Chapter 31, Problem 47P

(a)

To determine

To find:The reaction for the decay of ${}^{13!}\text{I}$ .

Explanation of Solution

According to Appendix B, ${}_{53}^{131}$ I decays by beta decay.

  ${}_{53}^{131}\text{I}\to {}_{54}^{131}\text{Xe}+{\mathrm{\beta }}^{-}+\overline{\mathrm{v}}$

(b)

To determine

To find:Duration required to ingest ingested

Answer to Problem 47P

  $27\text{days}$

Explanation of Solution

Given data:

Half −life of ${}_{53}^{131}\text{I}$ is 8.0 d

Final nuclei number is 10% of initial number, $\mathrm{N}=0.10{\mathrm{N}}_0$ .

Formula used:

The number of nuclei present after time t

  $\mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda }\mathrm{t}}$

Calculation:

The number of nuclei present is

  $\mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda }\mathrm{t}}$

After rearranging the above equation

  $\begin{array}{c}\mathrm{t}=-\frac{\ln \mathrm{N}/{\mathrm{N}}_0}{\mathrm{\lambda }}\\ =-\frac{{\mathrm{T}}_{1/2}\ln \mathrm{N}/{\mathrm{N}}_0}{0.693}\\ =-\frac{(8.0\text{d})\ln 0.10}{0.693}\\ =26.58\text{d}\\ \approx 27\text{d}\end{array}$

Conclusion:

  $27\text{d}$

(c)

To determine

To find:The mass of iodine

Answer to Problem 47P

  $8\times {10}^{-12}\text{kg}$

Explanation of Solution

Given data:

Absorbed amount is $1\text{mCi}\text{=}\text{1}\times {\text{10}}^{-3}\text{Ci}$

  $1\text{Ci}=3.70\times {10}^{10}\text{decays}/\text{s}$

Formula used:

The activity is given by $\mathrm{\Delta }\mathrm{N}/\mathrm{\Delta }\mathrm{t}=\mathrm{\lambda }\mathrm{N}$ .

Calculation:

(c) This can be used to find the number of nuclei, and then the mass can be found.

  $\mathrm{\Delta }\mathrm{N}/\mathrm{\Delta }\mathrm{t}=\mathrm{\lambda }\mathrm{N}$

Final number of nuclei is

  $\begin{array}{l}\mathrm{N}=\frac{\mathrm{\Delta }\mathrm{N}/\mathrm{\Delta }\mathrm{t}}{\mathrm{\lambda }}\\ =\frac{\left({\mathrm{T}}_{1/2}\right)(\mathrm{\Delta }\mathrm{N}/\mathrm{\Delta }\mathrm{t})}{0.693}\\ =\frac{(8.0\text{d})(86,400\text{s}/\text{d})\left(1\times {10}^{-3}\text{Ci}\right)\left(3.70\times {10}^{10}\text{decays}/\text{s}\right)}{0.693}\\ =3.69\times {10}^{13}\text{ nuclei }\end{array}$

The mass of iodine is

  $\mathrm{m}=3.69\times {10}^{13}\text{ nuclei }\left(\frac{0.131\text{kg}}{6.02\times {10}^{23}\text{nuclei}}\right)=8\times {10}^{-12}\text{kg}$

Conclusion:

The mass of iodine is $8\times {10}^{-12}\text{kg}$