Chapter 14, Problem 126AP

The activity of a radioactive sample is the number of nuclear disintegrations per second, which is equal to the first-order rate constant times the number of radioactive nuclei present. The fundamental unit of radioactivity is the curie (Ci). where 1 Ci corresponds to exactly $3.70\times {10}^{10}$ disintegrations per second. This decay rate is equivalent to that of 1 g of radium-226. Calculate the rate constant and half-life for the radium decay. Starting with 1.0 g of the radium sample, what is the activity after 500 yr? The molar mass of Ra-226 is 226.03 g/mol.

Interpretation Introduction

Interpretation:

The rate constant and half-life for radium decay is to be calculated. Also, the activity of the radium sample after $500\text{ yr}$ is to be determined.

Concept introduction:

Rate constant for a reaction is the proportionality constant, which relates the rate of reaction and the concentration of reactants in the reaction.

Half-life is the time required by a substance to reduce by half of its original quantity. Half-life for a substance can be calculated as follows:

$t_{1/2}=\frac{0.693}{k{\left[A\right]}_0}$

Here, $t_{1/2}$ is half-life of the substance, $k$ is the rate constant for the decomposition reaction of the substance, and ${\left[A\right]}_0$ is the initial concentration of reactant A.

Answer to Problem 126AP

Solution: Rate constant and half-life for radium decay is $1.4\times {10}^{-11}{\text{ s}}^{-1}$ and $5.0\times {10}^{10}\text{ s}$, respectively.

$3.035\times {10}^{10}\text{nuclear }\text{disintegrations}/\text{s}$.

Explanation of Solution

Given information: A $1.0\text{ g}$ of radium-$226$ sample disintegrates and its molar mass is $226.03\text{ g}/\text{mol}$.

To determine the rate constant for the radium decay, the number of radium nuclei in $1.0g$ of sample is determined as follows:

$\begin{array}{c}n=\left(1.0\cancel{\text{g Ra}}\right)\times \left(\frac{\text{1}\cancel{\text{mol}}\text{Ra}}{\text{226}\text{.03}\cancel{\text{g Ra}}}\right)\times \left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{Ra nuclei}}{\text{1 }\cancel{\text{mol}}\text{Ra}}\right)\\ =2.7\times {10}^{21}\text{ Ra nuclei}\end{array}$

Now, calculate rate constant from the activity and the number of nuclei as follows:

Activity $=kN$

Here, $k$ is the rate constant and $N$ is the number of nuclei.

Rearrange the equation to determine therate constant.

$k=\frac{\text{activity}}{N}$

Substitute values of activity and number of nuclei in the above expression as,

$\begin{array}{c}k=\left(\frac{3.70\times {10}^{10}\text{nuclear disintegrations}}{2.7\times {10}^{21}\text{ nuclei}}\right)\\ =1.4\times {10}^{-11}{\text{s}}^{-1}\end{array}$

The half-life of radium is determined as follows:

$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{k}$

Substitute the value of rate constant in the above equation,

$\begin{array}{c}{t}_{1/2}=\left(\frac{0.693}{1.4\times {10}^{-11}}\right)\\ =5.0\times {10}^{10}\text{s}\end{array}$

The activity of radium after $500\text{ yr}$ is as follows:

$\left(500\cancel{\text{yr}}\right)\times \left(\frac{365\cancel{\text{days}}}{1\text{yr}}\right)\times \left(\frac{24\cancel{\text{h}}}{1\cancel{\text{day}}}\right)\times \left(\frac{3600\text{s}}{1\cancel{\text{h}}}\right)=1.58\times {10}^{10}\text{s}$

Now, by using the first-order integrated rate law, the number of nuclei remaining after $500$ yearsis calculated as follows:

$\ln \frac{N_t}{N_o}=-kt$

Here, N_t is the number of nuclei at the given time, $N_o$ is the number of nuclei present in the element sample without any disintegration, k is the rate constant, and t is time for disintegration.

Substitute values of $k$, $t$, and $N_o$ in the above expression,

$\begin{array}{c}\ln \left(\frac{N_t}{2.7\times {10}^{21}}\right)=-\left(1.4\times {10}^{-11}1/\cancel{s}\right)\left(1.58\times {10}^{10}\cancel{s}\right)\\ \left(\frac{N_t}{2.7\times {10}^{21}}\right)=e^{-0.22}\\ N_t=2.168\times {10}^{21}\text{ Ra}\text{nuclei}\end{array}$

Now, the activity of radium sample after $500\text{ yr}$ is calculated as follows:

$\begin{array}{c}\text{Activity}=kN\\ =\left(1.4\times {10}^{-11}{\text{s}}^{-1}\right)\left(2.168\times {10}^{21}\text{nuclei}\right)\\ =3.035\times {10}^{10}\text{nuclear }\text{disintegrations}/\text{s}\end{array}$

Conclusion

The rate constant and half-life for the radium decay is $1.4\times {10}^{-11}{\text{ s}}^{-1}$ and $5.0\times {10}^{10}\text{ s}$, respectively. The activity of the radium sample after $500\text{ yr}$ is $3.035\times {10}^{10}\text{nuclear }\text{disintegrations}/\text{s}$.