# Question #b3120

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that a radioactive isotope's **nuclear half-life**, **half** of the atoms present in a sample of said isotope to undergo radioactive decay.

This means that if you start with

#A_0 * 1/2 = A_0/2^color(red)(1) -># after#color(red)(1)# half-life#A_0/2 * 1/2 = A_0/4 = A_0/2^color(red)(2) -># after#color(red)(2)# half-lives#A_0/4 * 1/2 = A_0/8 = A_0/2^color(red)(3) -># after#color(red)(3)# half-lives#A_0/8 * 1/2 = A_0/16 = A_0/2^color(red)(4) -># after#color(red)(4)# half-lives

#vdots#

and so on. You can thus say that the number of radioactive isotopes that **remain undecayed** after a period of time

#A_t = A_0 * (1/2)^color(red)(n)#

Here **number of half-lives** that pass in the given time

#color(red)(n) = t/t_"1/2"#

In your case, you know that you start with

You can thus say that

#12 = 100 * (1/2)^color(red)(n)#

Rearrange to get

#12/100 = (1/2)^color(red)(n)#

This will be equivalent to

#ln(12/100) = ln[(1/2)^color(red)(n)]#

which gets you

#color(red)(n) = ln(12/100)/ln(1/2) = 3.06#

So, you know that **half-lives** must pass in order for the number of radioactive isotopes to go from

You now have

#t = color(red)(n) * t_"1/2"#

and since you know that

#t_"1/2" = "1000 years"#

you can say that the rock is

#"age of rock" = 3.06 * "1000 years" = color(darkgreen)(ul(color(black)("3060 years")))#

I'll leave the answer rounded to three **sig figs**, but keep in mind that your values *do not* justify this value.