Chapter 30, Problem 68GP

(a)

To determine

To find: Activity of 1 g of $\text{I}$

Answer to Problem 68GP

The activity of 1 g of $\text{I}$ is $R=4.6\times {10}^{15}\text{decays/s}$

Explanation of Solution

Given:

Mass of $\text{I}$ , $m=1.0\text{ g}$

Half-life is $T_{\frac{1}{2}}=8.02\text{ days}=8.02\times 86400\text{s}=692928\text{ s}$

Formula used:

The activity can be given by: $R=\lambda N$

Calculation:

Firstly, we need to calculate the number of nuclei as follows:

  $\begin{array}{l}N=\frac{m{N}_A}{M}\\ N=\frac{\left(1.0\text{ g}\right)\left(6.022\times {10}^{23}\text{nuclei/mol}\right)}{\left(131\text{ g/mol}\right)}\\ N=4.6\times {10}^{21}\text{nuclei}\end{array}$

Now, the activity can be calculated as follows:

  $\begin{array}{l}R=\lambda N\\ R=\left(\frac{0.693}{T_{\frac{1}{2}}}\right)N\\ R=\left(\frac{0.693}{692928\text{ s}}\right)\left(4.6\times {10}^{21}\text{nuclei}\right)\\ R=4.6\times {10}^{15}\text{decays/s}\end{array}$

Conclusion:

The activity of 1 g of $\text{I}$ is $R=4.6\times {10}^{15}\text{decays/s}$

(b)

To determine

To find: Activity of 1 g of $\text{U}$

Answer to Problem 68GP

The activity of 1 g of $\text{U}$ is $R=1.24\times {10}^4\text{decays/s}$

Explanation of Solution

Given:

Mass of $\text{U}$ , $m=1.0\text{ g}$

Half-life is $T_{\frac{1}{2}}=4.47\times {10}^9\text{yr}=4.47\times {10}^9\times 86400\text{s}=1.41\times {10}^{17}\text{s}$

Calculation:

Firstly, we need to calculate the number of nuclei as follows:

  $\begin{array}{l}N=\frac{m{N}_A}{M}\\ N=\frac{\left(1.0\text{ g}\right)\left(6.022\times {10}^{23}\text{nuclei/mol}\right)}{\left(\text{238 g/mol}\right)}\\ N=2.53\times {10}^{21}\text{nuclei}\end{array}$

Now, the activity can be calculated as follows:

  $\begin{array}{l}R=\lambda N\\ R=\left(\frac{0.693}{T_{\frac{1}{2}}}\right)N\\ R=\left(\frac{0.693}{1.41\times {10}^{17}\text{s}}\right)\left(2.53\times {10}^{21}\text{nuclei}\right)\\ R=1.24\times {10}^4\text{decays/s}\end{array}$

Conclusion:

The activity of 1 g of $\text{U}$ is $R=1.24\times {10}^4\text{decays/s}$