Question

Question: The nuclide ${}^{\mathbf{131}}\mathbf{I}$undergoes beta decay with a half-life of $\mathbf{8}\mathbf{.}\mathbf{041}$days. Large quantities of this nuclide were released into the environment in the Chernobyl accident. A victim of radiation poisoning has absorbed of $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{g}\left(5.0\mathrm{\mu g}\right)$.

1. Compute the activity in becquerels, of the data-custom-editor="chemistry" ${}^{\mathbf{131}}\mathbf{I}$in this person, taking the atomic mass of the nuclide to equal $\mathbf{131}{\mathbf{gmol}}^{\mathbf{-}\mathbf{1}}$.
2. Compute the radiation absorbed dose, in milligrays caused by this nuclide during the first second after its ingestion. Assume that beta particles emitted by data-custom-editor="chemistry" ${}^{\mathbf{131}}\mathbf{I}$have an average kinetic energy of $\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{MeV}$, that all of this energy is deposited within the victim’s body and that victim weighs $\mathbf{69}\mathbf{}\mathbf{kg}$.
3. Is this dose likely to be lethal? Remember that the activity of${}^{\mathbf{131}}\mathbf{I}$diminishes as it decays.

Short answer

1. The total activity $=2.09x{10}^{10}\mathrm{Bq}$
   
2. The radiation absorbed dose is $0.019\mathrm{mGy}{\mathrm{s}}^{-1}$
   

3. The dose is lethal.

Step-by-step solution

Decay constant of I131

We can calculate the decay constant from the half-life of with the help of the following expression:

$$\begin{gathered} \mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}} \\ \mathrm{k}=\frac{0.693}{8.401}{\mathrm{days}}^{-1} \\ \mathrm{k}=0.082{\mathrm{days}}^{-1} \\ \mathrm{k}=\frac{0.082}{24\times 60\times 60}{\mathrm{s}}^{-1} \\ \mathrm{k}=9.5\times {10}^{-7}{\mathrm{s}}^{-1} \end{gathered}$$

Total activity

Amount of ${}^{131}\mathrm{I}$deposited in the body $=5.0\times {10}^{-6}\mathrm{g}$

Number of atoms present $\frac{5.0\times {10}^{-6}\mathrm{g}}{131{\mathrm{gmol}}^{-1}}\times \frac{6.023\times {10}^{23}\mathrm{atoms}}{1\mathrm{mol}}=2.2\times {10}^{16}\mathrm{atoms}$

$\begin{array}{rcl}\mathrm{A}& =& \mathrm{kN}\\ & =& 9.5\times {10}^7\times 2.2\times {10}^{16}{\mathrm{s}}^{-1}\\ & =& 2.09\times {10}^{10}{\mathrm{s}}^{-1}\end{array}$

Therefore, the total activity is $\begin{array}{rcl}2.09\times {10}^{10}{\mathrm{s}}^{-1}& =& 2.09\times {10}^{10}\mathrm{Bq}\end{array}$

Radiation absorbed dose

Total number of disintegration per second $=2.09\times {10}^{10}{s}^{-1}$

Average kinetic energy of each beta particle$=0.40MeV$

The total energy deposited per second$2.09\times {10}^{10}{s}^{-1}\times 0.40\mathrm{MeV}=0.83\times {10}^{10}\mathrm{MeV}{\mathrm{s}}^{-1}$

Total energy deposited per second in Joule$0.83\times {10}^{10}\mathrm{MeV}{\mathrm{s}}^{-1}\times 1.602\times {10}^{13}=1.328\times {10}^3{\mathrm{Js}}^{-1}$

Total energy deposited per kglocalid="1660932575842" $=\frac{1.328x{10}^3\mathrm{J}}{69\mathrm{kg}}=1.9\mathrm{x}{10}^{-5}{\mathrm{Jkg}}^{-1}{\mathrm{s}}^{-1}$

Now,

$\begin{array}{l}1{\mathrm{Jkg}}^{-1}=1\mathrm{Gy}\\ 1000{\mathrm{Jkg}}^{-1}=1\mathrm{mGy}\\ 1.9\times {10}^{-5}\mathrm{J}{\mathrm{kg}}^{-1}=0.019\mathrm{mGy}\end{array}$

The radiation absorbed dose is $0.019{\mathrm{mGys}}^{-1}$

Lethality of the dose

Energy deposited in the human body per day $2.09\times {10}^{10}{s}^{-1}\times 24\times 3600\times 0.40\mathrm{MeV}=7.2\times {10}^{14}\mathrm{MeV}{\mathrm{day}}^{-1}$

Energy deposited per kg per day$=\frac{7.2\times {10}^{14}\mathrm{MeVx}1.602\mathrm{x}{10}^{-13}\mathrm{J}}{69\mathrm{kg}}=1.66{\mathrm{Gyday}}^{-1}$

A dose of 5Gy is 50% lethal. As we can roughly estimate that the energy deposited in a week will be almost 11.68 Gy which will be lethal.