An ancient wooden club is found that contains 79 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.
mass of Carbon = 79 gm
No of Carbon atoms = 79*6.02e+23/12
when the tree is cut or dead
number of C-14 in the wood No = 79*6.02e+23 /12 * 1.3e-12 = 51.52 e+11
activity of Carbon at the time it was cut =c No ( c - C-14 decay const. = 3.8394e-12 /s )
suppose a time of t is elapsed from the time it was cut
current activity c Nc = c No exp (-0.693t/ T1/2 ) ( C-14 half life T1/2 = 5730 Yrs
= 7 decays /s
51.52e+11 *3.8394e-12 exp( -t/T1/2) = 7
t = 5952.27 Yrs
An ancient wooden club is found that contains 79 g of carbon and has an activity...
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