Samarium-153 has a half-life of 2 days.How much of this radioisotope remains after 10 days?

Question

Quizplus · Accepted Answer

After 10 days, there will have been 5 half-lives of Samarium-153 (10/2). 
Using the formula for half-life decay, we can calculate how much of the radioisotope remains:
Amount remaining = initial amount x (1/2)^(number of half-lives)
Amount remaining = 100% x (1/2)^5
Amount remaining = 100% x 0.03125
Amount remaining = 3.125%
Therefore, the best choice is D, about 3%.

Samarium-153 Has a Half-Life of 2 Days