Atomic Spectroscopy for Metal Analysis - Fundamentals of Environmental Sampling and Analysis - page 223

in atomic absorption or emission spectroscopy is a relatively low energy excitation
source, because temperature is often around 2000-3000 K, depending on the type
and velocity of the fuel gas. For example, the maximum temperatures that can be
achieved for air-acetylene and nitrous oxide-acetylene are 2250 and 2955 C,
respectively. The second common type of excitation source used in atomic
absorption is flameless graphite furnace. When a potential difference of a few volts
is applied on a graphite boat that holds a sample, the graphite rod behaves as
an electric resistor and is heated. The heating process is accomplished through a
four-stage cycle of a programmed temperature profile: (a) drying (100 C);
(b) decomposition (400 C); (c) atomization (2000 C); and (d) pyrolysis (cleaning,
2300 C). Drying is needed to avoid splashing of sample when a sudden temperature
increase is employed. Note that the atomization reactions in furnace are similar to
the processes in flame as described in Figure 9.1.

The Boltzmann Equation: Understanding the Temperature Effect
and the Sensitivity Difference Between Absorption and Emission
Spectrometry

Temperature has a profound effect on the ratio between the excited atoms and
unexcited (ground state) atoms in the flame or flameless atomizer. The quantitative
description of this effect is the Boltzmann equation:

N
N ¼

P
P e ðE EÞ=kT

ð9 : 3Þ

where N * and N are the number of atoms in an excited state and the ground state,
respectively, k is the Boltzmann constant (1

K), T is the temperature in
kelvin, and E E is the energy difference between the excited state and the ground
state. The P* and P are the statistical weights of the excited and ground states,
respectively, which can be determined by the number of states having equal energy
at each quantum level n (Section 8.1.2). In a hydrogen atom, for example, there are
P ¼ 2 ways that an atom can exist at the n ¼ 1 energy level (1s 2 ), and P ¼ 8 ways
that an atom can arrange itself at the n ¼ 2 energy level (2s 2 ,2p 6 ). The example
below illustrates the use and implications of Boltzmann equation.

: 3810 23 J =

EXAMPLE 9.1. Determine the ratio of sodium (Na) atoms in the 3p (excited state) to the
number in the 3s (ground state) at 2600 and 2610 K. Assume an average wavelength of 5893 ˚
for the 3p ! 3s transition. There are two quantum states in the 3s level and six in the 3p,
hence P =P ¼ 6=2 ¼ 3.

SOLUTION: Recall from Chapter 8 that the wavelength (l) and energy are related by:
E ¼ hc=l where h ¼Planck constant (6:6210 34 Js) and c ¼speed of light (310 8 m=s).
The wavelength needs to be converted into metric units in meters. Since 1m ¼ 10 10 ˚ ,
5893 A ¼ 5:89310 7 m.

E ¼ E E ¼ hc=l ¼ 6:6210 34 310 8 =ð5:89310 7 Þ

3710 19 J

¼ 3 :