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Learning Objectives

By the end of this chapter, you should be able to:

Introduction

Chapter 9 developed the general machinery for any central potential: separation into radial and angular parts, and the universal quantization of orbital angular momentum. This chapter specializes that machinery to the single most important central potential in atomic physics — the Coulomb attraction between an electron and a proton — and solves it for the hydrogen atom, the only atom for which the Schrödinger equation can be solved exactly in closed form. The result reproduces (and explains, rather than assumes) the energy levels first found empirically in atomic spectra and postulated ad hoc in the 1913 Bohr model, while revealing a far richer structure — angular momentum, spatial probability distributions, and a fourth quantum number, electron spin, with no classical counterpart at all.

10.1Energy Levels and Hydrogenic Structure

The Radial Equation and Energy Quantization

For the hydrogen atom, the central potential is the Coulomb attraction between the electron (charge e-e) and the proton (charge +e+e),

V(r)=e24πϵ0r.V(r) = -\frac{e^2}{4\pi\epsilon_0\, r}.

Substituting this V(r)V(r) into the radial equation obtained from the separation ψ=R(r)Y(θ,ϕ)\psi = R(r)Y(\theta,\phi) (Chapter 9), and requiring R(r)R(r) to be normalizable (i.e., to decay rather than blow up as rr\to\infty, and to remain finite at r=0r=0), restricts the allowed energies to exactly the same discrete set found by Bohr in 1913 from an ad hoc semiclassical model:

En=mee48ϵ02h21n2=13.6 eVn2,n=1,2,3,.E_n = -\frac{m_ee^4}{8\epsilon_0^2h^2}\,\frac{1}{n^2} = -\frac{13.6\ \text{eV}}{n^2}, \qquad n = 1, 2, 3,\ldots.

This agreement is a triumph for the Schrödinger equation — it reproduces a result that matched atomic spectroscopy to remarkable precision — but the derivation and interpretation are entirely different from Bohr’s. Bohr postulated that the electron moves on definite circular orbits, with angular momentum quantized as L=nL = n\hbar by assumption. The Schrödinger treatment makes no such assumption about orbits at all; instead, nn emerges purely as an index counting the normalizable solutions of the radial equation, the electron has no well-defined trajectory, and — as shown below — the ground state (n=1n=1) actually has zero orbital angular momentum, in direct contradiction to Bohr’s L=nL=n\hbar. The numerical agreement in EnE_n is, in this sense, a coincidence specific to the particular form of the 1/r1/r Coulomb potential, not a sign that the Bohr picture was substantially correct.

The mechanism behind the discreteness is worth making explicit, since the same pattern recurs throughout quantum mechanics. Solving the radial equation for a trial (negative) energy EE produces a function that falls off as eκre^{-\kappa r} at large rr, with κ=2meE/\kappa = \sqrt{-2m_eE}/\hbar, multiplied by an infinite series in powers of rr; for almost any choice of EE, that series itself grows fast enough that the full product still diverges as rr\to\infty, exactly the unnormalizable blow-up that ruled out most trial energies in the finite square well of Chapter 8. Demanding a normalizable wave function forces the series to terminate after a finite number of terms — collapsing it from an infinite series into a finite polynomial — and this termination happens only when 1/κa01/\kappa a_0 lands exactly on a positive integer, which is precisely the principal quantum number nn. The same termination condition also caps the orbital quantum number: the polynomial’s degree fixes a nonnegative integer nr=0,1,2,n_r = 0, 1, 2,\ldots (the number of interior radial nodes, discussed further below) related to nn and \ell by n=nr++1n = n_r + \ell + 1, so that nr0n_r\ge0 immediately forces n1\ell \le n-1 — and, for the ground state n=1n=1, forces nr==0n_r=\ell=0 exactly, which is the direct algebraic reason the ground state must have zero orbital angular momentum, not merely an empirical fact to be taken on faith.

The distance between Bohr’s picture and Schrödinger’s is easiest to see by running them side by side, as in Figure 10.1. Both reproduce the same EnE_n and therefore the same emission lines, but only one of them places the electron on an orbit.

Screenshot of the Models of the Hydrogen Atom simulation

Figure 10.1:Successive models of the hydrogen atom — Bohr, de Broglie, Schrödinger, and their predecessors — each firing photons at an atom and producing a spectrum. The models that agree on EnE_n disagree entirely about where the electron is.

Interactive simulation: Models of the Hydrogen Atom

Historical Context: Bohr’s 1913 Model and Its Limits

It is worth pausing on why a model built on postulates that turned out to be simply wrong — definite orbits, angular momentum quantized as L=nL=n\hbar — nonetheless earned Niels Bohr the 1922 Nobel Prize and remains the picture most people first encounter. In 1913, thirteen years before Schrödinger’s equation existed, Bohr combined classical circular-orbit mechanics with a single ad hoc quantization rule and reproduced the empirical Rydberg formula for hydrogen’s spectral lines essentially exactly, including its dependence on nuclear charge for hydrogenic ions (below) — a stunning success for a model with no derivation behind its central assumption. The de Broglie standing-wave argument of Chapter 7 later supplied a retroactive justification for L=nL=n\hbar, which is part of why the Bohr model survived as long as it did before being fully superseded.

The model’s failures, however, were just as instructive as its success. Extending Bohr’s orbit-quantization scheme to helium — even with the refinements (elliptical orbits, relativistic orbit precession) added by Arnold Sommerfeld through the 1910s — never produced a correct ionization energy or a stable ground-state configuration for a two-electron atom; the semiclassical machinery simply had no consistent way to handle two mutually interacting orbiting electrons. The model also could not predict which transitions between levels actually occur (the selection rules developed later in this chapter), could not account for the relative brightness of spectral lines, and — as already emphasized above — gets the ground state’s angular momentum flatly wrong: Bohr’s n=1n=1 orbit carries L=L=\hbar, while the true ground state has L=0L=0 and no orbit, well-defined trajectory, or definite radius at all, only the probability cloud shown in Figure 10.1. These failures, especially the inability to extend the model consistently beyond hydrogen, were a central motivation for the fully quantum-mechanical treatment developed in Chapters 89 and specialized to hydrogen in this chapter — a treatment that, unlike Bohr’s, generalizes cleanly to helium and every other atom in Chapter 11.

Hydrogenic Ions: Scaling with Nuclear Charge

Everything derived above specializes the general central-potential machinery of Chapter 9 to a nuclear charge of exactly +e+e. The same radial equation applies unchanged, with only e2Ze2e^2 \to Ze^2 in the Coulomb potential, to any hydrogenic (hydrogen-like) ion: a single electron bound to a nucleus of charge +Ze+Ze, such as singly ionized helium He+\text{He}^+ (Z=2Z=2) or doubly ionized lithium Li2+\text{Li}^{2+} (Z=3Z=3). Repeating the normalizability argument above with the rescaled potential gives

En(Z)=Z213.6 eVn2,an(Z)=n2Za0,E_n(Z) = -Z^2\,\frac{13.6\ \text{eV}}{n^2}, \qquad a_n(Z) = \frac{n^2}{Z}\,a_0,

where an(Z)a_n(Z) is the characteristic Bohr length scale of the nn-th shell. It is useful for comparing orbital sizes, but it is not the exact radial-probability peak for every (n,)(n,\ell) state; those peak positions also depend on \ell. Both dependences make direct physical sense: a larger nuclear charge attracts the electron more strongly, pulling its orbit inward (radius shrinks as 1/Z1/Z) while binding it more tightly — and because the potential energy itself scales as ZZ while the resulting spatial compression compounds that scaling once more, the binding energy grows as Z2Z^2, not simply ZZ. This Z2Z^2 scaling reappears with direct experimental consequence in Chapter 11, where it underlies Moseley’s law for the energies of characteristic X-ray transitions in multi-electron atoms.

Worked Example: The Ionization Energy and Orbital Size of He⁺

Find the ground-state ionization energy and the most probable electron–nucleus separation for He+\text{He}^+ (Z=2Z=2), a one-electron ion, and compare both to hydrogen’s.

E1(He+)=Z2(13.6 eV)=(2)2(13.6 eV)=54.4 eV,E_1(\text{He}^+) = -Z^2(13.6\ \text{eV}) = -(2)^2(13.6\ \text{eV}) = -54.4\ \text{eV},

so the ionization energy — the energy required to remove the electron entirely — is 54.4 eV54.4\ \text{eV}, exactly four times hydrogen’s 13.6 eV13.6\ \text{eV}, matching the measured ionization energy of He+\text{He}^+. The orbital size scales the other way,

a1(He+)=a0Z=0.0529 nm2=0.0265 nm,a_1(\text{He}^+) = \frac{a_0}{Z} = \frac{0.0529\ \text{nm}}{2} = 0.0265\ \text{nm},

half of hydrogen’s Bohr radius: the electron in He+\text{He}^+ is bound four times more tightly and orbits, on average, twice as close to the nucleus as the electron in neutral hydrogen — both consequences of the same doubled nuclear charge, entering the energy quadratically and the size linearly (inversely).

Quantum Numbers and Degeneracy in Hydrogen

Solving the full three-dimensional problem gives states labeled by the same three quantum numbers introduced in Chapter 9nn, \ell, mm_\ell — but with a further restriction, specific to the 1/r1/r Coulomb potential, tying \ell to nn:

n=1,2,3,,=0,1,,n1,m=,,.n = 1, 2, 3, \ldots, \qquad \ell = 0, 1, \ldots, n-1, \qquad m_\ell = -\ell, \ldots, \ell.

Because the energy EnE_n depends only on nn — not on \ell or mm_\ell — every state sharing a given nn is degenerate (equal in energy), regardless of its orbital angular momentum. Counting the total number of (,m)(\ell, m_\ell) combinations for a given nn gives n2n^2 degenerate states (before accounting for electron spin, discussed below): for example, n=2n=2 admits =0\ell=0 (one state, m=0m_\ell=0) and =1\ell=1 (three states, m=1,0,1m_\ell=-1,0,1), for 1+3=4=221+3=4=2^2 states total. This \ell-independence of the energy is itself notable — it does not hold for multi-electron atoms (Chapter 11), where the energy depends on \ell as well as nn, and is a special feature of the pure 1/r1/r Coulomb potential (technically, a signature of a hidden extra symmetry, beyond ordinary rotational symmetry, unique to the 1/r1/r potential).

The Hidden Symmetry Behind ℓ-Independent Degeneracy

Ordinary three-dimensional rotational symmetry guarantees that energy cannot depend on mm_\ell, since no direction in space is special — but it says nothing about \ell. A generic central potential’s energy levels do depend on \ell, and indeed they do for every atom with more than one electron (Chapter 11). Hydrogen’s extra, “accidental” \ell-independence traces back to a conserved quantity unique to the exact 1/r1/r potential: the classical Laplace–Runge–Lenz vector,

A=p×Lmee24πϵ0r^,\vec A = \vec p\times\vec L - \frac{m_ee^2}{4\pi\epsilon_0}\,\hat r,

which points from the focus of a classical Kepler orbit toward its perihelion and is constant in time only for a force that falls off as exactly 1/r21/r^2 — not 1/r2.011/r^{2.01}, not 1/r1.991/r^{1.99}. Quantum mechanically, A\vec A becomes an operator that commutes with the Hamiltonian and connects states of different \ell at the same nn, enlarging the symmetry governing hydrogen from ordinary three-dimensional rotations, SO(3)SO(3), to a larger four-dimensional rotation group, SO(4)SO(4), whose representations turn out to be exactly the n2n^2-fold degenerate multiplets observed. This is the same underlying fact about the 1/r21/r^2 force law that makes classical Kepler orbits close on themselves without precessing — the classical and quantum “accidents” are one and the same.

10.2Wave Functions, Spectra, and Spin

Wave Functions and Probability Distributions

The full wave functions, ψnm(r,θ,ϕ)=Rn(r)Ym(θ,ϕ)\psi_{n\ell m_\ell}(r,\theta,\phi) = R_{n\ell}(r)\,Y_{\ell m_\ell}(\theta,\phi), have structure worth examining qualitatively even without their explicit algebraic form. The ground state, ψ100\psi_{100}, is spherically symmetric (=0\ell=0, so Y00Y_{00} is constant) and decays exponentially, R10(r)er/a0R_{10}(r) \propto e^{-r/a_0}, where

a0=4πϵ02mee2=0.0529 nma_0 = \frac{4\pi\epsilon_0\hbar^2}{m_ee^2} = 0.0529\ \text{nm}

is the Bohr radius — reappearing here not as the radius of a Bohr orbit but as the natural length scale over which the ground-state probability density falls off. The radial probability distribution, P(r)=r2Rn(r)2P(r) = r^2|R_{n\ell}(r)|^2 (the probability per unit rr of finding the electron at distance rr from the nucleus, obtained by integrating ψ2|\psi|^2 over all angles at fixed rr), peaks at r=a0r = a_0 for the ground state — the most probable electron-nucleus distance in hydrogen’s ground state is exactly the Bohr radius, even though the electron has zero orbital angular momentum and hence, unlike in the Bohr picture, is not “orbiting” in any classical sense.

Written out in full, with proper normalization, the lowest few radial wave functions are

R10(r)=2a03/2er/a0,R20(r)=122a03/2(2ra0)er/2a0,R21(r)=126a03/2ra0er/2a0,R_{10}(r) = \frac{2}{a_0^{3/2}}\,e^{-r/a_0}, \qquad R_{20}(r) = \frac{1}{2\sqrt2\,a_0^{3/2}}\left(2-\frac{r}{a_0}\right)e^{-r/2a_0}, \qquad R_{21}(r) = \frac{1}{2\sqrt6\,a_0^{3/2}}\,\frac{r}{a_0}\,e^{-r/2a_0},

each verified against the standard hydrogen wave-function table (e.g. OpenStax Vol. 3, §8.2). The pattern generalizes: every Rn(r)R_{n\ell}(r) is an exponential er/na0e^{-r/na_0} (decay rate set only by nn) multiplied by a polynomial in rr of degree n1n-1, and the number of interior zeros of that polynomial — the radial nodes counted below — is exactly nr=n1n_r = n-\ell-1, the same integer introduced above via n=nr++1n=n_r+\ell+1. R10R_{10} (nr=0n_r=0) is a bare exponential with no node; R20R_{20} (nr=1n_r=1) has one node, at r=2a0r=2a_0, exactly where its parenthetical factor vanishes; R21R_{21} (nr=0n_r=0, since its higher \ell “uses up” the node budget at fixed n=2n=2) is a bare power of rr times an exponential, with no node at all despite sharing n=2n=2 with R20R_{20}.

Figure 10.2 plots the resulting P(r)=r2Rn(r)2P(r)=r^2|R_{n\ell}(r)|^2 for six low-lying states, making the node-counting rule directly visible rather than merely asserted: each panel’s dotted vertical lines mark its nodes, and the count in every panel matches n1n-\ell-1 exactly — zero for 1s1s, 2p2p, and 3d3d; one for 2s2s and 3p3p; two for 3s3s.

Six panels of radial probability distribution P(r) versus r over the Bohr radius, for the 1s, 2s, 2p, 3s, 3p, and 3d states of hydrogen, with dotted lines marking the radial nodes.

Figure 10.2:Radial probability distributions P(r)=r2Rn(r)2P(r)=r^2|R_{n\ell}(r)|^2 for the six lowest distinct (n,)(n,\ell) combinations. Each state’s node count matches n1n-\ell-1 exactly, and every curve for a given nn extends, on average, farther from the origin as nn increases — the size of the atom really does grow with nn, just not along a sharp Bohr orbit. Computed directly from the closed-form Rn(r)R_{n\ell}(r) formulas; see scripts/figures/.

More generally, Rn(r)R_{n\ell}(r) has n1n-\ell-1 radial nodes (points, other than r=0r=0 and r=r=\infty, where the probability density vanishes), and the angular functions Ym(θ,ϕ)Y_{\ell m_\ell}(\theta,\phi) have angular nodes (nodal planes or cones) whose count and shape depend on \ell and mm_\ell — giving rise to the familiar ss (spherical), pp (dumbbell-shaped, with a single nodal plane through the origin), and dd-orbital shapes used throughout chemistry (Chapter 12) to describe electron distributions in atoms and molecules.

The Effective Potential and the Centrifugal Barrier

Chapter 9 showed that the radial equation for any central potential can be recast as an effective one-dimensional Schrödinger equation for u(r)=rR(r)u(r)=rR(r), governed by an effective potential

Veff(r)=V(r)+2(+1)2mer2=e24πϵ0r+2(+1)2mer2,V_{\text{eff}}(r) = V(r) + \frac{\hbar^2\,\ell(\ell+1)}{2m_er^2} = -\frac{e^2}{4\pi\epsilon_0 r} + \frac{\hbar^2\,\ell(\ell+1)}{2m_er^2},

the attractive Coulomb term plus the repulsive centrifugal barrier introduced there. For =0\ell=0 (ss-states), Veff(r)=V(r)V_{\text{eff}}(r)=V(r): nothing but the bare attractive well, all the way in to r=0r=0. This is why only =0\ell=0 states can have a nonzero probability density ψ(0)2|\psi(0)|^2 at the nucleus. The radial probability P(r)=r2R(r)2P(r)=r^2|R(r)|^2, however, is zero at r=0r=0 for every state because the spherical volume element contributes the factor r2r^2; accordingly, the 1s1s, 2s2s, and 3s3s curves in Figure 10.2 all rise from zero at the origin even though their wave functions themselves need not. For >0\ell>0, the centrifugal term (1/r2\propto 1/r^2) dominates the attractive term (1/r\propto 1/r) at small rr and produces a genuine barrier that pushes the wave function away from the nucleus; the larger \ell is, the taller that barrier and the farther out the first significant probability appears, exactly as seen in the figure, where 2p2p, 3p3p, and 3d3d all start at zero and rise more gradually than their same-nn ss-state counterparts, and where 3d3d — the largest \ell at n=3n=3 — peaks farthest from the origin of the three n=3n=3 curves. This \ell-dependent “penetration” toward the nucleus is purely qualitative for hydrogen’s single electron, but it becomes quantitatively essential in Chapter 11, where it explains why an ss-electron in a multi-electron atom is attracted more strongly by (and screens other electrons from) the nuclear charge than a pp- or dd-electron of the same nn — the mechanism that breaks hydrogen’s accidental \ell-independent degeneracy once more than one electron is present.

Spectral Series and the Energy-Level Diagram

Every property established so far — discrete EnE_n, the selection rule developed below — can be assembled into a single picture: the ladder of allowed energies together with the transitions between them that are actually observed. Figure 10.3 draws that ladder for n=1n=1 through 5, together with the three most commonly tabulated spectral series, each named for whoever first catalogued it and each defined by a common lower level: the Lyman series (nf=1n_f=1, entirely in the ultraviolet), the Balmer series (nf=2n_f=2, spanning the visible), and the Paschen series (nf=3n_f=3, infrared). Because the spacing between successive EnE_n shrinks rapidly as nn grows (it falls off as roughly 1/n31/n^3 near the top of the ladder), the levels bunch up as they approach E=0E=0, and each series converges to a series limit: the shortest wavelength (highest photon energy) the series can produce, corresponding to a transition from nin_i\to\infty down to the series’ fixed nfn_f, and equal to the energy 13.6 eV/nf213.6\ \text{eV}/n_f^2 needed to ionize the atom starting from level nfn_f. Beyond that limit the spectrum stops being a set of discrete lines at all, since a photoionized electron (E>0E>0) is no longer confined to a discrete spectrum and can carry away any leftover energy continuously.

Hydrogen energy levels for n equals 1 through 5 on a compressed vertical scale, with clusters of arrows showing the Lyman series converging on n=1, the Balmer series converging on n=2, and the Paschen series converging on n=3.

Figure 10.3:Hydrogen energy levels n=1n=15 (vertical position compressed for legibility; labels give the real En=13.6 eV/n2E_n=-13.6\ \text{eV}/n^2), with the Lyman, Balmer, and Paschen series drawn as clusters of downward transitions converging on nf=1n_f=1, 2, and 3 respectively. Each series’ shortest-wavelength member is its series limit, reached only as nin_i\to\infty. Computed directly from EnE_n; see scripts/figures/.

Worked Example: The Balmer Series and the Visible Spectrum

The Rydberg formula is simply En=13.6 eV/n2E_n=-13.6\ \text{eV}/n^2 rewritten as a wavelength via hc/λ=EniEnfhc/\lambda = |E_{n_i}-E_{n_f}|:

1λ=R(1nf21ni2),R=1.097×107 m1.\frac{1}{\lambda} = R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right), \qquad R = 1.097\times10^7\ \text{m}^{-1}.

Find (a) the wavelength of Hα\text{H}_\alpha, the first (longest-wavelength) line of the Balmer series (ni=3nf=2n_i=3\to n_f=2), and (b) the Balmer series limit.

(a)

1λ=(1.097×107 m1)(1419)=1.524×106 m1λ=656.3 nm,\frac{1}{\lambda} = (1.097\times10^7\ \text{m}^{-1})\left(\frac14-\frac19\right) = 1.524\times10^6\ \text{m}^{-1} \quad\Longrightarrow\quad \lambda = 656.3\ \text{nm},

a deep red line — the Hα\text{H}_\alpha line used to image glowing hydrogen gas in nebulae and to trace the rotation of spiral galaxies.

(b) As nin_i\to\infty, 1/ni201/n_i^2\to0, so

1λlimit=R4=2.743×106 m1λlimit=364.6 nm,\frac{1}{\lambda_{\text{limit}}} = \frac{R}{4} = 2.743\times10^6\ \text{m}^{-1} \quad\Longrightarrow\quad \lambda_{\text{limit}} = 364.6\ \text{nm},

just past the violet edge of the visible spectrum, in the near ultraviolet. Every Balmer line falls between these two wavelengths, 364.6 nm<λ656.3 nm364.6\ \text{nm} < \lambda \le 656.3\ \text{nm} — which is why nearly the whole series, unlike the entirely ultraviolet Lyman series or the entirely infrared Paschen series, is visible to the human eye, and why Johann Balmer found the first four of these lines by curve-fitting alone in 1885, three decades before Bohr’s model — or Schrödinger’s equation — existed to explain them.

Electron Spin

By the mid-1920s, several pieces of spectroscopic evidence — most directly, the splitting of atomic beams passing through an inhomogeneous magnetic field — showed that the three quantum numbers (n,,m)(n,\ell,m_\ell) do not fully specify an electron’s state. In the Stern–Gerlach experiment (1922), a beam of (electrically neutral) silver atoms was passed through a strongly inhomogeneous magnetic field and allowed to strike a detector screen. A classical magnetic dipole, oriented randomly, should be deflected by an amount depending continuously on its orientation, producing a single smeared-out band on the screen. Instead, the beam split into exactly two discrete spots, symmetric about the undeflected position — direct evidence of space quantization (Chapter 9) applied to a new, previously unsuspected degree of freedom, since the outermost electron in a silver atom happens to be in an ss-state (=0\ell=0, hence zero orbital angular momentum and no orbital magnetic moment to produce any deflection at all), so the observed splitting could not be due to orbital angular momentum.

The resolution, proposed by Samuel Goudsmit and George Uhlenbeck (1925), is that the electron possesses an intrinsic angular momentum, spin, S\vec S, with no classical counterpart (it is not literally the electron spinning on its axis — such a picture leads to internal-consistency and speed-of-rotation contradictions and should be regarded purely as a suggestive name), quantized exactly as orbital angular momentum is, but with a spin quantum number restricted to the single value s=12s=\tfrac12:

S=s(s+1)=32,Sz=ms,ms=12,+12.S = \sqrt{s(s+1)}\,\hbar = \frac{\sqrt3}{2}\hbar, \qquad S_z = m_s\hbar, \qquad m_s = -\tfrac12, +\tfrac12.

The two allowed values of msm_s — “spin up” and “spin down” — account exactly for the two Stern–Gerlach spots. A complete specification of an electron’s state in hydrogen therefore requires four quantum numbers, (n,,m,ms)(n,\ell,m_\ell,m_s), and the count of degenerate states for a given nn becomes 2n22n^2 rather than n2n^2, the factor of 2 from the two spin orientations — a result central to the structure of the periodic table in Chapter 11.

10.3Magnetic Moments and Selection Rules

Magnetic Moments

Both orbital and spin angular momentum give the electron a magnetic dipole moment, since a circulating (or intrinsically “spinning”) charge behaves as a small current loop. The orbital magnetic moment is

μL=e2meL,μL,z=mμB,\vec\mu_L = -\frac{e}{2m_e}\vec L, \qquad \mu_{L,z} = -m_\ell\mu_B,

where μBe/2me=9.274×1024 J/T\mu_B \equiv e\hbar/2m_e = 9.274\times10^{-24}\ \text{J/T} is the Bohr magneton, a natural unit of atomic magnetic moment. The spin magnetic moment has an analogous form but with an extra numerical factor (the electron’s gg-factor, gs2g_s \approx 2, itself a prediction of relativistic quantum theory beyond the scope of the nonrelativistic Schrödinger equation used here):

μS,z=gsmsμB2msμB.\mu_{S,z} = -g_s\, m_s\, \mu_B \approx -2m_s\mu_B.

These magnetic moments are what couple to an external magnetic field in the Stern–Gerlach experiment (producing the observed splitting and deflection) and, coupling to each other and to nuclear magnetic moments, produce the fine and hyperfine splittings observed in high-resolution atomic spectra.

The Stern–Gerlach Force and a Numeric Deflection

A uniform magnetic field exerts a torque on a magnetic dipole — producing the Larmor precession pictured in Chapter 9’s precessing-top figure — but exerts zero net force, because the interaction energy μB-\vec\mu\cdot\vec B does not depend on position when B\vec B itself does not. Deflecting the beam at all, as Stern and Gerlach needed to do to see anything on their screen, requires a spatially varying (inhomogeneous) field, whose gradient converts the dipole’s orientation into a genuine transverse force:

Fz=μzBzz,F_z = \mu_z\,\frac{\partial B_z}{\partial z},

pushing atoms with μz>0\mu_z>0 one way and μz<0\mu_z<0 the other. Because μz\mu_z takes only the two discrete values μS,z=±μB\mu_{S,z}=\pm\mu_B (from the spin-moment formula above, with ms=±12m_s=\pm\tfrac12 and gs2g_s\approx2) rather than a continuum of values, the beam splits into exactly two discrete trajectories instead of spreading into one continuous smear — the direct experimental signature of space quantization that motivated this section.

Worked Example: Beam Splitting in a Stern–Gerlach Magnet

A beam of silver atoms (m=1.79×1025 kgm=1.79\times10^{-25}\ \text{kg}) effuses from an oven at T=1000 KT=1000\ \text{K}, giving a typical beam speed v=3kBT/m480 m/sv=\sqrt{3k_BT/m}\approx 480\ \text{m/s}. The beam passes through a magnet of length L=0.50 mL=0.50\ \text{m} producing a field gradient Bz/z=10 T/m\partial B_z/\partial z = 10\ \text{T/m}, then drifts a further D=0.50 mD=0.50\ \text{m} to a detector screen. Find the separation between the two spots.

The force on each atom is

Fz=μzBzz=±(9.274×1024 J/T)(10 T/m)=±9.27×1023 N,F_z = \mu_z\frac{\partial B_z}{\partial z} = \pm(9.274\times10^{-24}\ \text{J/T})(10\ \text{T/m}) = \pm9.27\times10^{-23}\ \text{N},

giving a transverse acceleration a=Fz/m=±518 m/s2a = F_z/m = \pm518\ \text{m/s}^2. While inside the magnet, for a time t1=L/v=1.04×103 st_1=L/v=1.04\times10^{-3}\ \text{s}, each atom acquires a transverse displacement z1=12at12=0.28 mmz_1=\tfrac12at_1^2 = 0.28\ \text{mm} and a transverse velocity vz=at1=0.54 m/sv_z=at_1=0.54\ \text{m/s}; it then drifts (no further transverse force, but still moving at vzv_z) for an additional time t2=D/v=1.04×103 st_2=D/v=1.04\times10^{-3}\ \text{s}, adding z2=vzt2=0.56 mmz_2=v_zt_2=0.56\ \text{mm}. The one-sided deflection is z1+z20.84 mmz_1+z_2\approx0.84\ \text{mm}, so the two spin states land

Δz=2(z1+z2)1.7 mm\Delta z = 2(z_1+z_2) \approx 1.7\ \text{mm}

apart on the screen — comfortably resolvable, and the same order of magnitude as the splitting Stern and Gerlach actually measured in 1922 with a weaker gradient over a shorter path, which is what made their two-spot pattern (rather than one smeared band) unambiguous.

The Zeeman Effect

Chapter 9 introduced the (normal) Zeeman effect qualitatively: an external field B\vec B along zz adds an energy μLB=mμBB-\vec\mu_L\cdot\vec B = m_\ell\mu_B B to each otherwise-degenerate mm_\ell sublevel of a given (n,)(n,\ell). Including the spin moment as well, the total field-induced shift of a state (m,ms)(m_\ell,m_s) is

ΔE=(m+gsms)μBB(m+2ms)μBB,\Delta E = (m_\ell + g_sm_s)\,\mu_B B \approx (m_\ell + 2m_s)\,\mu_B B,

so that adjacent orbital sublevels are separated by exactly μBB\mu_BB. For electric-dipole transitions, the corresponding magnetic quantum-number rule is Δm=0,±1\Delta m_\ell=0,\pm1: the Δm=±1\Delta m_\ell=\pm1 transitions produce the two side components, while Δm=0\Delta m_\ell=0 produces the central component of the normal Zeeman triplet. Historically, it was the observed splitting pattern (sometimes the simple three-line “normal” Zeeman pattern predicted by mm_\ell alone, but more often a more complicated “anomalous” pattern explainable only once spin and its distinct gg-factor were included) that provided some of the earliest indirect evidence for electron spin, years before Stern and Gerlach identified its direct mechanical signature.

Worked Example: Zeeman Splitting Frequency in a 1 T Field

Estimate the frequency splitting between adjacent mm_\ell sublevels for a hydrogen atom in a B=1.0 TB=1.0\ \text{T} field (a typical laboratory electromagnet), and the resulting wavelength shift of the Hα\text{H}_\alpha line (λ=656.3 nm\lambda=656.3\ \text{nm}, from the worked example above).

ΔE=μBB=(9.274×1024 J/T)(1.0 T)=9.27×1024 J=5.79×105 eV,\Delta E = \mu_BB = (9.274\times10^{-24}\ \text{J/T})(1.0\ \text{T}) = 9.27\times10^{-24}\ \text{J} = 5.79\times10^{-5}\ \text{eV},
Δf=ΔEh=9.27×1024 J6.626×1034 Js=1.40×1010 Hz=14.0 GHz\Delta f = \frac{\Delta E}{h} = \frac{9.27\times10^{-24}\ \text{J}}{6.626\times10^{-34}\ \text{J}\cdot\text{s}} = 1.40\times10^{10}\ \text{Hz} = 14.0\ \text{GHz}

(the electron’s Larmor precession frequency in this field). Converting to a wavelength shift via Δλ(λ2/c)Δf|\Delta\lambda|\approx(\lambda^2/c)\,\Delta f,

Δλ(656.3×109 m)2(1.40×1010 Hz)3.00×108 m/s0.020 nm,\Delta\lambda \approx \frac{(656.3\times10^{-9}\ \text{m})^2(1.40\times10^{10}\ \text{Hz})}{3.00\times10^8\ \text{m/s}} \approx 0.020\ \text{nm},

about 30,00030{,}000 times smaller than the line’s own wavelength — invisible by eye, but well within reach of a laboratory grating with resolving power R=λ/Δλ3×104R=\lambda/\Delta\lambda\sim3\times10^4 (Chapter 5), which is how the Zeeman effect is actually resolved in the lab.

Fine Structure: Spin–Orbit Coupling

Even with no external field at all, the energies En=13.6 eV/n2E_n=-13.6\ \text{eV}/n^2 are not quite the whole story. A useful semiclassical picture is that, in the electron’s rest frame, the orbiting (positively charged) nucleus constitutes a circulating current that produces a magnetic field at the electron’s location. That internal field couples to the electron’s spin magnetic moment, with the necessary relativistic treatment—including Thomas precession—giving the interaction called spin–orbit coupling. Its strength depends on the relative orientation of L\vec L and S\vec S (equivalently, on the total angular momentum J=L+S\vec J = \vec L+\vec S). Spin–orbit coupling vanishes for =0\ell=0, but the full fine structure also includes relativistic kinetic-energy and Darwin corrections, so ss-states can receive fine-structure shifts as well. A complete treatment requires relativistic quantum mechanics beyond the nonrelativistic Schrödinger equation used throughout this book. What matters here is the size of the resulting fine-structure correction:

ΔEfineα2En,αe24πϵ0c1137.07.30×103,\Delta E_{\text{fine}} \sim \alpha^2\,|E_n|, \qquad \alpha \equiv \frac{e^2}{4\pi\epsilon_0\hbar c} \approx \frac{1}{137.0} \approx 7.30\times10^{-3},

where α\alpha, the dimensionless fine-structure constant, measures the intrinsic strength of the electromagnetic interaction and sets the size of essentially every relativistic correction in atomic physics. Because α25×105\alpha^2\approx5\times10^{-5}, fine structure shifts a level by only a few parts in 105 of En|E_n| itself — far too fine to see with a simple grating, but readily resolved with a high-resolution spectrometer, and the historical reason spectral lines that look perfectly sharp in an introductory demonstration reveal themselves, under sufficient magnification, to be closely spaced multiplets (hence the name, coined for exactly this kind of fine-toothed splitting of what earlier instruments saw as a single line).

Worked Example: Order-of-Magnitude Fine-Structure Splitting

Estimate the fine-structure splitting of hydrogen’s n=2n=2 level, and compare it to the actual measured splitting of the 2p2p levels, ΔEfine4.5×105 eV\Delta E_{\text{fine}}\approx4.5\times10^{-5}\ \text{eV} (equivalently, about 10.9 GHz10.9\ \text{GHz}).

E2=13.6 eV4=3.40 eVΔEfineα2E2=(7.30×103)2(3.40 eV)1.8×104 eV,E_2 = -\frac{13.6\ \text{eV}}{4} = -3.40\ \text{eV} \quad\Longrightarrow\quad \Delta E_{\text{fine}} \sim \alpha^2|E_2| = (7.30\times10^{-3})^2(3.40\ \text{eV}) \approx 1.8\times10^{-4}\ \text{eV},

within about a factor of four of the measured value — entirely appropriate for an order-of-magnitude estimate that drops the numerical factors (jj-dependence and other combinatorics supplied only by a full relativistic treatment) while correctly capturing the essential physical scale: fine structure is smaller than the gross level spacing En|E_n| by a factor of α25×105\alpha^2\approx5\times10^{-5}, the same small parameter that controls, order by order, essentially every relativistic correction in atomic physics.

The same coupling, applied to the proton’s magnetic moment rather than the electron’s, is the basis of magnetic resonance imaging, and Figure 10.4 is a working model of it. A static field along zz splits the two spin orientations by ΔE=gμNB\Delta E = g\mu_N B, which for a proton in a 1 T1\ \text{T} field falls in the radio band; a transverse radio-frequency field tuned to exactly that frequency drives transitions between them, and the resonance is sharp enough that a deliberate spatial gradient in BB makes the resonant frequency a map of position. Everything in the sequence — the splitting proportional to BB, the resonance condition hf=ΔEhf = \Delta E, the return to equilibrium afterwards — is this section’s physics with μB\mu_B replaced by the nuclear magneton.

Screenshot of the Simplified MRI simulation

Figure 10.4:Proton spins in a magnetic field: Zeeman splitting, resonant absorption of a radio-frequency photon, and the field gradient that turns the resonance into an image.

Interactive simulation: Simplified MRI

Selection Rules

Not every pair of hydrogen energy levels is connected by an observable spectral line. An electron making a transition between stationary states typically does so by emitting or absorbing a single photon, and conservation of angular momentum and parity restricts which transitions can occur via this single-photon (electric dipole) process. The principal orbital selection rule is

Δ=±1,\Delta \ell = \pm 1,

and the magnetic quantum number must obey Δm=0,±1\Delta m_\ell=0,\pm1; the electron’s spin projection is unchanged, Δms=0\Delta m_s=0. There is no similarly strict restriction on Δn\Delta n, but the electric-dipole transition must change parity, which the Δ=±1\Delta\ell=\pm1 rule already guarantees for hydrogen orbitals. Transitions violating these rules (e.g., 2s1s2s \to 1s, both =0\ell=0) are called forbidden transitions — not absolutely impossible, but strongly suppressed, occurring (if at all) only through much slower, higher-order processes. The selection rules are why, for instance, the observed hydrogen spectral series (Lyman, Balmer, Paschen, etc., corresponding to transitions ending on nf=1,2,3,n_f = 1, 2, 3,\ldots) show specific line patterns rather than a line for every conceivable pair of levels.

10.4Summary

10.5Problems

Solution to Exercise 10.1 #

For 212\to1, the Rydberg formula gives

1λ=R(114)=34(1.097×107 m1),λ=121.5 nm.\frac1\lambda=R\left(1-\frac14\right)=\frac34(1.097\times10^7\ \text{m}^{-1}),\qquad \lambda=121.5\ \text{nm}.

At the series limit, 1/λ=R1/\lambda=R, so λ=91.2 nm\lambda=91.2\ \text{nm}. Both lines are the leftmost cluster of downward arrows in Figure 10.3, all converging on nf=1n_f=1. Therefore, the first Lyman line is 121.5 nm121.5\ \text{nm} and the limit is 91.2 nm91.2\ \text{nm}; both are ultraviolet, so the entire series is invisible to the eye.

Solution to Exercise 10.2 #

For n=3n=3, =0,1,2\ell=0,1,2. The allowed sets are (0,0)(0,0); (1,1),(1,0),(1,1)(1,-1),(1,0),(1,1); and (2,2),(2,1),(2,0),(2,1),(2,2)(2,-2),(2,-1),(2,0),(2,1),(2,2). Their count is 1+3+5=9=n21+3+5=9=n^2. Including ms=±12m_s=\pm\tfrac12 doubles this count to 18. Therefore, the n=3n=3 level has nine spatial states and eighteen states when spin is included.

Solution to Exercise 10.3 #

The changes in \ell are: (a) 212\to1, so Δ=1\Delta\ell=-1 and allowed; (b) 000\to0, so forbidden; (c) 101\to0, so allowed; and (d) 101\to0, so allowed. Because no mm_\ell values are specified, this classification asks whether the orbital transition has any allowed magnetic-sublevel components; those components must also satisfy Δm=0,±1\Delta m_\ell=0,\pm1.

Energy levels grouped by orbital type s, p, and d at n equals 1, 2, and 3, with solid arrows for the three allowed transitions and a dashed crossed arrow for the forbidden 3s to 2s transition.

Figure 10.5:Grouping the levels by \ell makes the rule visual: an allowed arrow always moves one column over, while 3s2s3s\to2s tries to stay in the same column and is forbidden.

Therefore, only 3s2s3s\to2s is forbidden by the electric-dipole selection rule.

Solution to Exercise 10.4 #

For =2\ell=2,

L=(+1)=6.L=\sqrt{\ell(\ell+1)}\hbar=\sqrt6\hbar.

The largest magnitude of mm_\ell is 2, and μL,z=mμB=2μB|\mu_{L,z}|=|m_\ell|\mu_B=2\mu_B (the sign is opposite to that of mm_\ell for an electron). Therefore, the 3d3d electron has orbital angular momentum 6\sqrt6\hbar and maximum zz-component magnetic-moment magnitude 2μB2\mu_B.

Solution to Exercise 10.5 #

Silver has one unpaired outer 5s5s electron. An ss state has =0\ell=0, so it has no orbital magnetic moment that could obscure the result; the two-way splitting is therefore due cleanly to spin. Helium’s paired 1s21s^2 electrons have canceling spin moments. Therefore, silver was ideal because its single outer ss electron leaves an uncompensated spin moment, whereas helium has no net moment.

Solution to Exercise 10.6 #

Ignoring the positive normalization constant, P(r)=r2e2r/a0P(r)=r^2e^{-2r/a_0}. Differentiation gives

dPdr=2re2r/a02a0r2e2r/a0=2re2r/a0(1ra0).\frac{dP}{dr}=2re^{-2r/a_0}-\frac2{a_0}r^2e^{-2r/a_0}=2re^{-2r/a_0}\left(1-\frac r{a_0}\right).

For r>0r>0, the derivative vanishes at r=a0r=a_0, changing from positive to negative there — exactly the peak of the 1s1s curve in Figure 10.2. Therefore, the ground-state radial probability is largest at the Bohr radius r=a0r=a_0.

Solution to Exercise 10.7 #

Hydrogenic ionization energy scales as Z2Z^2 and the most-probable radius as a0/Za_0/Z. Thus

EI=32(13.6 eV)=122.4 eV,rmax=0.529 A˚3=0.176 A˚=0.0176 nm.E_I=3^2(13.6\ \text{eV})=122.4\ \text{eV},\qquad r_{\max}=\frac{0.529\ \text{\AA}}3=0.176\ \text{\AA}=0.0176\ \text{nm}.
Ground-state radial probability distributions for hydrogen and doubly ionized lithium, each normalized to its own peak, with the lithium curve peaking at one-third the radius of hydrogen's.

Figure 10.6:Both curves have the same shape, rescaled: replacing Z=1Z=1 with Z=3Z=3 compresses the peak radius by 1/Z1/Z while (not shown to the same vertical scale) the binding energy grows by Z2Z^2.

Therefore, Li2+\text{Li}^{2+} has a 122.4 eV122.4\ \text{eV} ionization energy and a 0.0176 nm0.0176\ \text{nm} most-probable radius: nine times hydrogen’s energy and one-third its radius.

Solution to Exercise 10.8 #

The force magnitude is F=μB(Bz/z)=(9.274×1024)(15)=1.39×1022 NF=\mu_B(\partial B_z/\partial z)=(9.274\times10^{-24})(15)=1.39\times10^{-22}\ \text{N}, so a=F/m=777 m/s2a=F/m=777\ \text{m/s}^2. The magnet time is t1=L/v=3.33×104 st_1=L/v=3.33\times10^{-4}\ \text{s} and the drift time is t2=D/v=6.67×104 st_2=D/v=6.67\times10^{-4}\ \text{s}. One component deflects by

z=12at12+(at1)t2=2.16×104 m=0.216 mm.z=\tfrac12at_1^2+(at_1)t_2=2.16\times10^{-4}\ \text{m}=0.216\ \text{mm}.
Schematic of a beam entering an inhomogeneous magnet and splitting into two straight paths that reach the screen separated by 0.432 millimeters, one for each spin projection.

Figure 10.7:The two spin states feel opposite forces inside the magnet, coast in straight lines afterward, and arrive at the screen separated by 2z2z. (The deflection is exaggerated for visibility; the real displacement is a fraction of a millimeter.)

Therefore, the two opposite spin components are separated by 2z=0.432 mm2z=0.432\ \text{mm}.

Solution to Exercise 10.9 #

Adjacent orbital sublevels differ by ΔE=μBB\Delta E=\mu_BB, so

ΔE=(5.79×105 eV/T)(0.50 T)=2.90×105 eV,\Delta E=(5.79\times10^{-5}\ \text{eV/T})(0.50\ \text{T})=2.90\times10^{-5}\ \text{eV},
Δf=ΔEh=7.00×109 Hz.\Delta f=\frac{\Delta E}{h}=7.00\times10^9\ \text{Hz}.

Finally Δλλ2Δf/c=(486.1×109 m)2(7.00×109 Hz)/(3.00×108 m/s)=0.00551 nm\Delta\lambda\simeq\lambda^2\Delta f/c=(486.1\times10^{-9}\ \text{m})^2(7.00\times10^9\ \text{Hz})/(3.00\times10^8\ \text{m/s})=0.00551\ \text{nm}.

A single spectral line at 486.1 nanometers shown splitting into three closely spaced lines separated by 0.00551 nanometers when a 0.50 tesla field is applied.

Figure 10.8:The field-free Hβ_\beta line becomes a triplet spaced by Δλ\Delta\lambda, one component for each mm_\ell sublevel the upper and lower states split into.

Therefore, the 0.50 T0.50\ \text{T} splitting is 2.90×105 eV2.90\times10^{-5}\ \text{eV}, 7.00 GHz7.00\ \text{GHz}, and 0.0055 nm0.0055\ \text{nm}: the energy and frequency splittings are exactly half the corresponding 1.0 T1.0\ \text{T} values (both scale purely with BB), but the wavelength shift is not simply half the worked example’s 0.020 nm0.020\ \text{nm}, since it also depends on λ2\lambda^2 and Hβ\text{H}_\beta’s wavelength differs from Hα\text{H}_\alpha’s.

Solution to Exercise 10.10 #

For n=3n=3, E3=13.6/9=1.51 eV|E_3|=13.6/9=1.51\ \text{eV} and α2=(1/137)2=5.33×105\alpha^2=(1/137)^2=5.33\times10^{-5}. Thus

ΔEfineα2E3=(5.33×105)(1.51 eV)=8.1×105 eV.\Delta E_\mathrm{fine}\sim\alpha^2|E_3|=(5.33\times10^{-5})(1.51\ \text{eV})=8.1\times10^{-5}\ \text{eV}.

This is 4/94/9 of the n=2n=2 estimate because En1/n2|E_n|\propto1/n^2. Therefore, the n=3n=3 fine splitting is of order 8×105 eV8\times10^{-5}\ \text{eV} and becomes harder to resolve as nn rises.