{"id":186319,"date":"2025-01-24T17:53:35","date_gmt":"2025-01-24T17:53:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186319"},"modified":"2025-01-24T17:53:38","modified_gmt":"2025-01-24T17:53:38","slug":"the-half-life-of-plutonium-239-is-24300-years","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/24\/the-half-life-of-plutonium-239-is-24300-years\/","title":{"rendered":"The half-life of plutonium-239 is 24,300 years"},"content":{"rendered":"\n
    \n
  1. 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 ?<\/li>\n\n\n\n
  2. The half-life of radon-222 is 3.8 days. How much of a 100 gram sample is left after 15.2 days?<\/li>\n<\/ol>\n\n\n\n

    The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

    Let’s address both questions step by step, using the half-life formula:<\/p>\n\n\n\n

    Formula for Half-Life:<\/h3>\n\n\n\n

    The general formula for the amount of a substance remaining after a certain period is: A=A0\u00d7(12)tTA = A_0 \\times \\left( \\frac{1}{2} \\right)^{\\frac{t}{T}}<\/p>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • AA is the amount remaining after time tt,<\/li>\n\n\n\n
    • A0A_0 is the initial amount,<\/li>\n\n\n\n
    • TT is the half-life of the substance,<\/li>\n\n\n\n
    • tt is the time elapsed.<\/li>\n<\/ul>\n\n\n\n

      1. Plutonium-239 Question:<\/h3>\n\n\n\n

      Given:<\/strong><\/p>\n\n\n\n

        \n
      • Initial mass (A0A_0) = 8 kg,<\/li>\n\n\n\n
      • Final mass (AA) = 1 kg,<\/li>\n\n\n\n
      • Half-life (TT) = 24,300 years.<\/li>\n<\/ul>\n\n\n\n

        We need to solve for tt, the time it will take for the mass to reduce from 8 kg to 1 kg.<\/p>\n\n\n\n

        Using the formula: 1=8\u00d7(12)t24,3001 = 8 \\times \\left( \\frac{1}{2} \\right)^{\\frac{t}{24,300}}<\/p>\n\n\n\n

        Solving for tt: 18=(12)t24,300\\frac{1}{8} = \\left( \\frac{1}{2} \\right)^{\\frac{t}{24,300}}<\/p>\n\n\n\n

        Taking the natural logarithm of both sides: ln\u2061(18)=t24,300\u00d7ln\u2061(12)\\ln \\left( \\frac{1}{8} \\right) = \\frac{t}{24,300} \\times \\ln \\left( \\frac{1}{2} \\right)<\/p>\n\n\n\n

        Since ln\u2061(12)=\u2212ln\u2061(2)\\ln \\left( \\frac{1}{2} \\right) = -\\ln(2), we have: ln\u2061(1)\u2212ln\u2061(8)=t24,300\u00d7(\u2212ln\u2061(2))\\ln(1) – \\ln(8) = \\frac{t}{24,300} \\times (-\\ln(2)) ln\u2061(8)\u22482.0794,ln\u2061(2)\u22480.6931\\ln(8) \\approx 2.0794, \\quad \\ln(2) \\approx 0.6931<\/p>\n\n\n\n

        Thus: \u22122.0794=t24,300\u00d7(\u22120.6931)-2.0794 = \\frac{t}{24,300} \\times (-0.6931)<\/p>\n\n\n\n

        Now, solve for tt: t=2.0794\u00d724,3000.6931\u224872,500 yearst = \\frac{2.0794 \\times 24,300}{0.6931} \\approx 72,500 \\text{ years}<\/p>\n\n\n\n

        Thus, it will take approximately 72,500 years<\/strong> for the plutonium-239 to reduce from 8 kg to 1 kg.<\/p>\n\n\n\n

        \n\n\n\n

        2. Radon-222 Question:<\/h3>\n\n\n\n

        Given:<\/strong><\/p>\n\n\n\n

          \n
        • Initial mass (A0A_0) = 100 grams,<\/li>\n\n\n\n
        • Half-life (TT) = 3.8 days,<\/li>\n\n\n\n
        • Time elapsed (tt) = 15.2 days.<\/li>\n<\/ul>\n\n\n\n

          We need to determine how much of the original 100-gram sample remains after 15.2 days.<\/p>\n\n\n\n

          Using the same half-life formula: A=100\u00d7(12)15.23.8A = 100 \\times \\left( \\frac{1}{2} \\right)^{\\frac{15.2}{3.8}}<\/p>\n\n\n\n

          First, calculate 15.23.8\\frac{15.2}{3.8}: 15.23.8=4\\frac{15.2}{3.8} = 4<\/p>\n\n\n\n

          Now: A=100\u00d7(12)4A = 100 \\times \\left( \\frac{1}{2} \\right)^4 A=100\u00d7116=6.25 gramsA = 100 \\times \\frac{1}{16} = 6.25 \\text{ grams}<\/p>\n\n\n\n

          Thus, after 15.2 days, 6.25 grams<\/strong> of the original 100 grams of radon-222 will remain.<\/p>\n\n\n\n

          \n\n\n\n

          Explanation:<\/h3>\n\n\n\n
            \n
          1. Plutonium-239<\/strong>:\n
              \n
            • The half-life of plutonium-239 is 24,300 years, meaning after each 24,300-year period, half of the initial quantity decays. To find how long it will take for 8 kg to reduce to 1 kg, we apply the half-life formula. The calculation shows that it takes approximately 72,500 years<\/strong> for the mass to decrease to 1 kg.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • Radon-222<\/strong>:\n
                \n
              • Radon-222 has a much shorter half-life of 3.8 days. After 4 half-lives (which equals 15.2 days), the amount of radon reduces by a factor of 116\\frac{1}{16}, or 6.25 grams remaining from the original 100 grams.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

                These calculations illustrate how substances decay exponentially over time, with each “half-life” reducing the remaining amount by half, which can be modeled using the formula above.<\/p>\n","protected":false},"excerpt":{"rendered":"

                The Correct Answer and Explanation is : Let’s address both questions step by step, using the half-life formula: Formula for Half-Life: The general formula for the amount of a substance remaining after a certain period is: A=A0\u00d7(12)tTA = A_0 \\times \\left( \\frac{1}{2} \\right)^{\\frac{t}{T}} Where: 1. Plutonium-239 Question: Given: We need to solve for tt, the […]<\/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-186319","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186319","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=186319"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186319\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186319"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186319"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186319"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}