Q. 5.75 - An Introduction to General, Organic, and Biological Chemistry

Question

If the amount of radioactive phosphorus-$32$, used to treat leukemia, in a sample decreases from $1.2 \mathrm{mg}$ to $0.30 \mathrm{mg}$ in $28.6$ days, what is the half-life of phosphorus-$32$? $(5.4)$

Step-by-Step Solution

Answer

$1 half - life period for P-32=\frac{28.6\text{ days }}{2\text{ half lives }}\times 1 half life.$

Phosphorus-$32$ has a half-life of $14.3$ days.

Step 1: Given information

Half-life: A radioisotope's half-life is the time it takes for half of a sample to decay. In  $28.6$ days, the $P-32$ sample drops from $1.2 mg$ to$0.30mg.$

To begin, figure out how many half-lives are needed to reduce the sample from$1.2mg$ to $0.30mg$ in $28.6$ days.

$1.2\mathrm{mg} P-32\stackrel{1\text{ half-life }}{⟶}0.6\mathrm{mg} P-32\stackrel{2\text{ half-lives }}{⟶}0.3\mathrm{mg} P-32$

In $28.6$ days, two half-lives are required to reduce the$P-32$ sample from $1.2 \mathrm{mg}$ to $0.30 \mathrm{mg}.$

Step 2: Explanation

Calculate the half-life of $P-32$:

The following are the equality and conversion factors:

$2 \mathrm{half} - \mathrm{lives} \mathrm{for} \mathrm{P}-32=28.6 \mathrm{days}$

$\frac{28.6\text{ days }}{2\text{ half - lives }} or \frac{2\text{ half - lives }}{28.6\text{ days }}$

Step 3: Conclusion

Therefore

$1 half - life period for P-32$$=\frac{28.6\text{ days }}{2\text{ half lives }}\times 1 half life.$

Phosphorus-$32$ has a half-life of $14.3$days.