{"id":1871,"date":"2020-01-22T19:02:28","date_gmt":"2020-01-23T01:02:28","guid":{"rendered":"https:\/\/www.crpa-acrp-bulletin.ca\/?p=1871"},"modified":"2020-02-21T11:19:28","modified_gmt":"2020-02-21T17:19:28","slug":"crpar-prep-january-2020-preparation-a-la-designation-aacrp-janvier-2020","status":"publish","type":"post","link":"https:\/\/www.crpa-acrp-bulletin.ca\/fr\/2020\/01\/22\/crpar-prep-january-2020-preparation-a-la-designation-aacrp-janvier-2020\/","title":{"rendered":"CRPA(R) Prep, January 2020 \/ Pr\u00e9paration \u00e0 la d\u00e9signation (A)ACRP, janvier 2020"},"content":{"rendered":"

\"\"In this section of the\u00a0Bulletin<\/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 we use on the CRPA(R) exam and perhaps convince more members to challenge the exam.<\/p>\n

The CRPA(R) designation is the highest level of competency recognized by CRPA at the Canadian level. To find out more, visit the\u00a0CRPA website<\/a>.<\/p>\n

If you already have your CRPA(R) designation, we invite you to\u00a0submit questions<\/a>\u00a0to earn points for your registration maintenance!<\/p>\n

Question from the last issue:<\/strong><\/h1>\n

We went back to some calculations\u2026<\/p>\n

A Cs-137 source has a current activity of 20.76 kBq. A total number of 579,605 counts are measured over a five-minute counting period. If the background count rate is 88.7 cpm, what is the detector efficiency for this counter?<\/p>\n

a. 0.056
\nb. 0.093
\nc. 0.113
\nd. 0.1<\/p>\n

Our last question was purely knowledge based, so if you didn\u2019t know the answer or didn\u2019t know where to find it in the provided set of regulations, you might have had to throw a dart!<\/p>\n

This question is for all you calculation lovers out there. With general knowledge of counting instruments and associated calculations, you can figure this one out in your sleep. If you have only basic knowledge of radiation protection fundamentals, you should still be able to work out the answer. Even using unit analysis, you might be able to get the job done.<\/p>\n

The CRPA Radiation Safety Professional Registration Process document (found here<\/a>) is a good resource. Section 4 includes a recommended reading list with many resources that would address this particular topic. My favourite resource for this kind of question would be Knoll\u2019s Radiation Detection and Measurement<\/em>.[1]<\/a><\/p>\n

However, even if you have (misguidedly) ignored the registration subcommittee\u2019s terrific advice to go through the recommended reading list, and you managed to register for the exam while having no idea what detector efficiency is, you can still take a stab at the question by inferring what is meant. Basically we want to know, how good is this fictional detector at detecting?<\/p>\n

If you did (sensibly) heed the subcommittee\u2019s advice to go through the recommended reading, you would know that the detector efficiency (in this case, specifically, the absolute efficiency, which you would have understood from the context of the question and information provided) is given in Knoll as:<\/p>\n

Absolute Efficiency = Number of pulses recorded\/Number of radiation quanta emitted by the source<\/p>\n

The number of pulses recorded is a gross count of 579,605 in 5 minutes, or 115,921 counts per minute (cpm). When we subtract the background count rate of 88.7 cpm, we get a net count of 115,832.3 cpm. (Interestingly, the background count rate is so small compared to the measured rate that failure to include it should yield the same answer.)<\/p>\n

The number of quanta is given as an activity (assuming one quanta per disintegration). Remember that the definition of a becquerel is one disintegration per second (dps). So we are actually given 20,760 dps.<\/p>\n

Efficiency is usually described in units of counts per second (cps)\/disintegrations per second (dps). So we need to divide our net measured count rate by 60 (seconds per minute) to go from cpm to cps to get our units to line up, which gives us 1,931 cps. Plug \u2019n\u2019 play time!<\/p>\n

Absolute Efficiency = 1,931 counts per second (cps)\/20,760 disintegrations per second (dps)<\/p>\n

Working that out gives us a detector efficiency of 0.093 (9.3%) so the answer is b.<\/p>\n

The question for next time:<\/strong><\/h1>\n

Let\u2019s stay in the measurement and counting realm, but make it a little more difficult.<\/p>\n

A gamma counter produces a measurement of 1,805 counts over a five-minute period. What is the count rate and associated standard deviation?<\/p>\n

a. 361 + 115 cpm
\nb. 361 + 75\u00a0cpm
\nc. 361 + 9 cpm
\nd. 361 + 11 cpm<\/p>\n

[1] <\/a>Knoll, G.F. (2000).\u00a0Radiation Detection and Measurement <\/em>(4th ed.). Hoboken, NJ: John Wiley & Sons.<\/p><\/div><\/div>\n

\n

Dans cette section du 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\u00e9sentes dans un examen d\u2019agr\u00e9ment et ainsi peut-\u00eatre convaincre davantage de membres de passer l\u2019examen.<\/p>\n

La d\u00e9signation (A)ACRP est le niveau de comp\u00e9tence le plus \u00e9lev\u00e9 reconnu par l\u2019ACRP au niveau canadien. Pour en savoir plus, visitez le site Web de l\u2019ACRP<\/a>.<\/p>\n

Si vous avez d\u00e9j\u00e0 votre d\u00e9signation (A)ACRP, nous vous invitons \u00e0 soumettre des questions<\/a> afin d\u2019obtenir des points pour le maintien de votre agr\u00e9ment!<\/p>\n

Question du dernier num\u00e9ro\u00a0:<\/strong><\/h1>\n

Nous \u00e9tions de retour aux calculs\u2026<\/p>\n

Une source de Cs-137 a pr\u00e9sentement une activit\u00e9 de 20,76 kBq. Un total de 579\u2009605 comptes a \u00e9t\u00e9 mesur\u00e9 sur une p\u00e9riode de comptage de 5 minutes. Si le taux de comptage du fond naturel de rayonnement est de 88,7 cpm, quelle est l\u2019efficacit\u00e9 du d\u00e9tecteur de ce compteur?<\/p>\n

a. 0,056
\nb. 0,093
\nc. 0,113
\nd. 0,1<\/p>\n

Cette question est enti\u00e8rement fond\u00e9e sur le savoir. Ainsi, si vous ne connaissez pas la r\u00e9ponse ou si vous ne savez pas o\u00f9 la trouver dans la r\u00e9glementation fournie, il vous faudra tirer \u00e0 pile ou face!<\/p>\n

Avec des connaissances g\u00e9n\u00e9rales sur les instruments de comptage et les calculs associ\u00e9s, les amateurs de calculs peuvent y r\u00e9pondre. M\u00eame si vous ne disposez que d\u2019une connaissance de base sur les principes fondamentaux en radioprotection, vous devriez \u00eatre en mesure de trouver la r\u00e9ponse. En outre, analyser les unit\u00e9s permettra de r\u00e9pondre \u00e0 la question.<\/p>\n

Le document \u00ab\u00a0Processus d\u2019agr\u00e9ment professionnel de l\u2019ACRP<\/a>\u00a0\u00bb est une bonne ressource. Sa section\u00a04 comprend une liste de lectures recommand\u00e9es avec plusieurs ressources abordant ce sujet en particulier. Ma ressource pr\u00e9f\u00e9r\u00e9e pour ce type de question est Radiation Detection and Measurement<\/em> de Knoll.[1]<\/a><\/p>\n

Cependant, m\u00eame si vous avez (malencontreusement) ignor\u00e9 les formidables conseils du sous-comit\u00e9 de l\u2019agr\u00e9ment de passer en revue la liste des lectures recommand\u00e9es et que vous vous \u00eates inscrit \u00e0 l\u2019examen sans savoir ce qu\u2019est l\u2019efficacit\u00e9 d\u2019un d\u00e9tecteur, vous pouvez tout de m\u00eame tenter votre chance en d\u00e9duisant ce dont il est question. Fondamentalement, nous voulons savoir \u00e0 quel point ce d\u00e9tecteur fictif est capable de d\u00e9tecter.<\/p>\n

Si vous avez (judicieusement) suivi les conseils du sous-comit\u00e9, vous sauriez que l\u2019efficacit\u00e9 du d\u00e9tecteur (c\u2019est-\u00e0-dire l\u2019efficacit\u00e9 absolue, selon le contexte de la question et les informations fournies) est donn\u00e9e par Knoll comme \u00e9tant\u00a0:<\/p>\n

Efficacit\u00e9 absolue = Nombre d\u2019impulsions enregistr\u00e9es\/Nombre de quanta de rayonnement \u00e9mis par la source<\/p>\n

Le nombre d\u2019impulsions enregistr\u00e9es est un compte brut de 579\u2009605 en 5 minutes, ou 115\u2009921 comptes par minutes (cpm). En soustrayant le taux de comptage du fond naturel de rayonnement qui est de 88,7 cpm, nous obtenons un compte net de 115\u2009832,3 cpm. (Fait int\u00e9ressant, le taux de comptage du fond naturel de rayonnement est si faible par rapport au taux mesur\u00e9 que le fait de ne pas l\u2019inclure donne la m\u00eame r\u00e9ponse.)<\/p>\n

Le nombre de quanta est donn\u00e9 comme une activit\u00e9 (en supposant un quanta par d\u00e9sint\u00e9gration). Rappelez-vous que la d\u00e9finition du becquerel est de 1 d\u00e9sint\u00e9gration par seconde (dps). Ainsi, nous avons en r\u00e9alit\u00e9 20\u2009760 dps.<\/p>\n

L\u2019efficacit\u00e9 est g\u00e9n\u00e9ralement d\u00e9crite en unit\u00e9s de comptes par seconde (cps)\/d\u00e9sint\u00e9gration par seconde (dps). Nous devons donc diviser le taux de comptage net mesur\u00e9 par 60 (secondes par minute) pour passer de cpm \u00e0 cps afin d\u2019avoir les m\u00eames unit\u00e9s, ce qui donne 1\u2009931 cps. C\u2019est le temps de calculer\u00a0!<\/p>\n

Efficacit\u00e9 absolue = 1\u2009931 comptes par seconds (cps)\/
\n20\u2009760 d\u00e9sint\u00e9grations par seconde (dps)<\/p>\n

Comme l\u2019efficacit\u00e9 du d\u00e9tecteur est de 0,093 (9,3\u00a0%), la r\u00e9ponse est b.<\/p>\n

La question pour le prochain num\u00e9ro\u00a0:<\/strong><\/h1>\n

Restons dans le domaine des mesures et du comptage, mais complexifions les choses.<\/p>\n

Un compteur gamma mesure 1\u2009805 comptes sur une p\u00e9riode de 5 minutes. Quels sont le taux de comptage et l\u2019\u00e9cart type associ\u00e9s?<\/p>\n

a. 361 + 115 cpm
\nb. 361 + 75 cpm
\nc. 361 + 9 cpm
\nd. 361 + 11 cpm<\/p>\n

[1]<\/a> Knoll, G.F. (2000). Radiation Detection and Measurement <\/em>(4e<\/sup> \u00e9d.). Hoboken, NJ : John Wiley & Sons.<\/p>\n<\/div><\/div><\/div>

<\/div>\n
\n

Christopher Malcolmson<\/h2>\n

\"\"<\/h2>\n

Christopher Malcolmson has been a health physicist at McMaster University since 2005. He 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. Malcolmson is the Registration Subcommittee exam coordinator. He is also a member of the International Radiation Protection Association\u2019s Commission on Publications.<\/p>\n

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 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":"

In this section of the 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 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-1871","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-chroniques-regulieres","category-regular-columns"],"yoast_head":"\nCRPA(R) Prep, January 2020 \/ Pr\u00e9paration \u00e0 la d\u00e9signation (A)ACRP, janvier 2020 - 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\/fr\/2020\/01\/22\/crpar-prep-january-2020-preparation-a-la-designation-aacrp-janvier-2020\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"CRPA(R) Prep, January 2020 \/ Pr\u00e9paration \u00e0 la d\u00e9signation (A)ACRP, janvier 2020 - 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. 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