{"id":202738,"date":"2025-03-18T05:09:05","date_gmt":"2025-03-18T05:09:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=202738"},"modified":"2025-03-18T05:09:08","modified_gmt":"2025-03-18T05:09:08","slug":"what-is-the-half-life-of-an-isotope-if-436-seconds-are-required-for-the-number-of-cpm-to-diminish-from-14227-to-3672","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/18\/what-is-the-half-life-of-an-isotope-if-436-seconds-are-required-for-the-number-of-cpm-to-diminish-from-14227-to-3672\/","title":{"rendered":"What is the half-life of an isotope if 436 seconds are required for the number of cpm to diminish from 14,227 to 3672"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">What is the half-life of an isotope if 436 seconds are required for the number of cpm to diminish from 14,227 to 3672 Show your calculations.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To determine the half-life (( t_{1\/2} )) of the isotope, we use the first-order decay formula:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>N = N_0 e^{-\\lambda t}<br>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( N_0 ) is the initial count rate (14,227 cpm),<\/li>\n\n\n\n<li>( N ) is the final count rate (3,672 cpm),<\/li>\n\n\n\n<li>( \\lambda ) is the decay constant,<\/li>\n\n\n\n<li>( t ) is the time elapsed (436 seconds),<\/li>\n\n\n\n<li>( e ) is the natural logarithm base (~2.718).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Solve for ( \\lambda )<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Taking the natural logarithm on both sides:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>\\ln \\left( \\frac{N}{N_0} \\right) = -\\lambda t<br>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Substituting the values:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>\\ln \\left( \\frac{3672}{14227} \\right) = -\\lambda (436)<br>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>\\ln (0.2581) = -\\lambda (436)<br>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>-1.354 = -\\lambda (436)<br>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>\\lambda = \\frac{1.354}{436}<br>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>\\lambda = 0.00311 \\text{ s}^{-1}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine Half-Life<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The relationship between the half-life and decay constant is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>t_{1\/2} = \\frac{\\ln 2}{\\lambda}<br>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>t_{1\/2} = \\frac{0.693}{0.00311}<br>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>t_{1\/2} = 223 \\text{ seconds}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300 Words)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The half-life of an isotope is the time required for half of a radioactive substance to decay. It is a key property in nuclear chemistry and radiometric dating.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In this problem, we determined the half-life using first-order decay kinetics. Radioactive decay follows the exponential law, meaning the quantity of a substance decreases at a rate proportional to its current amount. This is described by the equation ( N = N_0 e^{-\\lambda t} ), where ( \\lambda ) is the decay constant.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By rearranging and solving for ( \\lambda ), we found it to be 0.00311 s(^{-1}). The half-life is calculated using the relationship ( t_{1\/2} = \\frac{\\ln 2}{\\lambda} ), which gives 223 seconds.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This means that every 223 seconds, half of the radioactive atoms decay. Understanding half-life is crucial in fields such as medicine (radiotherapy), archaeology (carbon dating), and nuclear power.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is the half-life of an isotope if 436 seconds are required for the number of cpm to diminish from 14,227 to 3672 Show your calculations. The correct answer and explanation is : To determine the half-life (( t_{1\/2} )) of the isotope, we use the first-order decay formula: [N = N_0 e^{-\\lambda t}] where: [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-202738","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202738","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=202738"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202738\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202738"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202738"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202738"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}