Physical Chemistry , 1st ed.

sample was energetically stimulated so that it emitted light). Many of these

spectra were complicated, but for some reason the spectrum of hydrogen was

relatively simple (see Figure 9.5). Hydrogen was the lightest known element

and probably the simplest, a fact that could hardly have been missed in the at-

tempts to interpret its spectrum. In 1885, the Swiss mathematician Johann

Jakob Balmer showed that the positions of the lines of light from hydrogen in

the visible portion of the spectrum could be predicted by a simple arithmetic

expression:



1

R 

1

4



n

1

2  (9.16)

where is the wavelength of the light,nis an integer greater than 2, and Ris

some constant whose value is determined by measuring the wavelengths of the

lines. The simplicity of the equation is startling, and it inspired other scientists

to analyze the spectrum of hydrogen in other regions of the spectrum, like the

infrared and ultraviolet regions. Although several other people (Lyman,

Brackett, Paschen, Pfund) were to discover other simple progressions of lines

in the hydrogen spectrum, in 1890 Johannes Robert Rydberg successfully gen-

eralized the progressions into a single formula:



1

 ̃ RH

n

1

 22 n

1

 (^21)  (9.17)
where n 1 and n 2 are different integers,n 2 is less than n 1 , and RHis known as
the Rydberg constant. The variable ̃is the wavenumber of the light and has
units of inverse centimeters, or cm^1 , indicating the number of light waves per
centimeter.* Interestingly enough, thanks to the precision with which the hy-
drogen atom spectrum can be measured, the Rydberg constant is one of the
most accurately known physical constants: 109,737.315 cm^1.
Example 9.2
Determine the frequencies in cm^1 for the first three lines for the Brackett
series of the hydrogen atom, where n 2  4.
Solution
Ifn 2 4, the first three lines in the Brackett series will have n 1 5, 6, and

7. Using equation 9.17 above and substituting for RHand n 2 ,we get

 ̃ 109,737.315 

4

1

2 n

1

 (^21) cm
 1
Substituting 5, 6, and 7 for n 1 above, we calculate 2469 cm^1 (n 1 5), 3810
cm^1 (n 1 6), and 4619 cm^1 (n 1 7).
But the questions remained: Why was the hydrogen spectrum so simple?
And why did Rydberg’s equation work so well? Although it was tacitly under-
stood that hydrogen was the lightest and simplest atom, there was absolutely
no reason to assume that a sample of this matter would give off only certain
wavelengths of light. It didn’t matter that the spectra of other elements were a
little more complicated and could not be described by any straightforward
250 CHAPTER 9 Pre-Quantum Mechanics
*The formal SI unit for wavenumber is m^1 , but quantities with cm^1 units are more
common.