A piece of wood from the ruins of an ancient building was found to have a C14 activity of 12 disintegration per minute per gram of its carbon content. The C14 activity of the living wood is 16 disintegration per minute per gram. How long ago did the tree, from which the wooden sample came, di? Given half-life of C14 is 5760 years.

Question

Filo · Accepted Answer

Rate of disintegration in old wood sample of C-14 radioactive atms is

12

atoms per min gm. Initially rate of distintegration of

C - 14

when the tree was live

= 16

atoms per min per gm.

T_{1/2}

of C-14 nuclei

= 5760

years
According to radioactive decay law

= N_{0} e^{- λ t} or R = R_{0} e^{- λ t}

12 = 16 e^{- λ t} \Rightarrow e^{λ t} = \frac{16}{12}

lo g_{e} e^{λ t} = lo g_{e} \frac{4}{3}

λ t = lo g_{e} \frac{4}{3}

t = \frac{2.303 \times lo g _{10} \frac{4}{3}}{λ} (∵ λ = \frac{0.6931}{T _{1/2}})

= \frac{2.303 \times ( lo g 4 - lo g 3 ) \times 5760}{0.6931} ye a r

∴ t = \frac{2.303 ( 0.6060 - 0.4771 ) 5760}{0.6931} = 2391.20 ye a rs

Question Text	A piece of wood from the ruins of an ancient building was found to have a $C^{14}$ activity of $12$ disintegration per minute per gram of its carbon content. The $C^{14}$ activity of the living wood is $16$ disintegration per minute per gram. How long ago did the tree, from which the wooden sample came, di? Given half-life of $C^{14}$ is $5760$ years.
Answer Type	Text solution:1
Upvotes	158

Text solutionVerified

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