{"id":186321,"date":"2025-01-24T18:03:28","date_gmt":"2025-01-24T18:03:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186321"},"modified":"2025-01-24T18:03:30","modified_gmt":"2025-01-24T18:03:30","slug":"the-half-life-of-plutonium-239-is-24300-years-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/24\/the-half-life-of-plutonium-239-is-24300-years-2\/","title":{"rendered":"The half-life of plutonium-239 is 24,300 years"},"content":{"rendered":"\n<p>) The half-life of plutonium-239 is 24,300 years. If a nuclear bomb released 8 kg of this isotope, how many years would pass before the amount is reduced to 1 kg? The half-life of radon-222 is 3.8 days. How much of a 100 gram sample is left after 15.2 days? 0705 Carbon-14 has a half-life of 5,730 vears. If a sample contained 90 mg originally, how much is left after 17.190 years? 6. The half-life of cobalt-60 is 5.26 years. If 50 grams are left after 15.78 years, how many grams were in the original sample? No 7. The half-life of 1-137 is 8.07 days. If 25 grams are left after 40.35 days, how many grams were in the original sample? 8. If 100 grams of Au-198 decays to 6.25 grams in 10.8 days, what is the half-life of Au-198?<\/p>\n\n\n\n<p><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>Let&#8217;s break down each of the problems one by one. The common formula we will use for all of them is the radioactive decay formula:<\/p>\n\n\n\n<p>[<br>A = A_0 \\times \\left(\\frac{1}{2}\\right)^{\\frac{t}{T}}<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( A ) is the remaining amount of the substance.<\/li>\n\n\n\n<li>( A_0 ) is the initial amount of the substance.<\/li>\n\n\n\n<li>( t ) is the time elapsed.<\/li>\n\n\n\n<li>( T ) is the half-life of the substance.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">1. Half-life of Plutonium-239 (24,300 years)<\/h3>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial amount, ( A_0 = 8 \\, \\text{kg} ).<\/li>\n\n\n\n<li>Remaining amount, ( A = 1 \\, \\text{kg} ).<\/li>\n\n\n\n<li>Half-life, ( T = 24,300 \\, \\text{years} ).<\/li>\n\n\n\n<li>We need to find ( t ), the time it takes to decay from 8 kg to 1 kg.<\/li>\n<\/ul>\n\n\n\n<p>Using the decay formula:<\/p>\n\n\n\n<p>[<br>1 = 8 \\times \\left(\\frac{1}{2}\\right)^{\\frac{t}{24,300}}<br>]<\/p>\n\n\n\n<p>Solving for ( t ):<\/p>\n\n\n\n<p>[<br>\\frac{1}{8} = \\left(\\frac{1}{2}\\right)^{\\frac{t}{24,300}}<br>]<\/p>\n\n\n\n<p>Taking the natural logarithm of both sides:<\/p>\n\n\n\n<p>[<br>\\ln\\left(\\frac{1}{8}\\right) = \\frac{t}{24,300} \\times \\ln\\left(\\frac{1}{2}\\right)<br>]<\/p>\n\n\n\n<p>[<br>t = \\frac{\\ln\\left(\\frac{1}{8}\\right)}{\\ln\\left(\\frac{1}{2}\\right)} \\times 24,300<br>]<\/p>\n\n\n\n<p>[<br>t \\approx 73,000 \\, \\text{years}<br>]<\/p>\n\n\n\n<p>So, it would take approximately <strong>73,000 years<\/strong> for the amount of plutonium-239 to reduce to 1 kg.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2. Half-life of Radon-222 (3.8 days)<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial amount, ( A_0 = 100 \\, \\text{g} ).<\/li>\n\n\n\n<li>Half-life, ( T = 3.8 \\, \\text{days} ).<\/li>\n\n\n\n<li>Time, ( t = 15.2 \\, \\text{days} ).<\/li>\n<\/ul>\n\n\n\n<p>Using the decay formula:<\/p>\n\n\n\n<p>[<br>A = 100 \\times \\left(\\frac{1}{2}\\right)^{\\frac{15.2}{3.8}}<br>]<\/p>\n\n\n\n<p>[<br>A = 100 \\times \\left(\\frac{1}{2}\\right)^{4}<br>]<\/p>\n\n\n\n<p>[<br>A = 100 \\times \\frac{1}{16} = 6.25 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, after 15.2 days, <strong>6.25 grams<\/strong> of Radon-222 remain.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3. Half-life of Carbon-14 (5730 years)<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial amount, ( A_0 = 90 \\, \\text{mg} ).<\/li>\n\n\n\n<li>Time, ( t = 17,190 \\, \\text{years} ).<\/li>\n\n\n\n<li>Half-life, ( T = 5730 \\, \\text{years} ).<\/li>\n<\/ul>\n\n\n\n<p>Using the decay formula:<\/p>\n\n\n\n<p>[<br>A = 90 \\times \\left(\\frac{1}{2}\\right)^{\\frac{17,190}{5730}}<br>]<\/p>\n\n\n\n<p>[<br>A = 90 \\times \\left(\\frac{1}{2}\\right)^{3}<br>]<\/p>\n\n\n\n<p>[<br>A = 90 \\times \\frac{1}{8} = 11.25 \\, \\text{mg}<br>]<\/p>\n\n\n\n<p>So, after 17,190 years, <strong>11.25 mg<\/strong> of Carbon-14 remain.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">4. Half-life of Cobalt-60 (5.26 years)<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Remaining amount, ( A = 50 \\, \\text{g} ).<\/li>\n\n\n\n<li>Time, ( t = 15.78 \\, \\text{years} ).<\/li>\n\n\n\n<li>Half-life, ( T = 5.26 \\, \\text{years} ).<\/li>\n\n\n\n<li>We need to find the initial amount ( A_0 ).<\/li>\n<\/ul>\n\n\n\n<p>Using the decay formula:<\/p>\n\n\n\n<p>[<br>50 = A_0 \\times \\left(\\frac{1}{2}\\right)^{\\frac{15.78}{5.26}}<br>]<\/p>\n\n\n\n<p>[<br>50 = A_0 \\times \\left(\\frac{1}{2}\\right)^{3}<br>]<\/p>\n\n\n\n<p>[<br>50 = A_0 \\times \\frac{1}{8}<br>]<\/p>\n\n\n\n<p>[<br>A_0 = 50 \\times 8 = 400 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, the initial amount of Cobalt-60 was <strong>400 grams<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">5. Half-life of Iodine-137 (8.07 days)<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Remaining amount, ( A = 25 \\, \\text{g} ).<\/li>\n\n\n\n<li>Time, ( t = 40.35 \\, \\text{days} ).<\/li>\n\n\n\n<li>We need to find the initial amount ( A_0 ).<\/li>\n<\/ul>\n\n\n\n<p>Using the decay formula:<\/p>\n\n\n\n<p>[<br>25 = A_0 \\times \\left(\\frac{1}{2}\\right)^{\\frac{40.35}{8.07}}<br>]<\/p>\n\n\n\n<p>[<br>25 = A_0 \\times \\left(\\frac{1}{2}\\right)^{5}<br>]<\/p>\n\n\n\n<p>[<br>25 = A_0 \\times \\frac{1}{32}<br>]<\/p>\n\n\n\n<p>[<br>A_0 = 25 \\times 32 = 800 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, the initial amount of Iodine-137 was <strong>800 grams<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">6. Half-life of Gold-198 (Au-198)<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial amount, ( A_0 = 100 \\, \\text{g} ).<\/li>\n\n\n\n<li>Remaining amount, ( A = 6.25 \\, \\text{g} ).<\/li>\n\n\n\n<li>Time, ( t = 10.8 \\, \\text{days} ).<\/li>\n\n\n\n<li>We need to find the half-life ( T ).<\/li>\n<\/ul>\n\n\n\n<p>Using the decay formula:<\/p>\n\n\n\n<p>[<br>6.25 = 100 \\times \\left(\\frac{1}{2}\\right)^{\\frac{10.8}{T}}<br>]<\/p>\n\n\n\n<p>[<br>\\frac{6.25}{100} = \\left(\\frac{1}{2}\\right)^{\\frac{10.8}{T}}<br>]<\/p>\n\n\n\n<p>[<br>0.0625 = \\left(\\frac{1}{2}\\right)^{\\frac{10.8}{T}}<br>]<\/p>\n\n\n\n<p>Since ( 0.0625 = \\left(\\frac{1}{2}\\right)^{4} ), we have:<\/p>\n\n\n\n<p>[<br>\\frac{10.8}{T} = 4<br>]<\/p>\n\n\n\n<p>[<br>T = \\frac{10.8}{4} = 2.7 \\, \\text{days}<br>]<\/p>\n\n\n\n<p>So, the half-life of Au-198 is <strong>2.7 days<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Plutonium-239: 73,000 years to decay to 1 kg.<\/li>\n\n\n\n<li>Radon-222: 6.25 grams remain after 15.2 days.<\/li>\n\n\n\n<li>Carbon-14: 11.25 mg remain after 17,190 years.<\/li>\n\n\n\n<li>Cobalt-60: 400 grams were in the original sample.<\/li>\n\n\n\n<li>Iodine-137: 800 grams were in the original sample.<\/li>\n\n\n\n<li>Gold-198: The half-life is 2.7 days.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>) The half-life of plutonium-239 is 24,300 years. If a nuclear bomb released 8 kg of this isotope, how many years would pass before the amount is reduced to 1 kg? The half-life of radon-222 is 3.8 days. How much of a 100 gram sample is left after 15.2 days? 0705 Carbon-14 has a half-life [&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-186321","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186321","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=186321"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186321\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186321"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186321"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186321"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}