Chapter 18, Problem 96QRT

Interpretation Introduction

Interpretation:

Age of the charcoal has to be calculated.

Concept Introduction:

For a radioactive isotope, the relative instability is expressed in terms of half-life.  Half-life is the time required for half of the given quantity of the radioisotope to undergo radioactive decay.  Half-life of the radioactive isotope is independent of the quantity of the radioactive isotope.  Half-life of a radioactive isotope can be expressed as,

    $t_{1/2}=\frac{\text{0}\text{.693}}{k}--------------(1)$

Number of disintegrations per unit time is known as the activity of the sample.  The rate of radioactive decay can be expressed as,

    $\text{Rate}\text{of}\text{radioactive}\text{decay}\text{=}\text{activity}\text{=}A=kN$

Where,

    $A$ is the activity of sample.

    $N$ is the number of radioactive atoms.

    $k$ is the first-order rate constant or decay constant.

If initial activity of the sample is $A_0$ at $t_0$, then after time $t$, this will have a smaller activity $A$.  If the number of radioactive atoms present initially is $N_0$, then after time $t$, the number of radioactive atoms present will be $N$.

    $\begin{array}{l}\frac{A}{A_0}=\frac{kN}{k{N}_0}\\ \frac{A}{A_0}=\frac{N}{N_0}----------------(2)\end{array}$

$N/N_0$ can be calculated using the integrated rate equation.

    $\ln N=-kt+\ln N_0--------------(3)$

Rearranging equation (3) as shown below,

    $\ln \frac{N}{N_0}=-kt------------------(4)$

Considering equation (2), the equation (3) can be rewritten in terms of fraction of radioactive atoms that is present after time $t$ as,

    $\ln \frac{A}{A_0}=-kt------------------(5)$

Explanation of Solution

Mass of carbon present in $\text{1}\text{.14}\text{g}$ of ${\text{CaCO}}_{\text{3}}$ can be calculated using the conversion factors as shown below,

    $\begin{array}{l}\text{1}\text{.14}\text{g}{\text{CaCO}}_{\text{3}}\text{×}\frac{\text{1}\text{mole}{\text{CaCO}}_{\text{3}}}{\text{100}\text{.087}\text{g}{\text{CaCO}}_{\text{3}}}\text{×}\frac{\text{1}\text{mole}\text{C}}{\text{1 mole}{\text{CaCO}}_{\text{3}}}\text{×}\frac{\text{12}\text{.0107}\text{g}\text{C}}{\text{1}\text{mole}\text{C}}\\ =\frac{13.692198}{100.087}\text{g}\text{C}\\ =0.\text{1368}\text{g}\text{C}\end{array}$

Half-life of carbon-14 is $\text{5730}\text{years}$.  Decay rate constant can be calculated as,

    $\begin{array}{l}{t}_{1/2}=\frac{\text{0}\text{.693}}{k}\\ k=\frac{\text{0}\text{.693}}{t_{1/2}}\\ =\frac{0.693}{5730\text{years}}=1.20\times {10}^{-4}{\text{y}}^{-1}\end{array}$

Disintegration rate of carbon-14 is given as $2.17\times {10}^{-12}\text{Bq}$.  Conversion of this unit into disintegration per second as shown below,

    $2.17\times {10}^{-12}\text{Bq}\text{=}2.17\times {10}^{-2}\text{dps}$

The initial activity ($A_0$) of the radioactive sample is $2.17\times {10}^{-2}\text{dps}$.  From this, the final activity ($A$) of the radioactive sample can be calculated as shown below,

    $\begin{array}{l}A=\frac{2.17\times {10}^{-2}\text{dps}\times \frac{60\text{s}}{1\min }}{0.1368\text{g}\text{C}}\\ -9.517{\min }^{-1}{\text{g}}^{-1}\end{array}$

Age of the charcoal can be calculated as shown below,

    $\ln \frac{A}{A_0}=-kt$

    $\begin{array}{l}\ln \left(\frac{\text{9}\text{.517}\text{d}{\text{min}}^{\text{-1}}{\text{g}}^{\text{-1}}}{\text{15}\text{.3}\text{d}{\text{min}}^{\text{-1}}{\text{g}}^{\text{-1}}}\right)=-\left(1.20\times {10}^{-4}{\text{yr}}^{\text{-1}}\right)t\\ \ln \left(0.6220\right)=-\left(1.20\times {10}^{-4}{\text{yr}}^{\text{-1}}\right)t\\ -0.4748=-\left(1.20\times {10}^{-4}{\text{yr}}^{\text{-1}}\right)t\\ t=\frac{0.4748}{1.20\times {10}^{-4}{\text{yr}}^{\text{-1}}}\\ =0.3956\times {10}^4\text{yr}\\ \text{=}3.956\times {10}^3\text{yr}\end{array}$

Therefore, the age of the charcoal is estimated to be $3.956\times {10}^3\text{yr}$.