Question #e3336
1 Answer
Explanation:
The half-life of a radioactive nuclide,
So if you take
#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 = A_0 * (1/2)^color(red)(n)#
with
#color(red)(n) = t/t_"1/2"#
Here
In your case, the sample has an initial mass of
#A_t = "52 mg" * (1/2)^(t/"5730 years")#
To find the amount of carbon-14 that remains undecayed after
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.