"5 half-lives\\\\\n\n\\% decayed \\space is \\space 100 \\% -> 50 \\% -> 25 \\% -> 12.5 \\%-> 6.125\\% -> 3.0625 \\%\\\\\n\n\nSo \\space amount \\space left \\space is \\space 3.0625 \\%\\\\\n\n\nWe \\space can\\space use \\space also\\space 1st \\space order \\space kinetics t = (\\frac{1}{k}) \\ln(\\frac{a}{a-x})\\\\\n\nt = 5 *20.4 , k = \\frac{0.693}{20.4}, a= 100 , we \\space get \\space a-x = 3.0625\\\\"
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