Question

I have no idea so i'm screwed please help me
Beryllium-11 has a half-life of 13.81 s. What percent of the initial sample will remain aft er 30.0 s?

Answer 1

Answer:

Let N0 be the initial atoms of Be11

N0 / 2^1 = N0 / 2  Number of Be11 after 1 half-life

N0 / 2^2 = N0/ 4   Number of Be11 after 2 half-lives

30/13.81 = 2.17 half lives

N0 / 2^2.17 = N0 / 4.51 = .222 N0 or 22.2 % of atoms remain