{"id":258679,"date":"2025-07-18T08:48:26","date_gmt":"2025-07-18T08:48:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=258679"},"modified":"2025-07-18T08:48:28","modified_gmt":"2025-07-18T08:48:28","slug":"if-a-radioactive-isotope-has-a-half-life-of-2-5-what-fraction-of-the-original-amount-will-remain-after-10-years","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/18\/if-a-radioactive-isotope-has-a-half-life-of-2-5-what-fraction-of-the-original-amount-will-remain-after-10-years\/","title":{"rendered":"If a radioactive isotope has a half-life of 2.5 , what fraction of the original amount will remain after 10 years"},"content":{"rendered":"\n<p>If a radioactive isotope has a half-life of 2.5 , what fraction of the original amount will remain after 10 years? How many years will it take to reduce the original amount to only 10%?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To answer this question, we will use the concept of half-life, which is the time required for half of the material to decay. Here&#8217;s how we can approach each part of the question:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. Fraction Remaining After 10 Years<\/h3>\n\n\n\n<p>The general formula for the decay of a radioactive substance is given by: N(t)=N0(12)tT1\/2N(t) = N_0 \\left( \\frac{1}{2} \\right)^{\\frac{t}{T_{1\/2}}}N(t)=N0\u200b(21\u200b)T1\/2\u200bt\u200b<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>N(t)N(t)N(t) is the amount remaining after time ttt<\/li>\n\n\n\n<li>N0N_0N0\u200b is the initial amount<\/li>\n\n\n\n<li>T1\/2T_{1\/2}T1\/2\u200b is the half-life of the isotope<\/li>\n\n\n\n<li>ttt is the time that has passed<\/li>\n<\/ul>\n\n\n\n<p>For this question:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Half-life, T1\/2=2.5T_{1\/2} = 2.5T1\/2\u200b=2.5 years<\/li>\n\n\n\n<li>Time, t=10t = 10t=10 years<\/li>\n<\/ul>\n\n\n\n<p>We substitute the given values into the equation: N(10)=N0(12)102.5=N0(12)4N(10) = N_0 \\left( \\frac{1}{2} \\right)^{\\frac{10}{2.5}} = N_0 \\left( \\frac{1}{2} \\right)^4N(10)=N0\u200b(21\u200b)2.510\u200b=N0\u200b(21\u200b)4 N(10)=N0\u00d7116N(10) = N_0 \\times \\frac{1}{16}N(10)=N0\u200b\u00d7161\u200b<\/p>\n\n\n\n<p>So, after 10 years, only 116\\frac{1}{16}161\u200b of the original amount will remain.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. Time to Reduce the Amount to 10%<\/h3>\n\n\n\n<p>Now, we need to find the time it takes for the original amount to reduce to 10% of its initial value. We want to find the time ttt when N(t)=0.1\u00d7N0N(t) = 0.1 \\times N_0N(t)=0.1\u00d7N0\u200b. Using the same decay formula: 0.1N0=N0(12)tT1\/20.1 N_0 = N_0 \\left( \\frac{1}{2} \\right)^{\\frac{t}{T_{1\/2}}}0.1N0\u200b=N0\u200b(21\u200b)T1\/2\u200bt\u200b<\/p>\n\n\n\n<p>Divide both sides by N0N_0N0\u200b: 0.1=(12)t2.50.1 = \\left( \\frac{1}{2} \\right)^{\\frac{t}{2.5}}0.1=(21\u200b)2.5t\u200b<\/p>\n\n\n\n<p>Taking the natural logarithm (ln) of both sides: ln\u2061(0.1)=t2.5ln\u2061(12)\\ln(0.1) = \\frac{t}{2.5} \\ln\\left( \\frac{1}{2} \\right)ln(0.1)=2.5t\u200bln(21\u200b)<\/p>\n\n\n\n<p>Now, solve for ttt: ln\u2061(0.1)=\u22122.3026andln\u2061(12)=\u22120.6931\\ln(0.1) = -2.3026 \\quad \\text{and} \\quad \\ln\\left( \\frac{1}{2} \\right) = -0.6931ln(0.1)=\u22122.3026andln(21\u200b)=\u22120.6931 \u22122.3026=t2.5(\u22120.6931)-2.3026 = \\frac{t}{2.5} (-0.6931)\u22122.3026=2.5t\u200b(\u22120.6931) t=\u22122.3026\u00d72.5\u22120.6931=8.3\u2009yearst = \\frac{-2.3026 \\times 2.5}{-0.6931} = 8.3 \\, \\text{years}t=\u22120.6931\u22122.3026\u00d72.5\u200b=8.3years<\/p>\n\n\n\n<p>Thus, it will take approximately 8.3 years to reduce the original amount to 10%.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>After 10 years, 116\\frac{1}{16}161\u200b of the original isotope will remain.<\/li>\n\n\n\n<li>It will take about 8.3 years to reduce the original amount to only 10%.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-211.jpeg\" alt=\"\" class=\"wp-image-258680\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>If a radioactive isotope has a half-life of 2.5 , what fraction of the original amount will remain after 10 years? How many years will it take to reduce the original amount to only 10%? The Correct Answer and Explanation is: To answer this question, we will use the concept of half-life, which is the [&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-258679","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258679","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=258679"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258679\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258679"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258679"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258679"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}