Question

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?

Answer 1

2.4g

I'll explain in a minute.

Answer 2

Sorry, 2.04g

Answer 3

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\[\frac{dN}{N}=n^2dt \rightarrow \int\limits_{}^{}\frac{dN}{N}=\int\limits_{}{}n^2dt \rightarrow \ln N=n^2t + c\]

Answer 4

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

Answer 5

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:\[\ln \frac{1}{2}=n^2(24100)yr \rightarrow n^2=\frac{\ln 1/2}{24100yr}=-\frac{\ln 2}{24000}\]

Answer 6

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.

Answer 7

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

Answer 8

Okay?

Answer 9

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

Answer 10

thanks

Answer 11

np

Answer 12

Is this your first time here?

Answer 13

yes

Answer 14

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