10 mg of carbon from living matter produce 200 counts per minute due to a small proportion of the radioactive isotope carbon-14. A piece of ancient wood of mass 10 mg is found to give 50 counts per minute. Find the age of the wood assuming that carbon-14 content of the atmosphere has remained unchanged. The half-life of carbon-14 is 5700 years.

Solution

The amount of ¹⁴C in the atmosphere remains constant. A living matter represents the amount of C¹⁴ present in the atmosphere. Ancient wood has 50 counts per minute. 10 mg in the beginning, must have the same count rates as from living matter.

Thus, the age of the ancient wood is the time in which the ¹⁴C count rate has decreased from the initial value of 200 counts per minute from 10 mg to the final value of 50 counts per minute from 10 mg.

We know that half life (T_1/2) for carbon is 5700 years.

If n is no of half lives then,

N = N_{0} . {(\frac{1}{2})}^{n} 50 = 200 {(\frac{1}{2})}^{n} \frac{50}{200} = {(\frac{1}{2})}^{n}

\frac{1}{4} = {(\frac{1}{2})}^{n} {(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{n}

n = 2 half lives required time = 5700 \times 2 = 11400 years = 1.14 \times 10^{4} years .

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