All living things are radioactive due to the presence of 14 6 C which undergoes

Question

All living things are radioactive due to the presence of 14 6 C which undergoes beta decay, emitting an electron. The half-life of 14 6 C is 5730 y. In live specimens, the level of 14 6 C present is constant. After death, the decay of 14 6 C slowly reduces the amount of 14 6 C present. Each gram of a live specimen emits 15.3 electrons per minute. An animal bone found at an archaeological site emits 75 electrons per minute. Its mass is 13.2 g. How old (years) is the bone? (How long ago did the animal die?) Give the answer to 2 significant figures.

Explanation / Answer

Given that :

half-life of 14C6 , t1/2 = 5730 year

initial bone activity, Q0 = 15.3 e/g-min

mass of the bone, m = 13.2 g

bone activity, Q = (89 e/min) / (13.2 g) = 6.74 e/g-min

The age of the bone (in years) which will be given as :

using an equation, we have

Age = t1/2 - ln (Q / Q0) / ln 2

Age = (5730 year) - ln [(89 / 13.2) / 15.3] / ln 2

Age = 6774 years

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