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Suppose a bone from an ancient human contains 18.5% of the C-14 found in living organisms....

Suppose a bone from an ancient human contains 18.5% of the C-14 found in living organisms. How old Is the bone?
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Answer #1

The half life of C-14   = 5730 years

K   = 0.693/t1/2

     = 0.693/5730    = 0.00012094 years^-1

K   = 2.303/t log[A0]/[A]

0.00012094 = 2.303/t log100/18.5

0.00012094     = 2.303*0.7328/t

t                         =2.303*0.7328/0.00012094   = 13954years     round off 1.4*10^4years

1.4*10^4 years old the bone >>>>answer

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