Chapter 29, Problem 79P

(a)

To determine

The mass of ${}^{131}\text{I}$ in the sample with initial activity of $64.5\text{ mCi}$.

Answer to Problem 79P

The mass of ${}^{131}\text{I}$ in the sample with initial activity of $64.5\text{ mCi}$ is $5.2\times {10}^{-7}\text{ g}$.

Explanation of Solution

A sample containing Iodine-131 has a initial activity of $R_0=64.5\text{ mCi}$ and molar mass of Iodine-131 is $M=130.9061246\text{g}/\text{mol}$ and its half-life is $T_{1/2}=8.0252\text{ days}$ . The Avogadro’s number is $N_A=6.022\times {10}^{23}\text{nuclei}/\text{mol}$.

Write the formula for half-life

$T_{1/2}=\tau \ln 2$                                                                                             (I)

Here, $T_{1/2}$ is the half-life and $\tau$ is the time constant.

Write the formula for the number of nuclei

$N=\frac{m{N}_A}{M}$                                                                                             (II)

Here, $N$ is the number of nuclei, $m$ is the mass, $N_A$ is the Avogadro’s number and $M$ is the molar mass.

Write the formula for the activity

$R=\frac{N}{\tau }$                                                                                                (III)

Here, $R$ is the activity.

Conclusion:

Substitute equation (I) and (III) in (II) to solve for $m$

$\begin{array}{l}m=\frac{NM}{N_A}\\ m=\frac{R\tau M}{N_A}\\ m=\frac{R{T}_{1/2}M}{N_A\ln 2}\end{array}$

Substitute $64.5\text{ mCi}$ for $R$, $6.022\times {10}^{23}\text{nuclei}/\text{mol}$ for $N_A$, $8.0252\text{ days}$ for $T_{1/2}$ and $130.9061246\text{g}/\text{mol}$ for $M$ in the above equation to find the value of $m$. [Note: $1\text{ Ci}=\text{3}\text{.7}\times {\text{10}}^{10}\text{ Bq}$]

$\begin{array}{l}m=\frac{64.5\times {10}^{-3}\text{ Ci}\times \text{3}\text{.7}\times {\text{10}}^{10}\text{Bq}/\text{Ci}\times 8.0252\text{ days}\times 86400\text{s}/\text{d}\times 130.9061246\text{g}/\text{mol}}{6.022\times {10}^{23}\text{nuclei}/\text{mol}\times \ln 2}\\ m=5.2\times {10}^{-7}\text{ g}\end{array}$

Thus, the mass of ${}^{131}\text{I}$ in the sample with initial activity of $64.5\text{ mCi}$ is $5.2\times {10}^{-7}\text{ g}$.

(b)

To determine

The activity of ${}^{131}\text{I}$ after $4.5\text{ days}$.

Answer to Problem 79P

The activity of ${}^{131}\text{I}$ after $4.5\text{ days}$ is $44\text{ mCi}$.

Explanation of Solution

The initial activity of ${}^{131}\text{I}$ is $R_0=64.5\text{ mCi}$, then the after a period of time $t=4.5\text{ days}$ the activity $R$ is

Write the formula for the activity

$R=R_0{e}^{-t/\tau }$                                                                                           (IV)

Here, $R$ is the activity after time $t$, $R_0$ is the activity at $t=0$ and $t$ is the time.

Conclusion:

Substitute equation (I) in (IV) to solve for $R$

$R=R_0{e}^{-t\ln 2/T_{1/2}}$

Substitute $64.5\text{ mCi}$ for $R_0$, $8.0252\text{ days}$ for $T_{1/2}$ and $4.5\text{ days}$ for $t$ in the above equation to find the value of $R$

$\begin{array}{l}R=64.5\text{ mCi}\times e^{-4.5\text{ days}\times \ln 2/8.0252\text{ days}}\\ R=44\text{ mCi}\end{array}$

Thus, the activity of ${}^{131}\text{I}$ after $4.5\text{ days}$ is $44\text{ mCi}$.