Technopedia

The technologist’s encyclopedia

Atomic Physics

Atomic physics describes the structure and behavior of atoms: the electromagnetic bound states of nuclei and electrons. It explains the periodic table, spectral lines, chemical bonding, and the quantized energy levels that underpin chemistry, spectroscopy, and modern electronics. It is relevant whenever the discrete energy structure of matter matters: from laser design to semiconductor physics to astrophysical spectroscopy.

Authoritative sources

The hydrogen atom

Definition. The hydrogen atom is the simplest atom: one proton and one electron bound by the Coulomb potential. Its energy levels are En=−13.6 eV/n2E_n = -13.6 \text{ eV}/n^2 for n=1,2,3,…n = 1, 2, 3, \ldots.

Details

From the Bohr model to the Schrödinger equation

In 1913, Niels Bohr proposed that the electron in hydrogen occupies only certain allowed circular orbits with quantized angular momentum L=nℏL = n\hbar. This ad hoc quantization reproduced the observed 1/n21/n^2 energy spacing and explained the hydrogen spectrum. The Bohr model was a crucial stepping stone: it identified the correct energy formula and the role of angular momentum quantization, but it could not explain line intensities, fine structure, or multi-electron atoms.

The full quantum-mechanical treatment replaces orbits with wavefunctions. Solving the Schrödinger equation for the Coulomb potential V(r)=−e2/(4πε0r)V(r) = -e^2/(4\pi\varepsilon_0 r) yields the same energy eigenvalues:

En=−mee42(4πε0)2ℏ2⋅1n2=−13.6 eVn2E_n = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2} \cdot \frac{1}{n^2} = -\frac{13.6 \text{ eV}}{n^2}

The quantum states of hydrogen are labeled by four quantum numbers:

The Bohr radius a0=4πε0ℏ2/(mee2)≈0.529a_0 = 4\pi\varepsilon_0 \hbar^2/(m_e e^2) \approx 0.529 Å sets the size scale of the atom.

Energy-level diagram

The figure below shows the first four energy levels of hydrogen and the principal transitions:

Energy (eV) n=1 (−13.6 eV) n=2 (−3.4 eV) n=3 (−1.51 eV) n=4 (−0.85 eV) 0 eV (ionization) Lyman-α (121.6 nm) Hα (656 nm) Hβ (486 nm) Lyman-β (102.6 nm)

Worked example: Lyman-alpha

The Lyman-alpha line corresponds to the transition n=2→n=1n=2 \to n=1:

Eγ=E2−E1=−3.4−(−13.6)=10.2 eVE_{\gamma} = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \text{ eV}

Wavelength: λ=hc/E=1240/10.2≈121.6\lambda = hc/E = 1240/10.2 \approx 121.6 nm (ultraviolet).

Authoritative sources

Example problem

Compute the ionization energy of hydrogen from the n=2n=2 state (i.e., the energy needed to take the electron from n=2n=2 to n=∞n=\infty).

Solution

The ionization energy from level nn is Eion=0−En=13.6/n2E_{\text{ion}} = 0 - E_n = 13.6/n^2 eV.

For n=2n=2: Eion=13.6/4=3.4E_{\text{ion}} = 13.6/4 = 3.4 eV.

A photon of wavelength λ=1240/3.4≈365\lambda = 1240/3.4 \approx 365 nm (near-UV) can ionize a hydrogen atom in its first excited state.

Fine structure and spin-orbit coupling

Definition. Fine structure is the small splitting of hydrogen energy levels caused by relativistic corrections to the electron kinetic energy and by the coupling of the electron’s spin magnetic moment to the magnetic field it experiences in its own rest frame (spin-orbit coupling). The splitting is of order α2\alpha^2 relative to the main 13.6/n213.6/n^2 spacing, where α≈1/137\alpha \approx 1/137 is the fine-structure constant.

Details

Why the levels split

In the basic Schrödinger treatment, all states with the same nn but different ℓ\ell are degenerate (e.g., 2s and 2p have the same energy). In reality, three relativistic effects lift this degeneracy:

  1. Relativistic kinetic energy correction: the electron speed in inner orbits approaches cc, so the kinetic energy p2/2mp^2/2m must be replaced by the relativistic expression. This shifts all levels slightly.

  2. Spin-orbit coupling: in the electron’s rest frame, the orbiting proton constitutes a current loop, producing a magnetic field Bso\mathbf{B}_{\text{so}}. The electron’s spin magnetic moment μs=−gsμBS/ℏ\boldsymbol{\mu}_s = -g_s \mu_B \mathbf{S}/\hbar interacts with this field:

Hso=12me2c21rdVdr L⋅SH_{\text{so}} = \frac{1}{2m_e^2 c^2} \frac{1}{r}\frac{dV}{dr}\, \mathbf{L} \cdot \mathbf{S}

The factor of 1/21/2 is the Thomas precession correction.

  1. Darwin term: a contact interaction affecting only ss-states (ℓ=0\ell=0).

The total angular momentum J=L+S\mathbf{J} = \mathbf{L} + \mathbf{S} is the good quantum number. For a given ℓ\ell, the spin couples to give j=ℓ±1/2j = \ell \pm 1/2 (except ℓ=0\ell=0 where only j=1/2j=1/2).

The fine-structure formula

The combined correction to the energy is:

En,j=−13.6 eVn2[1+α2n(nj+12−34)]E_{n,j} = -\frac{13.6 \text{ eV}}{n^2}\left[1 + \frac{\alpha^2}{n}\left(\frac{n}{j+\tfrac{1}{2}} - \frac{3}{4}\right)\right]

For n=2n=2: the 2p3/22p_{3/2} level is slightly higher than the 2p1/22p_{1/2} level, and the 2s1/22s_{1/2} level is degenerate with 2p1/22p_{1/2} (to this order). The 2p3/2→2s1/22p_{3/2} \to 2s_{1/2} transition produces the famous 2.1 μm “fine-structure line” of hydrogen.

Diagram: fine-structure splitting of n=2

Energy 2p3/2 2p1/2 2s1/2 (degenerate with 2p1/2) 1s1/2 2.1 μm ΔE

Authoritative sources

Example problem

Estimate the fine-structure splitting of the n=2n=2 level in hydrogen. Use α≈1/137\alpha \approx 1/137 and the main energy E2=−3.4E_2 = -3.4 eV.

Solution

The fine-structure correction is of order α2∣En∣\alpha^2 |E_n|:

ΔE∼α2⋅3.4 eV=3.41372≈1.8×10−4 eV\Delta E \sim \alpha^2 \cdot 3.4 \text{ eV} = \frac{3.4}{137^2} \approx 1.8 \times 10^{-4} \text{ eV}

More precisely, the 2p3/22p_{3/2}–2p1/22p_{1/2} splitting is ΔE=α2⋅3.4/2≈9×10−5\Delta E = \alpha^2 \cdot 3.4/2 \approx 9 \times 10^{-5} eV, corresponding to a wavelength shift of about 0.01 nm in the 656 nm Hα line.

Multi-electron atoms and the periodic table

Definition. Multi-electron atoms are described by the Pauli exclusion principle and the Aufbau principle: electrons fill orbitals in order of increasing energy, with at most two (opposite-spin) electrons per spatial orbital.

Details

Why multi-electron atoms are harder

In hydrogen, the electron sees a pure −e2/r-e^2/r Coulomb potential and the problem is exactly solvable. In a multi-electron atom with nuclear charge ZeZe, each electron feels:

  1. The attractive Coulomb potential of the nucleus: −Ze2/(4πε0r)-Ze^2/(4\pi\varepsilon_0 r).
  2. The repulsive Coulomb potential of all other electrons: +e2/(4πε0rij)+e^2/(4\pi\varepsilon_0 r_{ij}).

The electron–electron repulsion terms make the Schrödinger equation unsolvable in closed form. The key physical insight is screening (or shielding): inner electrons partially cancel the nuclear charge, so an outer electron sees an effective nuclear charge Zeff<ZZ_{\text{eff}} < Z. This is why the energy of an orbital depends on both nn and ℓ\ell: orbitals with smaller ℓ\ell (more penetrating, e.g. s orbitals) spend more time near the nucleus and feel a larger ZeffZ_{\text{eff}}, lowering their energy relative to higher-ℓ\ell orbitals with the same nn.

The filling rules

Orbital-filling diagram

The diagram below shows the filling order for the first 36 elements (up to krypton), with the n+ℓn+\ell rule:

Orbital filling order (Madelung rule: increasing n+ℓ, then n) 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 2 2 6 2 6 2 10 6 2 Each color = one value of n+ℓ. Filling proceeds left-to-right, top-to-bottom. Exceptions: Cr (3d⁵4s¹), Cu (3d¹⁰4s¹), and several heavier elements deviate from this order.

The periodic table

The periodic table organizes elements by atomic number ZZ. Elements in the same column (group) have the same valence electron configuration, explaining chemical periodicity. The nucleus provides the positive charge +Ze+Ze that binds the electrons; its internal structure (protons, neutrons, strong force) is the subject of nuclear physics.

Example. Carbon (Z=6Z = 6): configuration 1s2 2s2 2p21s^2\, 2s^2\, 2p^2. The two 2p electrons occupy separate orbitals with parallel spins (Hund’s rule), giving carbon its tetravalent chemistry.

Authoritative sources

Example problem

Write the ground-state electron configuration of iron (Z=26Z = 26) and of chromium (Z=24Z = 24). Explain why chromium is an exception to the Madelung rule.

Solution

Iron (Z=26Z=26): Fill in order 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶. Configuration: [Ar] 3d6 4s2[\text{Ar}]\, 3d^6\, 4s^2.

Chromium (Z=24Z=24): The Madelung rule predicts [Ar] 3d4 4s2[\text{Ar}]\, 3d^4\, 4s^2, but the observed configuration is [Ar] 3d5 4s1[\text{Ar}]\, 3d^5\, 4s^1.

Reason: a half-filled d-subshell (d5d^5) has extra exchange-stabilization energy (all five d-electrons have parallel spins, maximizing the number of parallel-spin pairs). The energy gain from this exchange interaction exceeds the small energy cost of promoting one 4s electron into the 3d subshell.

Spectral lines

Definition. Spectral lines are the discrete frequencies of electromagnetic radiation emitted or absorbed when an electron transitions between quantized energy levels: hν=Ei−Efh\nu = E_i - E_f.

Details

Connection to the hydrogen energy levels

The Rydberg formula is a direct restatement of the hydrogen energy-level formula. Starting from En=−13.6/n2E_n = -13.6/n^2 eV, the photon energy for a transition ni→nfn_i \to n_f is:

hν=Eni−Enf=13.6(1nf2−1ni2) eVh\nu = E_{n_i} - E_{n_f} = 13.6\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{ eV}

Converting to wavelength λ=c/ν=hc/(hν)\lambda = c/\nu = hc/(h\nu):

1λ=RH(1nf2−1ni2),RH=13.6 eVhc≈1.097×107 m−1\frac{1}{\lambda} = R_H\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right), \quad R_H = \frac{13.6 \text{ eV}}{hc} \approx 1.097 \times 10^7 \text{ m}^{-1}

So the Rydberg constant is not an independent empirical parameter — it is fixed by the same physics that gives En=−13.6/n2E_n = -13.6/n^2.

Emission, absorption, and selection rules

Spectral-line diagram

The figure below shows the hydrogen emission spectrum (visible Balmer series) and the corresponding absorption spectrum:

Wavelength (nm) 400 450 500 550 600 Hδ 410 Hγ 434 Hβ 486 Hα 656 Hydrogen emission spectrum (Balmer series, visible) Absorption: dark lines at same wavelengths

Multi-electron spectral lines

In multi-electron atoms, the energy levels depend on both nn and ℓ\ell (due to screening), so the spectrum is richer. The sodium D-line at 589 nm arises from the 3p→3s3p \to 3s transition. The fine structure splits this into two closely spaced lines at 589.0 and 589.6 nm (the 3p1/23p_{1/2} and 3p3/23p_{3/2} upper levels).

Authoritative sources

Example problem

A spectral line is observed at 486.1 nm. Identify the transition in hydrogen that produces it. (Use En=−13.6/n2E_n = -13.6/n^2 eV and hc=1240hc = 1240 eV·nm.)

Solution

Photon energy: E=1240/486.1≈2.55E = 1240/486.1 \approx 2.55 eV.

We need 13.6(1/nf2−1/ni2)=2.5513.6(1/n_f^2 - 1/n_i^2) = 2.55, i.e. 1/nf2−1/ni2=0.18751/n_f^2 - 1/n_i^2 = 0.1875.

Try nf=2n_f = 2: 1/4−1/ni2=0.1875⇒1/ni2=0.0625⇒ni=41/4 - 1/n_i^2 = 0.1875 \Rightarrow 1/n_i^2 = 0.0625 \Rightarrow n_i = 4.

This is the Hβ line: the n=4→n=2n=4 \to n=2 transition (Balmer-beta).

The Zeeman effect

Definition. The Zeeman effect is the splitting of atomic spectral lines in the presence of an external magnetic field. It arises because the magnetic field lifts the degeneracy of the mℓm_\ell (and msm_s) sub-levels, producing multiple closely spaced lines instead of one.

Details

Mechanism

In the absence of a field, states with the same n,ℓn, \ell but different mℓm_\ell are degenerate. An external magnetic field B=Bz^\mathbf{B} = B\hat{z} couples to the orbital and spin magnetic moments:

HZ=μB(Lz+gsSz)Bℏ=μB(mℓ+gs ms) BH_Z = \mu_B (L_z + g_s S_z)\frac{B}{\hbar} = \mu_B (m_\ell + g_s\, m_s)\, B

where μB=eℏ/(2me)≈5.79×10−5\mu_B = e\hbar/(2m_e) \approx 5.79 \times 10^{-5} eV/T is the Bohr magneton and gs≈2g_s \approx 2.

Each sub-level shifts by ΔE=μB(mℓ+gsms)B\Delta E = \mu_B (m_\ell + g_s m_s) B. Since the selection rule Δmℓ=0,±1\Delta m_\ell = 0, \pm 1 still applies, a single spectral line splits into multiple components.

Normal vs. anomalous Zeeman effect

Diagram: normal Zeeman splitting

m=−1 m=0 m=+1 p m=0 s Spectrum: σ⁻ π σ⁺ B

The Stark effect (brief note)

An external electric field similarly splits or shifts energy levels (the Stark effect). For hydrogen, the linear Stark effect mixes states of the same nn but different parity (e.g., 2s and 2p), lifting the mℓm_\ell degeneracy. The Stark effect is generally smaller than the Zeeman effect for typical laboratory fields but is important in astrophysics (pressure broadening in stellar atmospheres).

Authoritative sources

Example problem

A hydrogen spectral line at 500 nm is observed in a magnetic field of B=1B = 1 T. Estimate the splitting between adjacent Zeeman components. (Use μB=5.79×10−5\mu_B = 5.79 \times 10^{-5} eV/T and hc=1240hc = 1240 eV·nm.)

Solution

Energy splitting: ΔE=μBB=5.79×10−5\Delta E = \mu_B B = 5.79 \times 10^{-5} eV.

Wavelength splitting: Δλ=λ2ΔE/(hc)=(500)2×5.79×10−5/1240≈0.0117\Delta\lambda = \lambda^2 \Delta E / (hc) = (500)^2 \times 5.79 \times 10^{-5} / 1240 \approx 0.0117 nm.

So the three components are separated by about 0.012 nm — resolvable with a high-resolution spectrometer but invisible to the naked eye.

Molecular bonds

Definition. Molecular bonds arise from the electromagnetic interaction between the electrons and nuclei of adjacent atoms. The dominant types are covalent (shared electrons), ionic (electron transfer), and metallic (delocalized electrons). This section gives a brief atomic-physics introduction; a full treatment belongs to molecular physics and quantum chemistry.

Details

The LCAO method

The Linear Combination of Atomic Orbitals (LCAO) method approximates molecular orbitals as superpositions of atomic orbitals on the constituent atoms. For a diatomic molecule AB:

ψ±=12(1±S)(ϕA±ϕB)\psi_{\pm} = \frac{1}{\sqrt{2(1 \pm S)}}(\phi_A \pm \phi_B)

where ϕA\phi_A and ϕB\phi_B are atomic orbitals on atoms A and B, and S=⟨ϕA∣ϕB⟩S = \langle \phi_A | \phi_B \rangle is the overlap integral. The ++ combination is the bonding orbital (lower energy, electron density concentrated between the nuclei); the −- combination is the antibonding orbital (higher energy, node between the nuclei).

Molecular orbital diagram for H₂

Molecular orbital diagram for H₂ 1s (HA) −13.6 eV 1s (HB) −13.6 eV σ(1s) bonding −16.3 eV σ*(1s) antibonding −8.0 eV ↑↓ Energy D₀ ≈ 4.5 eV

Types of bonds

Van der Waals interactions

The Lennard-Jones potential models weak van der Waals (dispersion) interactions between neutral atoms or molecules:

V(r)=4ε[(σr)12−(σr)6]V(r) = 4\varepsilon\left[\left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^6\right]

The r−12r^{-12} term models Pauli repulsion at short range; the r−6r^{-6} term is the London dispersion attraction (induced dipole–induced dipole).

Authoritative sources

Example problem

Two hydrogen atoms approach each other. Using the MO picture, explain why the bonding σ orbital is lower in energy than the atomic 1s orbital, while the antibonding σ* orbital is higher. Estimate the bond order of H₂.

Solution

In the LCAO picture, the bonding combination ψ+=(ϕA+ϕB)/2(1+S)\psi_+ = (\phi_A + \phi_B)/\sqrt{2(1+S)} has constructive interference between the nuclei: the electron density is enhanced in the internuclear region, lowering the potential energy (the electron is attracted to both nuclei simultaneously). The kinetic energy cost (from the spatial confinement) is smaller than the potential energy gain, so the net energy drops below the atomic level.

The antibonding combination ψ−=(ϕA−ϕB)/2(1−S)\psi_- = (\phi_A - \phi_B)/\sqrt{2(1-S)} has a node between the nuclei: electron density is excluded from the bonding region, and the electron is effectively repelled from the internuclear zone, raising the energy.

Bond order = (number of electrons in bonding MOs − number in antibonding MOs) / 2 = (2 − 0)/2 = 1.

A bond order of 1 corresponds to a single bond, consistent with the observed H–H bond.

Interactive simulation

Explore the hydrogen energy levels interactively: click on two levels to see the emitted or absorbed photon’s wavelength and color.

Quiz

Example problem

An electron in a hydrogen atom makes a transition from n=3n = 3 to n=1n = 1. Compute the wavelength of the emitted photon. (Use En=−13.6/n2E_n = -13.6/n^2 eV and hc=1240hc = 1240 eV·nm.)

Solution

Energy of the photon: E=E3−E1=−13.6/9−(−13.6/1)=13.6(1−1/9)=13.6×8/9≈12.09E = E_3 - E_1 = -13.6/9 - (-13.6/1) = 13.6(1 - 1/9) = 13.6 \times 8/9 \approx 12.09 eV.

Wavelength: λ=hc/E=1240/12.09≈102.6\lambda = hc/E = 1240/12.09 \approx 102.6 nm.

This is the Lyman-beta line, in the ultraviolet.