# Why is there a hydrogen emission spectrum?

Jun 24, 2017

Well, the same reason why any atom has an emission spectrum... because everything composed of atoms can absorb light, and consequently emit light upon relaxation of the electrons in the atoms. Atoms are particularly interesting in that they have discrete emission spectra, i.e. there is no line broadening, and the "peaks" are all lines. This is due to the quantization of the energy levels with respect to quantum numbers. Here, hydrogen atom energy levels are quantized to the principal quantum number $n = 1 , 2 , . . . , \infty$, in integer increments. The energies for hydrogen atom are given by

${E}_{n} = - \frac{\text{13.60569253 eV}}{n} ^ 2$

and energy transitions are thus given by the Rydberg equation

$\Delta E = {E}_{f} - {E}_{i} = - \text{13.60569253 eV} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

from initial state with energy ${E}_{i}$ to final state with energy ${E}_{f}$ with corresponding quantum numbers ${n}_{i}$ and ${n}_{f}$, respectively.

The corresponding wavelength-based Rydberg equation would be:

$\frac{1}{\lambda} = \frac{{E}_{f} - {E}_{i}}{h c} = - \frac{e \cdot \text{13.60569253 eV}}{h c} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

where:

• $h = 6.62607004 \times {10}^{- 34} \text{J"cdot"s}$ is Planck's constant.
• $c = 2.99792458 \times {10}^{8} \text{m/s}$ is the speed of light.
• $e = 1.60217662 \times {10}^{- 19} \text{J/eV}$ is based on the charge of an electron.

(usually you see the constant in front to be written as $109737.316$ ${\text{m}}^{- 1}$, the Rydberg constant in ${\text{m}}^{- 1}$.)

So one can find the wavelengths corresponding to each energy transition. Thus, we have linked the discrete wavelengths in the emission spectra to the quantization of energy levels.