Question

Question: The nuclide undergoes alpha decay with a half-life of. An atomic energy worker breathes in of , which lodged permanently in a lung.

1. Compute the activity in becquerels, of the ingested, taking the atomic mass of the nuclide to be .
2. Determine the radiation absorbed dose in milligrays, during the first year after its ingestion. Assume that alpha particles emitted by have an average kinetic energy of , that all of this energy is deposited within the worker’s body, and the worker weighs 60 kg.
3. Is this dose likely to be lethal?

Short answer

1. The total activity$=1.09\times {10}^{4}Bq$
2. The radiation absorbed dose is $48m{{\mathrm{Gy}}^{-}}^1$
3. The dose is not fatal.

Step-by-step solution

Decay constant   

The decay constant can be calculated from the half life as follows:

$$\begin{gathered} \mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}} \\ \mathrm{k}=\frac{0.693}{2.411\times {10}^4}year{s}^{-1} \\ \mathrm{k}=0.287\times {10}^{-4}year{s}^{-1} \\ \mathrm{k}=\frac{0.287\times {10}^{-4}}{365\times 24\times 60\times 60}{\mathrm{s}}^{-1} \\ \mathrm{k}=9.1\times {10}^{-13}{\mathrm{s}}^{-1} \end{gathered}$$

Total activity

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

Number of atoms present $\frac{5.0\times {10}^{-6}\mathrm{g}}{239{\mathrm{gmol}}^{-1}}\times \frac{6.023\times {10}^{23}\mathrm{atoms}}{1\mathrm{mol}}=1.2\times {10}^{16}\mathrm{atoms}$

$\begin{array}{rcl}\mathrm{A}& =& \mathrm{kN}\\ & =& 9.1\times {10}^{-13}\times 1.2\times {10}^{16}{\mathrm{s}}^{-1}\\ & =& 1.09\times {10}^4{\mathrm{s}}^{-1}\end{array}$

Therefore, the total activity $1.09\times {10}^{14}{s}^{-1}$

Radiation absorbed dose

Total number of disintegration per second$1.09\times {10}^4{s}^{-1}$

Average kinetic energy of each beta particlerole="math" localid="1660968665664" $=5.24MeV$

The total energy deposited per year$1.09\times {10}^4{s}^{-1}\times 365\times 24\times 3600\times 5.24\mathrm{MeV}=1.8\times {10}^{12}\mathrm{MeV}yea{r}^{-1}$

Total energy deposited per year in Joule$1.8\times {10}^{12}\mathrm{MeV}yea{r}^{-1}\times 1.602\times {10}^{13}=0.288Jyea{r}^{-1}$

Total energy deposited per kg$=\frac{0.288{\mathrm{Jyr}}^{-1}}{60\mathrm{kg}}=0.0048{\mathrm{Jkg}}^{-1}y{r}^{-1}$

Now,

$\begin{array}{l}1{\mathrm{Jkg}}^{-1}=1\mathrm{Gy}\\ 1000{\mathrm{Jkg}}^{-1}=1\mathrm{mGy}\\ 0.0048\mathrm{J}{\mathrm{kg}}^{-1}=4.8\mathrm{mGy}\end{array}$

The radiation absorbed dose is $4.8mGyy{r}^{-1}$