10.6 Atomic Spectra, the Hydrogen Atom, Many-Electron Atoms, and Term Symbols

Key Takeaways

  • Photon energy is E = h nu = hc/lambda; a 500 nm photon carries 3.97 x 10^-19 J, or 239 kJ per mole of photons.

  • Hydrogen line wavenumbers follow the Rydberg equation with R_H = 109,678 cm^-1; the Balmer red line (n = 3 to 2) is at 656 nm and the Lyman series limit at 91.2 nm.

  • Hydrogen-like energies are E_n = -13.6 Z^2/n^2 eV, so He+ has an ionization energy of 54.4 eV and Li2+ 122.4 eV.

  • Many-electron Schrodinger equations cannot be solved exactly because of electron-electron repulsion; the orbital approximation and self-consistent-field methods use antisymmetric Slater determinants that build in the Pauli principle.

  • An atomic term symbol is (2S+1)L_J; carbon's 2p2 ground term is 3P0, nitrogen's is 4S3/2, and oxygen's is 3P2.

Last updated: October 2026

10.6 Atomic Spectra, the Hydrogen Atom, Many-Electron Atoms, and Term Symbols

Quick Answer: The Physical Chemistry TOS quantum area lists three phenomena (blackbody radiation, the photoelectric effect, and line spectra), simple systems, term symbols, the polyelectronic wave equation, and diatomic energy levels. The Inorganic TOS adds electromagnetic radiation to its quantum-theory area. Section 10.5 covers blackbody radiation, the photoelectric effect, and particle-in-a-box and oscillator models. This section covers photon energy, hydrogen line spectra, the hydrogen-atom solution, many-electron atoms, and atomic term symbols.


Electromagnetic Radiation

c=λν,E=hν=hcλ=hcν~c = \lambda\nu, \qquad E = h\nu = \frac{hc}{\lambda} = hc\tilde{\nu}

with h=6.626×10−34h = 6.626 \times 10^{-34} J s and c=2.998×108c = 2.998 \times 10^8 m/s. For a 500 nm photon, E=(6.626×10−34)(2.998×108)500×10−9=3.97×10−19E = \frac{(6.626 \times 10^{-34})(2.998 \times 10^8)}{500 \times 10^{-9}} = 3.97 \times 10^{-19} J (2.48 eV). Per mole of photons that is 239 kJ, comparable to bond energies, which is why visible and ultraviolet light can drive photochemistry.

RegionWavelengthMolecular Process
Gamma raysbelow about 0.01 nmNuclear transitions
X-rays0.01-10 nmInner-shell electrons
Ultraviolet-visible10-780 nmValence electrons
Infrared0.78-1000 micrometresMolecular vibrations
Microwave1 mm-1 mMolecular rotations
Radioabove 1 mNuclear spin flips (NMR)

Hydrogen Line Spectra and the Bohr Model

Excited hydrogen emits only certain wavelengths. Rydberg fitted them with

ν~=1λ=RH(1n12−1n22),RH=109,678 cm−1\tilde{\nu} = \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad R_H = 109{,}678\text{ cm}^{-1}

SeriesLower Level n1n_1RegionSeries Limit
Lyman1Ultraviolet91.2 nm
Balmer2Visible364.7 nm
Paschen3Near infrared820 nm
Brackett4Infrared1459 nm

Worked example (Balmer, n = 4 to 2): ν~=109,678(14−116)=20,565 cm−1\tilde{\nu} = 109{,}678\left(\frac{1}{4} - \frac{1}{16}\right) = 20{,}565\text{ cm}^{-1}, so λ=486\lambda = 486 nm (blue-green). The n = 3 to 2 line is at 656 nm (red).

Bohr's model (1913) quantized angular momentum (mvr=nℏmvr = n\hbar). It gave En=−13.6 eVn2E_n = -\frac{13.6\text{ eV}}{n^2} and orbit radii rn=n2a0r_n = n^2a_0 with a0=52.9a_0 = 52.9 pm. It reproduces hydrogen-like spectra but fails for many-electron atoms and wrongly assigns the ground state definite orbits and nonzero angular momentum.


The Schrodinger Solution for Hydrogen-like Atoms

The wavefunction separates into radial and angular parts, ψnlml=Rnl(r) Ylml(θ,ϕ)\psi_{nlm_l} = R_{nl}(r)\,Y_{lm_l}(\theta,\phi), with energies

En=−Z2n2 (13.6 eV)E_n = -\frac{Z^2}{n^2}\,(13.6\text{ eV})

  • Energy depends only on nn in a one-electron atom, so each level is n2n^2-fold degenerate (2n22n^2 with spin). That degeneracy is lifted in many-electron atoms.
  • Ionization energies scale with Z2Z^2: H 13.6 eV, He+He^+ 54.4 eV, and Li2+Li^{2+} 122.4 eV.
  • The orbital has n−l−1n - l - 1 radial nodes and ll angular nodes.
  • The radial distribution function P(r)=r2R2P(r) = r^2R^2 gives the probability of finding the electron at distance rr. For the 1s orbital it peaks at r=a0/Zr = a_0/Z, although ψ2\psi^2 itself is greatest at the nucleus. The mean radius of the 1s orbital is 1.5 a01.5\,a_0.

Many-Electron Atoms

The helium Hamiltonian contains an electron-electron repulsion term, e2/4πε0r12e^2/4\pi\varepsilon_0 r_{12}. Because of it, the equation cannot be solved exactly.

  • Ignoring repulsion gives E=2(−54.4)=−108.8E = 2(-54.4) = -108.8 eV, far below the experimental −79.0-79.0 eV.
  • Variation method: any trial function gives an energy at or above the true value. Letting the effective nuclear charge vary gives Zeff=27/16=1.69Z_{eff} = 27/16 = 1.69 and E=−77.5E = -77.5 eV; the electrons partly screen each other.
  • Orbital approximation and the self-consistent field (Hartree-Fock): each electron moves in the average field of the others, and the orbitals are iterated until they stop changing.
  • Pauli principle: the total wavefunction must be antisymmetric when any two electrons are exchanged. A Slater determinant builds this in automatically, and it vanishes if two electrons share all four quantum numbers.
  • Shielding and penetration split ss, pp, dd, and ff energies within a shell. That gives the filling order covered in Section 2.2.

Russell-Saunders Term Symbols

For light atoms, combine orbital angular momenta into LL and spins into SS, then couple them to JJ:

2S+1LJ,L=0,1,2,3→S,P,D,F;J=L+S,…,∣L−S∣^{2S+1}L_J, \qquad L = 0, 1, 2, 3 \rightarrow S, P, D, F; \qquad J = L+S, \ldots, \lvert L-S\rvert

Hund's rules for the ground term: maximize SS, then maximize LL; then J=∣L−S∣J = \lvert L - S\rvert if the subshell is less than half full or J=L+SJ = L + S if it is more than half full.

Atom (Configuration)Terms ArisingGround Term
H, Na (s1s^1)2S^2S2S1/2^2S_{1/2}
C (2p22p^2)3P^3P, 1D^1D, 1S^1S3P0^3P_0
N (2p32p^3)4S^4S, 2D^2D, 2P^2P4S3/2^4S_{3/2}
O (2p42p^4)3P^3P, 1D^1D, 1S^1S3P2^3P_2
F (2p52p^5)2P^2P2P3/2^2P_{3/2}

Selection rules for one-electron atomic transitions: Δl=±1\Delta l = \pm 1, ΔS=0\Delta S = 0, ΔL=0,±1\Delta L = 0, \pm 1, and ΔJ=0,±1\Delta J = 0, \pm 1 (but not J=0J = 0 to J=0J = 0). The sodium D doublet shows spin-orbit coupling: 2P3/2→2S1/2^2P_{3/2} \rightarrow {}^2S_{1/2} at 589.0 nm and 2P1/2→2S1/2^2P_{1/2} \rightarrow {}^2S_{1/2} at 589.6 nm. Term symbols for dnd^n ions in ligand fields are covered in Section 3.3.

Loading diagram...
Hydrogen Emission Series
Test Your Knowledge

What is the wavelength of the hydrogen emission line for the transition from n = 4 to n = 2? (R_H = 109,678 cm^-1)

A

656 nm

B

434 nm

C

122 nm

D

486 nm

Test Your Knowledge

What is the ground-state term symbol of the oxygen atom (1s2 2s2 2p4)?

A

1S0

B

3P0

C

1D2

D

3P2

Test Your Knowledge

Why can the Schrodinger equation for the helium atom not be solved exactly?

A

The electron-electron repulsion term couples the motion of the two electrons

B

Helium's electrons have opposite spins and cancel each other

C

Helium's nucleus is too heavy for the non-relativistic equation to apply at all

D

Its two electrons are in different shells with different n values

Sections you finish are checked off in the contents.