{"id":1153,"date":"2018-11-07T18:52:45","date_gmt":"2018-11-08T00:52:45","guid":{"rendered":"https:\/\/www.crpa-acrp-bulletin.ca\/?p=1153"},"modified":"2020-02-18T20:18:33","modified_gmt":"2020-02-19T02:18:33","slug":"crpar-prep-october-2018-preparation-a-la-designation-aacrp-octobre-2018","status":"publish","type":"post","link":"https:\/\/www.crpa-acrp-bulletin.ca\/fr\/2018\/11\/07\/crpar-prep-october-2018-preparation-a-la-designation-aacrp-octobre-2018\/","title":{"rendered":"CRPA(R) Prep, October 2018 \/ Pr\u00e9paration \u00e0 la d\u00e9signation (A)ACRP, octobre 2018"},"content":{"rendered":"<div class='content-column one_half'><div style=\"padding-right:10px;\"><p><img loading=\"lazy\" decoding=\"async\" class=\"alignright wp-image-994\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/CRPAR-icon_cropped-233x300.png\" alt=\"\" width=\"150\" height=\"193\" srcset=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/CRPAR-icon_cropped-233x300.png 233w, https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/CRPAR-icon_cropped-768x987.png 768w, https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/CRPAR-icon_cropped-797x1024.png 797w, https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/CRPAR-icon_cropped-520x668.png 520w, https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/CRPAR-icon_cropped.png 1868w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><\/p>\n<p>For those of you who are new to this corner of the <em>Bulletin<\/em>, here is what we are trying to do in this section. We introduce a question or two similar to the questions on the CRPA(R) exam in each issue. In the next issue, we go through the solution and propose a new question.<\/p>\n<p>The intention is to give people an idea of the types of questions that appear on a CPRA(R) exam and maybe, just maybe, convince more members to challenge the exam.<\/p>\n<p>Do you already have your CRPA(R) designation? If so, <em>we invite you to\u00a0<\/em><a href=\"mailto:cmalcol@mcmaster.ca\"><em>submit questions<\/em><\/a>\u00a0to earn points for your registration maintenance!<\/p>\n<p>So let\u2019s take a look at the solution to the question from the last issue.<\/p>\n<h1>Question from the last issue:<\/h1>\n<p>Here is the question from the <a href=\"https:\/\/www.crpa-acrp-bulletin.ca\/2018\/09\/08\/crpar-prep-september-2018-preparation-a-la-designation-aacrp-septembre-2018\/\" target=\"_blank\" rel=\"noopener noreferrer\">last issue<\/a>:<\/p>\n<p style=\"padding-left: 30px;\">One tenth value layer (TVL) is equal to approximately how many half value layers (HVLs)?<\/p>\n<p><strong>Proposed solution:<\/strong><\/p>\n<p>As is often the case, there are many paths to get to the correct solution.<\/p>\n<p>First, we need to know what the terms \u201ctenth value layer\u201d (TVL) and \u201chalf value layer\u201d (HVL) mean. These terms relate to radiation shielding and how thick a certain material needs to be to reduce the amount of radiation to one-tenth or one-half the original amount.\u00a0 The terms \u201ctenth thickness\u201d or \u201chalf thickness\u201d can also be used.<\/p>\n<p>Okay, great! Now that we know the definitions of each, we can represent these relationships mathematically.<\/p>\n<p>For HVL:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1157\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image003.png\" alt=\"\" width=\"50\" height=\"41\" \/><\/p>\n<p>Where:<\/p>\n<ul>\n<li><em>I<\/em> is the resulting intensity of radiation<\/li>\n<li><em>I<\/em><sub>0<\/sub> is the incident (or original) intensity of radiation<\/li>\n<li><em>n<\/em> is the number of HVLs<\/li>\n<\/ul>\n<p>A similar relationship for the TVL is given as:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1158\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image007.png\" alt=\"\" width=\"59\" height=\"41\" \/><\/p>\n<p>Where:<\/p>\n<ul>\n<li><em>I<\/em> is the resulting intensity of radiation<\/li>\n<li><em>I<\/em><sub>0<\/sub> is the incident (or original) intensity of radiation<\/li>\n<li><em>n<\/em> is the number of TVLs<\/li>\n<\/ul>\n<p>So, if we have one TVL, then:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1159\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image011.png\" alt=\"\" width=\"58\" height=\"41\" \/><\/p>\n<p>If we substitute that result into the HVL relationship for <em>I\u00a0<\/em>\/\u00a0<em>I<\/em><sub>0\u00a0<\/sub>, we get:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1157\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image003.png\" alt=\"\" width=\"50\" height=\"41\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1160\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image019.png\" alt=\"\" width=\"56\" height=\"38\" \/><\/p>\n<p>So solving for <em>n<\/em> gives us:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1162\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image021.png\" alt=\"\" width=\"56\" height=\"20\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1163\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image023.png\" alt=\"\" width=\"109\" height=\"20\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1164\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image025.png\" alt=\"\" width=\"113\" height=\"20\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1166\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image028.png\" alt=\"\" width=\"116\" height=\"41\" \/><\/p>\n<p>So, one TVL is approximately 3.3 HVLs!<\/p>\n<p>I am a huge proponent of checking your work to make sure your answer makes sense. It\u2019s one thing to do the math all the way through and work things out, but I often mess up some small detail like dropping a minus sign or missing a factor of 10 in a conversion or something like that.<\/p>\n<p>So, let\u2019s take a look at our answer and make sure it makes sense. Keep in mind that if you can\u2019t remember how to do the math and solve for an exponent, you can often figure it out just by using some common sense\u2014no math needed!<\/p>\n<p>Here\u2019s how that would work for this example:<\/p>\n<ul>\n<li>If I have one HVL, I knock intensity by a half.<\/li>\n<li>If I have two HLVs, I cut it by half of a half, or a quarter.<\/li>\n<li>If I have three HLVs, it\u2019s half of a half of a half, which is an eighth of the original. Cut it by half again and you get a sixteenth.<\/li>\n<\/ul>\n<p>So, we know that one TVL (a tenth) is somewhere between three and four HVLs, which means our answer makes sense.<\/p>\n<h1>New question:<\/h1>\n<p>There are a lot of different types of questions on the exam. Not all of them require calculation.\u00a0 For example, the exam is multiple choice and often the answer can be determined through a process of elimination without having to actually calculate the exact answer.<\/p>\n<p>Here\u2019s a non-calculation question to think about. We\u2019ll provide the answer in the next issue.<\/p>\n<p style=\"padding-left: 30px;\">What is the typical range of dose measurements possible for a thermoluminescent dosimeter (TLD)?<\/p>\n<p style=\"padding-left: 30px;\">a. 0.01 to 0.1 mSv<br \/>\nb. 0.1 mSv to 10 mSv<br \/>\nc. 1 to 100 mSv<br \/>\nd. 10 to 100 mSv<\/p><\/div><\/div>\n<div class='content-column one_half last_column'><div style=\"padding-left:10px;\"><div style=\"padding: 0 0 0 20px; border-left: #dddddd 1px solid;\">\n<p>Pour ceux d\u2019entre vous qui ne sont pas des habitu\u00e9s de la pr\u00e9sente chronique du <em>Bulletin<\/em>, l\u2019objectif de cette section est de proposer, \u00e0 chaque num\u00e9ro, une ou deux questions similaires \u00e0 celles qui se trouvent dans un examen pour l\u2019agr\u00e9ment (A)ACRP. Les r\u00e9ponses sont publi\u00e9es dans le num\u00e9ro subs\u00e9quent, accompagn\u00e9es de nouvelles questions.<\/p>\n<p>L\u2019objectif est de donner une id\u00e9e du type de questions trouv\u00e9es dans l\u2019examen de l\u2019agr\u00e9ment (A)ACRP. Avec un peu de chance, nous encouragerons ainsi plus de membres \u00e0 passer l\u2019examen et \u00e0 devenir professionnel de la radioprotection agr\u00e9\u00e9 (PRPA).<\/p>\n<p>En outre, <em>nous vous invitons \u00e0 <a href=\"mailto:cmalcol@mcmaster.ca\">soumettre des questions<\/a><\/em> afin d\u2019obtenir des points pour le maintien de votre d\u00e9signation (A)ACRP si vous la d\u00e9tenez d\u00e9j\u00e0!<\/p>\n<p>Analysons donc la r\u00e9ponse \u00e0 la question parue dans le dernier num\u00e9ro.<\/p>\n<h1>Question du dernier num\u00e9ro\u00a0:<\/h1>\n<p>Voici la question du <a href=\"https:\/\/www.crpa-acrp-bulletin.ca\/2018\/09\/08\/crpar-prep-september-2018-preparation-a-la-designation-aacrp-septembre-2018\/\" target=\"_blank\" rel=\"noopener noreferrer\">dernier num\u00e9ro<\/a>\u00a0:<\/p>\n<p style=\"padding-left: 30px;\">Une couche d\u2019att\u00e9nuation au dixi\u00e8me (TVL) est \u00e9gale \u00e0 environ combien de couches de demi-att\u00e9nuation (HVL) ?<\/p>\n<p><strong>Solution propos\u00e9e\u00a0:<\/strong><\/p>\n<p>Comme c\u2019est souvent le cas, il existe diff\u00e9rentes fa\u00e7ons de r\u00e9soudre correctement cette question.<\/p>\n<p>D\u2019abord, nous avons besoin de savoir ce que signifient une \u00ab couche d\u2019att\u00e9nuation au dixi\u00e8me \u00bb (TVL) et une \u00ab couche de demi-att\u00e9nuation \u00bb (HVL). Ces termes ont trait au blindage des rayonnements et \u00e0 l\u2019\u00e9paisseur requise pour certains mat\u00e9riaux afin de r\u00e9duire la quantit\u00e9 de rayonnement \u00e0 un dixi\u00e8me ou \u00e0 la moiti\u00e9 de sa quantit\u00e9 initiale. Les expressions \u00ab dixi\u00e8me d\u2019\u00e9paisseur \u00bb et \u00ab demi-\u00e9paisseur \u00bb peuvent \u00e9galement \u00eatre utilis\u00e9es.<\/p>\n<p>Maintenant que nous connaissons les d\u00e9finitions de ces expressions, nous pouvons les repr\u00e9senter math\u00e9matiquement.<\/p>\n<p>Pour la HVL, nous avons :<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1157\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image003.png\" alt=\"\" width=\"50\" height=\"41\" \/><\/p>\n<p>o\u00f9 :<\/p>\n<ul>\n<li><em>I<\/em> est l\u2019intensit\u00e9 de rayonnement r\u00e9sultante<\/li>\n<li><em>I<\/em><sub>0<\/sub> est l\u2019intensit\u00e9 de rayonnement incidente (ou initiale)<\/li>\n<li><em>n<\/em> est le nombre de HVLs<\/li>\n<\/ul>\n<p>Une \u00e9quation similaire est attribu\u00e9e pour la TVL :<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1158\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image007.png\" alt=\"\" width=\"59\" height=\"41\" \/><\/p>\n<p>o\u00f9 :<\/p>\n<ul>\n<li><em>I<\/em> est l\u2019intensit\u00e9 de rayonnement r\u00e9sultante<\/li>\n<li><em>I<\/em><sub>0<\/sub> est l\u2019intensit\u00e9 de rayonnement incidente (ou initiale)<\/li>\n<li><em>n<\/em> est le nombre de HVLs<\/li>\n<\/ul>\n<p>Donc, pour une TVL, nous avons :<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1159\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image011.png\" alt=\"\" width=\"58\" height=\"41\" \/><\/p>\n<p>En substituant ce r\u00e9sultat \u00e0 l\u2019expression <em>I\u00a0<\/em>\/\u00a0<em>I<\/em><sub>0<\/sub>\u00a0dans l\u2019\u00e9quation de la CDA, nous obtenons :<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1157\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image003.png\" alt=\"\" width=\"50\" height=\"41\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1160\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image019.png\" alt=\"\" width=\"56\" height=\"38\" \/><\/p>\n<p>Le r\u00e9sultat pour <em>n<\/em> est :<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1162\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image021.png\" alt=\"\" width=\"56\" height=\"20\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1163\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image023.png\" alt=\"\" width=\"109\" height=\"20\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1164\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image025.png\" alt=\"\" width=\"113\" height=\"20\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1166\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/10\/image028.png\" alt=\"\" width=\"116\" height=\"41\" \/><\/p>\n<p>En r\u00e9sum\u00e9, une TVL est approximativement \u00e9quivalente \u00e0 3,3 HVLs.<\/p>\n<p>Je suis un fervent partisan de la v\u00e9rification des r\u00e9ponses afin de m\u2019assurer qu\u2019elles soient logiques. C\u2019est une chose de faire le calcul, mais j\u2019oublie parfois quelques petits d\u00e9tails comme un signe n\u00e9gatif ou un facteur de 10 dans une conversion ou quelque chose du genre.<\/p>\n<p>Examinons donc notre r\u00e9ponse pour nous assurer que tout fonctionne. Gardez en t\u00eate que si vous ne vous souvenez pas comment faire le calcul, vous pouvez tout de m\u00eame avoir une id\u00e9e de la r\u00e9ponse avec une dose de bon sens; nul besoin de calculer !<\/p>\n<p>Voici comment y parvenir pour cet exemple :<\/p>\n<ul>\n<li>Si j\u2019ai une HVL, j\u2019att\u00e9nue la moiti\u00e9 de l\u2019intensit\u00e9.<\/li>\n<li>Si j\u2019ai deux HVLs, j\u2019att\u00e9nue la moiti\u00e9 de la moiti\u00e9; il ne reste qu\u2019un quart.<\/li>\n<li>Si j\u2019ai trois HVLs, j\u2019att\u00e9nue donc la moiti\u00e9 de la moiti\u00e9 de la moiti\u00e9, ce qui laisse un huiti\u00e8me de l\u2019intensit\u00e9 initiale. Diminuons de moiti\u00e9 \u00e0 nouveau et il ne reste qu\u2019un seizi\u00e8me.<\/li>\n<\/ul>\n<p>Comme nous savons qu\u2019une TVL\u00a0est \u00e9quivalente \u00e0 entre trois et quatre HVLs, notre r\u00e9ponse est logique.<\/p>\n<h1>Nouvelle question\u00a0:<\/h1>\n<p>Diff\u00e9rents types de questions apparaissent dans l\u2019examen. Certaines n\u00e9cessitent des calculs et d\u2019autres sont des questions \u00e0 choix multiples pour lesquelles les r\u00e9ponses peuvent \u00eatre trouv\u00e9es par un processus d\u2019\u00e9limination sans avoir recours \u00e0 des calculs.<\/p>\n<p>Voici une question qui donne \u00e0 r\u00e9fl\u00e9chir et pour laquelle aucun calcul n\u2019est n\u00e9cessaire. La r\u00e9ponse sera d\u00e9voil\u00e9e dans le prochain num\u00e9ro du <em>Bulletin<\/em>.<\/p>\n<p style=\"padding-left: 30px;\">Quel est le domaine de mesure typique d\u2019un dosim\u00e8tre thermoluminescent (DTL)?<\/p>\n<p style=\"padding-left: 30px;\">a. 0,01 \u00e0 0,1 mSv<br \/>\nb. 0,1 \u00e0 10 mSv<br \/>\nc. 1 \u00e0 100 mSv<br \/>\nd. 10 \u00e0 100 mSv<\/p>\n<\/div><\/div><\/div><div class='clear_column'><\/div>\n<div style=\"padding: 10px 10px 0 10px; background-color: #eeeeee; border: #dddddd 2px solid;\">\n<h2><img loading=\"lazy\" decoding=\"async\" class=\"alignright wp-image-991\" src=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/Chris-Malcolmson_cropped.jpg\" alt=\"\" width=\"125\" height=\"125\" srcset=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/Chris-Malcolmson_cropped.jpg 165w, https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/Chris-Malcolmson_cropped-150x150.jpg 150w, https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/Chris-Malcolmson_cropped-160x160.jpg 160w\" sizes=\"auto, (max-width: 125px) 100vw, 125px\" \/>Christopher Malcolmson<\/h2>\n<p>Christopher Malcolmson has been a health physicist at McMaster University since 2005.\u00a0He received a BSc from McMaster in 2004 and an MSc in 2011. He completed his CRPA(R) in 2009, American Board of Health Physics certification in 2012, and National Registry of Radiation Protection Technologists exam in 2016.\u00a0Malcolmson is currently a member of the CRPA board of directors (director of professional development) and the Registration Subcommittee exam coordinator. He is also a member of the International Radiation Protection Association\u2019s Commission on Publications.<\/p>\n<p>Christopher Malcolmson est sp\u00e9cialiste en radioprotection \u00e0 l\u2019Universit\u00e9 McMaster depuis 2005. Il a obtenu un baccalaur\u00e9at en sciences de cette m\u00eame universit\u00e9 en 2004, suivi d\u2019une ma\u00eetrise en 2011. Il a obtenu sa certification (A)ACRP en 2009, celle de l\u2019American Board of Health Physics (conseil am\u00e9ricain des sp\u00e9cialistes en radioprotection) en 2012, et a termin\u00e9 l\u2019examen de la National Registry of Radiation Protection Technologists (Registre national des technologues en radioprotection) en 2016. Malcolmson est pr\u00e9sentement membre du conseil d\u2019administration de l\u2019ACRP (directeur du perfectionnement professionnel) et coordonnateur de l\u2019examen au sous-comit\u00e9 des inscriptions. Il est \u00e9galement membre de la commission des publications de l\u2019Association internationale pour la protection contre les radiations.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>For those of you who are new to this corner of the <em>Bulletin<\/em>, here is what we are trying to do in this section. We introduce a question or two similar to the questions on the CRPA(R) exam in each issue. \/ Pour ceux d\u2019entre vous qui ne sont pas des habitu\u00e9s de la pr\u00e9sente chronique du <em>Bulletin<\/em>, l\u2019objectif de cette section est de proposer, \u00e0 chaque num\u00e9ro, une ou deux questions similaires \u00e0 celles qui se trouvent dans un examen pour l\u2019agr\u00e9ment (A)ACRP.<\/p>\n","protected":false},"author":31,"featured_media":995,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"footnotes":""},"categories":[5,4],"tags":[],"class_list":["post-1153","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-chroniques-regulieres","category-regular-columns"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.2 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>CRPA(R) Prep, October 2018 \/ Pr\u00e9paration \u00e0 la d\u00e9signation (A)ACRP, octobre 2018 - CRPA-ACRP Bulletin<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.crpa-acrp-bulletin.ca\/2018\/11\/07\/crpar-prep-october-2018-preparation-a-la-designation-aacrp-octobre-2018\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"CRPA(R) Prep, October 2018 \/ Pr\u00e9paration \u00e0 la d\u00e9signation (A)ACRP, octobre 2018 - CRPA-ACRP Bulletin\" \/>\n<meta property=\"og:description\" content=\"For those of you who are new to this corner of the Bulletin, here is what we are trying to do in this section. We introduce a question or two similar to the questions on the CRPA(R) exam in each issue. \/ Pour ceux d\u2019entre vous qui ne sont pas des habitu\u00e9s de la pr\u00e9sente chronique du Bulletin, l\u2019objectif de cette section est de proposer, \u00e0 chaque num\u00e9ro, une ou deux questions similaires \u00e0 celles qui se trouvent dans un examen pour l\u2019agr\u00e9ment (A)ACRP.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.crpa-acrp-bulletin.ca\/2018\/11\/07\/crpar-prep-october-2018-preparation-a-la-designation-aacrp-octobre-2018\/\" \/>\n<meta property=\"og:site_name\" content=\"CRPA-ACRP Bulletin\" \/>\n<meta property=\"article:published_time\" content=\"2018-11-08T00:52:45+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2020-02-19T02:18:33+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.crpa-acrp-bulletin.ca\/wp-content\/uploads\/2018\/07\/CRPAR-icon-ftr.png\" \/>\n\t<meta property=\"og:image:width\" content=\"1200\" \/>\n\t<meta property=\"og:image:height\" content=\"600\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"Chris Malcolmson, Health Physicist, McMaster University, Former CRPA Director of Professional Development \/ Sp\u00e9cialiste en radioprotection, Universit\u00e9 McMaster, Ancien directeur du perfectionnement professionnel de l\u2019ACRP\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"\u00c9crit par\" \/>\n\t<meta name=\"twitter:data1\" content=\"Chris Malcolmson, Health Physicist, McMaster University, Former CRPA Director of Professional Development \/ Sp\u00e9cialiste en radioprotection, Universit\u00e9 McMaster, Ancien directeur du perfectionnement professionnel de l\u2019ACRP\" \/>\n\t<meta name=\"twitter:label2\" content=\"Dur\u00e9e de lecture estim\u00e9e\" \/>\n\t<meta name=\"twitter:data2\" content=\"8 minutes\" \/>\n<script type=\"application\/ld+json\" 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