How old is a bone in which the Carbon-14 in it has undergone 3 half-lives? A) 11,400 years old Eliminate B) 17,100 years old C) 34,200 years old D) 57,000 years old

For this problem, we use the formula for radioactive decay which is expressed as follows:

An = Aoe^-kt

where An is the amount left after time t, Ao is the initial amount and k is a constant.

We calculate as follows:

at hal-life equal to 5700,

An = Aoe^-kt

0.5 = e^-k (5700)

k = 1.22x10^-4

An = Aoe^-kt

1.5 = e^-1.22x10^-4 (t)

t = 3323 years