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-$32$?$(5.4)$

Short answer

$1half-lifeperiodforP-32=\frac{28.6\text{days}}{2\text{half lives}}\times 1halflife.$

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

Step-by-step solution

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 $1.2mg$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$0.30\mathrm{mg}.$

    Explanation

Calculate the half-life of$P-32$:

The following are the equality and conversion factors:

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

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

    Conclusion

Therefore

$1half-lifeperiodforP-32$$=\frac{28.6\text{days}}{2\text{half lives}}\times 1halflife.$

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