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Standard XII

Chemistry

Law of Radioactivity

An old sample...

Question

An old sample of wood from an archaeological excavation was found to have a radioactivity of about 140.8 counts per minute due to $^{14} C (t_{1 / 2} = 5730$ years) as against 281.6 counts per minute for that of a freshly cut piece of wood. What is the age of the old sample of wood (in millennia, 1 millennium= 1000 years)?

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Solution

Age of the wood sample = $t = \frac{1}{λ} l n \frac{initial activity}{final activity} = \frac{t_{1 / 2}}{l n 2} l n \frac{281.6}{140.8}$ $= \frac{5730}{l n 2} l n 2$ = $5730 = 5730 y e a r s$

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The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.

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Age of the wood sample = t=1λlninitial activityfinal activity=t1/2ln 2ln281.6140.8 =5730ln 2ln2 = 5730=5730 years

Age of the wood sample = $t = \frac{1}{λ} l n \frac{initial activity}{final activity} = \frac{t_{1 / 2}}{l n 2} l n \frac{281.6}{140.8}$ $= \frac{5730}{l n 2} l n 2$ = $5730 = 5730 y e a r s$