## Gamma Rays Attenuation

The total cross-section of interaction of a gamma rays with an atom is equal to the sum of all three mentioned partial cross-sections:**σ = σ _{f} + σ_{C} + σ_{p }**

**σ**_{f}– Photoelectric effect

**σ**_{C}– Compton scattering

**σ**_{p}– Pair production

Depending on the gamma ray energy and the absorber material, one of the three partial cross-sections may become much larger than the other two. At small values of gamma ray energy the** photoelectric effect** dominates. **Compton scattering** dominates at intermediate energies. The compton scattering also increases with decreasing atomic number of matter, therefore the interval of domination is wider for light nuclei. Finally, **electron-positron pair production** dominates at high energies.Based on the definition of interaction cross-section, the dependence of gamma rays intensity on thickness of absorber material can be derive. If **monoenergetic gamma rays** are collimated into a **narrow beam** and if the detector behind the material only detects the gamma rays that passed through that material without any kind of interaction with this material, then the dependence should be simple **exponential attenuation of gamma rays**. Each of these interactions removes the photon from the beam either by absorbtion or by scattering away from the detector direction. Therefore the interactions can be characterized by a fixed probability of occurance per unit path length in the absorber. The sum of these probabilities is called the **linear attenuation coefficient**:

**μ = τ _{(photoelectric)} + σ_{(Compton)} + κ_{(pair)}**

### Linear Attenuation Coefficient

The attenuation of gamma radiation can be then described by the following equation.

**I=I _{0}.e^{-μx}**

, where I is intensity after attenuation, I_{o} is incident intensity, μ is the linear attenuation coefficient (cm^{-1}), and physical thickness of absorber (cm).

The materials listed in the table beside are air, water and a different elements from carbon (*Z*=6) through to lead (*Z*=82) and their linear attenuation coefficients are given for three gamma ray energies. There are two main features of the linear attenuation coefficient:

- The linear attenuation coefficient increases as the atomic number of the absorber increases.
- The linear attenuation coefficient for all materials decreases with the energy of the gamma rays.

### Half Value Layer

The half value layer expresses the thickness of absorbing material needed for reduction of the incident radiation intensity by a **factor of two**. There are two main features of the half value layer:

- The
**half value layer**decreases as the atomic number of the absorber increases. For example 35 m of air is needed to reduce the intensity of a 100 keV gamma ray beam by a factor of two whereas just 0.12 mm of lead can do the same thing. - The
**half value layer**for all materials increases with the energy of the gamma rays. For example from 0.26 cm for iron at 100 keV to about 1.06 cm at 500 keV.

### Mass Attenuation Coefficient

When characterizing an absorbing material, we can use sometimes the mass attenuation coefficient. **The mass attenuation coefficient** is defined as the ratio of the linear attenuation coefficient and absorber density **(μ/ρ)**. The attenuation of gamma radiation can be then described by the following equation:

**I=I _{0}.e^{-(μ/ρ).ρl}**

, where ρ is the material density, (μ/ρ) is the mass attenuation coefficient and ρ.l is the mass thickness. The measurement unit used for the mass attenuation coefficient cm^{2}g^{-1}.For intermediate energies the Compton scattering dominates and different absorbers have approximately equal mass attenuation coefficients. This is due to the fact that cross section of Compton scattering is proportional to the Z (atomic number) and therefore the coefficient is proportional to the material density ρ. At small values of gamma ray energy or at high values of gamma ray energy, where the coefficient is proportional to higher powers of the atomic number Z (for photoelectric effect σ_{f} ~ Z^{5}; for pair production σ_{p} ~ Z^{2}), the attenuation coefficient μ is not a constant.

### Example:

How much water schielding do you require, if you want to reduce the intensity of a 500 keV **monoenergetic** gamma ray beam (**narrow beam**) to **1%** of its incident intensity? The half value layer for 500 keV gamma rays in water is 7.15 cm and the linear attenuation coefficient for 500 keV gamma rays in water is 0.097 cm^{-1}.The question is quite simple and can be described by following equation:If the half value layer for water is 7.15 cm, the linear attenuation coefficient is:Now we can use the exponential attenuation equation:thereforeSo the required thickness of water is about **47.5 cm**. This is relatively large thickness and it is caused by small atomic numbers of hydrogen and oxygen. If we calculate the same problem for **lead (Pb)**, we obtain the thickness **x=2.8cm**.

### Linear Attenuation Coefficients

**Table of Linear Attenuation Coefficients** (in cm-1) for a different materials at gamma ray energies of 100, 200 and 500 keV.

Absorber | 100 keV | 200 keV | 500 keV |

Air | 0.000195/cm | 0.000159/cm | 0.000112/cm |

Water | 0.167/cm | 0.136/cm | 0.097/cm |

Carbon | 0.335/cm | 0.274/cm | 0.196/cm |

Aluminium | 0.435/cm | 0.324/cm | 0.227/cm |

Iron | 2.72/cm | 1.09/cm | 0.655/cm |

Copper | 3.8/cm | 1.309/cm | 0.73/cm |

Lead | 59.7/cm | 10.15/cm | 1.64/cm |

### Half Value Layers

**Table of Half Value Layers** (in cm) for a different materials at gamma ray energies of 100, 200 and 500 keV.

Absorber | 100 keV | 200 keV | 500 keV |

Air | 3555 cm | 4359 cm | 6189 cm |

Water | 4.15 cm | 5.1 cm | 7.15 cm |

Carbon | 2.07 cm | 2.53 cm | 3.54 cm |

Aluminium | 1.59 cm | 2.14 cm | 3.05 cm |

Iron | 0.26 cm | 0.64 cm | 1.06 cm |

Copper | 0.18 cm | 0.53 cm | 0.95 cm |

Lead | 0.012 cm | 0.068 cm | 0.42 cm |

### Validity of Exponential Law

The exponential law will always describe the attenuation of the primary radiation by matter. If secondary particles are producedor if the primary radiation changes its energy or direction, then the effective attenuation will be much less. The radiation will penetrate more deeply into matter than ispredicted by the exponential law alone. The processmust be taken into account whenevaluating the effect of radiation shielding.