Calculating the Rydberg equation allows you to predict the position of the emission lines in the spectra of hydrogen and hydrogen-like atoms. Find more about this fundamental discovery of physics: learn how to calculate the Rydberg equation, the wavelength's dependence on the quantum numbers, and the different values of the Rydberg constant with our short article.

Introduction to the Rydberg equation: atomic spectra

Spectra are the fingerprints of an atom. If we obtain light from an atom, anywhere in the cosmos, we can also discover information on the chemical nature of the atom itself.

Atoms of the same type (element) are identical: they undergo the same excitation and de-excitation processes in which energy is either absorbed or released by electrons in the electronic cloud. Each element does this in its way, allowing us to identify it.

To these two processes, we associate two different types of spectra, emission and absorption. On these, you will see either colorful lines over a dark background (if the atoms are emitting light thanks to the jumps of electrons from higher to lower energetic levels) or black lines over a colorful rainbow-like band (the atom's electrons are absorbing some of the light at specific frequencies to raise their energy). The existence of these lines was gradually discovered in the early XVIII century, on top of previous knowledge about spectra. It was already clear that there was a correlation between chemistry and light.

With increased knowledge in physics and chemistry and better equipment, scientists discovered more about these lines until the accurate observations of the spectra of many gases, which ushered in quantum mechanics at the end of the XIX century. We credit one of these discoveries to Johannes Rydberg, who first discovered the existence of a mathematical relationship between the wavelength of those lines.

The Rydberg formula for hydrogen: how to calculate wavelength from energy levels

Rydberg discovered that the spectral lines of hydrogen were not placed randomly on the spectrum, but lay at specific intervals. Those intervals are better seen if we reason in terms of wavenumbers.

The wavenumber is nothing but the inverse of the wavelength: n=1/λn=1/\lambda. By observing that emission lines appeared in series, he worked out a relationship now known as Rydberg equation for the wavelength. The Rydberg equation for hydrogen is:

1λ=RH(1n221n12)\frac{1}{\lambda} = R_{\text{H}}\cdot\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)

Where:

  • λ\lambda — The wavelength of the spectral line;
  • RHR_{\text{H}} — The Rydberg constant for hydrogen;
  • n1n_1 — The principal quantum number of the initial state; and
  • n2n_2 — The principal quantum number at the end of the transition.

🙋 Do you want to know what precisely the principal quantum number is? Head to our quantum number calculator for a quick and straightforward answer!

The Rydberg equation for hydrogen allows you to calculate the wavelength of the various lines of many spectral series. In particular:

  • For n1=1n_1 =1 we calculate the Lyman series;
  • For n1=1n_1 =1 we calculate the Balmer series; and
  • For n1=1n_1 =1 we calculate the Paschen series.

In all cases n2n_2 varies from n1n_1 to infinity, assuming only integer values.

The Rydberg equation for hydrogen-like atoms: a change in the Rydberg constant

The Rydberg equation also applies to all hydrogen-like atoms, atoms with a single valence electron. We are talking of alkaline metals such as lithium, sodium, and potassium or single ionized alkaline earth metals, such as magnesium and strontium.

We are not changing how we calculate the wavelength from the energy levels,; we only modify the Rydberg constant:

R=RZ2R =R_{\infty}\cdot Z^2

Where:

  • RR_{\infty} — The Rydberg constant; and
  • ZZ — The atomic number.

🙋 How many Rydberg constants are out there? The answer is two. For hydrogen, the Rydberg constant has to consider the effect of the non-negligible ratio between an electron's and a proton's mass. For heavier atoms, the electrons' mass is ignored. The values of the constants are:

  • RH=1.09678×107 m1R_{\text{H}} = 1.09678\times10^7\ \text{m}^{-1}; and
  • R=1.09737×107 m1R_{\infty} = 1.09737\times10^7\ \text{m}^{-1}.

The Rydberg equation for a hydrogen-like atom is then:

1λ=R ⁣ ⁣Z2 ⁣ ⁣(1n221n12)\frac{1}{\lambda} = R_{\infty}\!\cdot\!Z^2\!\cdot\!\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)

In the advanced mode of our Rydberg equation calculator, you can find the energy and the frequency corresponding to each wavelength. Learn about the respective conversions at our frequency calculator and photon energy calculator.

Calculate the wavelength with the Rydberg equation: examples for the Lyman series of hydrogen

We use the Rydberg equation to calculate the first four terms of the Lyman series. The lines are four possible excited states of the hydrogen atom, with the electron starting from the lowest possible energy.

To find the position of the first line, substitute n1=1n_1=1 and n2=2n_2=2:

λ ⁣=( ⁣RH ⁣ ⁣(112122))1 ⁣= ⁣121.57 nm\begin{align*} \lambda\! &=\left(\! R_{\text{H}}\!\cdot\!\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\right)^{-1}\! \\ &=\! 121.57\ \text{nm} \end{align*}

Use our Rydberg equation calculator to find the rest of the Lyman series:

  • For n2=3n_2=3, λ=102.57 nm\lambda = 102.57\ \text{nm};
  • For n2=4n_2=4, λ=97.25 nm\lambda = 97.25\ \text{nm};
  • For n2=5n_2=5, λ=94.97 nm\lambda = 94.97\ \text{nm};
  • For n2=6n_2=6, λ=93.78 nm\lambda = 93.78\ \text{nm};

As you can see, the interval between wavelengths is reducing. The series will eventually converge at the value 91.18 nm91.18\ \text{nm}.

This number was crucial to confirm Bohr's model for atomic hydrogen correctness. Learn more about it at our Bohr model calculator.

Davide Borchia
Atomic number
Initial state
Final state
Wavelength
nm
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