Problem 7. Two bones of the same mass and from the same type of animal are compa
ID: 1684327 • Letter: P
Question
Problem 7. Two bones of the same mass and from the same type of animal are compared. The first bone is from an animal that has just died and the second bone is from an animal that has died an unkown time ago. The Carbon-14 activity in the second bone is 1/6 that of the carbon-14 activity in the first bone. How old is the second bone. (The half-life of Carbon-14 is 5730 years.) (Solve for ? first, from the known half-life, then find the time for the activity to be reduced by a factor of 6.)Explanation / Answer
This is a first order reactions So rate constant , k = 0.693 /T Where T = half life = 5730 years --> k = 0.693 / 5730 = 1.21*10^-4 year^-1 ----------------------------------------------------------------- We know that A = Ao e^(-kt) ---> (A/Ao ) = e^(-kt) ---> ln ( A/ Ao ) = -kt Ao = activity of first bone Where A = activity of the second bone = Ao / 6 t = time taken = ? k = rate constant = 1.21*10^-4 year^-1 Plug the values we get ln( Ao / 6 / Ao ) = -( 1.21*10^-4 year^-1 ) * t ----> t = 14808 years So the second bone is 14808 years older