# Formula Rydberg Formula for Hydrogen Atom Frequency    Principal quantum number    Rydberg frequency

## Frequency

Unit
Frequency of electromagnetic radiation in vacuum with which the H atom is irradiated. Frequency is related to the energy of the radiation as follows: $$W ~=~ h \, f$$.

## Principal quantum number

Unit
This is an integer indicating an energy level of the H atom. The electron in the H atom can occupy an energy state, which is described by $$n$$.

It is true: $$n ~\lt~ m$$, that is, the $$n$$th energy state is lower than the $$m$$th energy state. The electron in the H atom can be excited to the $$m$$th energy state.

## Upper principal quantum number

Unit
This is an integer indicating an energy level of the H atom. The electron in the H atom can occupy this energy state, which is described by $$m$$, by being excited into this energy state by a photon. After a short time, the electron falls back to the lower state $$n$$ and the H atom emits a photon in the process. The energy of this photon corresponds to the difference of the energy between $$m$$ and $$n$$ states.

## Rydberg frequency

Unit
Rydberg frequency for the hydrogen atom is$$R_{\text f} ~=~ c \, R ~=~ 3.289 \, 841 \, 95 \,\cdot\, 10^{15} \, \mathrm{Hz}$$ where $$c$$ is the speed of light and $$R$$ is the Rydberg constant for the H atom: $$R = 1.097 373 15 \,\cdot\, 10^7 \, \frac{1}{\mathrm m}$$.

Multiplying the Rydberg frequency by Planck's constant $$h$$ yields the energy required to remove the electron from the H atom (i.e., to ionize the H atom):$$R_{\text f} \, h ~=~ 2.17 \cdot 10^{-18} \, \mathrm{J} ~=~ 13.6 \, \mathrm{eV}$$