The Rydberg formula is a mathematical formula used to predict the wavelength of light resulting from an electron moving between energy levels of an atom.

When an electron changes from one atomic orbital to another, the electron's energy changes. When the electron changes from an orbital with high energy to a lower energy state, a photon of light is created. When the electron moves from low energy to a higher energy state, a photon of light is absorbed by the atom.

Each element has a distinct spectral fingerprint. When an element's gaseous state is heated, it will give off light. When this light is passed through a prism or diffraction grating, bright lines of different colors can be distinguished. Each element is slightly different from other elements. This discovery was the beginning of the study of spectroscopy.

### Rydberg Formula Equation

Johannes Rydberg was a Swedish physicist who attempted to find a mathematical relationship between one spectral line and the next of certain elements. He eventually discovered there was an integer relationship between the wavenumbers of successive lines.

His findings were combined with Bohr's model of the atom to give the formula:

1/λ = RZ^{2}(1/n_{1}^{2} - 1/n_{2}^{2})

where

λ is the wavelength of the photon (wavenumber = 1/wavelength)

R = Rydberg's constant (1.0973731568539(55) x 10^{7} m^{-1})

Z = atomic number of the atom

n_{1} and n_{2} are integers where n_{2} > n_{1}.

It was later found n_{2} and n_{1} were related to the principal quantum number or energy quantum number. This formula works very well for transitions between energy levels of a hydrogen atom with only one electron. For atoms with multiple electrons, this formula begins to break down and give results that are incorrect.

The reason for the inaccuracy is that the amount of screening for inner electrons for outer electron transitions varies. The equation is too simplistic to compensate for the differences.

The Rydberg formula may be applied to hydrogen to obtain its spectral lines. Setting n_{1} to 1 and running n_{2} from 2 to infinity yields the Lyman series. Other spectral series may also be determined:

n_{1} | n_{2} | Converges Toward | Name |

1 | 2 → ∞ | 91.13 nm (ultraviolet) | Lyman series |

2 | 3 → ∞ | 364.51 nm (visible light) | Balmer series |

3 | 4 → ∞ | 820.14 nm (infrared) | Paschen series |

4 | 5 → ∞ | 1458.03 nm (far infrared) | Brackett series |

5 | 6 → ∞ | 2278.17 nm (far infrared) | Pfund series |

6 | 7 → ∞ | 3280.56 nm (far infrared | Humphreys series |

For most problems, you'll deal with hydrogen so you can use the formula:

1/λ = R_{H}(1/n_{1}^{2} - 1/n_{2}^{2})

where R_{H} is Rydberg's constant, since the Z of hydrogen is 1.

### Rydberg Formula Worked Example Problem

Find the wavelength of the electromagnetic radiation that is emitted from an electron relaxes from n = 3 to n = 1.

To solve the problem, start with the Rydberg equation:

1/λ = R(1/n_{1}^{2} - 1/n_{2}^{2})

Now plug in the values, where n_{1} is 1 and n_{2} is 3. Use 1.9074 x 10^{7} m^{-1} for Rydberg's constant:

1/λ = (1.0974 x 10^{7})(1/1^{2} - 1/3^{2})

1/λ = (1.0974 x 10^{7})(1 - 1/9)

1/λ = 9754666.67 m^{-1}

1 = (9754666.67 m^{-1})λ

1 / 9754666.67 m^{-1} = λ

λ = 1.025 x 10^{-7} m

Note the formula gives a wavelength in meters using this value for Rydberg's constant. You'll often be asked to provide an answer in nanometers or Angstroms.