Question #e3336

1 Answer
Dec 27, 2017

Answer:

#"15.5 mg"#

Explanation:

The half-life of a radioactive nuclide, #t_"1/2"#, is simply a measure of the time needed for half of a given sample to undergo radioactive decay.

So if you take #A_0# to be the initial mass of carbon-14, you can say that this sample will be reduced to

  • #A_0 * 1/2 = A_0/2 = 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
    #vdots#

and so on.

Now, let's say that #A_t# represents the mass of carbon-14 that remains undecayed after a period of time #t#. You can say that #A_t# will be equal to

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

with

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

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

In your case, the sample has an initial mass of #"52 mg"# and the half-life of carbon-14 is equal to #5730 #years, so you can say that after a period of time #t# passes, the sample will be reduced to

#A_t = "52 mg" * (1/2)^(t/"5730 years")#

To find the amount of carbon-14 that remains undecayed after #"10,000"# years, simply plug in this value into the above equation.

You will have

#color(red)(n) = ("10,000" color(red)(cancel(color(black)("years"))))/(5730color(red)(cancel(color(black)("years"))))#

and

#A_t = "52 mg" * (1/2)^("10,000"/5730) = color(darkgreen)(ul(color(black)("15.5 mg")))#

The answer is rounded to the nearest tenth.