Page 227 - Essentials of physical chemistry

Basic Spectroscopy 189

Let us consolidate the key values to make formulas that are easy to remember.

2 27 2
h (6:6260693 10 erg s)
me 4p (9:1093826 10 g)(4:803204 10 g cm =s)
a 0 ¼ 2 ¼ 2 28 10 1=2 3=2 2
¼ 0:5291772 10 8 cm
4 2 28 10 1=2 3=2 4
me 4p (9:1093826 10 g)(4:803204 10 g cm =s)
¼ 13:6057 eV
2
2
¼
2 h 2(6:6260693 10 27 erg s) (1:60217653 10 12 erg=eV)
In the second step, we have used the conversion from erg to eV in the denominator. Since
1eV ¼ 1.60217653 10 19 J, we just multiply by 1.0 10 7 to put the constant into ergs.
Although we have already delved into several unit conversions, we need just one more formula to
be on our way to understanding spectroscopy. The next formula is by far and away the most useful
formula in spectroscopy. If you attend a research seminar and the speaker gives energies in
kilocalories=mole he=she is probably an older chemist, if the speaker uses kilojoule=mole energies
he=she is probably a younger chemist, but if the speaker is a physicist you will probably hear all the
energy values in electron volts. Thus, even though the CRC Handbook and other texts strive to use
only SI units, our recommendation is to get used to electron volt energies. In x-ray analysis later in
this chapter the units are usually in kiloelectron volt (thousands of electron volts), so there is a large
part of the scientific community using electron volts.
With a warning to the students, we have selected electron volts as the most useful energy unit to
relate spectroscopy experiments to theory in the sense that a student can imagine the physical units.
However, the physical constants are revaluated every three years or so which makes past research
papers subject to drift in the values of the constants. Around 1960, quantum chemists addressed this
h
problem and chose yet another set of units in which c ¼ ¼ m e ¼ q e ¼ 1 to simplify theoretical
equations in ‘‘atomic units,’’ so that the equations were expressed totally in the basic mathematical
units. In these units (used by quantum chemistry computer programs), a person only needs to know
the latest value of an energy unit called the hartree and the latest value of the Bohr radius (a 0 )to
convert computer results back to laboratory results. At present (2010), 1 hartree ¼ 27.2113845 eV
and a 0 ¼ 0.52917720859 10 8 cm. We will not use these units until Chapter 17 but you can see
that the Bohr formulas simplify further in these units:

2
2
2
4 2 2
Z me Z n h n
(0:5 hartree); a 0 :
E(n, Z) ¼ 2 ¼ r(n, Z) ¼ 2 ¼
n 2 n Z me Z
h
A VERY USEFUL FORMULA
Here, we present a very simple but powerful formula. This formula is so useful that you can sit in a
research seminar or lecture and do mental arithmetic on the spot and then make a very intelligent
comment such as ‘‘Yes, Professor, but the wavelength for that transition should be . . . ’’ Recall that
Planck realized that the energy of a light wave is proportional to the frequency of the wave and
evaluated the proportionality constant to be ‘‘h.’’ As mentioned above, the same number occurs in the
slope of the data for the photoelectric effect as analyzed by A. Einstein in 1905. The modern value is
6.6260693 10 27 erg s. Assume there are two energy levels in a molecular system such that the
hc
. This situation happens so
difference between the two levels is a quantum with energy DE ¼ hn ¼
l
often in spectroscopy that we can develop a useful shortcut formula if we assume the l value is in
angstroms (1.0 10 8 cm) and we always want the energy value in electron volts. Then we have
10
8
hc (6:6260693 10 34 J s)(2:99792458 10 m=s)(10 A ˚ =m)
DE (eV) ¼ ¼ 19
l (A ˚ ) (1:60217653 10 J=eV)(l (A ˚ ))