Isaac
Cosmic Collision Forges Galactic One Ring—in X-rays
Photon Interactions with Matter
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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

$\displaystyle{ -\frac{\mathrm{d} n}{\mathrm{d} x} = N \sigma n_0 \text{,} }$ (1)

where $\sigma$ is the absorption cross section of each atomic scattering center, and $N$ is the volume density of atomic scattering centers. (Note that 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

$\displaystyle{ n(x) = n_0 \mathrm{e}^{-\mu x} \text{.} }$ (2)

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

$\displaystyle{ \mu_\text{m} \equiv \frac{\mu}{\rho} = \frac{N \sigma}{\rho} = \frac{N_\text{A} \sigma}{\bar{A}} \text{,} }$ (3)

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:

$\displaystyle{ n(x) = n_0 \mathrm{e}^{-\mu_\text{m} x} \text{.} }$ (4)

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 $\sigma$ is the sum of the cross sections for each type of photon interaction:

$\displaystyle{ \sigma = \sigma_{\text{PE}} + \sigma_{\text{C}} + \sigma_{\text{PP}} \text{,} }$ (5)

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. 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 kinetic 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. That is remarkable!

These cross sections are implemented for a single photon in a Fortran program I've written (get it here).

Contents

  1. Photoelectric Effect
  2. Compton Scattering
  3. Pair Production

Photoelectric Effect

Elijah

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:

$\displaystyle{ \sigma_\text{PE} = 4 \sqrt{2} \alpha^4 \bar{Z}^5 \frac{8 \pi r_\text{e}^2}{3} \left( \frac{m_\text{e} c^2}{E_\gamma} \right)^{7/2} \text{,} }$ (6)

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 finally $E_\gamma$ is the kinetic 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.

Compton Scattering

Elijah

The interaction of photons with "nearly" free electrons is described by a number of mechanisms:

Figure 1 illustrates Compton scattering, where we assume that the target electron is "free" (i.e., not bound to a host atom). An incident photon of kinetic energy $E_\gamma$ "collides" with a free electron at rest, transferring some of its energy and momentum to the electron. The photon, now with energy $E'_\gamma$, scatters at an angle $\theta$, while the electron recoils at an angle $\phi$ with velocity $v = \beta c$.

The geometry of Compton scattering
Figure 1. The geometry of Compton scattering. Diagram courtesy of Eric Benton.

If we know the energy of the incident photon $E_\gamma$ and are able to experimentally detect the scattered photon and thereby determine the angle $\theta$, we can find the energy of the scattered photon $E'_\gamma$ by means of conservation of energy and conservation of momentum. Conservation of energy for the Compton scattering interaction illustrated in Figure 1 allows us to write

$\displaystyle{ E_\gamma + m_\text{e} c^2 = E'_\gamma + \frac{m_\text{e} c^2}{\sqrt{1 - \beta^2}} \text{,} }$ (7)

where $\beta \equiv v /c = \sqrt{1 - \left[ m_\text{e} c^2 / (E_\gamma - E'_\gamma + m_\text{e} c^2) \right]^2 }$ from special relativity. After a lot of algebra, you will first write down the differential cross section for Compton scattering, which is specified by the Klein-Nishina formula. Integrating this formula over all angles gives the cross section for the Compton interaction:

$\displaystyle{ \sigma_\text{C} = \frac{\pi r_\text{e}^2}{\bar{E}} \left[ \left( 1 - \frac{2 (\bar{E} + 1)}{\bar{E}^2} \right) \ln(2 \bar{E} + 1) + \frac{1}{2} + \frac{4}{\bar{E}} - \frac{1}{2 (2 \bar{E} + 1)^2} \right] \text{,} }$ (8)

where $\bar{E}$ effectively compares the incident photon energy (in MeV) to the rest mass-energy of the electron, or $E_\gamma / m_\text{e} c^2$.

Pair Production

Elijah

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:

$\displaystyle{ E_\gamma = T_\text{e} + m_\text{e} c^2 + T_\text{p} + m_\text{e} c^2 \text{,} }$ (9)

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)]:

$\displaystyle{ \sigma_\text{PP} = \frac{\pi r_\text{e}^2}{2} (1 - \beta^2) \left[ (3 - \beta^4) \ln \left( \frac{1 + \beta}{1 - \beta} \right) - 2 \beta (2 - \beta^2) \right] \text{,} }$ (10)

where $\beta \equiv v /c = \sqrt{1 - \left[ m_\text{e} c^2 / (E + m_\text{e} c^2) \right]^2 }$ 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

Elijah

I wish to thank Dr. Eric Benton for the nuclear and particle physics notes that were used as inspiration for this page.

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