# Statistical Thermodynamics and Rate Theories/Electronic energy

Atoms and molecules can hold different energy levels that correspond to different distributions in how the electron density is distributed around them. These are counted as electronic degrees of freedom.
For example, the simplest atom is the hydrogen atom, with 1 proton and 1 electron. The attraction between the proton and electron in the hydrogen atom is governed by Coulomb's Law,
| principal quantum number | n=1,2,3,... |
| angular momentum quantum number | $l=0,1,...,n-1$ |
| magnetic quantum number | $m_{l}=-l,-l+1,...,0,l-1,l$ |
| spin quantum number | $m_{s}=+{\frac {1}{2}},-{\frac {1}{2}}$ |
    ${\mathcal {V}}(r)=-{\frac {e^{2}}{4\pi \epsilon _{0}r}}$
Where $e$ is the elementary charge of an electron, $r$ is the distance between the proton and electron in the hydrogen atom, and $\epsilon _{o}$ is the vacuum permittivity constant. When the Schrödinger equation is solved for this potential, four quantum numbers result, n, l, m$_l$, and m$_s$. This means that the particles can only take on certain discrete values of energy. The electronic energy of the hydrogen atom depends only on the principal quantum number $n$ via the following equation,
    $\epsilon _{n}=-{\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{o}^{2}\hbar ^{2}}}{\frac {1}{n^{2}}}$
Where $m_{e}$ is the mass of an electron. This means that all sets of angular momentum, l, and magnetic quantum numbers, m$_l$, and m$_s$, with the same value of n are degenerate, as they do not appear in this solution to the Schrödinger equation. This quantization of allowed energies give rise to the electronic energy levels of the hydrogen atom. This equation can also be used to calculate the energy levels of hydrogen like species such as He$^+$. An atomic factor, $Z^{2}$, would be added in the numerator to account for the extra protons in the atom, which would pull the electrons in more than the hydrogen atom, and thereby increase the energy.

## Example

Calculation of the energy for the excitation of O$_2$ from the triplet to singlet electronic state.
Singlet oxygen comes from the electronic excitation of triple oxygen, the energy is experimentally 0.98 electron volts per mole of electrons. The figure below illustrates the different states of oxygen. $^1$Δ$_g$ and $^1$Σ$_g$$^+$ represents singlet oxygen as it has a spin state equal to one, the triplet degenerate state is represented by $^3$Σ$_g$$^-$, the transition is represented by $^1$Δ$_g$←$^3$Σ$_g$$^-$
The degeneracy is calculated for this state is calculated from the formula g$_{el}$= 2S+1 where S is the sum of the spin of the electrons. For singlet oxygen it has one spin state of -1/2 and a spin state of +1/2 giving the sum to be 0 and the degeneracy to be calculated by g$_{el}$= 2(0)+1 to give a degeneracy of 1. Triple oxygen has two electrons in the spin up state to give it a value of 1. Which gives the degeneracy of 3 for this state.
The experimental value for this transition is experimentally calculated to be 0.98 eV found on Singlet Oxygen. This is for a wavelength of 1270 nm. the conversion factor from eV (electron volts) to kJ/mol is given by multiplying the number of electron volts by 96 kJ/mol, for the transition of triplet degenerate state to the singlet energy state it gives a value of 94.56 kJ/mol.