Question

Using potassium–argon radiometric dating, they discovered that the tools contained 25 percent radioactive potassium-40. How old are the tools?

Answer 1

*The half-life for potassium-40 to argon-40 is 1.25 billion years.

Answer 2

hey?

Answer 3

hi

Answer 4

okay so is that your answer with the star on it?

Answer 5

no its like an extra hint that was on the question

Answer 6

oh ok well let me see what i can do

Answer 7

ok!

Answer 8

sorry i cant help u

Answer 9

Oh well then I give it a try! :)
Ever heard about the radioactive decay law?

Answer 10

The radioactive decay law can be written as a differential equation as:
\[\large -\frac{ dN }{ dT }=\lambda N\]With a solution being:
\[\Large N(t)=N_0 \times e ^{-\lambda \times t} \]
Here is \(t\) the time, \(N(T)\) the amount of nuclei left at the time \(t\), \(N_0\) the amount of nucleic to the time \(t=0\) and \(\lambda\) a decay constant.

Sometimes this constant is not given, but we instead have the "half life". Half life is the amount of time \(t\) there will go before the following expression is true:
\[\large N(t)= \frac{ N_0 }{ 2 }\]
That is how long it will take before a sample has decayed to half the amount of nuclei. We can relate this quantity to the decay constant with following expression:
\[\Large t_{1/2}=\frac{ \ln(2) }{ \lambda }\] Where \(t_{1/2}\) is the half life.

As we are not given the exact amount of nucleic at the start and end point, but how many percent that has decayed, we can use a trick for exponential functions. We are told that at the time \(t\) the tool contained 25% potassium-40 using this information we can write:
\[\Large N(t)=0.25 \times N_0\]
We can substitute this into the radioactive decay law and obtain:
\[\Large 0.25 \times N_0=N_0 \times e ^{-\lambda \times t}\] The \(N_0\) cancel on both sides and we get an expression independent of the start and end number of nucleic:
\[\Large 0.25=e^{- \lambda \times t}\]
The challenge is now to evaluate \(\lambda\). Depending on the information given (usually the half life), we can solve the equation for \(t\) and get the solution to your question.

If any questions feel free to write.

Answer 11

Oh sorry didn't see you actually wrote you got the half life. In that case you should evaluate the decay constant \(\lambda\) using the half life \(t_{1/2}\)