# Question #36005

##### 1 Answer

#### Answer:

#### 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"# thatremains undecayedafter a period of time#t# #A_0# is the initial mass of#""^14"C"# #color(red)(n)# represents thenumber of half-livesthat pass in a given time period#t#

Now, you know that your fossil contains

#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

#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 **half-lives** have passed from the moment the initial sample of

If you take

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

which, in your case, is equal to

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

Since **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 **years old**.

The answer is rounded to three **sig figs**.