{"id":12749,"date":"2016-09-30T16:19:00","date_gmt":"2016-09-30T16:19:00","guid":{"rendered":"http:\/\/sitepourvtc.com\/?page_id=12749"},"modified":"2022-10-18T10:58:11","modified_gmt":"2022-10-18T10:58:11","slug":"shielding-gamma-radiation","status":"publish","type":"page","link":"https:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/shielding-of-ionizing-radiation\/shielding-gamma-radiation\/","title":{"rendered":"Shielding of Gamma Radiation"},"content":{"rendered":"<div   id="8n9s8rpp6h"    class=\"su-quote su-quote-style-default\"><div   id="8n9s8rpp6h"    class=\"su-quote-inner su-u-clearfix su-u-trim\"><strong>Shielding of ionizing radiation<\/strong>\u00a0simply means having <strong>some material between the source of radiation and you<\/strong> (or some device) that will <strong>absorb the <a title=\"Radiation\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/\">radiation<\/a><\/strong>. Lead blocks or sheets are widely used as a <strong>gamma rays<\/strong> shield.\u00a0 The major advantage of the lead shield is its compactness due to its higher density.<\/div><\/div>\n<p>In short, effective shielding of gamma radiation is in most cases based on the use of materials with two following material properties:<\/p>\n<ul>\n<li><strong><span style=\"font-weight: 400;\">high-density of material<\/span><\/strong><\/li>\n<li><strong><span style=\"font-weight: 400;\">the high atomic number of material \u00a0(high Z materials)<\/span><\/strong><\/li>\n<\/ul>\n<p><strong><span style=\"font-weight: 400;\">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.\u00a0\u00a0<\/span><\/strong><\/p>\n<p><strong><span style=\"font-weight: 400;\">A lead is widely used as a gamma shield.\u00a0 The major advantage of the lead shield is its compactness due to its higher density. On the other hand, <a title=\"Depleted Uranium\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-fuel\/uranium\/depleted-uranium\/\">depleted uranium<\/a> is much more effective due to its higher Z.\u00a0\u00a0Depleted uranium shields in portable gamma-ray sources.\u00a0<\/span><\/strong><\/p>\n<p><strong><span style=\"font-weight: 400;\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2021\/03\/shielding-of-gamma-rays.png\"><img loading=\"lazy\" class=\"wp-image-31316 size-large aligncenter\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2021\/03\/shielding-of-gamma-rays-1024x268.png\" alt=\"shielding of gamma rays\" width=\"1024\" height=\"268\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2021\/03\/shielding-of-gamma-rays-1024x268.png 1024w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2021\/03\/shielding-of-gamma-rays-300x78.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2021\/03\/shielding-of-gamma-rays-768x201.png 768w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2021\/03\/shielding-of-gamma-rays.png 1158w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><\/a><\/span><\/strong><\/p>\n<p>In <a title=\"Nuclear Power Plant\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/\">nuclear power plants<\/a>,\u00a0shielding of a <a title=\"Reactor Core\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor-core\/\">reactor core<\/a> can be provided by materials of reactor pressure vessel, reactor internals (<a title=\"Neutron Reflector\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor\/neutron-reflector\/\">neutron reflector<\/a>). Also, heavy concrete is usually used to shield both <a title=\"Shielding of Neutron Radiation\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/shielding-neutron-radiation\/\">neutrons<\/a> and gamma radiation.<\/p>\n<p>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 transports from and into the <a title=\"Reactor Core\" href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor-core\/\">reactor core<\/a>.<\/p>\n<p>In general, gamma radiation shielding is more complex and difficult than <a title=\"Shielding of Alpha Radiation\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/shielding-of-ionizing-radiation\/shielding-of-alpha-radiation\/\">alpha<\/a> or <a title=\"Shielding of Beta Radiation \u2013 Electrons\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/shielding-of-ionizing-radiation\/shielding-beta-radiation\/\">beta radiation shielding<\/a>. To comprehensively understand how a gamma-ray loses its initial energy, how it can be attenuated, and how it can be shielded, we must have detailed knowledge of its interaction mechanisms.<\/p>\n<p>See also more theory: <a title=\"Interaction of Gamma Radiation with Matter\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/\">Interaction of Gamma Radiation with Matter<\/a><\/p>\n<p>See also calculator: <a title=\"Rad Pro Calculator\" href=\"http:\/\/www.radprocalculator.com\/Gamma.aspx\">Gamma activity to dose rate (with\/without shield)<\/a><\/p>\n<p>See also XCOM &#8211; photon cross-section DB: <a title=\"XCOM Database\" href=\"http:\/\/www.nist.gov\/pml\/data\/xcom\/\">XCOM: Photon Cross Sections Database<\/a><\/p>\n<h2>Characteristics of Gamma Rays \/ Radiation<\/h2>\n<p><strong>Key features of <a title=\"Gamma Rays \/ Gamma Radiation\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/gamma-ray\/\">gamma rays<\/a><\/strong> are summarized in the following few points:<\/p>\n<ul>\n<li>\n<figure id=\"attachment_11684\" aria-describedby=\"caption-attachment-11684\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/total_photon_attenuation.png\"><img loading=\"lazy\" class=\"size-medium wp-image-11684\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/total_photon_attenuation-300x217.png\" alt=\"Attenuation coefficients.\" width=\"300\" height=\"217\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/total_photon_attenuation-300x217.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/total_photon_attenuation.png 900w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11684\" class=\"wp-caption-text\">Total photon cross-sections.<br \/>Source: Wikimedia Commons<\/figcaption><\/figure>\n<p>Gamma rays are <strong>high-energy photons<\/strong> (about 10 000 times as much energy as the visible photons), the same <a title=\"Photon \u2013 Fundamental Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/\">photons<\/a> as the photons forming the visible range of the electromagnetic spectrum &#8211; light.<\/li>\n<li>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.<\/li>\n<li>Gamma rays ionize matter primarily via <strong>indirect ionization<\/strong>.<\/li>\n<li>Although many possible interactions are known, there are three key interaction mechanisms with the matter.\n<ul>\n<li><a title=\"Photoelectric Effect\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/photoelectric-effect\/\"><strong>Photoelectric effect<\/strong><\/a><\/li>\n<li><a title=\"Compton Scattering\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/compton-scattering\/\"><strong>Compton scattering<\/strong><\/a><\/li>\n<li><a title=\"Positron-Electron Pair Production\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/pair-production\/\"><strong>Pair production<\/strong><\/a><\/li>\n<\/ul>\n<\/li>\n<li>Gamma rays travel <strong>at the speed of light,<\/strong> and they can travel thousands of meters in the air before spending their energy.<\/li>\n<li>Since gamma radiation is very penetrating, it must be shielded by very dense materials, such as lead or uranium.<\/li>\n<li>The distinction between X-rays and gamma rays is not so simple and has changed in recent decades. \u00a0According to the currently valid definition, X-rays are emitted by electrons outside the nucleus, while <b>the nucleus emits gamma rays<\/b>.<\/li>\n<li>Gamma rays frequently <strong>accompany the emission<\/strong> of <a title=\"Alpha Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/alpha-particle\/\">alpha<\/a> and <a title=\"Beta Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/beta-particle\/\">beta radiation<\/a>.<\/li>\n<\/ul>\n<div   id="8n9s8rpp6h"    class=\"collapsible-container\">\n<h2>Gamma Rays Attenuetion<\/h2>\n<p>The total cross-section of the <a title=\"Interaction of Gamma Radiation with Matter\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/\">interaction of gamma rays<\/a> with an atom is equal to the sum of all three mentioned partial cross-sections:<strong>\u03c3 = \u03c3<sub>f<\/sub> + \u03c3<sub>C<\/sub> + \u03c3<sub>p\u00a0<\/sub><\/strong><\/p>\n<ul>\n<li><strong>\u03c3<sub>f<\/sub> &#8211; Photoelectric effect<\/strong><\/li>\n<\/ul>\n<ul>\n<li><strong>\u03c3<sub>C<\/sub> &#8211; Compton scattering<\/strong><\/li>\n<\/ul>\n<ul>\n<li><strong>\u03c3<sub>p<\/sub> &#8211; Pair production<\/strong><\/li>\n<\/ul>\n<p>One of the three partial cross-sections may become much larger than the other two depending on the gamma-ray energy and the absorber material. At small values of gamma-ray energy, the<a title=\"Photoelectric Effect\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/photoelectric-effect\/\"><strong> photoelectric effect<\/strong><\/a> dominates. <a title=\"Compton Scattering\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/compton-scattering\/\"><strong>Compton scattering<\/strong><\/a> 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, <a title=\"Positron-Electron Pair Production\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/pair-production\/\"><strong>electron-positron pair production<\/strong><\/a> dominates at high energies.<\/p>\n<p>Based on the definition of interaction cross-section, the dependence of gamma rays intensity on the thickness of absorber material can be derived. If <strong>monoenergetic gamma rays<\/strong> are collimated into a <strong>narrow beam<\/strong> and if the detector behind the material only detects the gamma rays that passed through that material without any interaction, then the dependence should be simple <strong>exponential attenuation gamma rays<\/strong>. Each interaction removes the photon from the beam either by absorption or scattering away from the detector direction. Therefore the interactions can be characterized by a fixed probability of occurrence per unit path length in the absorber. The sum of these probabilities is called the <strong>linear attenuation coefficient<\/strong>:<\/p>\n<p style=\"text-align: center;\"><strong>\u03bc = \u03c4<sub>(photoelectric)<\/sub> + \u00a0\u03c3<sub>(Compton)<\/sub> + \u03ba<sub>(pair)<\/sub><\/strong><\/p>\n<figure id=\"attachment_11791\" aria-describedby=\"caption-attachment-11791\" style=\"width: 560px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/attenuation.png\"><img loading=\"lazy\" class=\"size-full wp-image-11791\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/attenuation.png\" alt=\"Gamma rays attuenuation\" width=\"570\" height=\"357\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/attenuation.png 570w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/attenuation-300x188.png 300w\" sizes=\"(max-width: 570px) 100vw, 570px\" \/><\/a><figcaption id=\"caption-attachment-11791\" class=\"wp-caption-text\">The relative importance of various processes of gamma radiation interaction with matter.<\/figcaption><\/figure>\n<h3>Linear Attenuation Coefficient<\/h3>\n<p>The following equation can then describe the attenuation of gamma radiation.<\/p>\n<p style=\"text-align: center;\"><strong>I=I<sub>0<\/sub>.e<sup>-\u03bcx<\/sup><\/strong><\/p>\n<p>where I is intensity after attenuation, \u00a0I<sub>o<\/sub> is incident intensity, \u00a0\u03bc is the linear attenuation coefficient (cm<sup>-1<\/sup>), and the physical thickness of absorber (cm).<\/p>\n<figure id=\"attachment_11792\" aria-describedby=\"caption-attachment-11792\" style=\"width: 290px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/exponential-attenuation.png\"><img loading=\"lazy\" class=\"size-medium wp-image-11792\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/exponential-attenuation-300x217.png\" alt=\"Attenuation\" width=\"300\" height=\"217\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/exponential-attenuation-300x217.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/exponential-attenuation.png 305w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11792\" class=\"wp-caption-text\">Dependence of gamma radiation intensity on absorber thickness<\/figcaption><\/figure>\n<p>The materials listed in the table beside are air, water, and different elements from carbon (<i>Z<\/i>=6) to lead (<i>Z<\/i>=82). Their linear attenuation coefficients are given for three gamma-ray energies. There are two main features of the linear attenuation coefficient:<\/p>\n<ul>\n<li>The linear attenuation coefficient increases as the atomic number of the absorber increases.<\/li>\n<li>The linear attenuation coefficient for all materials decreases with the energy of the gamma rays.<\/li>\n<\/ul>\n<h2>Half Value Layer<\/h2>\n<p>The half-value layer expresses the thickness of absorbing material needed to reduce the incident radiation intensity by a <strong>factor of two<\/strong>. There are two main features of the half-value layer:<\/p>\n<ul>\n<li>The <b>half-value layer<\/b> 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.<\/li>\n<li>The <b>half-value layer<\/b> 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.<\/li>\n<\/ul>\n<h3>Example<\/h3>\n<p><span style=\"color: #333333;\">How much water shielding do you require if you want to reduce the intensity of a 500 keV <strong>monoenergetic<\/strong> gamma-ray beam (<strong>narrow beam<\/strong>) to <strong>1%<\/strong> 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<sup>-1<\/sup>. <\/span><span style=\"color: #333333;\">The question is quite simple and can be described by the following equation:<\/span><span style=\"color: #333333;\"><img class=\"mathtex-equation-editor aligncenter\" src=\"http:\/\/chart.apis.google.com\/chart?cht=tx&amp;chl=I(x)%3D%5Cfrac%7BI_%7B0%7D%7D%7B100%7D%2C%5C%3B%5C%3B%20when%5C%3B%20x%20%3D%3F%20\" alt=\"I(x)=\\frac{I_{0}}{100},\\;\\; when\\; x =? \" align=\"absmiddle\" \/><\/span><span style=\"color: #333333;\">If the half-value layer for water is 7.15 cm, the linear attenuation coefficient is:<\/span><span style=\"color: #333333;\"><img class=\"mathtex-equation-editor aligncenter\" src=\"http:\/\/chart.apis.google.com\/chart?cht=tx&amp;chl=%5Cmu%3D%5Cfrac%7Bln2%7D%7B7.15%7D%3D0.097cm%5E%7B-1%7D\" alt=\"\\mu=\\frac{ln2}{7.15}=0.097cm^{-1}\" align=\"absmiddle\" \/><\/span><span style=\"color: #333333;\">Now we can use the exponential attenuation equation:<\/span><span style=\"color: #333333;\"><img class=\"mathtex-equation-editor aligncenter\" src=\"http:\/\/chart.apis.google.com\/chart?cht=tx&amp;chl=I(x)%3DI_0%5C%3Bexp%5C%3B(-%5Cmu%20x)\" alt=\"I(x)=I_0\\;exp\\;(-\\mu x)\" align=\"absmiddle\" \/><\/span><span style=\"color: #333333;\"><img class=\"mathtex-equation-editor aligncenter\" src=\"http:\/\/chart.apis.google.com\/chart?cht=tx&amp;chl=%5Cfrac%7BI_0%7D%7B100%7D%3DI_0%5C%3Bexp%5C%3B(-0.097%20x)\" alt=\"\\frac{I_0}{100}=I_0\\;exp\\;(-0.097 x)\" align=\"absmiddle\" \/><\/span><span style=\"color: #333333;\">therefore<\/span><span style=\"color: #333333;\"><img class=\"mathtex-equation-editor aligncenter\" src=\"http:\/\/chart.apis.google.com\/chart?cht=tx&amp;chl=%5Cfrac%7B1%7D%7B100%7D%3D%5C%3Bexp%5C%3B(-0.097%20x)\" alt=\"\\frac{1}{100}=\\;exp\\;(-0.097 x)\" align=\"absmiddle\" \/><\/span><span style=\"color: #333333;\"><img class=\"mathtex-equation-editor aligncenter\" src=\"http:\/\/chart.apis.google.com\/chart?cht=tx&amp;chl=ln%5Cfrac%7B1%7D%7B100%7D%3D-ln%5C%3B100%3D-0.097%20x\" alt=\"ln\\frac{1}{100}=-ln\\;100=-0.097 x\" align=\"absmiddle\" \/><\/span><span style=\"color: #333333;\"><img class=\"mathtex-equation-editor aligncenter\" src=\"http:\/\/chart.apis.google.com\/chart?cht=tx&amp;chl=x%3D%5Cfrac%7Bln100%7D%7B%7B0.097%7D%7D%3D47.47%5C%3Bcm\" alt=\"x=\\frac{ln100}{{0.097}}=47.47\\;cm\" align=\"absmiddle\" \/><\/span><span style=\"color: #333333;\">So the required thickness of water is about <strong>47.5 cm<\/strong>.\u00a0\u00a0This is a relatively large thickness, and it is caused by small atomic numbers of hydrogen and oxygen. If we calculate the same problem for <strong>lead (Pb)<\/strong>, we obtain the thickness <strong>x=2.8cm<\/strong>.<\/span><\/p>\n<p>Linear Attenuation Coefficients<\/p>\n<p><strong>Table of Linear Attenuation Coefficients<\/strong> (in cm-1) for different materials at gamma-ray energies of 100, 200, and 500 keV.<\/p>\n<table rules=\"rows\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\">Absorber<\/td>\n<td style=\"text-align: center;\">100 keV<\/td>\n<td style=\"text-align: center;\">200 keV<\/td>\n<td style=\"text-align: center;\">500 keV<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Air<\/td>\n<td style=\"text-align: center;\">0.0195\/m<\/td>\n<td style=\"text-align: center;\">0.0159\/m<\/td>\n<td style=\"text-align: center;\">0.0112\/m<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Water<\/td>\n<td style=\"text-align: center;\">0.167\/cm<\/td>\n<td style=\"text-align: center;\">0.136\/cm<\/td>\n<td style=\"text-align: center;\">0.097\/cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Carbon<\/td>\n<td style=\"text-align: center;\">0.335\/cm<\/td>\n<td style=\"text-align: center;\">0.274\/cm<\/td>\n<td style=\"text-align: center;\">0.196\/cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Aluminium<\/td>\n<td style=\"text-align: center;\">0.435\/cm<\/td>\n<td style=\"text-align: center;\">0.324\/cm<\/td>\n<td style=\"text-align: center;\">0.227\/cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Iron<\/td>\n<td style=\"text-align: center;\">2.72\/cm<\/td>\n<td style=\"text-align: center;\">1.09\/cm<\/td>\n<td style=\"text-align: center;\">0.655\/cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Copper<\/td>\n<td style=\"text-align: center;\">3.8\/cm<\/td>\n<td style=\"text-align: center;\">1.309\/cm<\/td>\n<td style=\"text-align: center;\">0.73\/cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Lead<\/td>\n<td style=\"text-align: center;\">59.7\/cm<\/td>\n<td style=\"text-align: center;\">10.15\/cm<\/td>\n<td style=\"text-align: center;\">1.64\/cm<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Half Value Layers<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/09\/half-value-layers.png\"><img loading=\"lazy\" class=\"aligncenter size-medium wp-image-12757\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/09\/half-value-layers-291x300.png\" alt=\"half value layer\" width=\"291\" height=\"300\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/09\/half-value-layers-291x300.png 291w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/09\/half-value-layers.png 306w\" sizes=\"(max-width: 291px) 100vw, 291px\" \/><\/a><div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:15px\"><\/div>\n<p>The half-value layer expresses the thickness of absorbing material needed to reduce the incident radiation intensity by a factor of two. With a half-value layer, it is easy to perform simple calculations.<br \/>\nSource: www.nde-ed.org<\/p>\n<div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:15px\"><\/div>\n<p><strong>Table of Half Value Layers<\/strong> (in cm) for different materials at gamma-ray energies of 100, 200, and 500 keV.<\/p>\n<table rules=\"rows\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\">Absorber<\/td>\n<td style=\"text-align: center;\">100 keV<\/td>\n<td style=\"text-align: center;\">200 keV<\/td>\n<td style=\"text-align: center;\">500 keV<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Air<\/td>\n<td style=\"text-align: center;\">3555 cm<\/td>\n<td style=\"text-align: center;\">4359 cm<\/td>\n<td style=\"text-align: center;\">6189 cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Water<\/td>\n<td style=\"text-align: center;\">4.15 cm<\/td>\n<td style=\"text-align: center;\">5.1 cm<\/td>\n<td style=\"text-align: center;\">7.15 cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Carbon<\/td>\n<td style=\"text-align: center;\">2.07 cm<\/td>\n<td style=\"text-align: center;\">2.53 cm<\/td>\n<td style=\"text-align: center;\">3.54 cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Aluminium<\/td>\n<td style=\"text-align: center;\">1.59 cm<\/td>\n<td style=\"text-align: center;\">2.14 cm<\/td>\n<td style=\"text-align: center;\">3.05 cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Iron<\/td>\n<td style=\"text-align: center;\">0.26 cm<\/td>\n<td style=\"text-align: center;\">0.64 cm<\/td>\n<td style=\"text-align: center;\">1.06 cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Copper<\/td>\n<td style=\"text-align: center;\">0.18 cm<\/td>\n<td style=\"text-align: center;\">0.53 cm<\/td>\n<td style=\"text-align: center;\">0.95 cm<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">Lead<\/td>\n<td style=\"text-align: center;\">\u00a00.012 cm<\/td>\n<td style=\"text-align: center;\">\u00a00.068 cm<\/td>\n<td style=\"text-align: center;\">\u00a00.42 cm<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<h2>Mass Attenuation Coefficient<\/h2>\n<p>When characterizing an absorbing material, we can sometimes use the mass attenuation coefficient. \u00a0<strong>The mass attenuation coefficient<\/strong> is defined as the ratio of the linear attenuation coefficient and absorber density <strong>(\u03bc\/\u03c1)<\/strong>. The following equation can then describe the attenuation of gamma radiation:<\/p>\n<p><strong>I=I<sub>0<\/sub>.e<sup>-(\u03bc\/\u03c1).\u03c1l<\/sup><\/strong><\/p>\n<p>, where \u03c1 is the material density, (\u03bc\/\u03c1) is the mass attenuation coefficient, and \u03c1.l is the mass thickness. The measurement unit was used for the mass attenuation coefficient cm<sup>2<\/sup>g<sup>-1<\/sup>.<\/p>\n<p>For intermediate energies, the Compton scattering dominates, and different absorbers have approximately equal mass attenuation coefficients. This is because the cross-section of Compton scattering is proportional to the Z (atomic number), and therefore the coefficient is proportional to the material density \u03c1. At small gamma-ray energy values or high gamma-ray energy values, the coefficient is proportional to higher powers of the atomic number Z (for photoelectric effect \u03c3<sub>f<\/sub> ~ Z<sup>5<\/sup>; for pair production \u03c3<sub>p<\/sub> ~ Z<sup>2<\/sup>), the attenuation coefficient \u03bc is not a constant.<\/p>\n<h2>Validity of Exponential Law<\/h2>\n<p>The exponential law will always describe the attenuation of the primary radiation by matter. If secondary particles are produced, or the primary radiation changes its energy or direction, the effective attenuation will be much less. The radiation will penetrate more deeply into matter than is predicted by the exponential law alone. The process must be taken into account when evaluating the effect of radiation shielding.<\/p>\n<figure id=\"attachment_11803\" aria-describedby=\"caption-attachment-11803\" style=\"width: 290px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/secondary_length.png\"><img loading=\"lazy\" class=\"size-medium wp-image-11803\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/secondary_length-300x165.png\" alt=\"Example of build-up of secondary particles. Strongly depends on character and parameters of primary particles. \" width=\"300\" height=\"165\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/secondary_length-300x165.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/secondary_length.png 698w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11803\" class=\"wp-caption-text\">Example of a buildup of secondary particles. Strongly depends on the character and parameters of primary particles.<\/figcaption><\/figure>\n<div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Calculation of Shielded Dose Rate in Sieverts from Contaminated Surface<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">\n<h2>Calculation of Shielded Dose Rate in Sieverts from Contaminated Surface<\/h2>\n<p>Assume a surface in which <strong>1.0 Ci of 137Cs contaminates.<\/strong> Assume that this <strong>contaminant<\/strong> can be approximated by the <strong>point isotropic source,<\/strong> which contains <strong>1.0 Ci of <sup>137<\/sup>Cs<\/strong>, with a <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radioactive-decay\/radioactive-decay-law\/half-life\/\">half-life<\/a> of <strong>30.2 years<\/strong>. Note that the relationship between half-life and the amount of a <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/atom-properties-of-atoms\/radionuclide-radioisotope\/\">radionuclide<\/a> required to give an activity of <a href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/radiation-protection\/units-of-radioactivity\/curie-unit-of-radioactivity\/\">one curie<\/a> is shown below. This amount of material can be calculated using \u03bb, which is the <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radioactive-decay\/radioactive-decay-law\/decay-constant\/\">decay constant<\/a> of certain nuclide:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/07\/Curie-Unit-of-Activity.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-24886\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/07\/Curie-Unit-of-Activity.png\" alt=\"Curie - Unit of Activity\" width=\"378\" height=\"61\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/07\/Curie-Unit-of-Activity.png 378w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/07\/Curie-Unit-of-Activity-300x48.png 300w\" sizes=\"(max-width: 378px) 100vw, 378px\" \/><\/a><\/p>\n<p>About 94.6 percent decays by <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radioactive-decay\/beta-decay-beta-radioactivity\/\">beta emission<\/a> to a metastable <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/atom-properties-of-atoms\/nuclides\/isomers\/\">nuclear isomer<\/a> of barium: barium-137m. The main photon peak of Ba-137m is <strong>662 keV<\/strong>. For this calculation, assume that all decays go through this channel.<\/p>\n<p><strong>Calculate the primary photon dose rate<\/strong>, in sieverts per hour (Sv.h<sup>-1<\/sup>), at the outer surface of a 5 cm thick lead shield. Then <strong>calculate the<\/strong> <strong>equivalent and effective dose rates<\/strong> for two cases.<\/p>\n<ol>\n<li>Assume that this external radiation field penetrates <strong>uniformly <\/strong>through the whole body. That means: Calculate the <strong>effective whole-body dose rate<\/strong>.<\/li>\n<li>Assume that this external radiation field penetrates <strong>only lungs<\/strong>, and the other organs are completely shielded. That means: Calculate the <strong>effective dose rate<\/strong>.<\/li>\n<\/ol>\n<p>Note that, primary photon dose rate neglects all secondary particles. Assume that the effective distance of the source from the dose point is <strong>10 cm<\/strong>. We shall also assume that the dose point is soft tissue and can reasonably be simulated by water, and we use the mass-energy absorption coefficient for water.<\/p>\n<p>See also: <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/gamma-ray-attenuation\/\">Gamma Ray Attenuation<\/a><\/p>\n<p>See also: <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/shielding-of-ionizing-radiation\/shielding-gamma-radiation\/\">Shielding of Gamma Rays<\/a><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The primary photon dose rate is <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/gamma-ray-attenuation\/\">attenuated exponentially<\/a>, and the dose rate from primary photons, taking account of the shield, is given by:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/dose-rate-calculation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-25304\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/dose-rate-calculation.png\" alt=\"dose rate calculation\" width=\"671\" height=\"307\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/dose-rate-calculation.png 671w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/dose-rate-calculation-300x137.png 300w\" sizes=\"(max-width: 671px) 100vw, 671px\" \/><\/a><\/p>\n<p>As can be seen, we do not account for the buildup of secondary radiation. If secondary particles are produced, or the primary radiation changes its energy or direction, the effective attenuation will be much less. \u00a0This 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.<\/p>\n<p>To calculate the <strong>absorbed dose rate<\/strong>, we have to use the formula:<\/p>\n<ul>\n<li>k = 5.76 x 10<sup>-7<\/sup><\/li>\n<li>S = 3.7 x 10<sup>10<\/sup> s<sup>-1<\/sup><\/li>\n<li>E = 0.662 MeV<\/li>\n<li>\u03bc<sub>t<\/sub>\/\u03c1 = <sup>\u00a0<\/sup>0.0326 cm<sup>2<\/sup>\/g (values are available at NIST)<\/li>\n<li>\u03bc = \u00a01.289 cm<sup>-1<\/sup> (values are available at NIST)<\/li>\n<li>D = 5 cm<\/li>\n<li>r = 10 cm<\/li>\n<\/ul>\n<p><strong>Result:<\/strong><\/p>\n<p>The resulting absorbed dose rate in grays per hour is then:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/absorbed-dose-rate-gray-calculation-1.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-25319\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/absorbed-dose-rate-gray-calculation-1.png\" alt=\"absorbed dose rate - gray - calculation\" width=\"551\" height=\"153\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/absorbed-dose-rate-gray-calculation-1.png 551w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/absorbed-dose-rate-gray-calculation-1-300x83.png 300w\" sizes=\"(max-width: 551px) 100vw, 551px\" \/><\/a><\/p>\n<p><strong>1) Uniform irradiation<\/strong><\/p>\n<p>Since the <a href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/radiation-protection\/equivalent-dose\/radiation-weighting-factor\/\">radiation weighting factor<\/a> for <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/gamma-ray\/\">gamma rays<\/a> 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 = H<sub>T<\/sub>) from the absorbed dose rate as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/11\/calculation-effective-dose-uniform.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-25478\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/11\/calculation-effective-dose-uniform.png\" alt=\"calculation - effective dose - uniform\" width=\"467\" height=\"182\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/11\/calculation-effective-dose-uniform.png 467w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/11\/calculation-effective-dose-uniform-300x117.png 300w\" sizes=\"(max-width: 467px) 100vw, 467px\" \/><\/a><\/p>\n<p><strong>2) Partial irradiation<\/strong><\/p>\n<p>In this case, we assume partial irradiation of lungs only. Thus, we have to use the <strong>tissue weighting factor<\/strong>, which is equal to <strong>w<sub>T<\/sub> = 0.12<\/strong>. The radiation weighting factor for gamma rays is equal to one. As a result, we can calculate the effective dose rate as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/11\/calculation-effective-dose-non-uniform.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-25480\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/11\/calculation-effective-dose-non-uniform.png\" alt=\"calculation - effective dose - non-uniform\" width=\"495\" height=\"178\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/11\/calculation-effective-dose-non-uniform.png 495w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/11\/calculation-effective-dose-non-uniform-300x108.png 300w\" sizes=\"(max-width: 495px) 100vw, 495px\" \/><\/a><\/p>\n<p>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.<\/p>\n<p>If we want to account for the buildup of secondary radiation, then we have to include the buildup factor. The <strong>extended formula<\/strong> for the dose rate is then:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/absorbed-dose-rate-gray-calculation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-25303\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/absorbed-dose-rate-gray-calculation.png\" alt=\"absorbed dose rate - gray\" width=\"693\" height=\"158\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/absorbed-dose-rate-gray-calculation.png 693w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2018\/10\/absorbed-dose-rate-gray-calculation-300x68.png 300w\" sizes=\"(max-width: 693px) 100vw, 693px\" \/><\/a><\/p>\n<\/div><\/div>\n<h2>Buildup Factors for Gamma Rays Shielding<\/h2>\n<p>The <strong>buildup factor<\/strong> is a correction factor that considers the influence of the scattered radiation plus any <strong>secondary particles<\/strong> in the medium during shielding calculations. If we want to account for the buildup of secondary radiation, then we have to include the <strong>buildup factor<\/strong>. The buildup factor is then a multiplicative factor that accounts for the response to the un-collided photons 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 un-collided dose.<\/p>\n<p>The <strong>extended formula<\/strong> for the dose rate calculation is:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2019\/01\/Buildup-Factor-in-Dose-Rate-Calculation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-25983\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2019\/01\/Buildup-Factor-in-Dose-Rate-Calculation.png\" alt=\"Buildup Factor\" width=\"689\" height=\"348\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2019\/01\/Buildup-Factor-in-Dose-Rate-Calculation.png 689w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2019\/01\/Buildup-Factor-in-Dose-Rate-Calculation-300x152.png 300w\" sizes=\"(max-width: 689px) 100vw, 689px\" \/><\/a><\/p>\n<p>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).<\/p>\n<\/div>\n<div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:20px 0;border-width:2px;border-color:#999999\"><\/div>\n<div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-33 lgc-tablet-grid-33 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\">\n<h2>See previous:<\/h2>\n<p>Shielding of Positrons<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/shielding-of-ionizing-radiation\/shielding-of-positrons\/\" class=\"su-button su-button-style-flat\" style=\"color:#606060;background-color:#ffffff;border-color:#cccccc;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px\" target=\"_self\"><span style=\"color:#606060;padding:7px 20px;font-size:16px;line-height:24px;border-color:#ffffff;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px;text-shadow:0px 0px 0px #000000;-moz-text-shadow:0px 0px 0px #000000;-webkit-text-shadow:0px 0px 0px #000000\"><i class=\"sui sui-link\" style=\"font-size:16px;color:#5d5d5d\"><\/i> <\/span><\/a><\/p><\/div><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-33 lgc-tablet-grid-33 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\">\n<h2>See above:<\/h2>\n<p>Shielding of Ionizing Radiation<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/shielding-of-ionizing-radiation\/\" class=\"su-button su-button-style-flat\" style=\"color:#606060;background-color:#ffffff;border-color:#cccccc;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px\" target=\"_self\"><span style=\"color:#606060;padding:7px 20px;font-size:16px;line-height:24px;border-color:#ffffff;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px;text-shadow:0px 0px 0px #000000;-moz-text-shadow:0px 0px 0px #000000;-webkit-text-shadow:0px 0px 0px #000000\"><i class=\"sui sui-link\" style=\"font-size:16px;color:#5d5d5d\"><\/i> <\/span><\/a><\/p><\/div><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-33 lgc-tablet-grid-33 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\">\n<h2>See next:<\/h2>\n<p>Shielding of Neutrons<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/shielding-neutron-radiation\/\" class=\"su-button su-button-style-flat\" style=\"color:#606060;background-color:#ffffff;border-color:#cccccc;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px\" target=\"_self\"><span style=\"color:#606060;padding:7px 20px;font-size:16px;line-height:24px;border-color:#ffffff;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px;text-shadow:0px 0px 0px #000000;-moz-text-shadow:0px 0px 0px #000000;-webkit-text-shadow:0px 0px 0px #000000\"><i class=\"sui sui-link\" style=\"font-size:16px;color:#5d5d5d\"><\/i> <\/span><\/a><\/p><\/div><\/div>\n","protected":false},"excerpt":{"rendered":"<p>In short, effective shielding of gamma radiation is in most cases based on the use of materials with two following material properties: high-density of material the high atomic number of material \u00a0(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 &#8230; <a title=\"Shielding of Gamma Radiation\" class=\"read-more\" href=\"https:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/shielding-of-ionizing-radiation\/shielding-gamma-radiation\/\" aria-label=\"More on Shielding of Gamma Radiation\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":12705,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"generate_page_header":""},"_links":{"self":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/12749"}],"collection":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/comments?post=12749"}],"version-history":[{"count":5,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/12749\/revisions"}],"predecessor-version":[{"id":32676,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/12749\/revisions\/32676"}],"up":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/12705"}],"wp:attachment":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/media?parent=12749"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}