{"id":11775,"date":"2016-03-19T20:19:01","date_gmt":"2016-03-19T20:19:01","guid":{"rendered":"http:\/\/sitepourvtc.com\/?page_id=11775"},"modified":"2022-10-14T07:46:33","modified_gmt":"2022-10-14T07:46:33","slug":"gamma-ray","status":"publish","type":"page","link":"https:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/gamma-ray\/","title":{"rendered":"Gamma Rays \/ Gamma Radiation"},"content":{"rendered":"<div   id="8n9s8rpp6h"    class=\"sue-shadow-wrap sue-content-wrap sue-shadow-inline-no\"><div   id="8n9s8rpp6h"    class=\"sue-shadow sue-shadow-style-vertical\">\n<div   id="8n9s8rpp6h"    class=\"sue-panel\" data-url=\"\" data-target=\"self\" style=\"background-color:#ffffff;color:#333333;border-radius:0px;-moz-border-radius:0px;-webkit-border-radius:0px;box-shadow:none;-moz-box-shadow:none;-webkit-box-shadow:none;border:1px solid #cccccc\"><div   id="8n9s8rpp6h"    class=\"sue-panel-content sue-content-wrap\" style=\"padding:15px;text-align:left\">\n<div class=\"su-heading su-heading-style-modern-1-dark su-heading-align-left\" id=\"\" style=\"font-size:13px;margin-bottom:20px\"><div   id="8n9s8rpp6h"    class=\"su-heading-inner\">\n<h2>Article Summary &amp; FAQs<\/h2>\n<\/div><\/div>\n<h3>What is gamma radiation?<\/h3>\n<p><em><strong>Gamma rays<\/strong>, also known as\u00a0<strong>gamma radiation<\/strong>, refer to electromagnetic radiation (no rest mass, no charge) with very high energies. Gamma rays are high-energy <a title=\"Photon \u2013 Fundamental Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/\">photons<\/a> with very short wavelengths and thus very high frequency. Since the gamma rays are in substance only very high-energy photons, they are very penetrating matter and are thus biologically hazardous. Gamma rays can travel thousands of feet in the air and can easily pass through the human body.<\/em><\/p>\n<h3>Key Facts<\/h3>\n<p><strong><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radioactive-decay\/gamma-decay-gamma-radioactivity\/\">Gamma Decay<\/a><\/strong><\/p>\n<p><em><strong>Gamma rays<\/strong>\u00a0are emitted by\u00a0<a title=\"Nuclear Stability\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/nuclear-stability\/\">unstable\u00a0nuclei<\/a> 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 <strong>accompanies other\u00a0<a title=\"Forms of Ionizing Radiation\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/forms-ionizing-radiation\/\">forms of decay<\/a><\/strong>, such as alpha or beta decay. The process of <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radioactive-decay\/gamma-decay-gamma-radioactivity\/isomeric-transition\/\">isomeric transition<\/a> is similar to any gamma emission but differs in that it involves the nuclei\u2019s intermediate metastable excited state(s).<\/em><\/p>\n<p><strong><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/gamma-ray\/gamma-rays-x-rays-difference-distinction\/\">Gamma Rays vs. X-rays<\/a><\/strong><\/p>\n<p><em>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>.<\/em><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/\"><strong>Interaction of Gamma Rays with Matter<\/strong><\/a><\/p>\n<p><em>Gamma rays ionize matter primarily via\u00a0<strong>indirect ionization<\/strong>. Although many possible interactions are known, there are <strong>three key interaction mechanisms <\/strong>with the matter.<\/em><\/p>\n<ul>\n<li><em><strong><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/photoelectric-effect\/\">Photoelectric effect<\/a>.\u00a0<\/strong>The photoelectric effect dominates\u00a0<strong>at low-energies of gamma rays<\/strong>.\u00a0The photoelectric effect leads to\u00a0<strong>the emission of photoelectrons<\/strong>\u00a0from matter when light (<a title=\"Photon \u2013 Fundamental Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/\">photons<\/a>) shines upon them.<\/em><\/li>\n<li><em><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/compton-scattering\/\"><strong>Compton scattering<\/strong><\/a>.\u00a0Compton scattering dominates\u00a0<strong>at intermediate energies.\u00a0<\/strong>Photons undergo a wavelength shift called\u00a0<strong>the Compton shift.<\/strong><\/em><\/li>\n<li><em><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/pair-production\/\"><strong>Pair production<\/strong><\/a>. For electron-positron pair production to occur, the electromagnetic energy of the photon must be above <strong>threshold energy<\/strong>, 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 equals <strong>1.02MeV (2 x 0.511MeV)<\/strong>\u00a0because the rest mass of a single electron is equivalent to 0.511MeV of energy.<\/em><\/li>\n<\/ul>\n<p><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/gamma-ray-attenuation\/\"><strong>Gamma Rays Attenuation<\/strong><\/a><\/p>\n<p><em>The following equation can then describe the attenuation of gamma radiation. <\/em><em><strong>I=I<sub>0<\/sub>.e<sup>-\u03bcx <\/sup><\/strong><\/em><em>\u00a0where I is intensity after attenuation, \u00a0I<sub>o<\/sub>\u00a0is incident intensity, \u00a0\u03bc is the linear attenuation coefficient (cm<sup>-1<\/sup>), and the physical thickness of absorber (cm).<\/em><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/shielding-of-ionizing-radiation\/shielding-gamma-radiation\/\"><strong>Shielding of Gamma Radiation<\/strong><\/a><\/p>\n<p><em>In short, effective shielding of gamma radiation is in most cases based on the use of materials with two following material properties:<\/em><\/p>\n<ul>\n<li><em>high-density of material.\u00a0<\/em><\/li>\n<li><em>the high atomic number of material \u00a0(high Z materials)<\/em><\/li>\n<\/ul>\n<p><em>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. Lead is widely used as a gamma shield.<\/em><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/radiation-detection\/gamma-spectroscopy\/cobalt-60-spectrum\/\"><strong>Example &#8211; Cobalt-60 Decay<\/strong><\/a><\/p>\n<p><em>Cobalt-60\u00a0\u00a0is an artificial\u00a0<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/atom-properties-of-atoms\/radionuclide-radioisotope\/\">radioactive isotope<\/a>\u00a0of cobalt with a\u00a0<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radioactive-decay\/radioactive-decay-law\/half-life\/\">half-life<\/a>\u00a0of\u00a05.2747 years. It is synthetically produced by neutron activation of cobalt-59 in\u00a0<a href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/what-is-nuclear-reactor\/\">nuclear reactors<\/a>. Cobalt-60 is a common calibration source found in many laboratories. The gamma spectrum has two significant peaks, one at 1173.2 keV and another at 1332.5 keV.\u00a0<\/em><\/p>\n<div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:10px 0;border-width:2px;border-color:#999999\"><\/div>\n<div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\">\n<div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-modern-light 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>Fear of Radiation \u2013 Is it rational?<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><strong>Fear of Radiation \u2013 Is it rational?<\/strong><\/p>\n<p><strong>Radiation is all around us.<\/strong>\u00a0We are continually exposed to\u00a0<a href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/radiation-protection\/sources-of-radiation\/natural-background-radiation\/\">natural background radiation<\/a>,\u00a0and it seems to be without any problem. Yes,\u00a0<strong>high doses<\/strong> of ionizing radiation are harmful and potentially lethal to living beings, but these doses must be high. Moreover, what is not harmful in high doses? Even a high amount of water can be lethal to living beings.<\/p>\n<p>The truth about<strong>\u00a0low-dose radiation<\/strong> health effects still needs to be found. It is not exactly known whether these low doses of radiation are detrimental or beneficial (and where is the threshold).<\/p>\n<p>But finally, if you compare risks, which arise from the existence of the radiation, natural or artificial, with risks that arise from everyday life, you must conclude that<strong>\u00a0fear of radiation is irrational<\/strong>. Humans are often inconsistent in our treatment of perceived risks. Even though two situations may have similar risks, people will find one permissible and another unjustifiably dangerous.<\/p>\n<p>See also:\u00a0<a href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/radiation-protection\/fear-of-radiation-is-it-rational\/\"><strong>Fear of Radiation \u2013 Is it rational?<\/strong><\/a><\/p>\n<\/div><\/div>\n<div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-modern-light 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>How can be gamma rays detected?<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">\n<strong>How can gamma rays be detected?<\/strong><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/gamma-ray\/\">Gamma rays<\/a> have very little trouble penetrating the metal walls of the detector.<strong>\u00a0Detectors<\/strong> may be categorized according to sensitive materials and methods that can be utilized to make a measurement:<\/p>\n<ul>\n<li><a href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/radiation-detection\/ga搜索引擎优化us-ionization-detector\/\"><strong>Ga搜索引擎优化us Ionization Detectors<\/strong><\/a><\/li>\n<li><a href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/radiation-detection\/scintillation-counter-scintillation-detector\/\"><strong>Scintillation Detectors<\/strong><\/a><\/li>\n<li><a href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/radiation-detection\/semiconductor-detectors\/\"><strong>Semiconductor Detectors<\/strong><\/a><\/li>\n<\/ul>\n<\/div><\/div>\n<div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-modern-light 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>What is gamma spectroscopy?<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">\n<strong>What is gamma spectroscopy?<\/strong><\/p>\n<p>In general,\u00a0<a href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/radiation-detection\/gamma-spectroscopy\/\"><strong>gamma spectroscopy<\/strong><\/a> studies the energy spectra of gamma-ray sources, such as in the nuclear industry, geochemical investigation, and astrophysics. Spectroscopes, or spectrometers, are sophisticated devices designed to measure the spectral power distribution of a source. The incident radiation generates a signal that allows determining the energy of the incident particle.<\/p>\n<\/div><\/div>\n<\/div>\n<\/div><\/div>\n<\/div><\/div>\n<div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:10px 0;border-width:2px;border-color:#999999\"><\/div>\n<div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:40px\"><\/div>\n<div class=\"su-heading su-heading-style-modern-1-dark su-heading-align-left\" id=\"\" style=\"font-size:13px;margin-bottom:20px\"><div   id="8n9s8rpp6h"    class=\"su-heading-inner\">\n<h2>Gamma Rays \/ Gamma Radiation<\/h2>\n<\/div><\/div>\n<p><strong>Gamma rays<\/strong>, also known as <strong>gamma radiation<\/strong>, refer to electromagnetic radiation (no rest mass, no charge) with very high energies. Gamma rays are high-energy <a title=\"Photon \u2013 Fundamental Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/\">photons<\/a> with very short wavelengths and thus very high frequency. Since the gamma rays are in substance only very high-energy photons, they are very penetrating matter and are thus biologically hazardous. Gamma rays can travel thousands of feet in the air and can easily pass through the human body.<\/p>\n<p><strong>Gamma rays<\/strong> are emitted by <a title=\"Nuclear Stability\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/nuclear-stability\/\">unstable\u00a0nuclei<\/a> 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 <strong>accompanies other <a title=\"Forms of Ionizing Radiation\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/forms-ionizing-radiation\/\">forms of decay<\/a><\/strong>, such as alpha or beta decay.<\/p>\n<figure id=\"attachment_11840\" aria-describedby=\"caption-attachment-11840\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Barium-137-radionuclide.png\"><img loading=\"lazy\" class=\"size-medium wp-image-11840\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Barium-137-radionuclide-300x300.png\" alt=\"Barium-137m is a product of a common fission product - Caesium - 137. The main gamma ray of Barium-137m is 661keV photon.\" width=\"300\" height=\"300\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Barium-137-radionuclide-300x300.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Barium-137-radionuclide-150x150.png 150w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Barium-137-radionuclide-66x66.png 66w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Barium-137-radionuclide.png 640w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11840\" class=\"wp-caption-text\">Barium-137m is a product of a common fission product &#8211; Caesium &#8211; 137. The main gamma-ray of Barium-137m is 661keV photon.<\/figcaption><\/figure>\n<p>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. <strong>Natural sources of gamma rays<\/strong> on Earth are, among other things, gamma rays from naturally occurring radionuclides, particularly potassium-40. \u00a0Potassium-40 is a radioactive isotope of potassium that has a very long half-life of 1.251\u00d710<sup>9<\/sup> 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 a natural source of gamma rays.<\/p>\n<p>See also: <a href=\"https:\/\/www.radiation-dosimetry.org\/what-is-gamma-dosimetry-gamma-dosimeter-definition\/\">Gamma Dosimetry<\/a><\/p>\n<div   id="8n9s8rpp6h"    class=\"collapsible-container\">\n<h2>Discovery of Gamma Rays<\/h2>\n<figure id=\"attachment_11798\" aria-describedby=\"caption-attachment-11798\" style=\"width: 158px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/becquerel.png\"><img loading=\"lazy\" class=\"size-full wp-image-11798\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/becquerel.png\" alt=\"Antoine Henri Becquerel\" width=\"168\" height=\"238\" \/><\/a><figcaption id=\"caption-attachment-11798\" class=\"wp-caption-text\">Antoine Henri Becquerel<\/figcaption><\/figure>\n<p>Gamma rays were discovered shortly after the discovery of X-rays. In 1896, French scientist <strong>Henri Becquerel<\/strong> discovered that uranium minerals could expose a photographic plate through another material. Becquerel presumed that uranium emitted some invisible light similar to X-rays, which W.C.Roentgen recently discovered. He called it \u201c<strong>metallic phosphorescence<\/strong>\u201d. Henri Becquerel had found <strong>gamma radiation<\/strong> being emitted by radioisotope <sup>226<\/sup>Ra (radium), which is part of the Uranium-series of uranium decay chain.<br \/>\nGamma rays were first thought to be particles with mass, for example, extremely energetic <a title=\"Beta Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/beta-particle\/\">beta particles<\/a>. This opinion failed because a magnetic field could not deflect this radiation, which indicated they had no charge. In 1914, gamma rays were observed to be reflected from crystal surfaces, proving they must be<strong> electromagnetic radiation<\/strong> with higher energy (higher frequency and shorter wavelengths).<\/p>\n<h2>Characteristics of Gamma Rays \/ Radiation<\/h2>\n<p><strong>Key features of gamma rays<\/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 photons 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><strong>Photoelectric effect<\/strong><\/li>\n<li><strong>Compton scattering<\/strong><\/li>\n<li><strong>Pair production<\/strong><\/li>\n<\/ul>\n<\/li>\n<li>\n<figure id=\"attachment_11707\" aria-describedby=\"caption-attachment-11707\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/cloud-chamber-2.png\"><img loading=\"lazy\" class=\"size-medium wp-image-11707\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/cloud-chamber-2-300x167.png\" alt=\"Comparison of particles in a cloud chamber. Source: wikipedia.org\" width=\"300\" height=\"167\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/cloud-chamber-2-300x167.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/cloud-chamber-2.png 922w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11707\" class=\"wp-caption-text\">Comparison of particles in a cloud chamber. Source: wikipedia.org<\/figcaption><\/figure>\n<p>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=\"su-accordion su-u-trim\"><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>Image: The relative importance of various processes of gamma radiation interactions with matter.<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">\n<figure id=\"attachment_11791\" aria-describedby=\"caption-attachment-11791\" style=\"width: 290px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/attenuation.png\"><img loading=\"lazy\" class=\"size-medium wp-image-11791\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/attenuation-300x188.png\" alt=\"Gamma rays attuenuation\" width=\"300\" height=\"188\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/attenuation-300x188.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/attenuation-320x202.png 320w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/attenuation.png 570w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/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<\/div><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div>\n<h2>Photoelectric Effect<\/h2>\n<p>See also : <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/photoelectric-effect\/\">Photoelectric effect<\/a><\/p>\n<ul>\n<li>The photoelectric effect dominates <strong>at low-energies of gamma rays<\/strong>.<\/li>\n<li>The photoelectric effect leads to <strong>the emission of photoelectrons<\/strong> from matter when light (<a title=\"Photon \u2013 Fundamental Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/\">photons<\/a>) shines upon them.<\/li>\n<li>The maximum energy an electron can receive in any one interaction is <strong><i>h\u03bd<\/i>.<\/strong><\/li>\n<li>The photoelectric effect only emits electrons if the photon reaches or exceeds <strong>threshold energy<\/strong>.<\/li>\n<li>A free-electron (e.g.,, from an atomic cloud) cannot absorb the entire energy of the incident photon. This is a result of the need to conserve both momentum and energy.<\/li>\n<li>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.<\/li>\n<\/ul>\n<div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:10px\"><\/div>\n<h3>Definition of Photoelectric effect<\/h3>\n<p>In the <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/photoelectric-effect\/\">photoelectric effect<\/a>, a photon undergoes an interaction with an electron that is bound in an atom.<\/p>\n<figure id=\"attachment_11855\" aria-describedby=\"caption-attachment-11855\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photoelectric-Effect-Potassium.jpg\"><img loading=\"lazy\" class=\"size-medium wp-image-11855\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photoelectric-Effect-Potassium-300x169.jpg\" alt=\"Photoelectric effect with photons from visible spectrum on potassium plate - threshold energy - 2eV\" width=\"300\" height=\"169\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photoelectric-Effect-Potassium-300x169.jpg 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photoelectric-Effect-Potassium.jpg 1024w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11855\" class=\"wp-caption-text\">Photoelectric effect with photons from visible spectrum on potassium plate &#8211; threshold energy &#8211; 2eV<\/figcaption><\/figure>\n<p>The incident photon completely disappears in this interaction, and the atom ejects an energetic photoelectron from one <strong>of its bound shells<\/strong>. The kinetic energy of the ejected photoelectron (E<sub>e<\/sub>) is equal to the incident photon energy (h\u03bd) minus the <a title=\"Binding Energy\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/binding-energy\/\">binding energy<\/a> of the photoelectron in its original shell (E<sub>b<\/sub>).<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\"><strong>E<sub>e<\/sub>=h\u03bd-E<sub>b<\/sub><\/strong><\/p>\n<p>Therefore, photoelectrons are only emitted by the photoelectric effect if the photon reaches or exceeds <strong>threshold energy<\/strong> &#8211; the electron\u2019s binding energy &#8211; <strong>the material\u2019s work function<\/strong>. For gamma rays with energies of more than hundreds keV, the photoelectron carries off the majority of the incident photon energy &#8211; h\u03bd.<\/p>\n<figure id=\"attachment_11817\" aria-describedby=\"caption-attachment-11817\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photoelectric_effect_2.jpg\"><img loading=\"lazy\" class=\"size-medium wp-image-11817\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photoelectric_effect_2-300x170.jpg\" alt=\"Gamma absorption by an atom. Source: laradioactivite.com\/\" width=\"300\" height=\"170\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photoelectric_effect_2-300x170.jpg 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photoelectric_effect_2.jpg 500w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11817\" class=\"wp-caption-text\">Gamma absorption by an atom.<br \/>Source: laradioactivite.com\/<\/figcaption><\/figure>\n<p>Following a photoelectric interaction, an ionized absorber atom is created with <strong>a vacancy in one of its bound shells<\/strong>. An electron from a shell will quickly fill this vacancy with lower binding energy (other shells) or capture 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 <strong>characteristic X-rays<\/strong> can also be 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.<\/p>\n<p>&nbsp;<\/p>\n<div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div>\n<h3>Cross-Sections of Photoelectric Effect<\/h3>\n<p><strong>At small values of gamma-ray energy, the photoelectric effect dominates<\/strong>. The mechanism is also enhanced for materials of high atomic number Z. It is not simple to derive an analytic expression for the probability of photoelectric absorption of gamma-ray per atom comprehensive ranges of gamma-ray energies. The probability of photoelectric absorption per unit mass is approximately proportional to:<\/p>\n<p style=\"text-align: center;\"><strong>\u03c4<sub>(photoelectric)<\/sub> = constant x Z<sup>N<\/sup>\/E<sup>3.5<\/sup><\/strong><\/p>\n<p>where <strong>Z<\/strong> is the atomic number, the exponent <strong>n<\/strong> varies between 4 and 5. <strong>E<\/strong> is the energy of the incident photon. The proportionality to higher powers of the atomic number Z is the main reason for using high Z materials, such as lead or depleted uranium in gamma-ray shields.<\/p>\n<figure id=\"attachment_11683\" aria-describedby=\"caption-attachment-11683\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/photoelectric_effect.png\"><img loading=\"lazy\" class=\"size-medium wp-image-11683\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/photoelectric_effect-300x214.png\" alt=\"Cross section of photoelectric effect.\" width=\"300\" height=\"214\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/photoelectric_effect-300x214.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/photoelectric_effect.png 828w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11683\" class=\"wp-caption-text\">Cross-section of the photoelectric effect.<\/figcaption><\/figure>\n<p>Although the probability of the photoelectric absorption of gamma photon decreases, in general, with increasing photon energy, there are <strong>sharp discontinuities<\/strong> in the cross-section curve. These are called <strong>\u201cabsorption edges\u201d<\/strong> and correspond to the binding energies of electrons from an atom\u2019s bound shells. For photons with the energy just above the edge, the photon energy is sufficient to undergo the photoelectric interaction with an electron from the 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 electrons from K-shell is energetically impossible, and therefore the probability drops abruptly. These edges also occur at binding energies of electrons from other shells (L, M, N &#8230;).<\/p>\n<h2>Compton Scattering<\/h2>\n<h3>Key characteristics of Compton Scattering<\/h3>\n<ul>\n<li>Compton scattering dominates <strong>at intermediate energies.<\/strong><\/li>\n<li>It is the scattering of photons <strong>by atomic electrons \u00a0<\/strong><\/li>\n<li>Photons undergo a wavelength shift called <strong>the Compton shift.<\/strong><\/li>\n<li>The energy transferred to the recoil electron can <strong>vary from zero to a large fraction<\/strong> of the incident gamma-ray energy.<\/li>\n<\/ul>\n<h3>Definition of Compton Scattering<\/h3>\n<p>Compton scattering is the inelastic or nonclassical scattering of a photon (which may be an X-ray or gamma-ray <a title=\"Photon \u2013 Fundamental Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/\">photon<\/a>) by a charged particle, usually an electron. In Compton scattering, the incident gamma-ray photon is deflected through an angle \u0398 with regard to its original direction. This deflection decreases the photon\u2019s frequency\u2019s energy (decrease in photon\u2019s frequency) and is called <strong>the Compton effect<\/strong>. The photon transfers a portion of its energy to <strong>the recoil electron<\/strong>. The energy transferred to the recoil electron can vary from zero to a large fraction of the incident gamma-ray energy because all scattering angles are possible. A. H.Compton observed the Compton scattering<strong>\u00a0in 1923<\/strong> at Washington University in St. Louis. Compton earned <strong>the Nobel Prize in Physics in 1927<\/strong> for this new understanding of the particle nature of photons.<\/p>\n<h3>Compton Scattering Formula<\/h3>\n<p>The Compton formula was published in 1923 in the Physical Review. Compton explained that the particle-like momentum of photons causes the X-ray shift. Compton scattering formula is the mathematical relationship between the shift in wavelength and the scattering angle of the X-rays. In the case of Compton scattering, the photon of frequency <i>f<\/i> collides with an electron at rest. The photon bounces off the electron upon collision, giving up some of its initial energy (given by Planck\u2019s formula E=hf). While the electron gains momentum (mass x velocity), the <strong>photon cannot lower its velocity<\/strong>. As a result of momentum conservation law, the photon must lower its momentum given by:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photon-momentum-formula.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-11869\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photon-momentum-formula.png\" alt=\"As a result of momentum conservetion law, the photon must lower its momentum given by this formula.\" width=\"177\" height=\"59\" \/><\/a><\/p>\n<p>So the decrease in photon\u2019s momentum must be translated into a <strong>decrease in frequency<\/strong> (increase in wavelength \u0394<b>\u03bb = \u03bb\u2019 &#8211; \u03bb<\/b>). The shift of the wavelength increased with scattering angle according to <strong>the Compton formula<\/strong>:<\/p>\n<p><strong><strong><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-Scattering-Formula.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-11870\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-Scattering-Formula.png\" alt=\"The shift of the wavelength increased with scattering angle according to the Compton formula\" width=\"236\" height=\"64\" \/><\/a><\/strong><\/strong><\/p>\n<figure id=\"attachment_11831\" aria-describedby=\"caption-attachment-11831\" style=\"width: 332px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/compton-scattering.gif\"><img loading=\"lazy\" class=\" wp-image-11831\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/compton-scattering.gif\" alt=\"Compton Scattering\" width=\"342\" height=\"247\" \/><\/a><figcaption id=\"caption-attachment-11831\" class=\"wp-caption-text\">In Compton scattering, the incident gamma-ray photon is deflected through an angle \u0398 with regard to its original direction. This deflection decreases the photon\u2019s frequency\u2019s energy (decrease in photon\u2019s frequency) and is called the Compton effect.<br \/>Source: hyperphysics.phy-astr.gsu.edu<\/figcaption><\/figure>\n<p>where<\/p>\n<p><b>\u03bb<\/b> is the initial wavelength of the photon<\/p>\n<p><b>\u03bb\u2019<\/b> is the wavelength after scattering,<\/p>\n<p><b>h <\/b>is the Planck constant = 6.626 x 10<sup>-34<\/sup> J.s<\/p>\n<p><b>m<\/b><b><sub>e<\/sub><\/b> is the electron rest mass (0.511 MeV)<\/p>\n<p><b>c<\/b> is the speed of light<\/p>\n<p><b>\u0398<\/b> is the scattering angle.<\/p>\n<p>The minimum change in wavelength (<i>\u03bb\u2032<\/i> \u2212 <i>\u03bb<\/i>) for the photon occurs when \u0398 = 0\u00b0 (cos(\u0398)=1) and is at least zero. The maximum change in wavelength (<i>\u03bb\u2032<\/i> \u2212 <i>\u03bb<\/i>) for the photon occurs when \u0398 = 180\u00b0 (cos(\u0398)=-1). In this case, the photon transfers to the electron as much momentum as possible. The maximum change in wavelength can be derived from the Compton formula:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-length.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-11867\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-length.png\" alt=\"The maximum change in wavelength can be derived from Compton formula. Compton length\" width=\"561\" height=\"78\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-length.png 561w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-length-300x42.png 300w\" sizes=\"(max-width: 561px) 100vw, 561px\" \/><\/a><\/p>\n<p>The quantity h\/m<sub>e<\/sub>c is known as the <strong>Compton wavelength<\/strong> of the electron and is equal to <strong>2.43\u00d710<sup>\u221212<\/sup> m<\/strong>.<\/p>\n<h3>Compton Scattering &#8211; Cross-Sections<\/h3>\n<p>The probability of Compton scattering per interaction with an atom increases linearly with atomic number Z because it depends on the number of electrons available for scattering in the target atom.<strong> The Klein-Nishina formula describes the angular distribution of photons scattered from a single free electron<\/strong>:<\/p>\n<p><strong><strong><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Klein-Nishina-Formula.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-11868\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Klein-Nishina-Formula.png\" alt=\"The angular distribution of photons scattered from a single free electron is described by the Klein-Nishina formula\" width=\"547\" height=\"85\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Klein-Nishina-Formula.png 547w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Klein-Nishina-Formula-300x47.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Klein-Nishina-Formula-540x85.png 540w\" sizes=\"(max-width: 547px) 100vw, 547px\" \/><\/a><\/strong><\/strong><\/p>\n<p>where \u03b5 = E<sub>0<\/sub>\/m<sub>e<\/sub>c<sup>2<\/sup> and r<sub>0<\/sub> is the &#8220;classical radius of the electron&#8221; equal to about 2.8 x 10<sup>-13<\/sup> cm. The formula gives the probability of scattering a photon into the solid angle element d\u03a9 = 2\u03c0 sin \u0398 d\u0398 when the incident energy is E<sub>0<\/sub>.<\/p>\n<figure id=\"attachment_11828\" aria-describedby=\"caption-attachment-11828\" style=\"width: 407px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-scattering-experiment.gif\"><img loading=\"lazy\" class=\"size-full wp-image-11828\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-scattering-experiment.gif\" alt=\"Compton scattering experiment\" width=\"417\" height=\"544\" \/><\/a><figcaption id=\"caption-attachment-11828\" class=\"wp-caption-text\">The wavelength change in such scattering depends only upon the angle of scattering for a given target particle.<br \/>Source: hyperphysics.phy-astr.gsu.edu\/<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_11686\" aria-describedby=\"caption-attachment-11686\" style=\"width: 290px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/Compton-scattering.png\"><img loading=\"lazy\" class=\"size-medium wp-image-11686\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/Compton-scattering-300x213.png\" alt=\"compton scattering\" width=\"300\" height=\"213\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/Compton-scattering-300x214.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/Compton-scattering.png 828w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11686\" class=\"wp-caption-text\">Cross-section of Compton scattering of photons by atomic electrons.<\/figcaption><\/figure>\n<figure id=\"attachment_11827\" aria-describedby=\"caption-attachment-11827\" style=\"width: 290px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton_angle_distribution.jpg\"><img loading=\"lazy\" class=\"size-medium wp-image-11827\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton_angle_distribution-300x209.jpg\" alt=\"Compton scattering - Angle distribution\" width=\"300\" height=\"209\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton_angle_distribution-300x209.jpg 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton_angle_distribution.jpg 648w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11827\" class=\"wp-caption-text\">Energies of a photon at 500 keV and an electron after Compton scattering.<br \/>Source: wikipedia.org<\/figcaption><\/figure>\n<div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div>\n<h3>Compton Edge<\/h3>\n<p>In spectrophotometry, <strong>the Compton edge<\/strong> is a feature of the spectrograph that results from the Compton scattering in the scintillator or detector. This feature is due to photons that undergo Compton scattering with a scattering angle of 180\u00b0 and then escape the detector. When a gamma-ray scatters off the detector and escapes, only a fraction of its initial energy can be deposited in the sensitive layer of the detector. It depends on the scattering angle of the photon, how much energy will be deposited in the detector. This leads to a spectrum of energies. The Compton edge energy corresponds to the full backscattered photon<strong>.<\/strong><\/p>\n<div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:10px\"><\/div>\n<h3>Inverse Compton Scattering<\/h3>\n<p><strong>Inverse Compton scattering<\/strong> is the scattering of low-energy photons to high energies by relativistic electrons. Relativistic electrons can boost the energy of low-energy photons by a potentially enormous amount (even gamma rays can be produced). This phenomenon is very important in astrophysics.<\/p>\n<figure id=\"attachment_11833\" aria-describedby=\"caption-attachment-11833\" style=\"width: 290px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-edge.gif\"><img loading=\"lazy\" class=\"size-medium wp-image-11833\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-edge-300x245.gif\" alt=\"Compton edge of 60Co on gamma spectrometer Na(Tl).\" width=\"300\" height=\"245\" \/><\/a><figcaption id=\"caption-attachment-11833\" class=\"wp-caption-text\">Compton edge of 60Co on gamma spectrometer Na(Tl).<\/figcaption><\/figure>\n<div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div>\n<figure id=\"attachment_11834\" aria-describedby=\"caption-attachment-11834\" style=\"width: 228px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Inverse-Compton-scattering.gif\"><img loading=\"lazy\" class=\"size-medium wp-image-11834\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Inverse-Compton-scattering-238x300.gif\" alt=\"Inverse Compton scattering\" width=\"238\" height=\"300\" \/><\/a><figcaption id=\"caption-attachment-11834\" class=\"wp-caption-text\">source: venables.asu.edu<\/figcaption><\/figure>\n<h2>Positron-Electron Pair Production<\/h2>\n<p>In general, <strong>pair production<\/strong> is a phenomenon of nature where <strong>energy is directly converted to matter<\/strong>. The phenomenon of pair production can be view two different ways. One way is <strong>a particle and anti-particle,<\/strong> and the other is <strong>a particle and a hole<\/strong>. The first way can be represented by the formation of <strong>electron and positron<\/strong>\u00a0from a packet of electromagnetic energy (<strong>high energy photon &#8211; gamma-ray<\/strong>) traveling through matter. \u00a0It is one of the possible ways in which gamma rays interact with matter. <strong>At high energies, this interaction dominates<\/strong>.<\/p>\n<p>For electron-positron pair production to occur, the electromagnetic energy of the photon must be above <strong>threshold energy<\/strong>, 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 equals <strong>1.02MeV (2 x 0.511MeV)<\/strong> because the rest mass of a single electron is equivalent to 0.511MeV of energy.<\/p>\n<p>If the original photon\u2019s 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.<\/p>\n<p>The presence of <strong>an electric field of a heavy atom<\/strong> such as lead or uranium<strong> is essential to satisfy the conservation of momentum and energy<\/strong>.\u00a0 The atomic nucleus must receive some momentum to satisfy both conservations of momentum and energy. Therefore a <a title=\"Photon \u2013 Fundamental Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/\">photon<\/a> pair production <strong>in free space cannot occur<\/strong>.<\/p>\n<p>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 both particles and the complete conversion of their rest mass back to pure energy (according to the E=mc<sup>2<\/sup> formula) in the form of two oppositely directed 0.511 MeV gamma rays (photons). The pair production phenomenon is therefore connected with<strong> the creation and destruction of matter<\/strong> in one reaction.<div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:10px\"><\/div>\n<h3>Positron-Electron Pair Production &#8211; Cross-Section<\/h3>\n<p>The probability of pair production, characterized by cross-section, is a <strong>very complicated function based on quantum mechanics<\/strong>. In general, the cross-section increases approximately with the square of atomic number <strong>(\u03c3<sub>p<\/sub> ~ Z<sup>2<\/sup>)<\/strong> and increases with photon energy, but this dependence is much more complex.<a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/Pair-production-in-chamber.jpg\"><img loading=\"lazy\" class=\"aligncenter size-medium wp-image-11706\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/Pair-production-in-chamber-216x300.jpg\" alt=\"\" width=\"216\" height=\"300\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/Pair-production-in-chamber-216x300.jpg 216w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/Pair-production-in-chamber.jpg 324w\" sizes=\"(max-width: 216px) 100vw, 216px\" \/><\/a><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div>\n<figure id=\"attachment_11685\" aria-describedby=\"caption-attachment-11685\" style=\"width: 290px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/pair_production.png\"><img loading=\"lazy\" class=\"size-medium wp-image-11685\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/pair_production-300x214.png\" alt=\"Pair production in nuclear field and electron field.\" width=\"300\" height=\"214\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/pair_production-300x214.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/03\/pair_production.png 829w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-11685\" class=\"wp-caption-text\">Cross-section of pair production in the nuclear field and electron field.<\/figcaption><\/figure>\n<h2>Gamma Rays Attenuation<\/h2>\n<p>The total cross-section of the interaction of gamma rays with an atom is equal to the sum of all three mentioned partial cross-sections:<\/p>\n<p><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<strong> photoelectric effect<\/strong> dominates. <strong>Compton scattering<\/strong> 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, <strong>electron-positron pair production<\/strong> 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 of these interactions removes the photon from the beam either by absorption or by 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 the 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<h3>Half Value Layer<\/h3>\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>Mass Attenuation Coefficient<\/h3>\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 style=\"text-align: center;\"><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). 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<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><\/p>\n<p><span style=\"color: #333333;\">The question is quite simple and can be described by the following equation:<\/span><\/p>\n<p><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><\/p>\n<p><span style=\"color: #333333;\">If the half-value layer for water is 7.15 cm, the linear attenuation coefficient is:<\/span><\/p>\n<p><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><\/p>\n<p><span style=\"color: #333333;\">Now we can use the exponential attenuation equation:<\/span><\/p>\n<p><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><\/p>\n<p><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><\/p>\n<p><span style=\"color: #333333;\">therefore<\/span><\/p>\n<p><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><\/p>\n<p><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><\/p>\n<p><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><\/p>\n<p><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><strong>Linear Attenuation Coefficients<\/strong><\/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><\/p>\n<p><strong>Half Value Layers<\/strong><\/p>\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<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<br \/>\npredicted by the exponential law alone. The process must be taken into account when<br \/>\nevaluating 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 the build-up 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>Spoiler title<\/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, which is <strong>contaminated by 1.0 Ci of <sup>137<\/sup>Cs<\/strong><strong>.<\/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 has 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 build-up 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 build-up 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 build-up of secondary radiation, then we have to include the build-up 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>Build-up Factors for Gamma Rays Shielding<\/h2>\n<p>The <strong>build-up 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 build-up of secondary radiation, then we have to include the <strong>build-up factor<\/strong>. The build-up 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 build-up 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 build-up 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<a href=\"\" 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><\/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>Photon<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/\" 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<a href=\"\" 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><\/div><\/div>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Gamma rays, also known as gamma radiation, refer to electromagnetic radiation (no rest mass, no charge) with 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 very high-energy photons, they are very penetrating matter and are thus biologically hazardous. &#8230; <a title=\"Gamma Rays \/ Gamma Radiation\" class=\"read-more\" href=\"https:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/gamma-ray\/\" aria-label=\"More on Gamma Rays \/ Gamma Radiation\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":11764,"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\/11775"}],"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=11775"}],"version-history":[{"count":16,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/11775\/revisions"}],"predecessor-version":[{"id":32492,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/11775\/revisions\/32492"}],"up":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/11764"}],"wp:attachment":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/media?parent=11775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}