Chapter 31, Problem 69GP

To determine

To Calculate: The initial activity of the ${\mathrm{T}}_{\mathrm{C}}$ sample if the whole-body dose cannot exceed $50\text{mrem}$ .

Answer to Problem 69GP

  $3\text{mCi}$

Explanation of Solution

Given:

Mass of the patient is, $\mathrm{m}=70\text{kg}$

Formula used:

Effective dose (in rem) is $=\mathrm{d}\mathrm{o}\mathrm{s}\mathrm{e}\left(\text{in rad}\right)\times \mathrm{Q}\mathrm{F}$

Activity, $\mathrm{A}=\mathrm{\lambda }\mathrm{N}$

Total number of nuclei decayed:

  $\mathrm{N}=\frac{\text{Effective dose}}{\left(\frac{1}{2}\right)\left(\text{Energy of each gamma ray}\right)}$

Where,

  $\mathrm{N}$ = number of nuclei decayed

  $\mathrm{\lambda }$ = decay constant

  $\mathrm{Q}\mathrm{F}$ = quality factor of gamma radiation

Calculation:

Effective dose (in rem) is

  $\begin{array}{l}=\mathrm{d}\mathrm{o}\mathrm{s}\mathrm{e}\left(\text{in rad}\right)\times \mathrm{Q}\mathrm{F}\\ =50\times {10}^{-3}\times {10}^{-2}\times 70\times \left(\frac{1}{1.60\times {10}^{-19}}\right)\\ =2.19\times {10}^{17}\text{eV}\end{array}$

Total number of nuclei decayed:

  $\mathrm{N}=\frac{\text{Effective dose}}{\left(\frac{1}{2}\right)\left(\text{Energy of each gamma ray}\right)}$

  $\begin{array}{l}\mathrm{N}=\left(2.19\times {10}^{17}\right)\left(\frac{1}{140\times {10}^3}\right)\left(2\right)\\ \mathrm{N}=3.13\times {10}^{12}\text{nuclei}\end{array}$

Activity can be given as:

  $\begin{array}{l}\mathrm{A}=\mathrm{\lambda }\mathrm{N}\\ \mathrm{A}=\left(\frac{0.693}{{\mathrm{T}}_{1/2}}\right)\mathrm{N}\\ \mathrm{A}=\left(\frac{0.693}{6}\right)\left(\frac{1}{3600}\right)\left(3.13\times {10}^{12}\right)\left(\frac{1}{3.7\times {10}^{10}}\right)\\ \mathrm{A}=3\times {10}^{-3}\text{Ci}\\ \mathrm{A}=\text{3}\text{mCi}\end{array}$

Conclusion:

  $\text{N = 3}\text{mCi}$ is the initial activity of the ${\mathrm{T}}_{\mathrm{C}}$ sample if the whole-body dose cannot exceed $50\text{mrem}$ .