Chapter 11, Problem 11.1P

To determine

The absorbed dose to the lung during 13-week period and during the 1-year period immediately following inhalation.

Answer to Problem 11.1P

In 13 weeks absorbed dose is $35\text{mGy}$ and in 1 year is $40\text{mGy}$ .

Explanation of Solution

Given:

Mean radioactivity concentration, $C=3.3{\text{MBq/m}}^3$ (in $2\text{hour}$ )

Formula used The effective half-life $T_{\text{E}}$ is

  $T_E=\frac{T_R\times T_B}{T_R+T_B}$

The effective clearance rate constant for these particles is

  ${\lambda }_E=\frac{0.693}{T_E}$

The cumulated activity for all four compartments after $t\text{day}$

  $\text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(t\right)\text{=}\frac{{\text{A}}_{s1}(0)}{{\lambda }_{E_1}}\left(1-e^{-{\lambda }_{1}t}\right)+\frac{{\text{A}}_{s2}(0)}{{\lambda }_{E_2}}\left(1-e^{-{\lambda }_{2}t}\right)+\frac{{\text{A}}_{s3}(0)}{{\lambda }_{E_3}}\left(1-e^{-{\lambda }_{3}t}\right)+\frac{{\text{A}}_{s3}(0)}{{\lambda }_{E_4}}\left(1-e^{-{\lambda }_{4}t}\right)$

The dose from the activity in the lung is given

  $\text{D}\left(\text{13wk}\right)\text{=}\text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(\text{13wk}\right)\text{Bq}\cdot \text{d}\text{×}\text{DCF,}\frac{\text{Gy}}{\text{Bq}\cdot \text{d}}$

Calculation:

Since no particle size is given, the ICRP default value of $1\text{μm}$ AMAD particles is used. The person inhales approximately $\text{10}{\text{m}}^3$ of air in an 8-hour day.

The inhaled activity:

  $\text{2}\text{hrs}\times 3.3\frac{\text{MEq}}{{\text{m}}^3}\times \frac{1\times {10}^6\text{Bq}}{\text{MBq}}\times \frac{10{\text{m}}^3}{8\text{hr}}=8.25\times {10}^6\text{Bq}$ is the total inhaled activity

The elimination constant for each of the compartments is calculated next.

ICRP 26 give the following depositions of $\text{1}\text{μm}$

AMAD particles:

	Deposition	Fraction in lungs	Total activity	Activity in lungs
N.P.	30%	0	$8.25\times {10}^6\text{Bq}$	0
T.B.	$8\%$	0.08	$8.25\times {10}^6\text{Bq}$	$6.6\times {10}^5\text{Bq}$
P	$25\%$	0.25	$8.25\times {10}^6\text{Bq}$	$2.1\times {10}^6\text{Bq}$
			Total deposited in lung	$2.7\times {10}^6\text{Bq}$

The effective clearance rate constant for these particles is

  ${\lambda }_E=\frac{0.693}{T_E}$

For 60 percent of the particles deposited in the P region , whose biological half --life $T_{\text{B}}=50$ days , and radiological half-life $T_{\text{R}}=87\text{days}$

The effective half-life $T_{\text{E}}$ is

  $T_E=\frac{T_R\times T_B}{T_R+T_B}$

  $=\frac{87\text{days}\times 50\text{days}}{87\text{days}+50\text{days}}=32\text{days}$

  ${\lambda }_{\text{E}}=\frac{0.693}{T_{1/2}}=\frac{0.693}{32\text{day}}=0.022{\text{d}}^{-1}$

Using ICRP 26 for following deposition of inhaled $1\text{μm}$ AMAD particles.

Region	% cleared	As(0), Bq	$T_B,{\text{d}}^{-1}$	$T_E,{\text{d}}^{\text{-1}}$	${\lambda }_E,{\text{d}}^{\text{-1}}$
TB	50	$0.5\times 6.6\times {10}^5$	$0.01$	$0.01$	69.3
TB.	50	$0.5\times 6.6\times {10}^5$	$0.2$	$0.2$	3.47
P	40	$0.4\times 2.61\times {10}^6$	$1$	$1$	0.693
P	60	$0.6\times 2.1\times {10}^6$	50	32	0.022

The cumulated activity for all four compartments after $t_1=91\text{day}$

  $\text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(t_1\right)\text{=}\frac{{\text{A}}_{s1}(0)}{{\lambda }_{E_1}}\left(1-e^{-{\lambda }_1{t}_1}\right)+\frac{{\text{A}}_{s2}(0)}{{\lambda }_{E_2}}\left(1-e^{-{\lambda }_2{t}_1}\right)+\frac{{\text{A}}_{s3}(0)}{{\lambda }_{E_3}}\left(1-e^{-{\lambda }_3{t}_1}\right)+\frac{{\text{A}}_{s3}(0)}{{\lambda }_{E_4}}\left(1-e^{-{\lambda }_4{t}_1}\right)$

  $\begin{array}{l} \text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(13\text{ weeks}\right)=\frac{(0.5)\times 6.6\times {10}^5\text{Bq}}{69.3{\text{d}}^{-1}}\left(1-e^{-69.3\times 91}\right)+\frac{(0.5)\times 6.6\times {10}^5\text{Bq}}{3.47{\text{d}}^{-\iota }}\left(1-e^{-3.47\times 91}\right)+\\ +\frac{(0.4)\times 2.1\times {10}^6\text{Bq}}{0.693{\text{d}}^{-1}}\left(1-e^{-0.693\times 91}\right)+\frac{0.6\times 2.1\times {10}^6\text{Bq}}{0.022{\text{d}}^{-1}}\left(1-e^{-0.022\times 91}\right)\end{array}$

  $$

  $\text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(13\text{weeks}\right)=5.1\times {10}^7\text{Bq}\cdot \text{d}$

The cumulated activity for all four compartments after $t=365\text{day}$

  $\text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(t_2\right)\text{=}\frac{{\text{A}}_{s1}(0)}{{\lambda }_{E_1}}\left(1-e^{-{\lambda }_1{t}_2}\right)+\frac{{\text{A}}_{s2}(0)}{{\lambda }_{E_2}}\left(1-e^{-{\lambda }_2{t}_2}\right)+\frac{{\text{A}}_{s3}(0)}{{\lambda }_{E_3}}\left(1-e^{-{\lambda }_3{t}_2}\right)+\frac{{\text{A}}_{s3}(0)}{{\lambda }_{E_4}}\left(1-e^{-{\lambda }_4{t}_2}\right)$

  $\begin{array}{l} \text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(1\text{year}\right)=\frac{(0.5)\times 6.6\times {10}^5\text{Bq}}{69.3{\text{d}}^{-1}}\left(1-e^{-69.3\times 365}\right)+\frac{(0.5)\times 6.6\times {10}^5\text{Bq}}{3.47{\text{d}}^{-\iota }}\left(1-e^{-3.47\times 365}\right)+\\ +\frac{(0.4)\times 2.1\times {10}^6\text{Bq}}{0.693{\text{d}}^{-1}}\left(1-e^{-0.693\times 365}\right)+\frac{0.6\times 2.1\times {10}^6\text{Bq}}{0.022{\text{d}}^{-1}}\left(1-e^{-0.022\times 365}\right)\end{array}$

  $\text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(1\text{year}\right)=5.9\times {10}^7\text{Bq}\cdot \text{d}$

Sulfur 35 emits a single beta whose mean energy is $\text{0}\text{.049}\text{MeV}$ , and no gammas. Using a lung weight of 1 kg (Appendix C), the DCF is calculated

  $\text{DCF}\text{=}\frac{\text{mGy}}{\text{Bq}\cdot \text{d}}=\frac{1\frac{\text{t}}{\text{s}\cdot \text{Bq}}\times 8.64\times {10}^4\frac{\text{s}}{\text{d}}\times 4.9\times {10}^{-2}\frac{\text{MeV}}{\text{t}}\times 1.6\times {10}^{-13}\frac{\text{J}}{\text{MeV}}}{1\text{kg}\times 1\frac{\text{J}/\text{kg}}{\text{Gy}}\times \frac{\text{1}\text{Gy}}{{10}^3\text{mGy}}}$

  $\text{DCF}=6.8\times {10}^{-7}\frac{\text{mGy}}{\text{Bq}\cdot \text{d}}$

Therefore

The dose from the activity in the lung is given

  $\text{D}\left(\text{13wk}\right)\text{=}\text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(\text{13wk}\right)\text{Bq}\cdot \text{d}\text{×}\text{DCF,}\frac{\text{Gy}}{\text{Bq}\cdot \text{d}}$

  $D\left(13\text{wk}\right)=5.1\times {10}^7\text{Bq}\cdot \text{d}\times 6.8\times {10}^{-7}\frac{\text{mGy}}{\text{Bq}\cdot \text{d}}=35\text{mGy}$

And,

  $\text{D}\left(1\text{yr}\right)\text{=}\text{<mover accent="true"><mo>A</mo><mo>˜</mo></mover>}\left(1\text{yr}\right)\text{Bq}\cdot \text{d}\text{×}\text{DCF,}\frac{\text{Gy}}{\text{Bq}\cdot \text{d}}$

  $D\left(1\text{yr}\right)=5.9\times {10}^7\text{Bq}\cdot \text{d}\times 6.8\times {10}^{-7}\frac{\text{mGy}}{\text{Bq}\cdot \text{d}}=40\text{mGy}$

Conclusion:

The absorbed dose in 13 weeks is $35\text{mGy}$ and in 1 year is $40\text{mGy}$ .