Chapter 6, Problem 6.27P

To determine

To Calculate:The absorbed dose to the liver.

Answer to Problem 6.27P

The absorbed dose to the liver is $0.7154\text{rad}$

Explanation of Solution

Given:

Activity of ${}^{99m}Tc$ -labeled sulfur colloid is $3\text{mci}\text{(111}\text{MBq)}$

The deposited percentage of the injectate is $60\%$

Formula used:

  $\begin{array}{l}D=\tilde{A}S\\ \tilde{A}=1.44A_0{T}_e\\ T_e=\frac{T_p\times T_b}{T_p+T_b}\\ D=1.44A_0{T}_{e}S\end{array}$

Where,

  $D$ - Absorbed dose

  $\tilde{A}$ -The cumulated activity

  $A_0$ - Administered activity

  $T_e$ - Effective half-life

  $T_p$ - Physical half-life ( $t_{1/2}$ - half-life)

  $T_b$ - Biological half-life

  $S$ - Absorbed dose per unit cumulated activity

[Values of $S$ for human phantoms can be found in Medical Internal Radiation Dose (MIRD) pamphlets published by the Society of Nuclear Medicine]

In ${}^{99m}Tc$ decay,

  $\begin{array}{l}S\left(livertoliver\right)=4.6\times {10}^{-5}\frac{\text{rad}}{\text{μ}\text{ci}\text{h}}\\ t_{1/2}(T_p)=6\text{h}\end{array}$

Calculation:

  $A_0=3000\mu ci\times 0.6=1800\mu ci$

Absorb dose to the liver [assuming $T_b\to \infty$ (there is no biological clearance)],

  $\begin{array}{l}D=1.44A_0{T}_{e}S\\ D=1.44\times 1800\text{μ}\text{ci}\times 6h\times 4.6\times {10}^{-5}\frac{\text{rad}}{\text{μ}\text{ci}\text{h}}\\ D=0.7154\text{rad}\end{array}$

Conclusion:

The absorbed dose to the liver = $0.7154\text{rad}$

To determine

To Calculate:The absorbed dose to the spleen.

Answer to Problem 6.27P

The absorbed dose to the liver is $2.5661\text{rad}$

Explanation of Solution

Given:

Activity of ${}^{99m}Tc$ -labeled sulfur colloid is $3mci(111MBq)$

The deposited percentage of the inject is $30\%$ .

Formula used:

  $\begin{array}{l}D=\tilde{A}S\\ \tilde{A}=1.44A_0{T}_e\\ T_e=\frac{T_p\times T_b}{T_p+T_b}\\ D=1.44A_0{T}_{e}S\end{array}$

Where,

  $D$ - Absorbed dose

  $\tilde{A}$ -The cumulated activity

  $A_0$ - Administered activity

  $T_e$ - Effective half-life

  $T_p$ - Physical half-life ( $t_{1/2}$ - half-life)

  $T_b$ - Biological half-life

  $S$ - Absorbed dose per unit cumulated activity

[Values of $S$ for human phantoms can be found in Medical Internal Radiation Dose (MIRD) pamphlets published by the Society of Nuclear Medicine]

In ${}^{99m}Tc$ decay,

  $\begin{array}{l}S\left(spleentospleen\right)=3.3\times {10}^{-4}\frac{rad}{\mu cih}\\ t_{1/2}(T_p)=6h\end{array}$

Calculation:

  $A_0=3000\text{μ}\text{ci×0}\text{.3=900}\text{μ}\text{ci}$

Absorb dose to the spleen [assuming $T_b\to \infty$ (there is no biological clearance)],

  $\begin{array}{l}D=1.44A_0{T}_{e}S\\ D=1.44\times 900\text{μ}\text{ci}\times 6h\times 3.3\times {10}^{-4}\frac{\text{rad}}{\text{μ}\text{ci}\text{h}}\\ D=2.5661\text{rad}\end{array}$

Conclusion:

The absorbed dose to the spleen = $2.5661\text{rad}$

To determine

To Calculate:The absorbed dose to the red marrow.

Answer to Problem 6.27P

The absorbed dose to the red marrow is $0.0804\text{rad}$

Explanation of Solution

Given:

Activity of ${}^{99m}Tc$ -labeled sulfur colloid is $\text{3}\text{mci}\text{(111}\text{MBq)}$

The deposited percentage of the inject is $10\%$ .

Formula used:

  $\begin{array}{l}D=\tilde{A}S\\ \tilde{A}=1.44A_0{T}_e\\ T_e=\frac{T_p\times T_b}{T_p+T_b}\\ D=1.44A_0{T}_{e}S\end{array}$

Where,

  $D$ - Absorbed dose

  $\tilde{A}$ -The cumulated activity

  $A_0$ - Administered activity

  $T_e$ - Effective half-life

  $T_p$ - Physical half-life ( $t_{1/2}$ - half-life)

  $T_b$ - Biological half-life

  $S$ - Absorbed dose per unit cumulated activity

[Values of $S$ for human phantoms can be found in Medical Internal Radiation Dose (MIRD) pamphlets published by the Society of Nuclear Medicine]

In ${}^{99m}Tc$ decay,

  $\begin{array}{l}S\left(redmarrowtoredmarrow\right)=3.1\times {10}^{-5}\frac{\text{rad}}{\text{μ}\text{ci}\text{h}}\\ t_{1/2}(T_p)=6\text{h}\end{array}$

Calculation:

  $A_0=3000\mu ci\times 0.1=300\mu ci$

Absorb dose to the red marrow [assuming $T_b\to \infty$ (there is no biological clearance)],

  $\begin{array}{l}D=1.44A_0{T}_{e}S\\ D=1.44\times 300\text{μ}\text{ci}\times 6h\times 3.1\times {10}^{-5}\frac{\text{rad}}{\text{μ}\text{ci}\text{h}}\\ D=0.0804\text{rad}\end{array}$

Conclusion:

The absorbed dose to the red marrow = $0.0804\text{rad}$