An ancient wooden club is found that contains 77 g of carbon and has an activity of 7.0 decays per second.
Determine its age assuming that in living trees the ratio of 14C/12C atoms is about 1.3×10^-12
T1/2 of carbon is 5700 years
Decay constant,
= ln 2/T1/20.693/5700=1.215×10^-4 yr^-1
Initial activity,A0= N14C
A0=1.215×10^-4×[{77 g×
(1 mol/12g)×6.023×10^23} ×(1.3×10^-12)
A0=0.6104×10^9 decay/ year
As N12C is 3.864×10^24 nuclei
Number of 14C in the sample of living tree is
N14= 1.3×10^-12×3.864×10^24=5.0232×10^12 nuclei
N=N0×e^-t
N= N0e^-t (1 year=3.16×10^7 sec)
e^(-t) =(7×3.16×10^7)/(1.215×10^-4×5.023×10^12)
-1.215×10^-4×t= ln(0.3624)= -1.0149
t= 1.0149/1.215×10^-4= 0.8353×10^4 years
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