{"id":1615,"date":"2019-09-19T09:13:56","date_gmt":"2019-09-19T15:13:56","guid":{"rendered":"https:\/\/www.crpa-acrp-bulletin.ca\/?p=1615"},"modified":"2020-02-18T20:53:57","modified_gmt":"2020-02-19T02:53:57","slug":"crpar-prep-september-2019-preparation-a-la-designation-aacrp-septembre-2019","status":"publish","type":"post","link":"https:\/\/www.crpa-acrp-bulletin.ca\/fr\/2019\/09\/19\/crpar-prep-september-2019-preparation-a-la-designation-aacrp-septembre-2019\/","title":{"rendered":"CRPA(R) Prep, September 2019 \/ Pr\u00e9paration \u00e0 la d\u00e9signation (A)ACRP, septembre 2019"},"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>In this section of the <em>Bulletin<\/em>, we introduce a question or two similar to the questions on the CRPA(R) exam. In the next issue, we will provide the solution. The intention is to give people an idea of the types of questions that we use on the CRPA(R) exam and perhaps convince more members to challenge the exam.<\/p>\n<p>If you already have your CRPA(R) designation, we invite you to\u00a0<a href=\"mailto:cmalcol@mcmaster.ca\">submit questions<\/a> to earn points for your registration maintenance!<\/p>\n<h1>Question from the last issue:<\/h1>\n<p>What is the net count rate in counts per minute (cpm) given the following information?<\/p>\n<ul>\n<li>Removable activity 3 Bq\/cm<sup>2<\/sup><\/li>\n<li>Area wiped 100 cm<sup>2<\/sup><\/li>\n<li>Collection factor for the wipe 10%<\/li>\n<li>Instrument efficiency 50%<\/li>\n<\/ul>\n<p>a. 15 cpm<br \/>\nb. 900 cpm<br \/>\nc. 90 cpm<br \/>\nd. 150 cpm<\/p>\n<p><strong>Proposed solution:<\/strong><\/p>\n<p>This is a straight-up calculation question, and experienced radiation safety people should be able to figure out the answer fairly quickly. That being said, a number of situations could make this question challenging. Maybe it\u2019s been a while since you\u2019ve had to do a calculation like this, or maybe your facility doesn\u2019t deal with a lot of loose contamination or indirect checking of surfaces, or maybe you just don\u2019t know how to do the calculation. If any of these apply, this is a perfect opportunity to employ an age-old exam technique: unit analysis.<\/p>\n<p>We want to get from Bq\/cm<sup>2<\/sup> to counts per minute (cpm). Let\u2019s keep that in mind while we figure out what is going on.<\/p>\n<p>We have wipe that we used to sample an area of 100 cm<sup>2<\/sup>. The area is contaminated with 3 Bq\/cm<sup>2<\/sup> (which we assume is uniformly distributed over the area). We are given a collection efficiency of 10%, which is the percentage of the surface contamination that is transferred to the wipe (and is a good rule of thumb in the absence of empirical data). Finally, we have the instrument efficiency, which gives us how many cpm we see on the instrument for every disintegration per minute (dpm)\u00a0in the sample for the geometry of this particular situation.<\/p>\n<p>If we think about what is going on, we can use unit analysis to make sure we are on the right track for this calculation. Basically, we can calculate the activity captured on the wipe and then calculate the reading expected on our measurement device for that activity.<\/p>\n<p>To calculate the activity on the wipe (in Bq collected), we multiply the removable contamination (in Bq\/cm<sup>2<\/sup>) by the area wiped (in cm<sup>2<\/sup>), then we multiply that result by the collection factor (in Bq collected per Bq removable contamination).<\/p>\n<p>activity on wipe<br \/>\n= removable contamination \u00d7 area wiped \u00d7 collection factor<br \/>\n= 3 Bq\/cm<sup>2<\/sup>\u00a0\u00d7 100 cm<sup>2<\/sup>\u00a0\u00d7 10% (Bq\/Bq)<br \/>\n= 30 Bq<\/p>\n<p>So, we have 30 Bq on the wipe.<\/p>\n<p>Note that the area units cancel and you are left with Bq. We are looking for a result in cpm. The efficiency for the counting system is given by cpm\/dpm. So we need to convert our sample activity to dpm. By definition, the Bq is disintegration per second (dps), so dpm is simply dps multiplied by 60 (60 seconds in 1 minute).<\/p>\n<p>To calculate the resulting count rate, we multiply the activity on wipe (in Bq) by the counting efficiency (in cpm\/dpm), then by 60 (dpm\/Bq).<\/p>\n<p>count rate result<br \/>\n= activity on wipe \u00d7 counting efficiency \u00d7 60 dpm\/Bq<br \/>\n= 30 Bq \u00d7 50% cpm\/dpm \u00d7 60 dpm\/Bq<br \/>\n= 900 cpm<\/p>\n<p>The Bq and dpm cancel, so you are left with an answer in cpm. The answer is b., 900 cpm.<\/p>\n<h1>The question for next time:<\/h1>\n<p>What is not an exemption under the regulations?<\/p>\n<p>a. A Canadian Nuclear Safety Commission inspector possessing a cesium-137 check source<\/p>\n<p>b. A company shipping a replacement source for an industrial radiographer<\/p>\n<p>c. An individual possessing less than an exemption quantity of iodine-125<\/p>\n<p>d. Naturally occurring radioactive material not used in development of nuclear power<\/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>Dans cette section du <em>Bulletin<\/em>, nous pr\u00e9sentons une ou deux questions similaires \u00e0 celles qui se trouvent dans un examen d\u2019agr\u00e9ment (A)ACRP, puis nous publions la solution dans le num\u00e9ro suivant. Le but est de donner aux gens une id\u00e9e du type de questions pr\u00e9sent\u00e9es lors d\u2019un examen d\u2019agr\u00e9ment, voire de convaincre davantage de membres de passer l\u2019examen.<\/p>\n<p>Si vous avez d\u00e9j\u00e0 votre d\u00e9signation (A)ACRP, nous vous invitons \u00e0 <a href=\"mailto:cmalcol@mcmaster.ca\">soumettre des questions<\/a> afin d\u2019obtenir des points pour le maintien de votre agr\u00e9ment!<\/p>\n<h1>Question du dernier num\u00e9ro :<\/h1>\n<p>Quel est le taux de comptage net en comptes par minute (cpm) quand on conna\u00eet les donn\u00e9es suivantes?<\/p>\n<ul>\n<li>Activit\u00e9 non fix\u00e9e 3\u00a0Bq\/cm\u00b2<\/li>\n<li>Surface \u00e9chantillonn\u00e9e 100\u00a0cm\u00b2<\/li>\n<li>Facteur de r\u00e9tention du frottis 10\u00a0%<\/li>\n<li>Efficacit\u00e9 de l\u2019instrument 50\u00a0%<\/li>\n<\/ul>\n<p>a. 15 cpm<br \/>\nb. 900 cpm<br \/>\nc. 90 cpm<br \/>\nd. 150 cpm<\/p>\n<p><strong>Solution propos\u00e9e\u00a0:<\/strong><\/p>\n<p>Il s\u2019agit d\u2019une question de calcul directe que les experts en radioprotection devraient pouvoir r\u00e9soudre assez rapidement. Malgr\u00e9 tout, certaines situations pourraient transformer cette question en d\u00e9fi\u00a0: cela fait peut-\u00eatre un moment que vous n\u2019avez pas fait ce calcul ou peut-\u00eatre ne savez-vous pas comment l\u2019effectuer. Il se peut aussi que votre installation ne g\u00e8re pas beaucoup de contamination non fix\u00e9e, ou encore n\u2019effectue pas de v\u00e9rification indirecte des surfaces.<\/p>\n<p>Si l\u2019une ou l\u2019autre de ces situations s\u2019applique, c\u2019est l\u2019occasion id\u00e9ale d\u2019utiliser la bonne vieille technique d\u2019analyse des unit\u00e9s.<\/p>\n<p>Nous voulons passer de Bq\/cm<sup>2<\/sup> \u00e0 comptes par minute (cpm). Gardons cela \u00e0 l\u2019esprit pendant que nous r\u00e9fl\u00e9chissons \u00e0 ce qui se passe.<\/p>\n<p>Nous disposons d\u2019un frottis utilis\u00e9 pour \u00e9chantillonner une surface de 100\u00a0cm<sup>2<\/sup>. La zone est contamin\u00e9e par 3\u00a0Bq\/cm<sup>2<\/sup> (suppos\u00e9ment r\u00e9partis uniform\u00e9ment sur la surface). Nous avons une efficacit\u00e9 de r\u00e9tention du frottis de 10\u00a0%, ce qui correspond au pourcentage de la contamination de surface transf\u00e9r\u00e9e sur le frottis (ce qui constitue une bonne approximation lorsqu\u2019il y a absence de donn\u00e9es empiriques). Enfin, l\u2019efficacit\u00e9 de l\u2019instrument nous indique combien de cpm nous voyons sur l\u2019instrument pour chaque d\u00e9sint\u00e9gration par minute (dpm) de l\u2019\u00e9chantillon pour la g\u00e9om\u00e9trie de cette situation particuli\u00e8re.<\/p>\n<p>Si nous r\u00e9fl\u00e9chissons \u00e0 ce qui se passe, nous pouvons utiliser l\u2019analyse des unit\u00e9s pour nous assurer que nous sommes sur la bonne voie pour ce calcul. Fondamentalement, nous pouvons calculer l\u2019activit\u00e9 retenue par le frottis, puis calculer la lecture attendue sur notre appareil de mesure pour cette activit\u00e9 donn\u00e9e.<\/p>\n<p>Pour calculer l\u2019activit\u00e9 sur le frottis (en Bq retenu), nous multiplions la contamination non fix\u00e9e (en Bq\/cm<sup>2<\/sup>) par la surface \u00e9chantillonn\u00e9e (en cm<sup>2<\/sup>), et nous multiplions le r\u00e9sultat par le facteur de r\u00e9tention du frottis (en Bq retenu, par contamination non fix\u00e9e en Bq).<\/p>\n<p>Activit\u00e9 sur le frottis<br \/>\n= contamination non fix\u00e9e \u00d7 surface de frottis \u00d7 facteur de r\u00e9tention<br \/>\n= 3 Bq\/cm<sup>2<\/sup>\u00a0\u00d7 100 cm<sup>2<\/sup> \u00d7 10 % (Bq\/Bq)<br \/>\n= 30 Bq<\/p>\n<p>Nous avons donc 30\u00a0Bq sur le frottis.<\/p>\n<p>Notons que les unit\u00e9s de surfaces s\u2019annulent et qu\u2019il ne reste que le Bq. Nous cherchons des r\u00e9sultats en cpm. L\u2019efficacit\u00e9 du syst\u00e8me de comptage est exprim\u00e9e en cpm\/dpm. Nous devons donc convertir l\u2019activit\u00e9 de notre \u00e9chantillon en dpm. Par d\u00e9finition, le\u00a0Bq est en d\u00e9sint\u00e9gration par seconde (dps). Ainsi, pour obtenir le nombre de dpm, il faut multiplier le nombre de dps par 60 (secondes en 1 minute).<\/p>\n<p>Pour calculer le compte r\u00e9sultant,<\/p>\n<p>nous multiplions l\u2019activit\u00e9 sur le frottis (en Bq) par l\u2019efficacit\u00e9 de comptage (en cpm\/dpm), puis par 60 (dpm\/Bq).<\/p>\n<p>r\u00e9sultat du taux de compte<br \/>\n= activit\u00e9 sur le frottis \u00d7 efficacit\u00e9 de comptage \u00d7 60 dpm\/Bq<br \/>\n= 30 Bq \u00d7 50 % cpm\/dpm \u00d7 60 dpm\/Bq<br \/>\n= 900 cpm<\/p>\n<p>Comme les Bq et les dpm s\u2019annulent, il ne reste que le cpm. La r\u00e9ponse est B) 900 CPM.<\/p>\n<h1><strong>Voici la question du prochain num\u00e9ro<\/strong> :<\/h1>\n<p>Qu\u2019est-ce qui n\u2019est pas une exemption r\u00e9glementaire?<\/p>\n<p>a. Un inspecteur de la Commission canadienne de s\u00fbret\u00e9 nucl\u00e9aire poss\u00e9dant une source de contr\u00f4le de c\u00e9sium 137.<\/p>\n<p>b. Une compagnie exp\u00e9diant une source de remplacement pour un radiographe industriel.<\/p>\n<p>c. Un individu poss\u00e9dant moins d\u2019une quantit\u00e9 d\u2019exemption d\u2019iode 125.<\/p>\n<p>d. Une mati\u00e8re radioactive naturelle non utilis\u00e9e dans la g\u00e9n\u00e9ration d\u2019\u00e9nergie nucl\u00e9aire.<\/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>In this section of the <em>Bulletin<\/em>, we introduce a question or two similar to the questions on the CRPA(R) exam. In the next issue, we will provide the solution. \/ Dans cette section du <em>Bulletin<\/em>, nous proposons une ou deux questions similaires \u00e0 celles qui se trouvent dans un examen pour l\u2019agr\u00e9ment (A)ACRP. Dans la publication suivante, nous fournissons la solution.<\/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-1615","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, September 2019 \/ Pr\u00e9paration \u00e0 la d\u00e9signation (A)ACRP, septembre 2019 - 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\/2019\/09\/19\/crpar-prep-september-2019-preparation-a-la-designation-aacrp-septembre-2019\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"CRPA(R) Prep, September 2019 \/ Pr\u00e9paration \u00e0 la d\u00e9signation (A)ACRP, septembre 2019 - CRPA-ACRP Bulletin\" \/>\n<meta property=\"og:description\" content=\"In this section of the Bulletin, we introduce a question or two similar to the questions on the CRPA(R) exam. In the next issue, we will provide the solution. \/ Dans cette section du Bulletin, nous proposons une ou deux questions similaires \u00e0 celles qui se trouvent dans un examen pour l\u2019agr\u00e9ment (A)ACRP. Dans la publication suivante, nous fournissons la solution.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.crpa-acrp-bulletin.ca\/2019\/09\/19\/crpar-prep-september-2019-preparation-a-la-designation-aacrp-septembre-2019\/\" \/>\n<meta property=\"og:site_name\" content=\"CRPA-ACRP Bulletin\" \/>\n<meta property=\"article:published_time\" content=\"2019-09-19T15:13:56+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2020-02-19T02:53:57+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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2019\",\"datePublished\":\"2019-09-19T15:13:56+00:00\",\"dateModified\":\"2020-02-19T02:53:57+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/www.crpa-acrp-bulletin.ca\\\/2019\\\/09\\\/19\\\/crpar-prep-september-2019-preparation-a-la-designation-aacrp-septembre-2019\\\/\"},\"wordCount\":1618,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/www.crpa-acrp-bulletin.ca\\\/#organization\"},\"image\":{\"@id\":\"https:\\\/\\\/www.crpa-acrp-bulletin.ca\\\/2019\\\/09\\\/19\\\/crpar-prep-september-2019-preparation-a-la-designation-aacrp-septembre-2019\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/www.crpa-acrp-bulletin.ca\\\/wp-content\\\/uploads\\\/2018\\\/07\\\/CRPAR-icon-ftr.png\",\"articleSection\":[\"Chroniques r\u00e9guli\u00e8res\",\"Regular 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