Posted by Moe on .
The radioactive isotope of Carcon (^14C)has a half life of 5730. What is the decay constant of ^14C? If we start with a sample of 1000 carcon nuclei, how many will be left in 22920 years time?
22920 years is exactly 4 half-lives, so the number of nuclei is 2^4 = 16 times less. That will leave 1000/16 = 67 C14 atoms.
The relationship of the decay constant (k) and the half-life (T1/2) is:
k = 0.693/(T1/2)
so k = 1.21*10^-4 years-1