Radioactive decay of carbon 14 years since death c14 atoms remaining per 1.0 x 108 c12 atoms 0 10,000 5,700 5,000 11,400 2,500 17,100 1,250 22,800 625 28,500 312 34,200 156 39,900 78 45,600 30 51,300 20 57,000 10 62,700 5 68,400 2 how old is a bone in which the carbon-14 in it has undergone 3 half-lives? a. 11,400 b.17,100 c. 34,200 d. 57,000 hurry i have a test

:Answer

1.0/5

7

MissPhiladelphia   Ambitious

17.2K answers

145.1M people helped

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 = A

c is the answer : )               your welcome

thinking it would be d not too sure but i think it's right