Question

Potassium-40 has a half-life of approximately 1.25 billion years. Approximately how many years will pass before a sample of potassium-40 contains one-eighth the original amount of parent isotope?

Answer 1

You know the radioactive decay law?

Answer 2

No, please explain.

Answer 3

Right. Well isotopes decay due to a exponential function. The differential equation that should be solved is:
\[\LARGE -\frac{ dN }{ N }= \lambda dt\]
The differential question has the solution:
\[\LARGE N(t)=N _{0}e ^{-\lambda*t}\]
Where the half-life is given by:
\[\LARGE t _{1/2}=\frac{ \ln(2) }{\lambda }\]

Answer 4

Now lets set up the problem: we want to know how much time there has went where there are 1/8 of the sample left. So the problem most be:
\[\LARGE N(t)=\frac{ 1 }{ 8 }N _{0}\]
The radioactive decay law then become:
\[\LARGE \frac{ 1 }{ 8 }N _{0}=N_{0}e ^{-\lambda*t}\]
The N_0 cancel on both sides:
\[\LARGE \frac{ 1 }{ 8 }=e ^{-\lambda*t}\]
We substitute the expression for the half-life:
\[\LARGE \frac{ 1 }{ 8 }=e ^{\frac{ \ln(2) }{ t _{1/2}}*t}\]
Solve for t and you have the answer.

Answer 5

@Yairenis

Do you know what "half life" means?

Answer 6

Woops a minus sign is missing in the last expression.

Answer 7

Anyway think I'm gonna solve for you, just to make this one an example for future work:
\[\LARGE \frac{ 1 }{ 8 }=e ^{-\frac{ \ln(2) }{ t _{1/2} }*t}\]
\[\LARGE \ln \left( \frac{ 1 }{ 8 } \right)=-\frac{ \ln(2) }{ t _{1/2} }*t\]
\[\LARGE t=-\frac{ \ln \left( \frac{ 1 }{ 8 } \right) }{ \frac{ \ln \left( 2 \right) }{ t _{1/2} } }\]

Answer 8

Don't worry about the minus sign... the natural logarithm to a number below 1 is negative, so the time is going to get positive.

Answer 9

So the answer would be 3.75 billion?

Answer 10

Yes please

Answer 11

Yea I get the same.

Answer 12

Then thank you for your help, Frostbite

Answer 13

No problem at all. Hope the outline was understandable?

Answer 14

Yes

Answer 15

Perfect.