Formula Rydberg Formula for Hydrogen Wavelength Principal quantum number Upper principal quantum number Rydberg constant
$$\lambda ~=~ \frac{1}{R \, \left( \frac{1}{n^2} - \frac{1}{m^2} \right)}$$ $$\lambda ~=~ \frac{1}{R \, \left( \frac{1}{n^2} - \frac{1}{m^2} \right)}$$
Wavelength
$$ \lambda $$ Unit $$ \mathrm{m} $$ Wavelength of electromagnetic radiation in vacuum with which the H atom is irradiated. The wavelength is related to the energy of the radiation as follows: \( W ~=~ h \, \frac{c}{\lambda} \). Thus, from the radiation of wavelength \( \lambda \) absorbed by the H atom, the energy absorbed by the H atom can be calculated.
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 $$ - $$ Upper principal quantum number 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 constant
$$ R $$ Unit $$ \frac{1}{\mathrm m} $$ Rydberg constant for the H atom is: \( R = 1.097 373 15 \,\cdot\, 10^7 \, \frac{1}{\mathrm m} \). It is expressed in units of wavenumber and multiplied by Planck's constant \(h\) and the speed of light \(c\), it gives energy necessary to remove the electron from the H atom (i.e., to ionize the H atom):$$ R \, h \, c ~=~ 2.17 \cdot 10^{-18} \, \mathrm{J} ~=~ 13.6 \, \mathrm{eV} $$