Question

A piece of bone of an animal from a ruin is found to have $^{14}C$ activity of 12 disintegration per minute per gram of its carbon content. The $^{14}C$ activity of a living animal is 16 disintegrations per minute per gram. How long ago nearly did the animal die? Given half life of $^{14}C$ is ( ${t_{1/2}} = 5760\,{\text{years}}$)
A. 1672 years
B. 2391 years
C. 3291 years
D. 4453 years

Answer 1

Hint-
We need to find the time elapsed. The decay rate equation gives the relation connecting the number of disintegration and the time elapsed. The decay rate equation is given as
$N = {N_0}{e^{ - \lambda t}}$
Where, $N$ is the number of disintegrations, ${N_0}$ is the total number of molecules present initially, $\lambda$ is the decay constant or disintegration constant and $t$ is the time elapsed.
From this we can find the value of time in terms of disintegration constant. The relation between half life and decay constant is given as
$\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$
Using this value we can calculate the time elapsed.

Complete step by step answer:
According to the law of radioactive decay the number of disintegrations will be proportional to the total number of nuclei in the sample.
The decay rate equation is given as
$N = {N_0}{e^{ - \lambda t}}$ ………………….(1)
Where, $N$ is the number of disintegrations, ${N_0}$ is the total number of molecules present initially, $\lambda$ is the decay constant or disintegration constant and $t$ is the time elapsed.
In the given question, the initial number of disintegration is given as 16.
The final disintegration per minute is 12.
We are asked to find the time elapsed.
Let us substitute the given values in the radioactive decay equation.
$N = {N_0}{e^{ - \lambda t}}$
$\rightarrow \dfrac{N}{{{N_0}}} = {e^{ - \lambda t}}$
Let us take logarithm on both sides of the equation,
$\rightarrow \ln \dfrac{N}{{{N_0}}} = - \lambda t$.....................(2)
Since, we know that if we take the logarithm of exponential both get canceled out.
$\therefore \ln \,{e^{ - \lambda t}} = - \lambda t$
According to the logarithm rule for division.
$\ln \dfrac{a}{b} = \ln \,a - \ln \,b$
Using this we can write equation (2) as
$\ln N - \ln \,{N_0} = - \lambda t$
Taking, minus sign to the left hand side we get
$\ln {N_o} - \ln \,N = \lambda t$
On substituting the value of $N$ and ${N_0}$ we get
$\ln \,16 - \ln \,12 = \lambda t$
$\rightarrow \ln \dfrac{{16}}{{12}} = \lambda t$
$\rightarrow 0.2876 = \lambda t$
$\therefore t = \dfrac{{0.2876}}{\lambda }$......................(3)
Half life is the time taken to reduce the initial number of nuclei to half. The relation between half life and decay constant is given as
$\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$
Half life is given in the question as
${t_{1/2}} = 5760\,{\text{years}}$
On substituting this value we get
$\lambda = \dfrac{{0.693}}{{5760}} = 1.203 \times {10^{ - 4}}$
Let us substitute this value of decay constant in equation 1
$\therefore t = \dfrac{{0.2876}}{{1.203 \times {{10}^{ - 4}}}} = 2390\,{\text{years}}$

So, the correct answer is option B.

Note:
Formulas to remember –
The radioactive decay equation.
$N = {N_0}{e^{ - \lambda t}}$
Where, $N$ is the number of disintegrations, ${N_0}$ is the total number of molecules present initially, $\lambda$ is the decay constant or disintegration constant and $t$ is the time elapsed.
The relation between half-life and decay constant is given as
$\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$