The radioisotope Phosphorus-32 is used in leukemia therapy. The half-life of this isotope os 14.26 days. Which equation determines the percent of an initial amount of the isotope remaining after t days?A= 14.26 (1/2)^100/tA= 14.26(1/2)^t/100A= 100(1/2)^14.26/tA= 100(1/2)^t/14.26
Accepted Solution
A:
the remaining value= the initial value × (1/2)^(t/T), t is the number of years, T is half life. So D is the right answer.