If 18g of a radioactive substance are present initially and 8 yr later only 9 g remain, how much of the substance will be present after 20 yr?

Answer 1

clearly the half-life is 8 years. So, after t years, the remaining fraction is

18*2^(-t/8)

So, plug in t=20

Answer 2

half gone in 8 years?

x = Xi e^-kt

.5 = e^-8k

ln .5 = -.693 = - 8 k
so
k = .08664

x = 18 e^-.08664 t

when t = 20
x = 3.18 grams