# Spectral series' formula of a given atom (other than hydrogen-like)?

The hydrogen spectral series is given by the Rydberg formula:

The energy differences between levels in the Bohr model, and hence the wavelengths of emitted/absorbed photons, is given by the Rydberg formula:

$${1 \over \lambda} = R \left( {1 \over (n^\prime)^2} - {1 \over n^2} \right) \qquad \left( R = 1.097373 \times 10^7 \ \mathrm{m}^{-1} \right)$$ where $n$ is the upper energy level, $n'$ is the lower energy level, and $R$ is the Rydberg constant.

There is a similar formula for every hydrogen-like atom.

Question: Is there a theoretic formula for the spectral series of a given atom (other than hydrogen-like)? Else, why, what are the difficulties?

## 1 Answer

There are, in general, no closed form solutions (aka formulas) for the spectra of multi-electron atoms. There are reasonably precise formulas for special cases, like approximate values of x-ray transitions from inner shell electrons, though. Unlike in case of hydrogen and Rydberg atoms, which can be treated as a non-relativistic one-body problems (i.e. for which the Schroedinger equation is a good approximation), no such simplification exists for atoms with Z>2 and more than one electron. Precise calculation of spectral properties of heavy atoms requires a fully relativistic treatment of a quantum mechanical many-body problem. The correct theoretical framework for that is quantum electrodynamics (QED), which is very complex and can only produce numerical results.

• Could you be a bit more precise about the complexity of QED? Is the QED equation "not solvable" for the helium or even for the hydrogen without simplification? What are the mathematical obstructions for finding solutions (no way to find solutions as Fourier series, analytic functions...)? – Sebastien Palcoux Aug 10 '14 at 22:52
• @SébastienPalcoux: not even the Schrodinger equation is soluble for atoms with more than one electron. However we can use numerical methods to construct very accurate approximations. CuriousOne mentions QFT because for heavy atoms the energy of the inner shell electrons is so high that their velocity becomes relativistic. Since Schrodinger's equation is non-relativistic it no longer gives an accurate description of the atom. Actually we can use relativistic QM, so strictly speaking QFT isn't needed. – John Rennie Aug 11 '14 at 9:41
• @SébastienPalcoux: there is no formula for the spectral series in multielectron atoms. The transition energies can only be calculated numerically. – John Rennie Aug 11 '14 at 10:30
• @SébastienPalcoux: in principle the spectrum is zero between peaks. In practice spectra always contain background light e.g. from black body radiation. – John Rennie Aug 11 '14 at 14:42
• @Sebastien Palcoux: I would agree with John Rennie about the absence of exact solutions for all but the most trivial case of the quantum mechanical two body problem. Curiously, nature has given us an enormous gift, because the hydrogen spectrum allowed Erwin Schroedinger to guess the equation named after him, which gave us an enormous insight into the structure of quantum mechanics. Without this one, highly symmetric case, we might have been guessing around a lot longer, before someone would have come up with the equation, which made everything that came thereafter easier to understand. – CuriousOne Aug 11 '14 at 22:18