Rydberg The Equation
Using the quantum mechanical resonance model for the electron cloud in atoms, we can
understand the absorption or emission of photons as resulting from the electron cloud
expanding out to a larger shell or contracting inward to a smaller one, respectively. The
difference in energy between the shells gives the energyand therefore the frequency
and wavelengthof the photon. We know that the energy level of a given shell can be
found in units of electron-volts from E = -13.6/n.
Notice, however, that a hydrogen atom cannot produce photons of every possible
wavelength. Since the principal quantum number n is an integer, and since the energy in a
shell depends on n, there are certain values of energy difference that the electron clouds
can never create. Its somewhat like the sound pitches that a piano can never produce
because they lie in between two adjacent keys. If we are dealing with photons coming
from a large mixture of atoms, there tends to be such a wide variety of possible emissions
that a complete spectrum of photons can be produced. This is, in fact, how the blackbody
radiation spectrum comes into being. If there is only one kind of atoms, however, an
emission spectrum results that consists only of a set of bright lines of frequencies
characteristic to the energy levels of that atom. The rest of the spectrum is completely
dark. The mixture of bright lines gives the excited element its distinct color, such as the
hydrogen blue that comes from the methane burning in a gas stove, or the green color of a
barium flame test, or the bright red-orange of a neon light. Broken up through a prism or
diffraction grating, the emission can spectrum be used to identify the presence of an
element uniquely. Helium was found in the spectrum of the sun (since its bright lines
matched nothing else known) before it was ever found on earth.
When this phenomenon was first observed, it was very difficult to explain. There were
several individuals who studied hydrogens emission spectrum, and their names have
come to be permanently associated with the band of lines they studied. Lyman studied a
set of lines in the UV portion of the spectrum. All of these result from electrons dropping
from higher levels down to the n = 1 (K) shell. Balmer studied lines in the visible portion,
all resulting from the electron falling down to n = 2 (L). We know that in each case, the
final value (which we will now call n' ) must be smaller than the initial value n where the
electron started. We can then use the Rydberg equation to solve for the wavelength of the
emitted light. The equation will be presented in an unusual way here, but one that is,
hopefully, easier. It will involve a value we will call the Rydberg wavelength, which is
just the wavelength of a 13.6eV photon: 91.1nm. It is therefore important to remember
that this equation yields values in nanometers: = 91.1 n n' / (n n' ).
A typical problem would read, Find the wavelength of the third
hydrogen Balmer line. Because it is a Balmer line, n' = 2 as taken from
the table at right. Then, since we seek the third line in that spectrum, n =
2 + 3 = 5. We plug n' = 2 and n = 5 into the Rydberg equation and get
= 434nm. The most common error by far is forgetting to square n
and n'.
name
Lyman
Balmer
Paschen
Brackett
Pfund
n'
1
2
3
4
5