Attenuation of Photons in Matter
Unlike energetic charged particles which have a finite range when traversing matter, photons of a given energy do not possess such a discrete range, but rather are absorbed exponentially as a function of distance in the absorbing medium. This phenomenon is referred to as attenuation. For a beam of $n_0$ monoenergetic photons, the rate of absorption of the photons as a function of depth in the absorber is
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$\displaystyle{ -\frac{\mathrm{d} n}{\mathrm{d} x} = N \sigma n_0 \text{,} }$ |
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where $\sigma$ is the absorption cross section of each atomic scattering center (or absorbing molecule), and $N$ is the volume density of atomic scattering centers. (The negative sign above represents absorption.) We define $\mu \equiv N \sigma$, where $\mu$ is referred to as the linear attenuation coefficient. We can then integrate Equation (1) with respect to distance $x$ to obtain
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$\displaystyle{ n(x) = n_0 \mathrm{e}^{-\mu x} \text{.} }$ |
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Similarly, if we consider depth $x$ in terms of not distance, but of areal density (in g/cm2), we can define the mass attenuation coefficient such that
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$\displaystyle{ \mu_\text{m} \equiv \frac{\mu}{\rho} = \frac{N \sigma}{\rho} = \frac{N_\text{A} \sigma}{\bar{A}} \text{,} }$ |
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where $\rho$ is the mass density (in g/cm3), $\bar{A}$ is the atomic molar mass (in g/mol) of the absorbing medium, and $N_\text{A}$ is Avogadro's number (in molecules/mol), respectively. Equation (2) now becomes:
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$\displaystyle{ n(x) = n_0 \mathrm{e}^{-\mu_\text{m} x} \text{.} }$ |
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Practical physics applications typically make use of Equation (4) wherein values of $\mu_\text{m}$ are tabulated.
Mechanisms of Photon Interaction with Matter
There are a number of different mechanisms by which photons can interact with matter. However, only three of these mechanisms—the photoelectric effect, Compton scattering, and electron/positron pair production—dominate, while the remainder usually make only a negligible contribution. The total absorption cross section $\sigma$ in Equation (3) is the sum of the cross sections for each type of photon interaction:
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$\displaystyle{ \sigma = \sigma_\text{PE} + \sigma_\text{C} + \sigma_\text{PP} \text{,} }$ |
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where $\sigma_\text{PE}$ is the cross section for photoelectric interactions, $\sigma_\text{C}$ is the cross section for Compton interactions, and $\sigma_\text{PP}$ is the cross section for pair production. (A cross section in nuclear physics can informally be read as an energy-dependent probability for interaction.) Since $\mu \propto \sigma$ from Equation (3), there is a corresponding mass attenuation coefficient for each cross section.
The cross sections given in Equations (6), (8), and (10) below are given in terms of the circular electron area $\pi r_\text{e}^2$ the photon "sees", scaled by a function that is dependent on the energy of the incident photon $E_\gamma$. In all cases $E_\gamma$ is compared to the rest mass-energy of the electron $m_\text{e} c^2$, suggesting a deep relationship between the two.
These cross sections are implemented in a Fortran program for a single photon. If you're not a fortranmann, then it is also here in JavaScript:
Contents
- Photoelectric Effect
- Compton Scattering
- Pair Production
Photoelectric Effect
Free electrons cannot fully absorb photons without violating conservation of energy and conservation of momentum, and for that reason, free electrons are not subject to the photoelectric effect. Indeed, the photoelectric effect should be thought of as interaction between an incident photon and an entire atom, not just one of its electrons.
Photoelectric interactions are more likely to eject electrons from the more tightly bound inner electron shells of the atom than the less tightly bound outer shells, since interactions involving inner shell electrons occur closer to the center of the atom where it is easier to transfer an incident photon's linear momentum to the atom. This is because nearly all the mass of the atom is concentrated in its nucleus.
The cross section for the photoelectric effect is:
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$\displaystyle{ \sigma_\text{PE} = 4 \sqrt{2} \alpha^4 \bar{Z}^5 \frac{8 \pi r_\text{e}^2}{3} \left( \frac{m_\text{e} c^2}{E_\gamma} \right)^{7/2} \text{,} }$ |
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where $\alpha$ is the fine-structure constant, $\bar{Z}$ is the atomic number of the absorbing medium, $r_\text{e}$ is the classical electron radius, the quantity $m_\text{e} c^2$ is the rest mass-energy of the electron, and $E_\gamma$ is the energy of the incident photon. Equation (6) should be applied if E > 0.1_dp .AND. E < 0.35_dp
, or where the photon energy is above the K-absorption edge. Please see the Notes for more details.
Pair Production
Pair production is essentially the reverse process of bremsstrahlung. When in close proximity to a heavy nucleus, a photon can be converted into an electron/positron pair. Conservation of energy for this process dictates
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$\displaystyle{ E_\gamma = T_\text{e} + m_\text{e} c^2 + T_\text{p} + m_\text{e} c^2 \text{,} }$ |
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where $T_\text{e}$ and $T_\text{p}$ are the kinetic energy of the electron and positron, respectively. Because the rest mass of the electron and the positron must come from the energy of the photon, there is a threshold energy of $2 m_\text{e} c^2$ = 1.022 MeV below which this process cannot occur.
The cross section for pair production originates from quantum electrodynamics, and is given by [R.J. Gould and G.P. Schréder, "Pair Production in Photon-Photon Collisions," Phys. Rev. 155, 1404 (1967)]:
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$\displaystyle{ \sigma_\text{PP} = \frac{\pi r_\text{e}^2}{2} (1 - \beta^2) f(\beta) \text{,} }$ |
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where the function $f(\beta)$ is equal to $(3 - \beta^4) \ln \left( \frac{1 + \beta}{1 - \beta} \right) - 2 \beta (2 - \beta^2)$. As before, the factor $\beta \equiv v /c = \sqrt{1 - \left[ m_\text{e} c^2 / (E + m_\text{e} c^2) \right]^2 }$ is from special relativity. In this case $E$ is actually the relativistic kinetic energy of the electron/positron in the center-of-mass (CM) frame of reference.
Acknowledgement
I wish to thank Dr. Eric Benton for the nuclear and particle physics notes that were used as inspiration for this page.