An archaeologist graduate student found a leg bone of a large animal during the building of a new science building. The bone had a carbon-14 decay rate of 13.6 disintegrations per minute per gram of carbon. Living organisms have a decay rate of 15.3 disintegrations per minute. How many years ago did death occur for the animal

981 .52 years

Explanation:

According to the first order kinetics for radioactive decay, the rate constant is given as

k = ㏑

making t subject formulae becomes

t= ㏑

but K = 0.693/ t1/2

using the half life,  t 1/2 for carbon 14 = 5730 years since not given

therefore k = 0.693/ 5730 y

1.2 x 10 ^-4y-1

Bringing back our equation becomes

t= ㏑

where initial activity = 15.3 disintegrations per minute per gram for living organsism

and final activity=  13.6 disintegrations per minute per gram for leg bone

t = 1/1.2 x 10^-4y-1 ㏑( 15.3/ 13.6)

=981 .52years

The age of the leg bone is 981.52 years

multiplying and dividing the two decimals and the eqations by the same number of the overalls

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