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Exponential decay and half life problem

  1. May 23, 2013 #1
    the half life of C14 is 5730 years. if a sample of C14 has a mass of 20 micrograms at time t = 0, how much is left after 2000 years?


    I learned from somewhere that these exponential decay and half life problems use the equation

    y = ab^t or y = a(1+r)^t

    where y = total, a = initial amount, b or r = growth rate and t = time


    I think I have to solve for b, but I am not sure which t to use, whether its t=0 or t = 2000. I know a is given to be 20mg, but is y also a...so I am not really sure how to go about solving this. The end answer is 20exp (-ln/5730)(2000)...but I have no idea how to get to that. Please help, I can't even find the start in this situation because the y and a are confusing me.
     
  2. jcsd
  3. May 23, 2013 #2

    SteamKing

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  4. May 24, 2013 #3

    HallsofIvy

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    The fact that a process has a "half life" means that there is a specific time, T, until, however much there was initially, there is half left.

    Suppose there is initially "C". That is A(0)= C. Then after time, T, we have A(T)= (1/2)C. After another time, T, so a total of 2T, we have half of that: [itex]A(2T)= (1/2)((1/2)C)= (1/2)^2C[/itex]. After yet another time T, so a total of 3T, we have half of that: [itex]A(3T)= (1/2((1/2)^2C)= (1/2)^3C[/itex]. Do you see the point? If the total time is nT, we have [itex]A(nT)= (1/2)^n C[/itex]. If we write the total time as t= nT, then n= t/T so [itex]A(t)= (1/2)^{t/T} C[/itex]. We are simply multiplying by 1/2, for every multiple of "1/2" life. And that works even when it is not an integer multiple: if t= T/3, t/T= 1/3 and we have [itex]A(T/3)= (1/2)^{1/3}C[/itex].

    You know that [itex]A(t)= (1/2)^{t/T}C[/itex] where "T" is the half life (so T= 5730 years) and "C" is the initial amount (so C= 20 [itex]\mu grams[/itex]). You want to find A(t) with t= 2000 years. Notice that t and T are both in "years" so t/T will be a diensionless ratio.


    (All exponentials are interchangeble. If we have [itex]y= a^x[/itex] and want to change it to base b, we can use the fact that exponential, base b, and logarithm, base b, are inverse: [itex]y= b^{log_b(a^x)}= b^{x log_b(a)}[/itex] so that [itex]a^x[/itex] becomes b to a multiple of x- that multiple being [itex]log_b(a)[/itex].)
     
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