## anonymous 5 years ago the half-life of plutonium-239 is 24,100 years what is its initial amount if after 1000 years its amount is 2.1 grams?

1. anonymous

2.4g I'll explain in a minute.

2. anonymous

Sorry, 2.04g

3. anonymous

This is exponential decay. The rate of growth or decay in something undergoing this kind of change is in proportion to itself. That is,$\frac{dN}{dt} =n^2N$where I've written n^2 as the constant of proportionality to emphasize it's magnitude (we can worry about signs later). You solve this type of equation as$\frac{dN}{N}=n^2dt \rightarrow \int\limits_{}^{}\frac{dN}{N}=\int\limits_{}{}n^2dt \rightarrow \ln N=n^2t + c$

4. anonymous

Exponentiate both sides to get$N=e^{n^2t+c}=e^ce^{n^2t}=Ce^{n^2t}$

5. anonymous

When t=0, the equation is N(0)=C, so C is the initial amount. Also, when t=24100 years, half of the initial substance is gone, so at this time,$\frac{1}{2}N(0)=N(0)e^{n^2(24,100)} \rightarrow \frac{1}{2}=e^{n^2(24,100)}$take the natural log of both sides to solve for n^2:$\ln \frac{1}{2}=n^2(24100)yr \rightarrow n^2=\frac{\ln 1/2}{24100yr}=-\frac{\ln 2}{24000}$

6. anonymous

So your equation for the amount of plutonium-239 at time t is:$N(t)=2.1 e^{-\frac{\ln 2}{24100yr}t} grams$where t is in years.

7. anonymous

For t=1000 years,$N(1000)=2.1 e^{-\ln 2/24100yr \times 1000yr}g \approx 2.04g$

8. anonymous

Okay?

9. anonymous

There's a typo. above -> 24000 should be 24100.

10. anonymous

thanks

11. anonymous

np

12. anonymous

Is this your first time here?

13. anonymous

yes

14. anonymous

We like to be thanked in fan points ;) There should be a link next to my name, "Become a fan".