Chapter 19, Problem 26A

Interpretation Introduction

Interpretation: Given its half-life, fraction of Technitium-99 that remains in a patient’s body after 2.0 days is to be determined.

Concept introduction:Half-life of a radioactive sample is the time taken by it to reduce its concentration to half. Formula used to calculate the fraction of sample left after half-lives is as follows:

  $\text{Fraction of sample left}=\frac{1}{2^n}$

Answer to Problem 26A

Fraction ofTechnitium-99 sample that remains in a patient’s body after 2.0 days = $\frac{1}{256}$ .

Explanation of Solution

Amount of Technitium-99 dosage $=100\mu g$

Half-life of Technitium-99 = $6hrs$

Number of hours in 2 days $=2\times 24hrs$

  $=48hrs$

Number of half-lives in 2 days= $\frac{48hrs}{6hrs}$

= $8$

Formula used to calculate the fraction of sample left after $n$ half-lives is

  $\text{Fraction of sample left}=\frac{1}{2^n}$

Fraction of Technitium-99 sample left $=\frac{1}{{\left(2\right)}^8}$

= $\frac{1}{256}$

Conclusion

Fraction ofTechnitium-99 sample that remains in a patient’s body after 2.0 days = $\frac{1}{256}$ .