{"id":38354,"date":"2025-06-26T09:27:01","date_gmt":"2025-06-26T09:27:01","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=38354"},"modified":"2025-06-26T09:27:03","modified_gmt":"2025-06-26T09:27:03","slug":"the-half-life-of-plutonium-239-is-24300-years","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-half-life-of-plutonium-239-is-24300-years\/","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? Use the law of radiation decay<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this, we use the <strong>law of radioactive decay<\/strong>, which is based on exponential decay. The formula is: A(t)=A0\u22c5(12)t\/TA(t) = A_0 \\cdot \\left(\\frac{1}{2}\\right)^{t\/T}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A(t)A(t) is the amount remaining after time tt<\/li>\n\n\n\n<li>A0A_0 is the initial amount<\/li>\n\n\n\n<li>TT is the half-life<\/li>\n\n\n\n<li>tt is the time that has passed<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A0=8A_0 = 8 kg<\/li>\n\n\n\n<li>A(t)=1A(t) = 1 kg<\/li>\n\n\n\n<li>T=24,300T = 24,300 years<\/li>\n<\/ul>\n\n\n\n<p>We want to solve for tt.<\/p>\n\n\n\n<p>Start by setting up the equation: 1=8\u22c5(12)t\/243001 = 8 \\cdot \\left(\\frac{1}{2}\\right)^{t \/ 24300}<\/p>\n\n\n\n<p>Divide both sides by 8: 18=(12)t\/24300\\frac{1}{8} = \\left(\\frac{1}{2}\\right)^{t \/ 24300}<\/p>\n\n\n\n<p>Now recognize that 18=(12)3\\frac{1}{8} = \\left(\\frac{1}{2}\\right)^3, so: (12)3=(12)t\/24300\\left(\\frac{1}{2}\\right)^3 = \\left(\\frac{1}{2}\\right)^{t \/ 24300}<\/p>\n\n\n\n<p>Since the bases are the same, equate the exponents: 3=t243003 = \\frac{t}{24300}<\/p>\n\n\n\n<p>Multiply both sides by 24,300: t=3\u22c524300=72,900&nbsp;yearst = 3 \\cdot 24300 = 72,900 \\text{ years}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>72,900 years<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>Radioactive decay follows an exponential pattern, where the amount of a radioactive substance is halved after each half-life. In this problem, plutonium-239 has a half-life of 24,300 years, meaning that every 24,300 years, only half of the substance remains. Starting with 8 kg and aiming to find when only 1 kg remains means the substance must go through three half-lives: from 8 to 4 kg, then to 2 kg, and finally to 1 kg. Each of those steps takes 24,300 years. Multiplying 3 half-lives by 24,300 years gives 72,900 years. This shows how long radioactive materials can persist in the environment and why their containment is a serious issue in nuclear safety and environmental science.<\/p>\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? Use the law of radiation decay The correct answer and explanation is: To solve this, we use the law of radioactive decay, which is based [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-38354","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38354","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=38354"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38354\/revisions"}],"predecessor-version":[{"id":38355,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38354\/revisions\/38355"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=38354"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=38354"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=38354"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}