{"id":270446,"date":"2025-07-25T07:41:17","date_gmt":"2025-07-25T07:41:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=270446"},"modified":"2025-07-25T07:41:20","modified_gmt":"2025-07-25T07:41:20","slug":"an-isotope-of-radon-gas-rn-222-undergoes-decay-with-a-half-life-of-3-8-days","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/25\/an-isotope-of-radon-gas-rn-222-undergoes-decay-with-a-half-life-of-3-8-days\/","title":{"rendered":"An isotope of radon gas, Rn-222, undergoes decay with a half-life of 3.8 days."},"content":{"rendered":"\n

An isotope of radon gas, Rn-222, undergoes decay with a half-life of 3.8 days. How long does it take for the amount of Rn-222 to be decreased to one-sixteenth of the original amount?<\/p>\n\n\n\n

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

To determine how long it takes for the amount of Rn-222 to decrease to one-sixteenth of the original amount, we can use the concept of half-life and apply it to the decay process.<\/p>\n\n\n\n

Step 1: Understanding the concept of half-life<\/h3>\n\n\n\n

The half-life of an isotope is the time it takes for half of the substance to decay. In this case, we are given that the half-life of Rn-222 is 3.8 days.<\/p>\n\n\n\n

Step 2: Using the decay formula<\/h3>\n\n\n\n

The general decay formula is: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

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

    \n
  • N(t)N(t)N(t) is the remaining amount of substance after time ttt.<\/li>\n\n\n\n
  • N0N_0N0\u200b is the initial amount of substance.<\/li>\n\n\n\n
  • T1\/2T_{1\/2}T1\/2\u200b is the half-life (3.8 days in this case).<\/li>\n\n\n\n
  • ttt is the time elapsed.<\/li>\n<\/ul>\n\n\n\n

    We want to find the time ttt when the remaining amount of radon is one-sixteenth of the original amount, meaning N(t)=116N0N(t) = \\frac{1}{16} N_0N(t)=161\u200bN0\u200b.<\/p>\n\n\n\n

    Step 3: Setting up the equation<\/h3>\n\n\n\n

    Substitute N(t)=116N0N(t) = \\frac{1}{16} N_0N(t)=161\u200bN0\u200b into the decay equation:116N0=N0(12)t3.8\\frac{1}{16} N_0 = N_0 \\left(\\frac{1}{2}\\right)^{\\frac{t}{3.8}}161\u200bN0\u200b=N0\u200b(21\u200b)3.8t\u200b<\/p>\n\n\n\n

    Step 4: Simplifying the equation<\/h3>\n\n\n\n

    Cancel N0N_0N0\u200b from both sides of the equation (assuming N0\u22600N_0 \\neq 0N0\u200b\ue020=0):116=(12)t3.8\\frac{1}{16} = \\left(\\frac{1}{2}\\right)^{\\frac{t}{3.8}}161\u200b=(21\u200b)3.8t\u200b<\/p>\n\n\n\n

    Now express 116\\frac{1}{16}161\u200b as (12)4\\left(\\frac{1}{2}\\right)^4(21\u200b)4:(12)4=(12)t3.8\\left(\\frac{1}{2}\\right)^4 = \\left(\\frac{1}{2}\\right)^{\\frac{t}{3.8}}(21\u200b)4=(21\u200b)3.8t\u200b<\/p>\n\n\n\n

    Step 5: Solving for ttt<\/h3>\n\n\n\n

    Since the bases are the same, we can set the exponents equal to each other:4=t3.84 = \\frac{t}{3.8}4=3.8t\u200b<\/p>\n\n\n\n

    Now solve for ttt:t=4\u00d73.8=15.2 dayst = 4 \\times 3.8 = 15.2 \\text{ days}t=4\u00d73.8=15.2 days<\/p>\n\n\n\n

    Conclusion:<\/h3>\n\n\n\n

    It will take 15.2 days<\/strong> for the amount of Rn-222 to decrease to one-sixteenth of the original amount.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    An isotope of radon gas, Rn-222, undergoes decay with a half-life of 3.8 days. How long does it take for the amount of Rn-222 to be decreased to one-sixteenth of the original amount? The Correct Answer and Explanation is: To determine how long it takes for the amount of Rn-222 to decrease to one-sixteenth of […]<\/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-270446","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/270446","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=270446"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/270446\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=270446"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=270446"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=270446"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}