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Question: A piece of bone of an animal from a ruin is found to have \({}^{14}C\) activity of \(12\) disintegra...

A piece of bone of an animal from a ruin is found to have 14C{}^{14}C activity of 1212 disintegration per minute per gm of its carbon content. The 14C{}^{14}C activity of a living animal is 1616 disintegration per minute per gm. Nearly, how long ago did the animal die?(Given half-life of 14C{}^{14}C is t12=5760years{{t}_{\dfrac{1}{2}}}=5760\,years)
A. 16721672 years
B. 23912391 years
C. 32913291 years
D. 44534453 years

Explanation

Solution

The concept used here is of radioactive decay. 14C{}^{14}C is a radioactive isotope of carbon and hence degenerates.
Half-life: It is the time period after which the material concentration is reduced to half of its initial concentration
t12=0.693α{{t}_{\dfrac{1}{2}}}=\dfrac{0.693}{\alpha } where α\alpha \,is rate constant of the reaction.
All radioactive decay is first order reaction

Complete step by step answer:
The first order decay equation isA=Aoeαt..........(1)A={{A}_{o}}{{e}^{-\alpha t}}..........\left( 1 \right)
where AA is final concentration or final condition disintegration per minute per gm,Ao{{A}_{o}}is initial concentration or initial condition disintegration per minute per gm and α\alpha \,is rate constant of the reaction.
In our case,the disintegration from animal bone ruin is final concentration whose age from the time being dead is to be found out
So A=12A=12disintegration per minute per gram
The initial concentration is disintegration from living animal
SoAo=16{{A}_{o}}=16disintegration per minute per gram
t12=5760years{{t}_{\dfrac{1}{2}}}=5760\,years
From this,
α=0.6935760.......(2)\alpha =\dfrac{0.693}{5760}.......\left( 2 \right)
Putting all these values in equation ,
12=16e0.693t576012=16{{e}^{-\dfrac{0.693t}{5760}}}
Taking loge{{\log }_{e}}both sides we get,
loge1216=0.6935760t\Rightarrow {{\log }_{e}}\dfrac{12}{16}=\dfrac{-0.693}{5760}t
loge43=0.6935760t\Rightarrow {{\log }_{e}}\dfrac{4}{3}=\dfrac{0.693}{5760}t

loge43×57600.693=t\Rightarrow {{\log }_{e}}\dfrac{4}{3}\times \dfrac{5760}{0.693}=t

t=2391yearst=2391\,years
So, on solving the value of time for which the animal has been dead is 2391years2391\,years.

The correct option for this will be B.

Note:
There are many types of reactions of different order.
Order determines the pace at which the reaction proceeds and the relation of time with concentration.
Many types of reactions exist like zero order reaction first order reaction and second order reaction.
For zero order half-life is dependent on initial concentration and for first order it is independent of it.