1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

  4. May 24, 2013 #3


    User Avatar
    Science Advisor

    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].)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted