## Basic principles of radiation protection

In radiation protection there are three ways how to protect people from identified radiation sources:

• Limiting Time. The amount of radiation exposure depends directly (linearly) on the time people spend near the source of radiation. The dose can be reduced by limiting exposure time.
• Distance. The amount of radiation exposure depends on the distance from the source of radiation. Similarly to a heat from a fire, if you are too close, the intensity of heat radiation is high and you can get burned. If you are at the right distance, you can withstand there without any problems and moreover it is comfortable. If you are too far from heat source, the insufficiency of heat can also hurt you. This analogy, in a certain sense, can be applied to radiation also from radiation sources.
• Shielding. Finally, if the source is too intensive and time or distance do not provide sufficient radiation protection, the shielding must be used. Radiation shielding usually consist of barriers of lead, concrete or water. There are many many materials, which can be used for radiation shielding, but there are many many situations in radiation protection. It highly depends on the type of radiation to be shielded, its energy and many other parametres. For example, even depleted uranium can be used as a good protection from gamma radiation, but on the other hand uranium is absolutely inappropriate shielding of neutron radiation.

Description of Gamma Rays

Gamma rays, also known as gamma radiation, refers to electromagnetic radiation (no rest mass, no charge) of a very high energies. Gamma rays are high-energy photons with very short wavelengths and thus very high frequency. Since the gamma rays are in substance only a very high-energy photons, they are very penetrating matter and are thus biologically hazardous. Gamma rays can travel thousands of feet in air and can easily pass through the human body.Gamma rays are emitted by unstable nuclei in their transition from a high energy state to a lower state known as gamma decay. In most practical laboratory sources, the excited nuclear states are created in the decay of a parent radionuclide, therefore a gamma decay typically accompanies other forms of decay, such as alpha or beta decay.Radiation and also gamma rays are all around us. In, around, and above the world we live in. It is a part of our natural world that has been here since the birth of our planet. Natural sources of gamma rays on Earth are inter alia gamma rays from naturally occurring radionuclides, particularly potassium-40.  Potassium-40 is a radioactive isotope of potassium which has a very long half-life of 1.251×109 years (comparable to the age of Earth). This isotope can be found in soil, water also in meat and bananas. This is not the only example of natural source of gamma rays.

## Characteristics of Gamma Rays / Radiation

Key features of gamma rays are summarized in following few points:

• Gamma rays are high-energy photons (about 10 000 times as much energy as the visible photons), the same photons as the photons forming the visible range of the electromagnetic spectrum – light.
• Photons (gamma rays and X-rays) can ionize atoms directly (despite they are electrically neutral) through the Photoelectric effect and the Compton effect, but secondary (indirect) ionization is much more significant.
• Gamma rays ionize matter primarily via indirect ionization.
• Although a large number of possible interactions are known, there are three key interaction mechanisms  with matter.
• Gamma rays travel at the speed of light and they can travel thousands of meters in air before spending their energy.
• Since the gamma radiation is very penetrating matter, it must be shielded by very dense materials, such as lead or uranium.
• 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.
• Gamma rays frequently accompany the emission of alpha and beta radiation.
Image: The relative importance of various processes of gamma radiation interactions with matter.

In short, effective shielding of gamma radiation is in most cases based on use of materials with two following material properties:

• high-density of material.
• high atomic number of material  (high Z materials)

However, low-density materials and low Z materials can be compensated with increased thickness, which is as significant as density and atomic number in shielding applications.

A lead is widely used as a gamma shield.  Major advantage of lead shield is in its compactness due to its higher density. On the other hand depleted uranium is much more effective due to its higher Z.  Depleted uranium is used for shielding in portable gamma ray sources.

In nuclear power plants shielding of a reactor core can be provided by materials of reactor pressure vessel, reactor internals (neutron reflector). Also heavy concrete is usually used to shield both neutrons and gamma radiation.

Although water is neither high density nor high Z material, it is commonly used as gamma shields. Water provides a radiation shielding of fuel assemblies in a spent fuel pool during storage or during transports from and into the reactor core.

In general, the gamma radiation shielding is more complex and difficult than the alpha or beta radiation shielding. In order to understand comprehensively the way how a gamma ray loses its initial energy, how can be attenuated and how can be shielded we must have detailed knowledge of the its interaction mechanisms.

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

## Gamma Rays Attenuetion

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 = σf + σC + σ

• σf – Photoelectric effect
• σC – Compton scattering
• σp – 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=I0.e-μx

, where I is intensity after attenuation,  Io 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.

## 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

The half value layer expresses the thickness of absorbing material needed for reduction of the incident radiation intensity by a factor of two. With half value layer it is easy to perform simple calculations.
Source: www.nde-ed.org

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

## 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=I0.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 cm2g-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 σf ~ Z5; for pair production σp ~ Z2), the attenuation coefficient μ is not a constant.

## 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.

Calculation of Shielded Dose Rate in Sieverts from Contaminated Surface

## Calculation of Shielded Dose Rate in Sieverts from Contaminated Surface

Assume a surface, which is contamined by 1.0 Ci of 137Cs. Assume that this contaminant can be aproximated by the point isotropic source which contains 1.0 Ci of 137Cs, which has a half-life of 30.2 years. Note that the relationship between half-life and the amount of a radionuclide required to give an activity of one curie is shown below. This amount of material can be calculated using λ, which is the decay constant of certain nuclide:

About 94.6 percent decays by beta emission to a metastable nuclear isomer of barium: barium-137m. The main photon peak of Ba-137m is 662 keV. For this calculation, assume that all decays go through this channel.

Calculate the primary photon dose rate, in sieverts per hour (Sv.h-1), at the outer surface of a 5 cm thick lead shield. Then calculate the equivalent and effective dose rates for two cases.

1. Assume that this external radiation field penetrates uniformly through the whole body. That means: Calculate the effective whole-body dose rate.
2. Assume that this external radiation field penetrates only lungs and the other organs are completely shielded. That means: Calculate the effective dose rate.

Note that, primary photon dose rate neglects all secondary particles. Assume that the effective distance of the source from the dose point is 10 cm. We shall also assume that the dose point is soft tissue and it can reasonably be simulated by water and we use the mass energy absorption coefficient for water.

Solution:

The primary photon dose rate is attenuated exponentially, and the dose rate from primary photons, taking account of the shield, is given by:

As can be seen, we do not account for the buildup of secondary radiation. If secondary particles are produced or if the primary radiation changes its energy or direction, then the effective attenuation will be much less.  This assumption generally underestimates the true dose rate, especially for thick shields and when the dose point is close to the shield surface, but this assumption simplifies all calculations. For this case the true dose rate (with the buildup of secondary radiation) will be more than two times higher.

To calculate the absorbed dose rate, we have to use in the formula:

• k = 5.76 x 10-7
• S = 3.7 x 1010 s-1
• E = 0.662 MeV
• μt/ρ =  0.0326 cm2/g (values are available at NIST)
• μ =  1.289 cm-1 (values are available at NIST)
• D = 5 cm
• r = 10 cm

Result:

The resulting absorbed dose rate in grays per hour is then:

Since the radiation weighting factor for gamma rays is equal to one and we have assumed the uniform radiation field (the tissue weighting factor is also equal to unity), we can directly calculate the equivalent dose rate and the effective dose rate (E = HT) from the absorbed dose rate as:

In this case we assume a partial irradiation of lungs only. Thus, we have to use the tissue weighting factor, which is equal to wT = 0.12. The radiation weighting factor for gamma rays is equal to one. As a result, we can calculate the effective dose rate as:

Note that, if one part of the body (e.g.,the lungs) receives a radiation dose, it represents a risk for a particularly damaging effect (e.g., lung cancer). If the same dose is given to another organ it represents a different risk factor.

If we want to account for the buildup of secondary radiation, then we have to include the buildup factor. The extended formula for the dose rate is then:

## Buildup Factors for Gamma Rays Shielding

The buildup factor is a correction factor that considers the influence of the scattered radiation plus any secondary particles in the medium during shielding calculations. If we want to account for the buildup of secondary radiation, then we have to include the buildup factor. The buildup factor is then a multiplicative factor which accounts for the response to the uncollided photons so as to include the contribution of the scattered photons. Thus, the buildup factor can be obtained as a ratio of the total dose to the response for uncollided dose.

The extended formula for the dose rate calculation is:

The ANSI/ANS-6.4.3-1991 Gamma-Ray Attenuation Coefficients and Buildup Factors for Engineering Materials Standard, contains derived gamma-ray attenuation coefficients and buildup factors for selected engineering materials and elements for use in shielding calculations (ANSI/ANS-6.1.1, 1991).

## See previous:

Shielding of Positrons