Chapter 13, Problem 13.22P

A worker accidentally ingested an unknown amount of ^σUCo activity. His body burden was measured by whole-body counting from day 1 until day 14 after the ingestion, with the following results:

The whole-body intake retention fraction (IRF) for ingested ⁶⁰Co is given as
$IRF\left(t\right)=0.5e^{-1.386t}+0.3e^{-0.1155t}+0.1e^{01155t}+0.1e^{-8.663\times {10}^{4}t}$
(a) Estimate the amount of ingested activity
(b) What was the committed dose from the ingested radiocobalt?

To determine

The amount of ingested activity.

Answer to Problem 13.22P

  $129.6\text{kBq}$

Explanation of Solution

Given info:

The body burden was measured by whole-body counting from day 1 until day 14 after the ingestion with the following results

Day: 1, 2, 3, 4, 7, 14

kBq: 75.3 63.4 54.3 49.2 42.0 31

The whole-body retention function for ingested ${}^{60}C$ is given as

  $Q(t)=0.5e^{-1.386t}+0.3e^{-0.01155t}+0.1e^{-0.01155t}+0.1e^{-0.00086t}$

Formula used:

The intake from a whole body measurement

  $A_{si}(0)=\frac{\text{measured}\text{activity}}{Q(t)}=\frac{A_{si}(t)}{Q(t)}$

Here, $Q(t\text{)}$ is the intake retention function gives the fraction of the intake remaining at time t days after intake.

Calculation:

So,

  $A_{si}(0)=\frac{\text{measured}\text{activity}}{0.5\times e^{-1.386t}+0.3e^{-0.1155t}+0.1e^{-0.01155t}+0.1e^{-8.663t}}$

For 1 day, we have

  $\begin{array}{l}{A}_{sl}(1)=\frac{75.3\text{kBq}}{0.5e^{-1.386\times 1}+0.3e^{-0.1155\times 1}+0.1e^{-0.01155\times 1}+0.1e^{-8.663\times {10}^{-4}}}\\ A_{sl}(1)=127.39\text{kBq}\end{array}$

Simlarly, intake from each of the whole body measurements

  $\begin{array}{l}{A}_{sl}(2)=135.78kBq\\ A_{sl}(3)=130.43kBq\\ A_{sl}(4)=127.42kBq\\ A_{sl}(7)=129.10kBq\\ A_{sl}(14)=127.32kBq\end{array}$

Thus mean,

  $\begin{array}{l}=\frac{127.39\text{kBq+}135.78kBq+130.43kBq+127.42kBq+129.10kBq+127.32kBq}{6}\\ =129.6\text{kBq}\end{array}$

To determine

What was the committed dose from the ingested radiocobalt?

Answer to Problem 13.22P

  $2.9\times {10}^{-3}\text{Sv}$

Explanation of Solution

Given info:

The body burden was measured by whole-body counting from day 1 until day 14 after the ingestion with the following results Day: 1, 2, 3, 4, 7, 14

kBq: 75.3 63.4 54.3 49.2 42.0 31

The whole-body retention function for ingested ${}^{60}C$ is given as

  $Q(t)=0.5e^{-1.386t}+0.3e^{-0.01155t}+0.1e^{-0.01155t}+0.1e^{-0.00086t}$

Formula used:

The cumulated activity,

  $\tilde{A}={\displaystyle \sum \frac{A_{si}(0)}{{\lambda }_{Ei}}}$

Substituting the values, we get

  $\begin{array}{l}\tilde{A}=129.6\times {10}^3\text{Bq}\left(\frac{0.5}{1.386{\text{d}}^{\text{-1}}}+\frac{0.3}{0.1155{\text{d}}^{\text{-1}}}+\frac{0.1}{0.1155{\text{d}}^{\text{-1}}}+\frac{0.1}{8.663\times {10}^{-4}{\text{d}}^{\text{-1}}}\right)\\ \tilde{A}=1.647\times {10}^7\text{Bq}\text{.d}\\ {\text{S}}_{co-60}\text{(body}\leftarrow \text{body)}=2.7\times {10}^{-5}\frac{\text{rad}}{\text{μCi}\text{.hr}}\times \frac{\text{0}\text{.01Gy}}{\text{rad}}\times \frac{\text{1μCi}}{3.7\times {10}^4\text{Bq}}\times 24\frac{\text{h}}{\text{d}}\times 1\frac{\text{Sv}}{\text{Gy}}\\ {\text{S}}_{co-60}\text{(body}\leftarrow \text{body)}=1.75\times {10}^{-10}\frac{\text{Sv}}{\text{Bq}\text{.d}}\end{array}$

The dose is given by

  $\begin{array}{l}H\text{= }\tilde{A}\text{Bq}\text{.d}\times S_{co-60}\text{(body}\leftarrow \text{body)}\frac{\text{Sv}}{\text{Bq}\text{.d}}\\ H=1.647\times {10}^7\text{Bq}\text{.d}\times 1.75\times {10}^{-10}\frac{\text{Sv}}{\text{Bq}\text{.d}}\\ H=2.9\times {10}^{-3}\text{Sv}\end{array}$

Here, $Q(t\text{)}$ is the intake retention function gives the fraction of the intake remaining at time t days after intake.

Calculation:

So,

  $A_{si}(0)=\frac{\text{measured}\text{activity}}{0.5\times e^{-1.386t}+0.3e^{-0.1155t}+0.1e^{-0.01155t}+0.1e^{-8.663t}}$

For 1 day, we have

  $\begin{array}{l}{A}_{sl}(1)=\frac{75.3\text{kBq}}{0.5e^{-1.386\times 1}+0.3e^{-0.1155\times 1}+0.1e^{-0.01155\times 1}+0.1e^{-8.663\times {10}^{-4}}}\\ A_{sl}(1)=127.39\text{kBq}\end{array}$

Simlarly, intake from each of the whole body measurements

  $\begin{array}{l}{A}_{sl}(2)=135.78kBq\\ A_{sl}(3)=130.43kBq\\ A_{sl}(4)=127.42kBq\\ A_{sl}(7)=129.10kBq\\ A_{sl}(14)=127.32kBq\end{array}$

Thus mean,

  $\begin{array}{l}=\frac{127.39\text{kBq+}135.78kBq+130.43kBq+127.42kBq+129.10kBq+127.32kBq}{6}\\ =129.6\text{kBq}\end{array}$