Photon Interactions on the Microscopic Scale

1. Photon Interaction Mechanisms

When a photon passes through a material, it may interact with the atoms of that material in several ways, depending on the photon energy and the atomic number \( Z \) of the absorber. These interactions can be divided into two broad categories: interactions with the nucleus and interactions with orbital electrons of the atoms in the absorber.

2. Interactions with the Nucleus

Photons can interact with the nuclei of the absorber atoms in the following ways:

Note: Photonuclear reactions and pair production are significant only at high photon energies (above several MeV) and typically occur in high-Z materials like lead or uranium.

3. Interactions with Orbital Electrons

Photons can also interact with the orbital electrons of the atoms in the absorber. The nature of these interactions depends on the energy of the photon and the binding energy of the electron:

4. Fate of the Photon after Interaction

After a photon interacts with an atom, there are two possible outcomes:

5. Example of Photon Interaction: Compton Scattering

Let's consider an example of Compton scattering. A photon with energy \( E_\gamma = 1 \, \text{MeV} \) interacts with a free electron, transferring part of its energy to the electron. The Compton scattering formula is given by:

\[ \lambda' - \lambda = \frac{h}{m_e c} \left( 1 - \cos \theta \right) \]

Where:

For a scattering angle \( \theta = 90^\circ \), the change in wavelength can be calculated, and the energy of the scattered photon can be determined from its new wavelength. This effect is one of the primary interactions for photons in the diagnostic energy range (e.g., X-rays).

6. Electron Vacancies and Auger Effect

In photon interactions, particularly with the inner shells of atoms, electron vacancies are created. These vacancies are typically filled by electrons from higher atomic shells, emitting characteristic X-rays (fluorescence photons). Alternatively, the vacancy can be filled by the emission of an Auger electron, a secondary electron that carries away the energy released during the transition.

The Auger effect occurs when an atom undergoes internal electron transitions, and the vacancy produced in the inner shell is filled by an electron from an outer shell, emitting an Auger electron instead of a photon. The cascade of Auger electrons can produce ionization densities that are comparable to those produced by alpha particles.

\[ \text{Auger Yield} = (1 - \omega) \]

Where \( \omega \) is the fluorescence yield, representing the probability that a vacancy will be filled by the emission of a photon instead of an Auger electron.

7. Fluorescence Yield and Auger Effect

The fluorescence yield (\( \omega \)) defines the probability that a vacancy in a given shell will be filled by the emission of a characteristic photon. The probability that the vacancy will be filled by an Auger electron is \( (1 - \omega) \). The yield depends on the atomic number of the material and the shell in question.

The fluorescence yield is illustrated in the graph below:

8. Compton Effect: Detailed Example

To better understand the Compton effect, let’s calculate the change in energy of the photon after scattering. Consider a photon with an energy \( E_\gamma = 1 \, \text{MeV} \) scattering off a free electron at an angle of \( 90^\circ \). According to the Compton scattering equation:

\[ \Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} \left( 1 - \cos \theta \right) \]

We know the following constants:

  • Planck’s constant, \( h = 6.626 \times 10^{-34} \, \text{J}\cdot\text{s} \)
  • Electron mass, \( m_e = 9.11 \times 10^{-31} \, \text{kg} \)
  • Speed of light, \( c = 3.0 \times 10^8 \, \text{m/s} \)
  • Incident photon energy, \( E_\gamma = 1 \, \text{MeV} = 1.602 \times 10^{-13} \, \text{J} \)

First, we need to calculate the initial wavelength \( \lambda \) of the incident photon:

\[ \lambda = \frac{h c}{E_\gamma} = \frac{(6.626 \times 10^{-34}) \times (3.0 \times 10^8)}{1.602 \times 10^{-13}} \approx 1.24 \times 10^{-12} \, \text{m} \]

For a scattering angle of \( 90^\circ \) (i.e., \( \theta = 90^\circ \), so \( \cos 90^\circ = 0 \)), we find:

\[ \Delta \lambda = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(3.0 \times 10^8)} \times (1 - 0) \approx 2.426 \times 10^{-12} \, \text{m} \]

The change in wavelength, \( \Delta \lambda \), represents the increase in wavelength (and thus the reduction in energy) of the scattered photon. To find the energy of the scattered photon, we use the relation:

\[ E'_\gamma = \frac{h c}{\lambda + \Delta \lambda} \]

Substituting the values:

\[ E'_\gamma = \frac{(6.626 \times 10^{-34}) \times (3.0 \times 10^8)}{1.24 \times 10^{-12} + 2.426 \times 10^{-12}} \approx 0.73 \, \text{MeV} \]

The scattered photon has an energy of approximately 0.73 MeV, and thus has lost energy due to the scattering event. This energy is transferred to the ejected electron, which is characteristic of the Compton effect.

9. Photoelectric Effect

The photoelectric effect occurs when a photon is absorbed by an atom, causing the ejection of an electron from one of its inner shells. The energy of the incident photon is used to overcome the binding energy \( E_B \) of the electron, and the remaining energy is given to the ejected electron as kinetic energy:

\[ E_\gamma = E_B + K.E. \]

Where \( E_\gamma \) is the energy of the photon, \( E_B \) is the binding energy of the ejected electron, and \( K.E. \) is the kinetic energy of the ejected electron.

For instance, consider a photon with energy \( E_\gamma = 100 \, \text{keV} \) interacting with a tungsten atom (where the binding energy of an inner shell electron is \( E_B = 70 \, \text{keV} \)). The kinetic energy of the ejected electron is:

\[ K.E. = E_\gamma - E_B = 100 \, \text{keV} - 70 \, \text{keV} = 30 \, \text{keV}. \]

This energy is released as the kinetic energy of the photoelectron.

10. Pair Production

Pair production occurs when a high-energy photon (with energy greater than 1.022 MeV) interacts with the electromagnetic field of a nucleus, creating an electron-positron pair. The energy of the photon is split into the rest mass energies of the electron and positron and the kinetic energy of the pair. The equation for pair production is:

\[ E_\gamma = 2m_e c^2 + K.E. \]

Where:

  • \( E_\gamma \) is the energy of the incident photon,
  • \( 2m_e c^2 \) is the combined rest mass energy of the electron and positron (since the rest mass of an electron or positron is \( m_e c^2 = 0.511 \, \text{MeV} \), the total rest mass energy is 1.022 MeV),
  • \( K.E. \) is the kinetic energy of the produced electron and positron.

For example, if a photon with energy \( E_\gamma = 3 \, \text{MeV} \) undergoes pair production, the energy left for the kinetic energy of the electron-positron pair is:

\[ K.E. = E_\gamma - 2m_e c^2 = 3 \, \text{MeV} - 1.022 \, \text{MeV} = 1.978 \, \text{MeV}. \]

This means that the electron and positron each receive approximately half of the remaining energy as kinetic energy.

11. Rayleigh Scattering

Rayleigh scattering occurs when a photon is scattered by the orbital electrons of an atom without energy loss. This means the photon changes direction but maintains its original energy. Rayleigh scattering is significant at lower photon energies and is more prevalent in low-Z materials, where the scattering cross-section is relatively high for low-energy photons. Rayleigh scattering is described by the following equation:

\[ \frac{d\sigma}{d\Omega} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{Rayleigh}} = \left(\frac{e^2}{4\pi\epsilon_0}\right)^2 \frac{1}{\sin^4 \left( \frac{\theta}{2} \right)}. \]

Where \( \frac{d\sigma}{d\Omega} \) is the differential scattering cross-section, \( \theta \) is the scattering angle, and \( e \) is the charge of the electron. This scattering has no energy loss, and the photon retains its energy, making it a purely elastic scattering event.

Conclusion

Photon interactions with matter involve a variety of processes, each with a characteristic energy dependence and interaction probability. These interactions—ranging from the photoelectric effect, Compton scattering, and pair production to Rayleigh scattering—play key roles in determining the attenuation of photon beams as they pass through materials. Understanding these processes is critical for applications in medical imaging, radiation therapy, and nuclear physics.

The balance between different interactions (such as the photoelectric effect at low photon energies, and Compton scattering at intermediate energies) depends on the material's atomic number and the photon energy. High-Z materials are more likely to undergo photoelectric absorption, while low-Z materials favor Compton scattering. This knowledge is fundamental in designing effective shielding, detectors, and radiation-based therapies.