Question #36005

1 Answer
Jul 25, 2017

Answer:

#"1060 years"#

Explanation:

Your tool of choice here will be the following equation

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

Here

  • #A_t# is the mass of #""^14"C"# that remains undecayed after a period of time #t#
  • #A_0# is the initial mass of #""^14"C"#
  • #color(red)(n)# represents the number of half-lives that pass in a given time period #t#

Now, you know that your fossil contains #88.0%# of the initial mass of #""^14"C"#, so you can say that you have

#A_t = 88.0/100 * A_0#

Plug this into the above equation to get

#88.0/100 * color(red)(cancel(color(black)(A_0))) = color(red)(cancel(color(black)(A_0))) * (1/2)^color(red)(n)#

#88.0/100 = (1/2)^color(red)(n)#

Rearrange to solve for #n#

#ln(88.0/100) = ln[(1/2)^color(red)(n)]#

You will end up wtih

#color(red)(n) * ln(1/2) = ln(88.0/100)#

which will get you

#color(red)(n) = ln(88.0/100)/ln(1/2) = 0.1844#

This tells you that #0.1844# half-lives have passed from the moment the initial sample of #""^14"C"# started to undergo radioactive decay to the moment you took your measurements.

If you take #t_"1/2"# to be the half-live of #""^14"C"#, you can say that you have

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

which, in your case, is equal to

#t = 0.1844 * t_"1/2"#

Since #""^14"C"# has a half-life of about #5730# years (see here), you can say that

#t = 0.1844 * "5730 years" = color(darkgreen)(ul(color(black)("1060 years")))#

have passed, i.e. the fossil is about #1060# years old.

The answer is rounded to three sig figs.