Half-Life of a Radioactive SubstanceDate: 4 Aug 1995 17:10:44 -0400 From: Margie Salaz Subject: Half-Life Dear Ken, Here is my question: The half-life of a radioactive substance is d days. If you begin with a sample weighing b g, how long will it be until c g remains? Here is what I did: I used the Decay Formula: N = No e^kt 1/2No = No e^dk 1/2 = e^dk (1/2)^1/d = (e^dk)^1/d = ek (1/2)^1/d = ek cg = bg (e^k)^t Cg/bg = bg/bg <(1/2)^1/d>^t c/b = (1/2)^1/d^t Here is where I began to get bogged down. Will you please see what I have done wrong here: c/b = (1/1/2^d)^t d c/b = 1/1/2^t d In c/b/In 1/2 = t The last line is the right answer, but I think my last 3 steps aren't in line with the final product. Will you help me? Thanks, Margie Salaz Cuba, NM Date: 7 Aug 1995 09:29:01 -0400 From: Dr. Ethan Subject: Re: Half-Life Well I am quite impressed - you made my job very easy. You knew exactly where your mistake was. I will start there. >cg = bg (e^k)^t >Cg/bg = bg/bg <(1/2)^1/d>^t >c/b = (1/2)^1/d^t Up to here you are fine. But here is where you make your error. Your next line would be correct if it was -d not 1/d. What you should do next is raise both sides to the d power. Then you have (c/b)^d = (1/2)^t Then take the ln (natural log) of both sides: you get ln [ (c/b)^d ] = ln [ (1/2)^t] which equals d ln[c/b] = t ln[1/2] So t= d * ln[c/b]/ln[1/2] Hope that helps. Ethan Doctor On Call |
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