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

## Frequency

`$$ f $$`Unit

`$$ \mathrm{Hz} = \frac{ 1 }{ \mathrm{s} } $$`

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

`$$ n $$`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

`$$ m $$`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

`$$ R_{\text f} $$`Unit

`$$ \mathrm{Hz} $$`

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):`$$ R_{\text f} \, h ~=~ 2.17 \cdot 10^{-18} \, \mathrm{J} ~=~ 13.6 \, \mathrm{eV} $$`