Chapter 29, Problem 19P

(a)

To determine

The half-life in seconds.

Answer to Problem 19P

The half-life is $6.95\times {10}^5\text{s}$.

Explanation of Solution

Given info: Half-life of $\text{I}$ is 8.04 days.

$1\text{day}=\left(24\right)\left(3600\text{s}\right)$

The half life is,

$\begin{array}{c}{t}_{1/2}=\left(8.04\text{days}\right)\frac{\left(24\right)\left(3600\text{s}\right)}{1\text{day}}\\ =6.95\times {10}^5\text{s}\end{array}$

Conclusion:

The half-life is $6.95\times {10}^5\text{s}$.

(b)

To determine

The decay constant.

Answer to Problem 19P

The decay constant is $9.97\times {10}^{-7}{\text{s}}^{-1}$.

Explanation of Solution

Given info: Half-life of $\text{I}$ is $6.95\times {10}^5\text{s}$.

Formula to calculate the decay constant is,

$\lambda =\frac{\ln 2}{t_{1/2}}$

Substitute $6.95\times {10}^5\text{s}$ for $t_{1/2}$ in the above equation to get $\lambda$.

$\begin{array}{c}\lambda =\frac{\ln 2}{6.95\times {10}^5\text{s}}\\ =9.97\times {10}^{-7}{\text{s}}^{-1}\end{array}$

Conclusion:

The decay constant is $9.97\times {10}^{-7}{\text{s}}^{-1}$.

(c)

To determine

The activity in SI unit.

Answer to Problem 19P

The activity in SI unit is $1.9\times {10}^4\text{Bq}$.

Explanation of Solution

Given info: Activity of $\text{I}$ is $0.500\text{μCi}$.

$1\text{Ci}=3.7\times {10}^{10}\text{Bq}$

The activity in SI unit is,

$\begin{array}{c}R=\left(0.500\times {10}^{-6}\text{Ci}\right)\left(\frac{3.7\times {10}^{10}\text{Bq}}{1\text{Ci}}\right)\\ =1.9\times {10}^4\text{Bq}\end{array}$

Conclusion:

The activity in SI unit is $1.9\times {10}^4\text{Bq}$.

(d)

To determine

The number of $\text{I}$.

Answer to Problem 19P

The number of $\text{I}$ is $1.9\times {10}^{10}$.

Explanation of Solution

Given info: Activity of $\text{I}$ is $0.500\text{μCi}$.

Formula to calculate the number of $\text{I}$ is,

$N=\frac{R}{\lambda }$

Substitute $0.500\text{μCi}$ for R and $9.97\times {10}^{-7}{\text{s}}^{-1}$ for $\lambda$ in the above equation to get R.

$\begin{array}{c}N=\frac{0.500\text{μCi}}{9.97\times {10}^{-7}{\text{s}}^{-1}}\\ =\frac{\left(0.500\times {10}^{-6}\text{Ci}\right)\left(\frac{3.7\times {10}^{10}\text{Bq}}{1\text{Ci}}\right)}{9.97\times {10}^{-7}{\text{s}}^{-1}}\\ =1.9\times {10}^{10}\end{array}$

Conclusion:

The number of $\text{I}$ is $1.9\times {10}^{10}$.

(e)

To determine

The number of half-lives completed and activity.

Answer to Problem 19P

The number of half-lives completed is 5.

The activity is $0.200\text{mCi}$.

Explanation of Solution

Section 1:

To determine: The number of half-lives completed.

Answer: The number of half-lives completed is 5.

Explanation:

Given info: Activity ( $R_0$) of $\text{I}$ at a given time is $6.40\text{mCi}$. Time elapsed is 40.2 days. Half-life of $\text{I}$ is 8.04 days.

Formula to calculate the number of half-lives is,

$n=\frac{t}{t_{1/2}}$

Substitute 40.2 days for t and 8.04 days for $t_{1/2}$ in the above equation to get n.

$\begin{array}{c}n=\frac{40.2\text{days}}{8.04\text{days}}\\ =5\end{array}$

The number of half-lives completed is 5.

Section 2:

To determine: The activity.

Answer: The activity is $0.200\text{mCi}$.

Explanation:

Given info: Activity ( $R_0$) of $\text{I}$ at a given time is $6.40\text{mCi}$. Time elapsed is 40.2 days. Half-life of $\text{I}$ is 8.04 days.

Formula to calculate the activity is,

$R=\frac{R_0}{2^n}$

Substitute $6.40\text{mCi}$ for $R_0$ and 5 for n in the above equation to get R.

$\begin{array}{c}R=\frac{6.40\text{mCi}}{2^5}\\ =0.200\text{mCi}\end{array}$

The activity is $0.200\text{mCi}$.

Conclusion:

The number of half-lives completed is 5.

The activity is $0.200\text{mCi}$