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