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The half-life of silicon-32 is 710 years. If 80 grams is present now, how much will be present ijn 200 years?

I used A(t)=Ae^kt to solve for the rate (k).

A(710)=1/2Ae^k(710) 1/2A=Ae^k(710) 1/2=e^k(710) ln1/2=k710 ln1/2/710=k k=

And this is where I'm confused. Am I even starting in the right place?

Ryan Burwell
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  • restate of what I've tried to make it more clear...A(710)=1/2Ae^k(710)... 1/2A=Ae^k(710)... 1/2=e^k(710)... ln1/2=k710... ln1/2/710=k... k= – Ryan Burwell Nov 06 '14 at 04:57
  • Kind of tough to read. Doing it your way (I would use a different but equivalent calculation) you should get $k-\frac{\ln(1/2)}{710}$. Now that you know $k$, you can use the formula to solve the how much after $200$ years problem. – André Nicolas Nov 06 '14 at 05:03
  • Your question is not clear. If the 80g have just been created, none of it would have decayed in 200 years as it stays for 710 years. If its creation was uniformly distributed over the past 710 years, then $\frac{510}{710}\times{80}g=57.46g (approx.)$ would be left today. How is this anything to do with logarithms? Do we have to take into account what it decays into? – ghosts_in_the_code Nov 06 '14 at 06:56

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Hint: $$m=m_02^{-t/T}\implies m=80\times2^{-200/710}=?$$

RE60K
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