — relativity is measurable by iron, structural by uranium. And one thing does not change: the level pattern. Every $n$ keeps its full $n^2$ -fold degeneracy, because scaling changes only the strength of the $1/r$ potential, never its shape — and the shape is what the hidden symmetry of Accidental Degeneracy and the SO(4) Symmetry cares about.

The scaling laws

The hydrogenic potential differs from hydrogen's in exactly one place. A nucleus of charge $+Ze$ pulls with

V(r) = -\frac{Ze^2}{4\pi\varepsilon_0\,r} \qquad (1)

where: $Z$ — the nuclear charge number (protons in the nucleus); $e$ — the elementary charge; $r$ — the electron–nucleus separation. Formally this is hydrogen's potential with $e^2 \to Ze^2$ : wherever the hydrogen solution carries the combination $e^2$ , the hydrogenic answer carries $Ze^2$ . That single promotion is the whole theory of this article. Chase the substitution through the energy formula of The Hydrogen Atom: Coulomb Radial Equation — its prefactor carries $(e^2)^2$ — and through the Bohr radius, which carries $e^2$ downstairs:

E_n = -\frac{Z^2}{n^2}\,\mathrm{Ry} = -13.6\,\frac{Z^2}{n^2}\ \mathrm{eV}, \qquad a_Z = \frac{a_0}{Z} \qquad (2)

where: $\mathrm{Ry} = 13.6057\ \mathrm{eV}$ — the Rydberg energy; $a_0 = 0.529\ \mathrm{\mathring{A}}$ — the Bohr radius; $n = 1, 2, 3, \dots$ Two factors of $Z$ in the energy, one inverse factor in the length: the ground state of $\mathrm{He}^+$ is bound four times as hard as hydrogen's in an orbital half the size. Derivation A below gets both laws from one change of variable, so you never have to trust the "count the $e^2$ 's" argument alone. The wavefunctions rescale just as rigidly. Because the radial coordinate enters the solution only through $r/a_Z = Zr/a_0$ , the hydrogenic state is the hydrogen state viewed at magnification $1/Z$ :

\psi^{(Z)}_{n\ell m}(\mathbf{r}) = Z^{3/2}\,\psi^{(Z=1)}_{n\ell m}(Z\mathbf{r}) \qquad (3)

where: $Z^{3/2}$ — the normalization factor forced by the volume element ( $\mathrm{d}^3r$ shrinks by $Z^3$ , and $\sqrt{Z^3} = Z^{3/2}$ ). Same nodes, same angular structure, same probability pattern — compressed. Expectation values inherit the compression: $\langle r\rangle_{1s} = \tfrac{3}{2}a_0/Z$ and $\langle 1/r\rangle_{1s} = Z/a_0$ , and via the virial theorem the potential energy $\langle V\rangle \propto Z\cdot Z = Z^2$ tracks the total energy exactly. The general moment machinery lives at Radial Probability Distribution and Expectation Values; here only the $Z$ -bookkeeping matters. Momentum tells the cautionary tale. Confinement to $a_0/Z$ means $p \sim Z\hbar/a_0$ , so the characteristic speed in the ground state is

v_1 \sim Z\alpha c, \qquad \alpha = \frac{e^2}{4\pi\varepsilon_0\hbar c} \approx \frac{1}{137.036} \qquad (4)

where: $\alpha$ — the fine-structure constant, hydrogen's speedometer reading from The Bohr Radius and Characteristic Atomic Scales, now multiplied by $Z$ . Relativistic corrections enter at relative order $(v/c)^2 = (Z\alpha)^2$ : five parts in $10^5$ for hydrogen, $3.6\%$ for hydrogen-like iron ( $Z\alpha = 0.19$ ), and no longer a correction at all for $\mathrm{U}^{91+}$ , where $Z\alpha = 0.67$ . The machinery of those corrections — fine structure, and ultimately the Dirac equation — belongs to the approximation-methods and relativistic chapters; this article only marks where the border lies.

which formula	when to reach for it
$E_n = -13.6\,Z^2/n^2\ \mathrm{eV}$	any level or ionization energy of a one-electron ion
$1/\lambda = Z^2 R\,(1/n_f^2 - 1/n_i^2)$	any spectral line; use the ion's reduced-mass $R$ for precision
$a_Z = a_0/Z$ , $\psi^{(Z)} = Z^{3/2}\psi^{(1)}(Z\mathbf r)$	sizes, overlaps, expectation values
$v_n \sim Z\alpha c/n$ , corrections $\sim (Z\alpha)^2$	deciding whether non-relativistic scaling is trustworthy
$E \approx -13.6\,Z_{\mathrm{eff}}^2/n^2\ \mathrm{eV}$	estimating many-electron atoms with a screened charge

Derivation AOne substitution solves every hydrogenic ion

The goal is both scaling laws — $E \propto Z^2$ and $a \propto 1/Z$ — derived honestly from the radial equation rather than by counting factors of $e^2$ in the final answers. The strategy is a single change of variable: guess that the natural length is $a_0/Z$ , rescale, and watch every trace of $Z$ vanish from the equation. Once the equation is $Z$ -free, its eigenvalues are hydrogen's forever, and $Z$ can only survive in the units — which is exactly what a scaling law means. Watch where the $Z^2$ assembles: one factor from the deeper potential, one from the tighter confinement.

Step 1. Start from the radial equation for $u(r) = rR(r)$ in the hydrogenic field — the equation built at The Radial Equation and Effective Potential and the Centrifugal Barrier, with the potential (1):

-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2u}{\mathrm{d}r^2} + \left[-\frac{Ze^2}{4\pi\varepsilon_0 r} + \frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]u = Eu.

Why: this is hydrogen's radial equation with $e^2 \to Ze^2$ — the centrifugal term carries no charge and is untouched.

Step 2. Rescale the length by the anticipated contraction: set $\rho = Zr/a_0$ , i.e. measure $r$ in units of $a_0/Z$ . Using $a_0 = 4\pi\varepsilon_0\hbar^2/me^2$ to express the Coulomb strength, every explicit $Z$ collects into one overall factor:

-\frac{\mathrm{d}^2u}{\mathrm{d}\rho^2} + \left[-\frac{2}{\rho} + \frac{\ell(\ell+1)}{\rho^2}\right]u = \frac{2m a_0^2}{\hbar^2 Z^2}\,E\,u.

Why: the bracket is now the dimensionless hydrogen problem — no $Z$ anywhere on the left. The knob has been absorbed into the units of length and the units of energy.

Step 3. Match eigenvalues. The dimensionless equation is exactly the one hydrogen solves, and its bound eigenvalues are $-1/n^2$ — the termination argument of The Hydrogen Atom: Coulomb Radial Equation, not re-run here. So

\frac{2m a_0^2}{\hbar^2 Z^2}\,E = -\frac{1}{n^2} \quad\Longrightarrow\quad E_n = -\frac{\hbar^2 Z^2}{2m a_0^2\,n^2} = -\frac{Z^2}{n^2}\,\mathrm{Ry}.

Why: the eigenvalue spectrum of a $Z$ -free equation cannot know about $Z$ ; the physical energy inherits $Z^2$ purely from the unit conversion.

Step 4. Read off the length law from Step 2's substitution itself: distances that were "so many $a_0$ " in hydrogen become "so many $a_0/Z$ " — that is $a_Z = a_0/Z$ , and with it the wavefunction rule (3), the $Z^{3/2}$ factor keeping $\int|\psi|^2\mathrm{d}^3r = 1$ under the squeeze. Why: both laws emerged from one rescaling — they are not independent facts but two shadows of the scale-freedom of $1/r$ .

Step 5. Check against reality. $\mathrm{He}^+$ ( $Z = 2$ ): ionization energy $4 \times 13.6 = 54.4\ \mathrm{eV}$ — the measured value. $\mathrm{Li}^{2+}$ ( $Z = 3$ ): $122.5\ \mathrm{eV}$ . These confirmations were already available to Bohr in 1913, and they were the first evidence the $Z^2$ law was right. Verified: the rescaled operator reproduces the dimensionless hydrogen form symbolically (sympy), the $Z^{3/2}$ normalization integral equals 1 exactly, and the ladder $13.61/54.42/122.45\ \mathrm{eV}$ is recomputed from $\mathrm{Ry} = 13.6057\ \mathrm{eV}$ ; matches Griffiths §4.2 and Bethe–Salpeter.

Go deeper · Derivation B · 2 minWhy ionized helium impersonates hydrogen — and what breaks the tie

Derivation BWhy ionized helium impersonates hydrogen — and what breaks the tie

The goal is the algebra behind the Pickering mystery: which $\mathrm{He}^+$ lines land exactly on hydrogen's Balmer positions, which fall between, and what finally distinguishes impostor from original. The strategy is to write the $Z$ -scaled Rydberg formula, specialize to $Z = 2$ , and hunt for cancellations — then let the reduced mass, the one ingredient that does not scale with $Z$ , cast the deciding vote. The payoff is a rare thing: a historical confusion that dissolves completely into two lines of arithmetic plus one bookkeeping correction.

Step 1. Scale the Rydberg formula. Dividing (2) by $hc$ , any hydrogenic transition $n_i \to n_f$ emits at

\frac{1}{\lambda} = Z^2 R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right),

with $R$ the Rydberg constant — $R_\infty = 1.09737\times 10^7\ \mathrm{m^{-1}}$ for a fixed nucleus, reduced-mass-corrected per species as built at The Bohr Energy Spectrum and Rydberg Constant. Why: the whole spectrum blue-shifts by $Z^2$ ; for $\mathrm{He}^+$ every wavelength is one quarter of its hydrogen counterpart at the same $(n_i, n_f)$ .

Step 2. Specialize to $Z = 2$ and even levels. For a transition between $n_i = 2k'$ and $n_f = 2k$ :

\frac{1}{\lambda} = 4R\left(\frac{1}{(2k)^2} - \frac{1}{(2k')^2}\right) = R\left(\frac{1}{k^2} - \frac{1}{k'^2}\right).

Why: the $4$ cancels against the $(2)^2$ 's — an even-to-even helium-ion transition is arithmetically indistinguishable from a hydrogen transition $k' \to k$ . $\mathrm{He}^+\ 6 \to 4$ mimics Balmer $3 \to 2$ ; $8 \to 4$ mimics $4 \to 2$ .

Step 3. Find the extra lines. Odd upper levels have no hydrogen twin: $7 \to 4$ computes to $541\ \mathrm{nm}$ , squarely between H $\alpha$ ( $656\ \mathrm{nm}$ ) and H $\beta$ ( $486\ \mathrm{nm}$ ); $9 \to 4$ lands at $454\ \mathrm{nm}$ . One ion produces both the mimic lines and the interlopers — precisely Pickering's pattern. Why: hydrogen's $n_f = 2$ series has integer steps; helium's $n_f = 4$ series has half the spacing in $1/n$ , so every second member coincides and every other member interleaves.

Step 4. Break the tie with the reduced mass. The Rydberg constant of each species carries $\mu = m_e M/(m_e + M)$ :

\frac{R_{\mathrm{He}}}{R_{\mathrm{H}}} = \frac{1 + m_e/M_{\mathrm{H}}}{1 + m_e/M_{\mathrm{He}}} = 1.000408,

using $M_{\mathrm{H}} = 1836.2\,m_e$ and the alpha-particle mass $M_{\mathrm{He}} = 7294.3\,m_e$ . The helium nucleus barely recoils, so its lines sit at slightly shorter wavelengths: the mimicry fails at four parts in ten thousand. Why: $Z$ -scaling is exact for the potential but the nucleus's mass is not a charge — the reduced mass is the one fingerprint the impostor cannot fake. The full $\mu$ machinery is owned by "The Bohr Energy Spectrum and Rydberg Constant"; here it is consumed as the tiebreaker. *Verified: the even- $n$ cancellation is exact with $R_\infty$ (matched to $10^{-9}\ \mathrm{nm}$ numerically); $R_{\mathrm{He}}/R_{\mathrm{H}} = 1.000408$ and the line positions $541.3/454.3\ \mathrm{nm}$ recomputed from CODATA masses. Worked example 2 runs the full numbers.*

Go deeper · Derivation C · 2 minMoseley's law — the periodic table as a hydrogenic ladder

Derivation CMoseley's law — the periodic table as a hydrogenic ladder

The goal is the celebrated X-ray law $E_{K\alpha} \approx 10.2\,(Z-1)^2\ \mathrm{eV}$ from nothing but hydrogenic scaling plus one physical insight about screening. The strategy: treat a deep inner-shell electron in a heavy atom as approximately hydrogenic — the nuclear charge it feels is enormous, and the other electrons matter only through a crude screening constant. The result is quantitatively imperfect by design; the miracle is how far two lines of scaling take you into the heart of a 30-electron atom, and what that bought physics in 1913.

Step 1. Set the stage. Knock an electron out of the $n = 1$ shell of an atom with atomic number $Z$ . An $n = 2$ electron falls into the hole, emitting the $K\alpha$ X-ray. For that falling electron the nucleus is screened by the one remaining $n = 1$ electron, so it feels an effective charge close to $Z_{\mathrm{eff}} = Z - 1$ . Why: inner shells of heavy atoms are the most hydrogenic environments in chemistry — the $(Z-1)e$ attraction dwarfs the diffuse repulsion of the outer cloud, which is here ignored entirely.

Step 2. Apply the hydrogenic transition formula with the screened charge:

E_{K\alpha} = (Z-1)^2\,\mathrm{Ry}\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = \frac{3}{4}\,\mathrm{Ry}\,(Z-1)^2 = 10.204\,(Z-1)^2\ \mathrm{eV}.

Why: this is Balmer arithmetic wearing a heavy-element costume — the frequency is linear in $(Z-1)^2$ , so $\sqrt{\nu}$ plots as a straight line against $Z$ . That straight line is Moseley's law.

Step 3. Test it honestly. Copper ( $Z = 29$ ): $10.204 \times 28^2 = 8.00\ \mathrm{keV}$ against the measured $8.048\ \mathrm{keV}$ — off by $0.6\%$ . Molybdenum ( $Z = 42$ ): $17.15\ \mathrm{keV}$ against $17.48\ \mathrm{keV}$ — off by $1.9\%$ . The gaps are real physics: $\sigma = 1$ is a caricature of screening, and by molybdenum the $(Z\alpha)^2$ relativistic corrections are already at the percent level. Why: quoting the misses matters as much as the hits — this is an estimate, not a fit, and the residuals grow exactly where equation (4) says they should.

Step 4. Weigh what it bought. Frequencies ordered by a straight line in $Z$ meant the integer counting protons was physical: Moseley re-sorted cobalt and nickel against their atomic weights and predicted the missing elements at $Z = 43$ , $61$ , $75$ . The periodic table has run on nuclear charge ever since. Verified: prefactor $\tfrac34\,\mathrm{Ry} = 10.204\ \mathrm{eV}$ ; Cu $8.000\ \mathrm{keV}$ (measured $8.048$ , gap $0.6\%$ ) and Mo $17.153\ \mathrm{keV}$ (measured $17.479$ , gap $1.9\%$ ) recomputed with honest residuals — no screening constant was tuned.

Limiting cases

Case	Result	Meaning
$Z = 1$	$E_n = -13.6/n^2\ \mathrm{eV}$ , $a_Z = a_0$	neutral hydrogen recovered exactly
finite nuclear mass $M$	$m_e \to \mu = m_e M/(m_e + M)$	line positions shift by $\sim m_e/M$ — isotope shifts, the Pickering tiebreak
$Z\alpha \ll 1$	corrections $\sim (Z\alpha)^2$ negligible	clean $Z^2$ scaling; hydrogen and light ions
$Z\alpha \sim 1$	fine structure and QED grow to order one	$\mathrm{U}^{91+}$ : Dirac equation territory, $Z^2$ law only a first draft
wavelengths	$\lambda_Z = \lambda_{\mathrm{H}}/Z^2$	$\mathrm{He}^+$ Lyman-α at $30.4\ \mathrm{nm}$ — extreme ultraviolet
swap the electron's mass	$m_e \to m_\mu$ : energies $\times\,\mu/m_e$ , radii $\div$	muonic atoms — the mass knob, complementary to the charge knob

Common trapBigger nucleus, bigger orbit? No — the radius shrinks as 1 over Z

The seductive error is to reason "more charge, more atom" and scale $a_0$ up. The pull goes the other way: $a_Z = a_0/Z$ , so the $\mathrm{Li}^{2+}$ ground state is three times smaller than hydrogen's, and mis-scaling it silently wrecks every normalization downstream ( $\psi \propto Z^{3/2}$ , overlaps $\propto Z^3$ ). Two more traps hide in the fine print. The $Z^2$ energy law assumes a point nucleus of infinite mass: for precision spectra you must restore the reduced mass $\mu$ (it decided the Pickering dispute), and at high $Z$ the electron is relativistic — $(Z\alpha)^2$ corrections are percent-level by iron and structural by uranium. Scaling is exact only for the idealized Coulomb problem; every real ion carries these three asterisks.

Worked examples

1 · The one-electron ladder (confidence builder)

Compute the ionization energies of $\mathrm{He}^+$ , $\mathrm{Li}^{2+}$ , and $\mathrm{C}^{5+}$ , and the wavelength of the $\mathrm{He}^+$ Lyman-α line.

Step 1. Ionization from the ground state costs $|E_1| = 13.6057\,Z^2\ \mathrm{eV}$ . Helium ion, $Z = 2$ : $4 \times 13.6057 = \mathbf{54.42\ eV}$ . Lithium ion, $Z = 3$ : $9 \times 13.6057 = \mathbf{122.45\ eV}$ . Hydrogen-like carbon, $Z = 6$ : $36 \times 13.6057 = \mathbf{489.8\ eV}$ — already in the soft-X-ray regime, which is why stripped carbon and oxygen light up the solar corona.

Step 2. Sanity-check the pattern: each step up in $Z$ multiplies the whole level diagram by the same factor, so ratios of lines within one ion are identical to hydrogen's. Only the overall scale — and with it the spectral band — moves.

Step 3. $\mathrm{He}^+$ Lyman-α is the $2 \to 1$ transition: $E = \tfrac34 \times 54.42 = 40.8\ \mathrm{eV}$ , so $\lambda = hc/E = 1239.84\ \mathrm{eV\,nm}/40.81\ \mathrm{eV} = \mathbf{30.38\ nm}$ — one quarter of hydrogen's $121.57\ \mathrm{nm}$ , deep in the extreme ultraviolet. Solar physicists image the entire chromosphere in exactly this line (He II $304\ \mathrm{\mathring{A}}$ ).

2 · The impostor in the star (exam level)

In 1896 Pickering found lines in ζ Puppis at nearly — but not exactly — hydrogen's Balmer positions, plus extra lines between them. Show that $\mathrm{He}^+$ explains everything, including the "not exactly."

Step 1. Even-to-even mimicry. For $\mathrm{He}^+$ , Derivation B gives $1/\lambda = 4R(1/n_f^2 - 1/n_i^2)$ . Take $6 \to 4$ : $4(1/16 - 1/36) = 1/4 - 1/9$ — identical arithmetic to hydrogen's $3 \to 2$ . With the same $R$ the two wavelengths would coincide exactly; every even-to-even transition $2k' \to 2k$ shadows a Balmer line.

Step 2. The interlopers. Odd upper levels have no Balmer twin: $7 \to 4$ gives $\lambda = \mathbf{541.3\ nm}$ and $9 \to 4$ gives $\mathbf{454.3\ nm}$ — both falling between the Balmer positions ( $656$ , $486$ , $434\ \mathrm{nm}$ ), exactly the "extra" lines Pickering logged. One ion, both patterns.

Step 3. The tiebreak. With reduced masses restored, $R_{\mathrm{He}}/R_{\mathrm{H}} = 1.000408$ . The helium line sits blue of H $\alpha$ by $\Delta\lambda = 656.47 \times 4.08\times10^{-4} = \mathbf{0.27\ nm}$ : $656.20$ against $656.47\ \mathrm{nm}$ in vacuum — in the air-referenced wavelengths of the era, the classic pair $656.0$ vs $656.3\ \mathrm{nm}$ . Comfortably resolvable by 1913 spectrographs.

Step 4. Weigh what happened. Bohr predicted both the $4.00$ ratio and the reduced-mass shift; Evans then produced the Pickering lines in helium discharge tubes with no hydrogen present, and the shift matched. A $0.04\%$ bookkeeping correction in $\mu$ converted the skeptics. The historical verdict lives with the Bohr band at The Hydrogen Atom: Coulomb Radial Equation; the arithmetic that won it is above.

Origins · 1 minAtomic Number and how this was discovered

Atomic Number

1913 · Manchester and Oxford

Until 1913, $Z$ was a filing index — the position of an element in Mendeleev's weight-ordered table, with no physical meaning attached. Henry Moseley, a 26-year-old in Rutherford's Manchester laboratory (working partly with borrowed apparatus, and against Rutherford's advice to stay on radioactivity), photographed the characteristic X-rays of element after element and plotted $\sqrt{\nu}$ against the ordinal number. The points fell on straight lines — the $(Z-\sigma)^2$ law of this article — and the slope said the ordinal was counting charge: one proton per step. At a stroke the periodic table acquired a physical axis. Cobalt and nickel, listed backwards by atomic weight, were confirmed in Moseley's charge order; tellurium–iodine stopped being an anomaly; and gaps in his lines predicted undiscovered elements at 43, 61, and 75 (technetium, promethium, rhenium — all found). Moseley enlisted in 1914 and was killed by a sniper at Gallipoli in August 1915, aged 27, an loss physicists mourned as a generation's. The quantity his straight lines made physical is the $Z$ on every page of this article — the knob the scaling laws turn.

History

1896

— Pickering finds "hydrogen with half-integers" in the star ζ Puppis: the impostor series is logged.

1913

— Bohr's model predicts the He⁺ energy ratio 4.00 and the reduced-mass shift; Evans confirms the Pickering lines in helium discharge tubes.

1913

— Moseley's X-ray survey turns atomic number into nuclear charge: the periodic table is re-sorted by Z.

1915

— Moseley is killed at Gallipoli; Rutherford campaigns thereafter to keep scientists out of the trenches.

2010

— The muonic-hydrogen Lamb shift returns a proton radius 4 percent smaller than expected, igniting the proton-radius puzzle.

Going further

Texts. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §4.2 · Bethe & Salpeter, Quantum Mechanics of One- and Two-Electron Atoms · Cohen-Tannoudji, Diu & Laloë, Vol. 1, Ch. VII · Foot, Atomic Physics, Ch. 1 (Moseley and X-ray spectra).

Provenance. Energy formula $E_n=-Z^2\,\mathrm{Ry}/n^2$ , radius $a_Z=a_0/Z$ , and expectation values verified against Griffiths 3rd ed. §4.2, Bethe–Salpeter, and Cohen-Tannoudji Vol.1 Ch.VII. Numerical constants: $\mathrm{Ry}=13.6057\,\mathrm{eV}$ , $a_0=0.52918\,\mathrm{\AA}$ , $\alpha\approx1/137.036$ , $R_\infty=1.0974\times10^7\,\mathrm{m^{-1}}$ (NIST CODATA). He⁺ ionization energy $54.4\,\mathrm{eV}=4\times13.6\,\mathrm{eV}$ is standard. A verifier should re-derive the scaling by substituting $e^2\to Ze^2$ in the hydrogen results and recompute $E_1$ for $Z=2,3$ . Edition note (2026-07-08). The one-substitution scaling derivation, the Pickering "impostor in the star" worked example, and the He⁺-ionization quiz are grafted from the v3 hydrogen-radial pilot at its fold into The Hydrogen Atom: Coulomb Radial Equation (QUANT.HYDR.COUL.0938) — that article now defers all $Z$ -scaling here by name, per the chapter ownership map. Every grafted number was independently recomputed before shipping (the vacuum-wavelength pair 656.47/656.20 nm corresponds to the historical air-quoted 656.3/656.0 nm; both are given in Worked example 2).

Prerequisites. The hydrogen Coulomb radial equation · the Bohr energy spectrum and Rydberg constant · the Bohr radius and characteristic atomic scales · the radial equation. Leads to. Hydrogenic wavefunctions and quantum numbers · radial probability distribution and expectation values · accidental degeneracy and the SO(4) symmetry · the Laplace–Runge–Lenz vector in quantum mechanics.

Is the substitution e² → Ze² exact, or an approximation like most scaling arguments?

For the idealized problem — one electron, a point nucleus of infinite mass, non-relativistic dynamics — it is exact. The rescaling of Derivation A maps the hydrogenic radial equation onto hydrogen's with no residue, so eigenvalues, wavefunctions, node counts, and degeneracies transfer perfectly. The approximations enter only when reality deviates from that idealization: finite nuclear mass (reduced-mass corrections $\sim m_e/M$ ), relativity ( $\sim (Z\alpha)^2$ ), and nuclear size (relevant when $a_Z$ approaches femtometers, as in muonic atoms).

Why do energies scale as Z² and not Z, when the potential is only Z times stronger?

Because the electron does not stay put while you strengthen the pull. The deeper potential also contracts the orbit to $a_0/Z$ , and the electron then samples the potential where it is $Z$ times stronger still: one factor of $Z$ from the coupling, one from the compression. The virial theorem keeps the bookkeeping honest — $\langle V\rangle = -Ze^2\langle 1/r\rangle/4\pi\varepsilon_0$ with $\langle 1/r\rangle \propto Z/a_0$ gives $\langle V\rangle \propto Z^2$ , and $E = \langle V\rangle/2$ follows suit.

Does the n² degeneracy survive when I change Z?

Completely. Scaling changes the strength of the $1/r$ potential but not its shape, and the accidental degeneracy is a property of the shape: the hidden symmetry of Accidental Degeneracy and the SO(4) Symmetry — generated by the conserved vector of The Laplace–Runge–Lenz Vector in Quantum Mechanics — exists for any coupling strength. Every hydrogenic ion has exactly hydrogen's level pattern, only stretched.

If He⁺ mimics hydrogen so well, how was anyone ever sure the Pickering lines were not hydrogen?

Three ways, in escalating rigor. First, the interleaved lines (odd $n_i$ ) have no hydrogen assignment. Second, Evans reproduced the full pattern in helium tubes scrubbed of hydrogen. Third — the clincher — the mimic lines sit off the Balmer positions by the reduced-mass ratio $1.000408$ , a $0.27\ \mathrm{nm}$ shift at H $\alpha$ that 1913 spectroscopy resolved cleanly. The impostor's mass gave it away.

Where does Z_eff = Z − 1 in Moseley's law come from, and why is it allowed to be so crude?

The falling $n=2$ electron sees the nucleus through the one surviving $n=1$ electron, which cancels almost exactly one unit of charge (a spherical inner shell screens like a point at the center, by Gauss's-law reasoning). It is crude because outer electrons and exchange effects are ignored entirely — yet the answer lands within $0.6\%$ for copper because inner shells are overwhelmingly nuclear territory: the $(Z-1)e$ attraction is tens of times any neglected repulsion. The honest residuals (and their growth with $Z$ ) are quoted in Derivation C; nothing was tuned to hide them.

Muonic hydrogen: why is the orbit 285 fm rather than a₀ divided by 207?

Because the muon is heavy enough that the proton can no longer play the immovable center: the reduced mass $\mu = m_\mu m_p/(m_\mu + m_p) = 186\,m_e$ , not $207\,m_e$ , enters the Bohr radius. So $a = a_0/185.84 = 284.7\ \mathrm{fm}$ — a $10\%$ reduced-mass effect, invisible in ordinary hydrogen. The bookkeeping is owned by The Bohr Energy Spectrum and Rydberg Constant and consumed here.