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. Potasium-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.
Before Albert Einstein, notably the German physicist Max Planck had prepared the way for the concept by explaining that objects that emit and absorb light do so only in amounts of energy that are quantized, that means every change of energy can occur only by certain particular discrete amounts and the object cannot change energy in any arbitrary way. The concept of modern photon came into general use after the physicist Arthur H. Compton demonstrated (1923) the corpuscular nature of X-rays. This was the validation that Einstein’s hypothesis that light itself is quantized.
The term photon comes from Greek phōtos, “light” and a photon is usually denoted by the symbol γ (gamma). The photons are also symbolized by hν (in chemistry and optical engineering), where h is Planck’s constant and the Greek letter ν (nu) is the photon’s 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.
Photons are gauge bosons for electromagnetism, having no electric charge or rest mass and one unit of spin. Common to all photons is the speed of light, the universal constant of physics. In empty space, the photon moves at c (the speed of light – 299 792 458 metres per second).
- The photoelectric effect dominates at low-energies of gamma rays.
- The photoelectric effect leads to the emission of photoelectrons from matter when light (photons) shines upon them.
- The maximum energy an electron can receive in any one interaction is hν.
- Electrons are only emitted by the photoelectric effect if photon reaches or exceeds a threshold energy.
- A free electron (e.g. from atomic cloud) cannot absorb entire energy of the incident photon. This is a result of the need to conserve both momentum and energy.
- The cross-section for the emission of n=1 (K-shell) photoelectrons is higher than that of n=2 (L-shell) photoelectrons. This is a result of the need to conserve momentum and energy.
Definition of Photoelectric effect
Therefore photoelectrons are only emitted by the photoelectric effect if photon reaches or exceeds a threshold energy – the binding energy of the electron – the work function of the material. For gamma rays with energies of more than hundreds keV, the photoelectron carries off the majority of the incident photon energy – hν.
Following a photoelectric interaction, an ionized absorber atom is created with a vacancy in one of its bound shells. This vacancy is will be quickly filled by an electron from a shell with a lower binding energy (other shells) or through capture of a free electron from the material. The rearrangement of electrons from other shells creates another vacancy, which, in turn, is filled by an electron from an even lower binding energy shell. Therefore a cascade of more characteristic X-rays can be also generated. The probability of characteristic x-ray emission decreases as the atomic number of the absorber decreases. Sometimes , the emission of an Auger electron occurs.
Cross-Sections of Photoelectric Effect
τ(photoelectric) = constant x ZN/E3.5
where Z is the atomic number, the exponent n varies between 4 and 5. E is the energy of the incident photon. The proportionality to higher powers of the atomic number Z is the main reason for using of high Z materials, such as lead or depleted uranium in gamma ray shields.
Although the probability of the photoelectric absorption of gamma photon decreases, in general, with increasing photon energy, there are sharp discontinuities in the cross-section curve. These are called “absoption edges” and they correspond to the binding energies of electrons from atom’s bound shells. For photons with the energy just above the edge, the photon energy is just sufficient to undergo the photoelectric interaction with electron from bound shell, let say K-shell. The probability of such interaction is just above this edge much greater than that of photons of energy slightly below this edge. For gamma photons below this edge the interaction with electron from K-shell in energetically impossible and therefore the probability drops abruptly. These edges occur also at binding energies of electrons from other shells (L, M, N …..).
Compton Scattering – Cross-Sections
where ε = E0/mec2 and r0 is the “classical radius of the electron” equal to about 2.8 x 10-13 cm. The formula gives the probability of scattering a photon into the solid angle element dΩ = 2π sin Θ dΘ when the incident energy is E0.
Inverse Compton Scattering
Positron-Electron Pair Production
In order for electron-positron pair production to occur, the electromagnetic energy of the photon must be above a threshold energy, which is equivalent to the rest mass of two electrons. The threshold energy (the total rest mass of produced particles) for electron-positron pair production is equal to 1.02MeV (2 x 0.511MeV) because the rest mass of a single electron is equivalent to 0.511MeV of energy.
If the original photon’s energy is greater than 1.02MeV, any energy above 1.02MeV is according to the conservation law split between the kinetic energy of motion of the two particles.
The presence of an electric field of a heavy atom such as lead or uranium is essential in order to satisfy conservation of momentum and energy. In order to satisfy both conservation of momentum and energy, the atomic nucleus must receive some momentum. Therefore a photon pair production in free space cannot occur.
Moreover, the positron is the anti-particle of the electron, so when a positron comes to rest, it interacts with another electron, resulting in the annihilation of the both particles and the complete conversion of their rest mass back to pure energy (according to the E=mc2 formula) in the form of two oppositely directed 0.511 MeV gamma rays (photons). The pair production phenomenon is therefore connected with creation and destruction of matter in one reaction.
Positron-Electron Pair Production – Cross-Section
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 = σf + σC + σp
- 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.
, 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.
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:
, 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.
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:
So 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
|Absorber||100 keV||200 keV||500 keV|
Half Value Layers
|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 produced
or 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 is
predicted by the exponential law alone. The processmust be taken into account when
evaluating the effect of radiation shielding.