# Does Rydberg's formula work for different orbitals?

Rydberg's formula is given as $\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { { n }_{ 1 }^{ 2 } } } -\frac { 1 }{ { { n }_{ 2 }^{ 2 } } } \right)$ where ${ n }_{ 2 }$ and ${ n }_{ 1 }$ are the principal quantum numbers of the orbitals of the electrons before and after its transition respectively. It is well known that the different orbitals in the same shell have different energies. Also, when an electron is brought down from a higher energy state to a lower energy state, it releases energy in the form of photons of suitable frequency which can be found by using the formula above. But, Rydberg's formula only works if the electron is transitioning between two different shells, and returns a value of 0 if used for different orbitals in the same shell which doesn't sound right. How can this be possible? Or is the energy difference between two orbitals of the same shell not enough to radiate a photon of tangible energy difference?

• When using the formula as written the assumption is made that there is no splitting of energy levels and for a given $n$ there is only one energy associated with it. – Farcher Feb 23 '16 at 10:09
• The Rydberg formula doesn't "know" anything about orbitals. It's basically a fit to the observed spectra of hydrogen-like atoms, without using any model for the microscopic structure that causes these series of spectral lines. If you want to talk about the atomic structure, it would probably be better to use modern terminology derived from quantum mechanics. – CuriousOne Feb 23 '16 at 10:10
• Bohr's atomic model for hydrogen was a great success because it explains why the Rydberg formula to works for hydrogen. Something more is needed to explain more complicated atoms. – Peter Diehr Feb 23 '16 at 11:25
• I don't feel that any of the above comments actually address the question. – lemon Feb 23 '16 at 11:27

The Rydberg formula only works for hydrogenic atoms, and in hydrogenic atoms all the orbitals with the same principal quantum number have (approximately) the same energy. The $2s$ and $2p$ have the same energy as do the $3s$, $3p$ and $3d$, and so on.

The Rydberg formula only works where the potential energy of the electron varies as $r^{-1}$. If we have more than one electron present then the electrons repel each other and they screen each other from the nucleus. As a result the potential is strictly speaking no longer even central, though to a good approximation we can treat the electron potential as central but no longer varying as $r^{-1}$ (more on this here if you're interested).

In hydrogenic atoms the different angular momentum states are only approximately of equal energy because relativistic effects cause a splitting. For example in hydrogen the $2s$ is slightly higher in energy than the $2p$, and this is known as the Lamb shift. However this is a tiny effect.

Rydberg's formula can be derived from scattering theory between an electron and a proton (using the known asymptotic behavior of the regular and irregular Coulomb functions. Using similar arguments, you can describe the scattering between an electron and an singly charged ion, e.g. He$^+$. As stated by John Rennie, the potential is not strictly a Coulomb potential anymore due to the other electron(s), but the idea is that this only matters at short distances between the electron and ion core. By matching the asymptotic solutions of the hydrogen atom to the short range solutions of the more-electron problem (which consists of superpositions of regular and irregular Coulomb functions), you effectively get a collisional phase shift. When proceeding and trying to derive the Rydberg formula, you'll see that the expression changes slightly:

$$\tilde{\nu}=\frac{E_\text{IP}}{hc}-\frac{\mathcal{R}_M}{(n-\mu_\ell)^2}$$

Here I slightly rewrite the Rydberg formula so that $\tilde{\nu}$ is the spectral position (energy) of the Rydberg state with principal quantum number $n$ and orbital angular momentum $\ell$. As you can see, the principle quantum number $n$ is replaced by $n-\mu_\ell$, the effective quantum number, where $\mu_\ell$ is the quantum defect which can be related to the collision phase shift. The magnitude of the quantum defect depends on the orbital angular momentum of the Rydberg electron. For small $\ell$, the electron "sees" a lot of the short-range potential and $\mu_\ell$ is relatively large, while for $\ell \gt 3$ the electron does not penetrate the core anymore and $\mu_\ell\approx 0$. Each value of $\ell$ defines a channel. The nice thing is that $\mu_\ell$ is a very smooth and almost constant function in the energy of the electron, even when the system is ionized.

Another thing that should be changed in the Rydberg formula is the Rydberg constant $\mathcal{R}_\infty$, as the Rydberg constant assumes an infinite mass of the nucleus, therefore you need to use the mass-corrected Rydberg constant for the system with mass $M$ this is defined as

$$\mathcal{R}_M = \mathcal{R}_\infty \frac{M-m_\text{e}}{M}$$

where $m_\text{e}$ is the electron mass.

Things can get complicated as different Rydberg channels may interact, resulting in shifts of the energy levels. These effects can be accounted for using multichannel quantum-defect theory (MQDT), that basically expands the single channel collision problem to more channels.

Even for simple molecules such as H$_2$ and ammonia, people have successfully applied this approach. See Ross, AIP Conf. Proc. 225, 73 (1991) for a nice introduction.