Question #b3120

1 Answer
Jun 8, 2017

Answer:

Here's what I got.

Explanation:

The idea here is that a radioactive isotope's nuclear half-life, #t_"1/2"#, tells you the time needed for half of the atoms present in a sample of said isotope to undergo radioactive decay.

This means that if you start with #A_0# radioactive isotope, you can say that you will end up 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 #t# is equal to

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

Here #color(red)(n)# represents the number of half-lives that pass in the given time #t#.

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

In your case, you know that you start with #100# radioactive isotopes and that #88# have already decayed to their daughter nuclides. This, of course, implies that you're left with #12# radioactive isotopes after an unknown time #t# passes.

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 #3.06# half-lives must pass in order for the number of radioactive isotopes to go from #100# to #12#.

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.