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7.5: Absorption Cross-Sections

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    107255
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    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.

    Readings

    1. Herzberg, G., Molecular Spectra and Molecular Structure: Infrared and Raman of Polyatomic Molecules. Prentice-Hall: New York, 1939; Vol. II, p. 261.
    2. McHale, J. L., Molecular Spectroscopy. 1st ed.; Prentice Hall: Upper Saddle River, NJ, 1999.

    7.5: Absorption Cross-Sections is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Andrei Tokmakoff via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.