Modern Physics · Chapter 21

Atomic Structure

The hydrogen atom is exactly solvable by quantum mechanics, yielding energy levels, orbitals, and selection rules that explain every spectral line ever observed.

PrerequisitesQuantummechanics(Ch.20)⋅Electrostatics—Coulombpotential(Ch.13Quantum mechanics (Ch. 20) \cdot Electrostatics — Coulomb potential (Ch. 13
Learning Goals
  • Calculate hydrogen energy levels and orbital radii using the Bohr model.
  • Identify the spectral series of hydrogen and compute photon wavelengths for transitions.
  • Label electron states with all four quantum numbers and determine orbital degeneracy.
  • Apply the Pauli exclusion principle to write ground-state electron configurations.
  • Explain how the Stern-Gerlach experiment reveals quantized spin and the two spin states.

21.1 The Bohr Model

Niels Bohr proposed in 1913 that the electron in hydrogen orbits the proton only at specific radii where the angular momentum is quantized: L = nℏ (n = 1, 2, 3, …). Setting the Coulomb attraction equal to centripetal force and imposing this condition yields discrete radii and energies:

rn=n2a0(a0=0.0529nm=Bohrradius)r_{n} = n^{2} a_{0} \qquad (a_{0} = 0.0529 nm = Bohr radius)(21.1)
En=−13.6eV/n2E_{n} = -13.6 eV / n^{2}(21.2)

The ground state (n = 1) has E₁ = −13.6 eV. The minus sign means the electron is bound — you must supply 13.6 eV to ionize hydrogen from the ground state. The Bohr model correctly predicts hydrogen's spectrum but fails for multi-electron atoms and cannot explain line intensities or fine structure. It was superseded by Schrödinger's equation — but its energy levels are exactly right for hydrogen.

Example 21.1 — Hydrogen Spectral Lines — the Balmer Series

Findthewavelengthofthephotonemittedwhenhydrogentransitionsfromn=3ton=2.ThisFind the wavelength of the photon emitted when hydrogen transitions from n=3 to n=2. This istheH−αline,theprominentredlineofhydrogenis the H-\alpha line, the prominent red line of hydrogen

Energies:E3=−13.6/9=−1.511eVE2=−13.6/4=−3.400eVE_{3} = -13.6/9 = -1.511 eV \qquad E_{2} = -13.6/4 = -3.400 eV
Photon energy:\DeltaE=E3−E2=−1.511−(−3.400)=1.889eV\DeltaE = E_{3} - E_{2} = -1.511 - (-3.400) = 1.889 eV
Wavelength:λ=hc/\DeltaE=(4.136×10−15eV⋯×3×108m/s)/1.889eV=\lambda = hc/\DeltaE = (4.136\times10^{-15} eV\cdots \times 3\times10^{8} m/s) / 1.889 eV = 657 nm (red ✓)
Figure 21.1. Hydrogen spectral series. The left panel shows the energy level diagram with transition arrows; the right panel shows the resulting spectral lines at their actual wavelengths. Toggle between the Lyman (UV), Balmer (visible), and Paschen (IR) series. Click a transition label to highlight it.
Definition 21.1 — Spectral Series of Hydrogen
Lymanseries:transitionston=1(ultraviolet,91–122nmLyman series: transitions to n=1 (ultraviolet, 91–122 nmBalmerseries:transitionston=2(visible,365–656nm)—theseriesvisibletothenakedBalmer series: transitions to n=2 (visible, 365–656 nm) — the series visible to the naked eyePaschenseries:transitionston=3(nearinfraredPaschen series: transitions to n=3 (near infraredGeneral:1/λ=RH(1/nf2−1/ni2),RH=1.097×107m−1(RydbergconstantGeneral: 1/\lambda = R_{H} (1/n_{f}^{2} - 1/n_{i}^{2}), R_{H} = 1.097\times10^{7} m^{-1} (Rydberg constant

21.2 Quantum Numbers and Orbitals

Solving the Schrödinger equation in 3D with the Coulomb potential gives four quantum numbers that completely characterize each electron state:

Definition 21.2 — The Four Quantum Numbers
n(principal):n=1,2,3,⋯Determinesenergy:En=−13.6/n2eV.Shellprincipal): n = 1, 2, 3, \cdots \qquad Determines energy: E_{n} = -13.6/n^{2} eV. Shellℓ(angularmomentum):ℓ=0,1,⋯ ,n−1.Subshell(s,p,d,fforℓ=0,1,2,3).Shapeoforbangular momentum): ℓ = 0, 1, \cdots, n-1. \qquad Subshell (s, p, d, f for ℓ = 0,1,2,3). Shape of orbtal.mℓm_ℓ(magnetic):mℓ=−ℓ,⋯ ,0,⋯ ,+ℓ.Orientationinspacemagnetic): m_ℓ = -ℓ, \cdots, 0, \cdots, +ℓ. \qquad Orientation in spacemsm_{s}(spin):ms=+12or−12.Intrinsicangularmomentumoftheelectronspin): m_{s} = +\frac{1}{2} or -\frac{1}{2}. \qquad Intrinsic angular momentum of the electron

The orbital shapes are striking: s orbitals are spherical, p orbitals have two lobes along an axis (p_x, p_y, p_z), d orbitals have four lobes. The probability density|ψ|² gives the region of space where the electron is likely to be found — not a definite orbit but a cloud.

21.3 The Pauli Exclusion Principle and the Periodic Table

Definition 21.3 — Pauli Exclusion Principle (1925)
Notwoelectronsinanatomcanhavethesamesetoffourquantumnumbers(n,ℓ,mℓ,msNo two electrons in an atom can have the same set of four quantum numbers (n, ℓ, m_ℓ, m_{s}. Each quantum state can hold at most one electron.

This principle, combined with the energy ordering of orbitals, explains the entire periodic table. Electrons fill the lowest available states (aufbau principle), with at most two per orbital (spin up and spin down). The filling order explains why:

n=1: 1s² (2 electrons) → Helium is noble (full shell)
n=2: 2s² 2p⁶ (8 electrons) → Neon is noble
n=3: 3s² 3p⁶ (8 electrons) → Argon is noble
Transition metals: 3d subshell fills after 4s (energy ordering crosses at n=3,4)

Chemical properties — valence, bonding, reactivity — follow from the outermost electrons and their quantum numbers. Quantum mechanics reduces chemistry to physics.

21.4 Electron Spin and the Stern-Gerlach Experiment

In 1922, Stern and Gerlach sent silver atoms through an inhomogeneous magnetic field and observed the beam split into exactly two components — not a continuous smear. This demonstrated that the angular momentum of the valence electron is quantized with only two possible projections: m_s = +½ and m_s = −½. This intrinsic angular momentum — spin — has no classical analogue. Its magnitude is |S| = ℏ√(s(s+1)) = ℏ(√3)/2 for spin-½ particles.

Example 21.2 — Electron Configuration of Iron

Writetheground−stateelectronconfigurationofiron(Z=26)andidentifythenumberofWrite the ground-state electron configuration of iron (Z = 26) and identify the number of unpaired electrons responsible for its magnetic properties.

Fill orbitals:1s22s22p63s23p64s23d6(26electronstotal1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{6} \qquad (26 electrons total
3d subshell:6electronsin5d−orbitals.ByHund′srule,maximizespin:↑↑↑↑↑+1paired=4unpaired6 electrons in 5 d-orbitals. By Hund's rule, maximize spin: ↑↑↑↑↑ + 1 paired = 4 unpaired
Magnetic moment:4unpairedelectrons→4Bohrmagnetonsofmagneticmoment→ironisferromagnetic4 unpaired electrons \to 4 Bohr magnetons of magnetic moment \to iron is ferromagnetic
Definition 21.4 — Common Traps
  • Orbitals are not planet-like orbits: they are probability amplitudes with quantized angular structure.
  • Quantum numbers have allowed ranges: l and m are constrained by n.
  • Pauli exclusion applies to full quantum states: no two electrons share all four quantum numbers.
  • Spectral lines come from energy differences: photons are emitted or absorbed during transitions.
Exercises — 21.1–21.4 Atomic Structure
1.
Whatistheenergyofthen=3energylevelinhydrogenWhat is the energy of the n = 3 energy level in hydrogen?
eV
Straightforward
2.
Findthewavelengthofthephotonemittedinthen=3→n=2transitionofhydrogenFind the wavelength of the photon emitted in the n = 3 \to n = 2 transition of hydrogen.
nm
Straightforward
3.
Howmanydistinctelectronstates(n,ℓ,mℓ,ms)existinthen=3shellHow many distinct electron states (n, ℓ, m_ℓ, m_{s}) exist in the n = 3 shell?
Intermediate
4.WritetheelectronconfigurationsforNa(Z=11)andCl(Z=17).ExplainintermsofquantumWrite the electron configurations for Na (Z=11) and Cl (Z=17). Explain in terms of quantum numbers why they react to form NaCl.
Intermediate
5.Whyisthe2s→1stransitioninhydrogenforbiddenbyelectricdipoleselectionrules?WhWhy is the 2s \to 1s transition in hydrogen forbidden by electric dipole selection rules? What happens instead? Derive the selection rule from the transition matrix element.
Challenging
Key Takeaways
  • Bohrmodel:En=−13.6/n2eV,rn=n2a0—correctenergies,wrongphysicalpictureBohr model: E_{n} = -13.6/n^{2} eV, r_{n} = n^{2}a_{0} — correct energies, wrong physical picture.
  • Fourquantumnumbers(n,ℓ,mℓ,ms)completelylabeleachelectronstateFour quantum numbers (n, ℓ, m_ℓ, m_{s}) completely label each electron state.
  • Pauli exclusion: no two electrons share all four quantum numbers — the basis of the periodic table.
  • Orbitalsareprobabilityclouds∣ψ∣2,notdefiniteorbitsOrbitals are probability clouds |\psi|^{2}, not definite orbits.
  • Spectrallines:\DeltaE=Ei−Ef=hc/λ—emissionwhenelectrondropstolowerlevelSpectral lines: \DeltaE = E_{i} - E_{f} = hc/\lambda — emission when electron drops to lower level.
  • Spinisintrinsicangularmomentumwithms=±12—revealedbyStern−Gerlach,noclassicalSpin is intrinsic angular momentum with m_{s} = \pm\frac{1}{2} — revealed by Stern-Gerlach, no classical analogue.