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An archaeologist finds 14C in a sample of material to be decaying at 20 Geiger Counter clicks per second. A modern equivalent sample of the same material decays at 160 clicks per second. The half-life of 14C is 5,730 years. How old is the sample

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Answer:

17188 years

Explanation:

Recall the formula;

0.693/t1/2 = 2.303/t log(No/N)

t1/2 = half life of 14C

No = initial activity of 14C

N = activity of 14C at time t

t = age of the sample

Substituting values

0.693/5730= 2.303/t log(160/ 20)

1.21 * 10^-4= 2.0798/t

t = 2.0798/1.21 * 10^-4

t = 17188 years

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