Chapter 11, Problem 11.9P

(a)

To determine

The dose to the thyroid due to an acute exposure of $1\text{Bq}{\text{.s/m}}^{\text{3}}$ of ${}^{\text{131}}\text{I}$ .

Answer to Problem 11.9P

  $D\left(\text{thyroid},{}^{\text{131}}\text{I}\right)=1.4\times {10}^{-10}\text{Gy}$

Explanation of Solution

Given:

Mass of thyroid is $m_t=20\text{g}$

Formula used:

The initial dose rate is

  ${\dot{D}}_0=\frac{q\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times E_E\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{m\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}$

Where,

  $m=$ Mass

  $q=$ Radioactivity

  $E_E=$ Energy per transition

Calculation:

According to Appendix C average person breaths 20 L per min. So, the light activity is:

  $\begin{array}{l}1\frac{\text{Bq}\text{.s}}{{\text{m}}^{\text{3}}}\times 20\frac{\text{L}}{\min }\times \frac{1\min }{60\sec }\times \frac{1{\text{m}}^{\text{3}}}{{10}^3\text{L}}\\ =3.33\times {10}^{-4}\text{Bq}\end{array}$

The below table is drawn referring ICRP 2:

	${}^{\text{131}}\text{I}$			${}^{\text{133}}\text{I}$
	$T_E,\text{d}$	${\lambda }_E,{\text{d}}^{\text{-1}}$	$E_E,\raisebox{1ex}{$\text{MeV}$}\!\left/ \!\raisebox{-1ex}{$\text{t}$}\right.$	$T_E,\text{d}$	${\lambda }_E,{\text{d}}^{\text{-1}}$	$E_E,\raisebox{1ex}{$\text{MeV}$}\!\left/ \!\raisebox{-1ex}{$\text{t}$}\right.$
Thyroid	$7.6$	$0.09$	$0.23$	$0.87$	$0.8$	$0.54$
Body	$7.6$	$0.09$	$0.44$	$0.87$	$0.8$	$0.54$

The dose is calculated as:

  $D=\frac{D_0}{{\lambda }_E}$

The initial dose rate for is

  ${\dot{D}}_0=\frac{q\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times E_E\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{m\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}$

For ${}^{\text{131}}\text{I}$ :

  $\begin{array}{l}{\dot{D}}_0=\frac{3.33\times {10}^{-4}\times 0.23\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times 0.23\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{0.02\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}\\ {\dot{D}}_0=1.22\times {10}^{-11}\left(\frac{\text{Gy}}{\text{day}}\right)\end{array}$

So, dose to the thyroid is:

  $\begin{array}{l}D\left(\text{thyroid},{}^{\text{131}}\text{I}\right)=\frac{{\dot{D}}_0}{{\lambda }_E}=\frac{1.22\times {10}^{-11}\left(\frac{\text{Gy}}{\text{day}}\right)}{0.09{\text{day}}^{-1}}\\ =1.35\times {10}^{-10}\text{Gy}\\ \cong 1.4\times {10}^{-10}\text{Gy}\end{array}$

Conclusion:

The dose to the thyroid due to an acute exposure of $1\text{Bq}{\text{.s/m}}^{\text{3}}$ of ${}^{\text{131}}\text{I}$ is $D\left(\text{thyroid},{}^{\text{131}}\text{I}\right)\cong 1.4\times {10}^{-10}\text{Gy}$ .

(b)

To determine

The dose to the thyroid due to an acute exposure of $1\text{Bq}{\text{.s/m}}^{\text{3}}$ of ${}^{\text{133}}\text{I}$ .

Answer to Problem 11.9P

  $3.6\times {10}^{-11}\text{Gy}$

Explanation of Solution

Given:

Mass of thyroid is $m_t=20\text{g}$

Formula used:

The initial dose rate is,

  ${\dot{D}}_0=\frac{q\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times E_E\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{m\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}$

Where,

  $m$ =Mass

  $q$ =Radioactivity

  $E_E$ =Energy per transition

Calculation:

The initial dose rate is,

  ${\dot{D}}_0=\frac{q\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times E_E\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{m\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}$

For ${}^{133}I$ :

  $\begin{array}{l}{\dot{D}}_0=\frac{3.33\times {10}^{-4}\times 0.23\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times 0.54\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{0.02\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}\\ {\dot{D}}_0=2.58\times {10}^{-13}\left(\frac{\text{Gy}}{\text{day}}\right)\end{array}$

So, dose to the thyroid is:

  $\begin{array}{l}D\left(\text{thyroid}{,}^{133}I\right)=\frac{{\dot{D}}_0}{{\lambda }_E}=\frac{2.58\times {10}^{-13}\left(\frac{\text{Gy}}{\text{day}}\right)}{0.09{\text{day}}^{-1}}\\ D\left(\text{thyroid}{,}^{133}I\right)=3.57\times {10}^{-11}\text{Gy}\\ \cong 3.6\times {10}^{-11}\text{Gy}\end{array}$

Conclusion:

The dose to the thyroid due to an acute exposure of $1\text{Bq}{\text{.s/m}}^{\text{3}}$ of ${}^{133}I$ is: $D\left(\text{thyroid}{,}^{133}I\right)\cong 3.6\times {10}^{-11}\text{Gy}$

(c)

To determine

The total body dose due to the protein bound iodine.

Answer to Problem 11.9P

  $D\left(\text{Body}{,}^{131}I\right)=2.5\times {10}^{-13}\text{Gy}$

Explanation of Solution

Given:

Mass of body is $m_b=70\text{kg}$

Formula used:

The initial dose rate is,

  ${\dot{D}}_0=\frac{q\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times E_E\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{m_b\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}$

Where,

  $m$ =Mass

  $q$ =Radioactivity

  $E_E$ =Energy per transition

Calculation:

The dose is calculated as:

  $D=\frac{D_0}{{\lambda }_E}$

The initial dose rate for is

  ${\dot{D}}_0=\frac{q\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times E_E\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{m_b\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}$

For ${}^{131}I$ :

  $\begin{array}{l}{\dot{D}}_0=\frac{3.33\times {10}^{-4}\times 0.77\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times 0.44\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{70\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}\\ {\dot{D}}_0=2.2\times {10}^{-14}\left(\frac{\text{Gy}}{\text{day}}\right)\end{array}$

So, dose to the thyroid is:

  $\begin{array}{l}D\left(\text{Body}{,}^{131}I\right)=\frac{{\dot{D}}_0}{{\lambda }_E}=\frac{2.2\times {10}^{-13}\left(\frac{\text{Gy}}{\text{day}}\right)}{0.09{\text{day}}^{-1}}\\ D\left(\text{Body}{,}^{131}I\right)=2.47\times {10}^{-13}\text{Gy}\\ \cong 2.5\times {10}^{-13}\text{Gy}\end{array}$

Conclusion:

The total body dose due to the protein bound iodine is $D\left(\text{Body},{}^{\text{131}}\text{I}\right)\cong 2.5\times {10}^{-13}\text{Gy}$

(d)

To determine

The effective dose for body from each isotope.

Answer to Problem 11.9P

  $D\left(\text{Body}{,}^{131}I\right)=5.3\times {10}^{-14}\text{Gy}$

Explanation of Solution

Given:

Mass of body is $m_b=70\text{kg}$

Formula used:

The initial dose rate is,

  ${\dot{D}}_0=\frac{q\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times E_E\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{m_b\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}$

Where,

  $m$ =Mass

  $q$ =Radioactivity

  $E_E$ =Energy per transition

Calculation:

The dose is calculated as:

  $D=\frac{D_0}{{\lambda }_E}$

The initial dose rate for is,

  ${\dot{D}}_0=\frac{q\left(\text{Bq}\right)\times 1\frac{\text{tps}}{\text{Bq}}\times E_E\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{m_b\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}$

For ${}^{\text{133}}\text{I}$ :

  $\begin{array}{l}{\dot{D}}_0=\frac{\left(3.33\times {10}^{-4}\times 0.77\right)\text{Bq}\times 1\frac{\text{tps}}{\text{Bq}}\times 0.84\left(\frac{\text{MeV}}{\text{t}}\right)\times 1.6\times {10}^{-13}\left(\frac{\text{J}}{\text{MeV}}\right)\times 8.64\times {10}^4\left(\frac{\text{s}}{\text{day}}\right)}{70\left(\text{kg}\right)\times 1\left(\frac{\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}{\text{Gy}}\right)}\\ {\dot{D}}_0=4.25\times {10}^{-14}\left(\frac{\text{Gy}}{\text{day}}\right)\end{array}$

So, dose to the thyroid is:

  $\begin{array}{l}D\left(\text{Body},{}^{\text{131}}\text{I}\right)=\frac{{\dot{D}}_0}{{\lambda }_E}=\frac{4.25\times {10}^{-14}\left(\frac{\text{Gy}}{\text{day}}\right)}{0.8{\text{day}}^{-1}}\\ D\left(\text{Body},{}^{\text{131}}\text{I}\right)=5.3\times {10}^{-14}\text{Gy}\end{array}$

Conclusion:

The effective dose for body from each isotope is $D\left(\text{Body}{,}^{131}I\right)=5.3\times {10}^{-14}\text{Gy}$