Phosphorus-32 has a half-life of 14.0 days. Starting with 8.00g of 32P, how many grams will remain after 70.0days ?

May 31, 2015

You'll be left with 0.25 g of phosphorus-32.

The nuclear half-life of a radioactive isotope expresses the time needed for a sample of that isotope to reach half of its initial value.

In your case, regardless with how much phosphorus-32 you start with, you'll be left with half of that initial mass after 14.0 days. After another 14 days, you'll be left with half of what you had after the first 14 days, which is equivalent to a quarter of the initial mass.

The original mass of the isotope will be halved for each passing of a half-life. In your case, you'll be left with

• 50% of the initial mass $\to$ after 1 half-life has passed;
• 25% of the initial mass $\to$ after 2 half-lives have passed;
• 12.5% of the initial mass $\to$ after 3 half-lives have passed;
• 6.25% of the initial mass $\to$ after 4 half-lives have passed;
• 3.125% of the initial mass $\to$ after 5 half-lives have passed.

and so on.

When the numbers allow it, such as you have in your problem, you can easily figure out how many half-lives have passed by dividing the total time passed by the half-life of the isotope

"number of half-lives" = (70.0cancel("days"))/(14.0cancel("days")) = 5

This means that you need to divide the initial mass by 2 for every half-life that passed, which is equivalent to having

$\frac{\text{remaining mass" = "initial mass"/(underbrace(2 * 2 * ... * 2)_(color(blue)("5 times"))) = "initial mass}}{2} ^ 5$

In your case, you'll be left with

m_"remaining" = "8.00 g"/2^5 = 8/32 = color(green)("0.250 g")