# 7.5: Absorption Cross-Sections


The rate of absorption induced by a monochromatic electromagnetic field is

$w _ {k \ell} ( \omega ) = \frac {\pi} {2 \hbar^{2}} \left| E _ {0} ( \omega ) \right|^{2} | \langle k | \hat {\varepsilon} \cdot \overline {\mu} | \left. \ell \rangle \right|^{2} \delta \left( \omega _ {k \ell} - \omega \right) \label{6.64}$

The rate is clearly dependent on the strength of the field. The variable that you can most easily measure is the intensity $$I$$, the energy flux through a unit area, which is the time-averaged value of the Poynting vector, $$S$$:

$S = \varepsilon _ {0} c^{2} ( \overline {E} \times \overline {B} ) \label{6.65}$

$I = \langle S \rangle = \frac {1} {2} \varepsilon _ {0} c E _ {0}^{2} \label{6.66}$

Using this we can write

$w _ {k \ell} = \frac {4 \pi} {3 \varepsilon _ {0} c h^{2}} I ( \omega ) | \langle k | \overline {\mu} | \ell \rangle |^{2} \delta \left( \omega _ {k \ell} - \omega \right) \label{6.67}$

where I have also made use of the uniform distribution of polarizations applicable to an isotropic field:

$\left| \overline {E} _ {0} \cdot \hat {x} \right| = \left| \overline {E} _ {0} \cdot \hat {y} \right| = \left| \overline {E} _ {0} \cdot \hat {z} \right| = \frac {1} {3} \left| E _ {0} \right|^{2}.$

Now let’s relate the rates of absorption to a quantity that is directly measured, an absorption cross section $$\alpha$$:

\begin{align} \alpha &= \frac {\text {total energy absorbed per unit time}} {\text {total incident intensity (energy/unit time/area)}} \label{6.68} \\[4pt] &= \frac {\hbar \omega w _ {k \ell}} {I} \end{align}

Note $$\alpha$$ has units of cm2 . The golden rule rate for absorption also gives the same rate for stimulated emission. Given two levels $$| m \rangle$$ and $$| n \rangle$$,

$w _ {n m} = w _ {m n}$

$\therefore \left( \alpha _ {A} \right) _ {n m} = \left( \alpha _ {S E} \right) _ {m n} \label{6.69}$

We can now use a phenomenological approach to calculate the change in the intensity of incident light, $$I$$, due to absorption and stimulated emission passing through a sample of length $$L$$. Given that we have a thermal distribution of identical non-interacting particles with quantum states such that the level $$| m \rangle$$ is higher in energy than $$| n \rangle$$:

$\frac {d I} {d x} = - N _ {n} \alpha _ {A} I + N _ {m} \alpha _ {s E} I \label{6.70}$

$\frac {d I} {I} = - \left( N _ {n} - N _ {m} \right) \alpha\, d x \label{6.71}$

Here $$N_n$$ and $$N_m$$ are population of the upper and lower states, but expressed as population densities (cm-3). Note that $$I$$ and $$α$$ are both functions of the frequency of the incident light. If $$N$$ is the molecular density,

$N _ {n} = N \left( \frac {e^{- \beta E _ {n}}} {Z} \right) \label{6.72}$

Integrating Equation \ref{6.71} over a path length $$L$$, we have

\begin{align} T &= \frac {I} {I _ {0}} \\[4pt] &= e^{- \Delta N \alpha L} \label{6.73} \\[4pt] &\approx e^{- N \alpha L} \end{align}

We see that the transmission of light through the sample decays exponentially as a function of path length.

$\Delta N = N _ {n} - N _ {m}$

is the thermal population difference between states. The second expression in Equation \ref{6.73} comes from the high-frequency approximation applicable to optical spectroscopy. Equation \ref{6.73} can also be written in terms of the familiar Beer–Lambert Law:

$A = - \log \frac {I} {I _ {0}} = \epsilon C L \label{6.74}$

where $$A$$ is the absorbance and $$C$$ is the sample concentration in mol L-1, which is related to the number density via Avagadro’s number $$N_A$$,

$C \left[ \operatorname {mol} L^{- 1} \right] = \frac {N \left[ c m^{- 3} \right]} {N _ {A}} \times 1,000 \label{6.75}$

In Equation \ref{6.74}, the characteristic molecular quantity that describes the sample’s ability to absorb the light is $$\epsilon$$, the molar decadic extinction coefficient, given in L mol-1 cm-1. With these units, we see that we can equate $$\epsilon$$ with the cross section as

$\epsilon = \frac {N _ {A} \alpha} {2303} \label{6.76}$

In the context of sample absorption characteristics, our use of the variable $$α$$ for cross section should not be confused with another use as an absorption coefficient with units of cm-1 that is equal to $$Nα$$ in Equation \ref{6.73}.

These relationships also allow us to obtain the magnitude of the transition dipole matrix element from absorption spectra by integrating over the absorption line shape:

\begin{align} \left| \mu _ {i f} \right|^{2} &= \frac {6 \varepsilon _ {0} \hbar^{2} 2303 c} {N _ {A} n} \int \frac {\varepsilon ( v )} {v} d v \label{6.77} \\[4pt] &= \left( 108.86\, L \,mol^{- 1}\, cm^{-1}D^{-2} \right)^{- 1} \int \frac {\varepsilon ( v )} {v} d v \end{align}

Here the absorption line shape is expressed in molar decadic units and the frequency in wavenumbers.