{"id":33857,"date":"2023-12-26T11:59:59","date_gmt":"2023-12-26T11:59:59","guid":{"rendered":"https:\/\/www.greatassignmenthelp.com\/questions\/?p=33857"},"modified":"2023-12-26T12:01:02","modified_gmt":"2023-12-26T12:01:02","slug":"half-life-formula","status":"publish","type":"post","link":"https:\/\/www.greatassignmenthelp.com\/questions\/half-life-formula\/","title":{"rendered":"What is Half-Life Formula?"},"content":{"rendered":"<p><strong><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: large;\"><u>Answer:<\/u><\/span><\/span><\/strong><b> <\/b><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">A half-life is the amount of time needed for half of an atomic nuclide in a given sample to react, or the amount of time needed for neutrons to decay into proton, electron, and antineutrino groups.<\/span><\/span> <span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">This is a mechanism that occurs in many different kinds of nuclear and chemical incidents, such as the decay of radioactive isotopes or nuclei.<\/span><\/span><\/p>\n<h2 id=\"1-the-detection-of-half-lives\"><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: large;\"><b>The Detection of Half-Lives<\/b><\/span><\/span><\/h2>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Ernest Rutherford, one of the leading scientists of his day, is credited with discovering half-lives as well as alpha and beta radiation. Alongside scientist Joseph John Thompson, Rutherford conducted complementary experiments that ultimately led to the discovery of electrons, placing him at the forefront of this momentous discovery. Research on radioactivity started when Rutherford realized the possibilities of what he was seeing. The difference between beta and alpha radiation was discovered by him two years later. Since samples of radioactive materials decayed by half in the same period, this led to his discovery of half-lives.<\/span><\/span><\/p>\n<h2 id=\"2-derivation-of-the-half-life-formula\"><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: large;\"><b>Derivation of the Half Life Formula<\/b><\/span><\/span><\/h2>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">We first begin with the exponential decay law, which is expressed as follows:<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">N(t) = N0e<\/span><\/span> <span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">\u2212\u03bbt<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Additionally, one needs to set t = T1\/2 and N (T1\/2) = \u00bd N0<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">N (T1\/2) = 1\/2N0 = N0e &#8211;<\/span><\/span> <span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">\u03bbT1\/2<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Next, calculate the logarithm by dividing by N0.<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">\u00bd = e<\/span><\/span> <span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">\u2212\u03bbt<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Thus, In (1\/2) = \u2212\u03bbT1\/2<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Now solving for T1\/2,<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">T1\/2 = \u22121\u03bb In (1\/2)<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">The &#8220;-1&#8221; can be raised to an exponent of the logarithm using the logarithmic rules. Ultimately, this provides<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">T1\/2 = In (2))\/\u03bb<\/span><\/span><\/p>\n<h2 id=\"3-the-half-life-of-certain-substances\"><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: large;\"><b>The Half-Life\u00a0of Certain Substances<\/b><\/span><\/span><\/h2>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Here are some examples of elements and their associated half-lives, as the half-life formula has been used to determine the half-lives of several isotopes:<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Silver-94 &#8211; 0.42 seconds<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Neutron &#8211; 10.2 min<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Iodine-131 &#8211; 8 days<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Carbon-14 &#8211; 5,730 years<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Plutonium-239 &#8211; 24,100 years <\/span><\/span><\/p>\n<h2 id=\"4-a-few-examples-of-the-half-life-formula\"><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: large;\"><b>A Few Examples of the Half-Life Formula<\/b><\/span><\/span><\/h2>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\"><b>Question:<\/b><\/span><\/span><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\"> If we were to calculate a certain radioactive substance&#8217;s half-life and use a decay constant of 0.003 per year, the result would be ____. <\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\"><b>Solution: <\/b><\/span><\/span><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">t1\/2 = ln (2)\/\u03bb<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">t1\/2 = 0.693 \/ 0.003 yearly<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">t1\/2 = 231 years<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">In other words, this indicates that the half-life of the material in question is 231 years\u00a0and that half of it will take that long to decompose. <\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\"><b>Question: <\/b><\/span><\/span><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Determine the decay constant value of a radioactive material with a half-life of 0.04 seconds.<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\"><b>Solution: <\/b><\/span><\/span><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Given that the substance&#8217;s half-life is t1\/2 = 0.04<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">To determine a substance&#8217;s half-life, apply the half-life formula.<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">t1\/2 = 0.693\/\u03bb<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">\u03bb= 0.693\/0.04<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">= 17.325<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Thus, 17.325 s-1 is the radioactive substance&#8217;s decay constant.<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\"><b>Question: <\/b><\/span><\/span><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">If the material contains 90% of the carbon-14 present in living tissue, then calculate the age of the Shroud of Turin.<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\"><b>Solution: <\/b><\/span><\/span><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">The original equation&#8217;s N(t)\/N0 equals 90%, or 0.90, which can be changed to 0.90= e(\u03bb*t).<\/span><\/span><b> <\/b><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Since the value of lambda is unknown, we can utilize the first half-life formula, which yields 0.693 over 5730 years since the half-life of carbon-14 is known. <\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Returning to the previous equation, we can see that ln 0.90 = -\u03bbt by calculating the natural logarithm of both sides. <\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">We would get t = ln 0.90\/\u03bb by rearranging this new equation to isolate t.<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Solving the two equations for time t and changing lambda to the appropriate equation <\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">t = ln 0.90\/ (0.693\/5730 years)<\/span><\/span><\/p>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">t = 871 years <\/span><\/span><\/p>\n<h2 id=\"5-what-are-the-applications-of-half-life\"><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: large;\"><b>What Are the Applications of Half-Life?<\/b><\/span><\/span><\/h2>\n<p><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Geologists and archaeologists and even physicians use this formula. Some of the notable applications of half-life are:<\/span><\/span><\/p>\n<ul>\n<li><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">The formula has been used in the medical industry, where injections of radioactive materials into patients are most likely to occur to establish the safe handling duration. According to an established protocol, a sample is considered safe once its radioactivity falls below detection limits, which happens every ten half-lives. <\/span><\/span><\/li>\n<li><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">Experts utilize the process of radioactive dating to ascertain the age of a variety of artifacts and other materials, including rocks and ancient things. The precise ages of rocks, other geological features, and man-made things on Earth are among the details it provides regarding their attributions. <\/span><\/span><\/li>\n<li><span style=\"font-family: Times New Roman, serif;\"><span style=\"font-size: medium;\">For organic entities to function, carbon is necessary. While both carbon-12 and carbon-13 are stable isotopes, carbon-12 is the most prevalent and may be found in almost all organic structures. Carbon-14, an unstable isotope generated in the atmosphere by radiation from space which is also found on Earth.<\/span><\/span><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Answer: A half-life is the amount of time needed for half of an atomic nuclide in a given sample to react, or the amount of [&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":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v23.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>What is Half-Life Formula?<\/title>\n<meta name=\"description\" content=\"A half-life is the amount of time needed for half of an atomic nuclide in a given sample to react, or the amount of time needed for neutrons to decay into proton, electron, and antineutrino groups.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.greatassignmenthelp.com\/questions\/half-life-formula\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"What is Half-Life Formula?\" \/>\n<meta property=\"og:description\" content=\"A half-life is the amount of time needed for half of an atomic nuclide in a given sample to react, or the amount of time needed for neutrons to decay into proton, electron, and antineutrino groups.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.greatassignmenthelp.com\/questions\/half-life-formula\/\" \/>\n<meta property=\"og:site_name\" content=\"Questions - 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