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anonymous
 5 years ago
the halflife of plutonium239 is 24,100 years what is its initial amount if after 1000 years its amount is 2.1 grams?
anonymous
 5 years ago
the halflife of plutonium239 is 24,100 years what is its initial amount if after 1000 years its amount is 2.1 grams?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02.4g I'll explain in a minute.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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\[\frac{dN}{N}=n^2dt \rightarrow \int\limits_{}^{}\frac{dN}{N}=\int\limits_{}{}n^2dt \rightarrow \ln N=n^2t + c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exponentiate both sides to get\[N=e^{n^2t+c}=e^ce^{n^2t}=Ce^{n^2t}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When 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:\[\ln \frac{1}{2}=n^2(24100)yr \rightarrow n^2=\frac{\ln 1/2}{24100yr}=\frac{\ln 2}{24000}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So your equation for the amount of plutonium239 at time t is:\[N(t)=2.1 e^{\frac{\ln 2}{24100yr}t} grams\]where t is in years.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For t=1000 years,\[N(1000)=2.1 e^{\ln 2/24100yr \times 1000yr}g \approx 2.04g\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a typo. above > 24000 should be 24100.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this your first time here?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We like to be thanked in fan points ;) There should be a link next to my name, "Become a fan".
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