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Standard X

Biology

Carbon Dating

A pieces of b...

Question

A pieces of bone an animals from a ruin is found to have $^{14} C$ activity of 12 disintegrations per minute per gm of its carbon content. The $^{14} C$ activity of of a living animal is 16 disintegration per minute per gm How long ago nearly did the animal die?
(Given half of life $^{14} C$ is $t_{1 / 2}$ =5760 years)

1672 years

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2391 years

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3291 years

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4453 years

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Solution

The correct option is B 2391 years
The initial activity of c -12 is $= A_{0} = 16$ dpm.
after a time of t years, activity reduces to A = 12 dpm.
from Radioactivity Law,
$A = A_{0} e^{- λ t}$
where $λ = \frac{l n_{2}}{T},$ T = Half life
$\Rightarrow 12 = 16 e^{- λ t}$
$\Rightarrow t = \frac{T (-)}{l n_{2}} l n (\frac{12}{16}) = T l o g_{2} (\frac{3}{4})$
So,
$t = 5760 l o g_{2} (\frac{4}{3}) ≃ 2390.6 y e a r s .$ (Ans)

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A pieces of bone an animals from a ruin is found to have 14C activity of 12 disintegrations per minute per gm of its carbon content. The14C activity of of a living animal is 16 disintegration per minute per gm How long ago nearly did the animal die?(Given half of life 14Cis t1/2=5760 years)

The correct option is B 2391 yearsThe initial activity of c -12 is =A0=16 dpm.after a time of t years, activity reduces to A = 12 dpm.from Radioactivity Law,A=A0e−λt where λ=ln2T, T = Half life⇒12=16e−λt⇒t=T(−)ln2ln(1216)=Tlog2(34)So,t=5760log2(43)≃2390.6years. (Ans)