Pulling it All Together: The Spectrum of the Hydrogen Atom

Let's take some of the concepts introduced above and work with them so we get a better understanding. We will just be dealing with the hydrogen atom. It's complicated enough, and the helium atom—with just two electrons—is so much more complicated that it would be impossible to describe here. We can use the Bohr model of the atom only for hydrogen.

The above image shows a continuous spectrum (top), an absorption spectrum (middle), and an emission spectrum (bottom) for some of the "Balmer lines" of hydrogen—those that lie within the visible part of the spectrum. The sketch to the right is a more-detailed version of Fig. 4.17 of your text, indicating additional possible transitions for an electron in a hydrogen atom. Notice that the energy jump of the electron from n = 3 to n = 2 results in a photon of wavelength 656.3 nm, or one that is in the red part of the spectrum. For an electron to jump from n = 2 to n = 3 would involve absorbing a photon of wavelength 656.3 nm. If the electron were at n = 2 and absorbed a photon having an energy of more than 3.4 electron volts (eV), it would have enough energy to escape the proton. If the proton were to capture a free electron, then a photon having a wavelength of 364.6 nm would be emitted. The line would be off to the left of the spectra shown above, deep in the violet part. Further explanation is given just below in these notes.

First of all, let's relate wavelength and frequency. Recall the relationship: c = wavelength * frequency. We know the speed of light, 2.9979 × 10⁸ m/s, and so if we have either the wavelength or the frequency, we can calculate the other value. For example (and let's stick with those electron jumps that result in photons in the visible part of the spectrum for now), the jump from energy level 3 to 2 results in a line at 656.3 nm. What is the frequency of this photon? How does the frequency of this photon compare to that of a photon with a wavelength of 364.6 nm?

In this comparison, we see that the photon with the shorter wavelength has a higher frequency; that is, more waves are going to pass per second by any given reference point. Note that the energy jump that resulted in the shorter wavelength was greater (more than 3.4 eVs) that the one with the wavelength of 656.3 nm (about 1.9 eVs). Do you see a pattern forming here? A wavelength of 656.3 nm is about 1.8 times 364.6 nm; the ratio of the frequencies (higher divided by lower) is about 1.8; 1.9 eVs times 1.8 is 3.4 eVs. They are all inter-related. You could do the same kind of quantitative analysis for each of the other wavelengths for the hydrogen lines. It does not matter whether these lines are seen in absorption, with the electrons absorbing photons of specific energies, or emission, with the electrons emitting photons of specific energies: it is the ratio of the energy jumps that is important.

Since we learned above that the energy of a photon is directly proportional to its frequency, we also now know that the photon with the shorter wavelength also has more energy. How do the energies actually compare? The energy is equal to Planck's constant times the frequency, E = hv. Planck's constant is 6.626 × 10^–34 joule * sec. Comparing energies using the frequencies calculated above we find:

E = (6.626 × 10^–34 joule * sec)(4.57 × 10¹⁴ per sec) = 3.03 × 10^–19 joules = 1.9 eVs
E = (6.626 × 10^–34 joule * sec)(8.22 × 10¹⁴ per sec) = 5.45 × 10^–19 joules = 3.4 eVs

where we've used the fact that 1 eV = 1.602 × 10^–19 joules. Do you see how all of this is interrelated? Try to follow the general steps taken in this mathematical analysis even if you do not understand it all. If we were to do similar calculations for the electron jumps that started or ended with n = 1, what is called the Lyman series, we would find out that these are photons with a great deal of energy. Since we see that they have shorter wavelengths, they must have higher frequencies, and higher frequencies mean higher energies. We can relate these to something more familiar to us: the ultraviolet radiation we receive from the Sun that causes premature aging and even cancer. UVA photons have wavelengths between 320 and 400 nm; UVB, 280-320 nm; and UVC, 180-280 nm. UVC radiation is extremely toxic to life. Fortunately for us, the Earth has developed an ozone layer that blocks this harmful radiation.

It is important to note that the hydrogen atom has these signature lines—this unique fingerprint—no matter what the temperature of the star is. For stars that are moving towards us or away from us, the wavelengths that we observe will be slightly different, but the pattern of lines will be the same. If we correct for the motion of the star, we will find that the wavelengths are exactly those that we measure in a laboratory here on Earth. Exactly. Without exception. The same can be said for every other element, for the ion of those elements, or for the isotopes of those elements. In each case, we see a unique fingerprint that identifies the element, ion, or isotope. Now, stars in real life are not made up of pure hydrogen, they all have helium, with a dash of carbon, nitrogen, oxygen, and other elements mixed in. What we observe, then, are these unique spectra all piled on top of each other. Unraveling the composition of the star is much like tracking down a criminal when there are 20 or 30 different sets of fingerprints on the weapon. Astronomers meticulously, with the help of computers (naturally) unravel the elements present in a star by matching the patterns, element by element, until as many as possible are identified.