Question

The half-life period of \[{}^{14}C\] is \[5370\,years\] . In a sample of dead trees, the proportion of \[{}^{14}C\] is found to be $60\% $ in comparison to living trees. Calculate the age of the sample.

Answer 1

Hint: Most of the radioactive decay reaction is first order as we discuss the chemical kinetics of such a reaction. So, for first order reactions there are two formulas which are used mostly one is the relation between half-life and rate constant and second relation is the relation of proportion and time.

Complete step-by-step answer:
We know that a first order reaction is when the rate of reaction depends upon the concentration of only one reactant and its rate expression can be written down as, $Rate\,\alpha \,[A]$ here $[A]$ is the concentration of reactant. Now for the first order reaction we have two relations, for the given question which says that for a sample of \[{}^{14}C\] half-life is \[5370\,years\] now by this information we got that the sample becomes half in concentration in this period of time. We can find out the value of rate constant (k) from the relation,
$k = \dfrac{{0.693}}{{\left( {{t_{\dfrac{1}{2}}}} \right)}}$
Putting value of half-life in the above expression, we get $k = \dfrac{{0.693}}{{5370\,year}} = \,0.00012905\,yea{r^{ - 1}}$
Now it was given in the question that proportion of \[{}^{14}C\] is found to be $60\% $ as compared to living trees it means that we can take the concentration of \[{}^{14}C\] in living trees as $100\% $ .
Now applying next relation which is expression of rate constant and concentration,
$T = \dfrac{{2.303}}{k}\,\log \left( {\dfrac{{{A_ \circ }}}{A}} \right)$ Here, ${A_ \circ }$ is the initial concentration while $A$ is the remaining concentration
As it was given that proportion was found to be $60\% $ it means this is given the remaining concentration. Now let’s put the values in the equation.
$T = \dfrac{{2.303}}{{0.00012905\,yea{r^{ - 1}}}}\,\log \left( {\dfrac{{100}}{{60}}} \right)$ = $\dfrac{{2.303}}{{0.00012905\,yea{r^{ - 1}}}}\,\log \,(1.66)$
$ = \,\dfrac{{2.303}}{{0.00012905\,yea{r^{ - 1}}}}\, \times \,0.220108$ $ = \,3928.9158\,years$
Above, we got the age of sample which is $3928.9158\,years$

Note: In the calculation of time or rate constant, there are chances of mistake so always keep in mind that the unit of rate constant for first order reaction should be ${(time)^{ - 1}}$ it may be in seconds or minutes or can be in years. Putting initial concentration always $100$ when the proportion is given in percentage. In the denominator we used to write the remaining concentration in many problems the concentration decreased. So putting the remaining concentration is important for right calculation.