Chapter 18, Problem 36QRT

Interpretation Introduction

Interpretation:

Age of the chariot and the year it was made has to be estimated if the $\text{C}$ activity is 11.2 disintegrations per minute per gram.

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,

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