What percent of the phosphorus-32 decays each day?

Question

The amount y (in grams) of the radioactive isotope phosphorus-32 remaining after t days is y=a(0.5)^t/14, where a is the initial amount (in grams). Round the answer to the nearest hundredth of a percent

Answer 1

In this equation, the value of "14" is the half life of 32-P in days.  That is to say, the half life for this nuclide is 14 days.

Fraction remaining = 0.5^n where n = number of half lives.  In this question they are looking for the fraction remaining after 1 day, so 1/14 = 0.0714 half lives.  Thus fraction remaining after 1 day = 0.5^0.0714 = 0.9517 or 95.17 % remains.

To get the % that decays after 1 day, it will be 100% - 95.17% = 4.83%.