**X-rays**, also known as **X-radiation**, refers to electromagnetic radiation (no rest mass, no charge) of high energies. X-rays are high-energy photons with short wavelengths and thus very high frequency. The radiation frequency is key parameter of all photons, because it determines the energy of a photon. Photons are categorized according to the energies from low-energy radio waves and infrared radiation, through visible light, to high-energy X-rays and gamma rays.

Most X-rays have a wavelength ranging from 0.01 to 10 nanometers (3×10^{16} Hz to 3×10^{19} Hz), corresponding to energies in the range 100 eV to 100 keV. X-ray wavelengths are shorter than those of UV rays and typically longer than those of gamma rays. The distinction between X-rays and gamma rays is not so simple and has changed in recent decades. According to the currently valid definition, **X-rays are emitted by electrons** outside the nucleus, while **gamma rays are emitted by the nucleus**.

## X-ray Attenuation

As the **high-energy photons** pass through material, their energy is decreasing. This is known as **attenuation**. The attenuation theory is valid for **X-rays** and **gamma rays** as well. It turns out that higher energy photons (hard X-rays) travel through tissue more easily than low-energy photons (i.e., the higher energy photons are less likely to interact with matter). Much of this effect is related to the photoelectric effect. The probability of photoelectric absorption is approximately proportional to (Z/E)^{3}, where Z is the atomic number of the tissue atom and E is the photon energy. As E gets larger, the likelihood of interaction drops rapidly. For higher energies, Compton scattering becomes dominant. Compton scattering is about constant for different energies although it slowly decreases at higher energies.

### Exponential Attenuation

Assume that **monoenergetic X-rays** are collimated into a **narrow beam** and the detector behind the material only detects the X-rays that passed through that material without any kind of interaction with this material, then the dependence should be simple **exponential attenuation of X-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)}**

## Linear Attenuation Coefficient – X-rays

The attenuation of X-rays 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 are air, water and a different elements from carbon (*Z*=6) through to lead (*Z*=82) and their linear attenuation coefficients are given for two X-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 X-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 X-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 X-rays. For example from 0.26 cm for iron at 100 keV to about 0.64 cm at 200 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 X-rays** 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 X-ray energy, where the coefficient is proportional to higher powers of the atomic number Z (for photoelectric effect σ_{f} ~ Z^{3}), the attenuation coefficient μ is not a constant.

See also calculator: Gamma activity to dose rate (with/without shield)

See also XCOM – photon cross-section DB: XCOM: Photon Cross Sections Database

## Example:

How much water schielding do you require, if you want to reduce the intensity of a 100 keV **monoenergetic** X-ray beam (**narrow beam**) to **1%** of its incident intensity? The half value layer for 100 keV X-rays in water is 4.15 cm and the linear attenuation coefficient for 100 keV X-rays in water is 0.167 cm^{-1}. The problem is quite simple and can be described by following equation:

If the half value layer for water is 4.15 cm, the linear attenuation coefficient is:Now we can use the exponential attenuation equation:

So the required thickness of water is about **27.58 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=0.077 cm**.

**Linear Attenuation Coefficients**

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

Absorber | 100 keV | 200 keV | 500 keV |

Air | 0.0195/m | 0.0159/m | 0.0112/m |

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 photon 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 process must be taken into account when evaluating the effect of radiation shielding.