MODULE 16_03
Practical Aspects of Light Absorption
We can go to the laboratory and conduct the following experiment. Place an absorbing specimen
normal to a beam of monochromatic radiation from a continuously operating source and measure the
intensity of the beam before (Ii) and after the beam has passed through the specimen.
l
It
Ii
Ignoring losses from reflection at the surfaces and scattering within the specimen, you find that the
absorber attenuates the light intensity. The quantity It falls exponentially with distance, l, through the
sample.
I t  I i e  l
(16.1)
where  (m ) is termed the absorption coefficient and characterizes the attenuation due to path length.
-1
If the sample consists of a solution of molecular absorbers in a transparent solvent, we can characterize
the attenuation factor per molecule. Thus with n absorbing molecules per m3 in the length l, we can
write
I t  I i e  nl
where m per molecule) is the molecular absorption coefficient.
2
(16.2)
Chemists tend to express
concentrations in terms of molarity, when
n  103 N A [ M ]
(16.3)
where [M] is the molarity of the absorber solution and NA is the Avogadro constant. In addition the
decadic base is used and so the molar attenuation is given by
I t  I i 10 [ M ]l
(16.4)
where  (M cm ) is the (molar decadic) extinction coefficient. Equation (16.4) is one way of
-1
-1
expressing the Beer-Lambert law. The extinction coefficient is the practical measure of the efficiency
1
with which a molecule, or a chromophoric entity, absorbs light. It varies from chromophore to
chromophore, and is a function of the wavelength of the light that is applied.
Another way of expressing equation (16.4) is in logarithmic form
I 
log  i    [ M ]l   A
 It 
where A is termed the absorbance of the absorbing entity (sample).
(16.5)
Figure 16.1 shows plots of the absorbance as a function of wavelength for solutions of benzene and
anthracene in hexane solvent. An absorption spectrophotometer was employed to obtain the plots. In
the experiment the concentration and the path length were fixed so the amplitude variations arises from
variation of the extinction coefficient with wavelength. The reason for this variation will be examined
later. 
1.0
0.6
anthracene
benzene
0.5
0.8
absorbance
absorbance
0.4
0.3
0.6
0.4
0.2
0.2
0.1
0.0
0.0
220
225
230
235
240
245
250
255
260
265
270
275
280
280
300
320
340
360
380
400
420
440
wavelength/nm
wavelength/nm
FIGURE 16.1
In many cases we, as photoscientists are interested in the quantity of light that is absorbed by a sample.
Ignoring scattering and reflections we can see that
Ii  It  I a
(16.6)
where Ia is the light absorbed in the sample during the exposure. Combining equation (16.4) with
(16.6) we obtain
I a  Ii  It  I i (1  10 [ M ]l )
 I i (1  10 A )
(16.7)
The quantities Ia and It can be expressed either as the total number of photons accumulated over the
exposure time (note that NA of photons = 1 einstein), or as a rate, in photons per second.
2
Since the absorption of one photon by a molecule generates one primary excited state, then the rate of
primary excited state production equals the rate of photon absorption.
d [ M *]
 I a  I i (1  10 A )
dt
(16.8)
in units of (molecules) per second.
The connection between theory and experiment
At this point it is useful to relate the experimental quantity  to the theoretical quantities
B( B f i  B f i ) and fi.
Consider a plane of area A at a position x with radiation incident on the plane from the left. All
photons within a length ct, (c = speed of light) and hence in a volume Act will pass through the
plane in the interval t. If the energy density of the field is U, then the total electromagnetic energy
passing through the plane in t is Uact.
The energy flux, F, is the energy per unit time per unit area and so
F  UAct / At  cU
(16.9)
In the frequency range  to +d the energy density is
dU  rad ( E )d
(16.10)
and so the energy flux in the same range is dF  c  rad ( E )d .
If we write dF  I ( )d where I is the radiation intensity, then I  c  rad .
Now let us consider the absorption that occurs within a slab of thickness dl. Let the total number of
molecules in the slab that can absorb radiation between  and d be n(dso the total number
density of absorbers is N   n( )d .
The rate at which a given molecule absorbs a photon
is ku l  B rad ( Eul ) . Each photon has energy h, and so the rate of change of energy density is

dU
 h k f i n( )d  n( )h B  rad ( E )d 
dt

The energy entering the slab at x from the left in dt is F(x)Adt, and the energy leaving the slab on the
right at x+dl is F(x+dl)Adt. The difference between these two quantities is the rate of change of
energy in the slab volume and is given by
Rearranging we get
d (UAdl )
 F ( x  dl ) A  F ( x) A
dt
(16.12)
dU F ( x  dl )  F ( x) dF


dt
dl
dl
(16.13)
3
dF dI
 d . By using equation (16.7) we can obtain
dl dl
n( )h
dI  n( )h B  rad dl  
BIdl
(16.14)
c
We also are aware that the reduction in light intensity by a beam on passing through a solution of
Remembering that I  c  rad , and therefore
length dl containing absorbing molecules of concentration [M] is
dI   ( )[ M ]Idl
(16.15)
 ( ) hn( ) B


c[ M ]
(16.16)
see
Integrating both sides with respect to  over all the frequencies in the radiation field leads to
 ( ) BhN BhN
(16.17)
   cA  c A
For typical absorption bands () is non-zero over a very narrow range of frequencies, so it is not too
much of an approximation to set  = fi. On doing this and putting J    ( ) d we have
J   h fi / c  N A B
(16.18)
In an earlier module we showed that the Einstein B coefficient is related to the transition dipole
moment according to
2
B   fi / 6 0
2
thus
J
 fi N A  fi
2
(16.19)
3 0 c
Thus we have established a connection between the integrated extinction coefficient (a measurable
quantity) and the theoretical quantities B and the transition dipole moment. The final thing to do is to
introduce the oscillator strength of a transition, f, which is a dimensionless quantity.
 4 me fi
f 
2
 3e
and combining this with equation (16.19) we obtain
 4m c 
f   e 20  J
 N Ae 

  fi

2
 6.257x1019 x( J / m 2 mol 1s 1 )
In practice, for allowed electric dipole transitions, f = 1; for forbidden transitions f<<1



4
(16.20)
(16.21)