Question

Gold (Au) has a synthetic isotope that is relatively unstable. After 25.5 minutes, a 128-gram sample has decayed to 2 g.

What is the half-life of this isotope?


8.5 minutes

2.25 minutes

1.5 minutes

4.25 minutes

Answer 1

x = Xi e^-kt
2 = 128 e^-25.5 k
0.015625 = e^-25.5 k
ln 0.015625 = -25.5 k
-4.15888 = -25.5 k
k = 0.163
so
1 = 2 e^-0.163 t for half life
0.5 = e^-0.163 t
ln 0.5 = -0.693 = -0.163 t
t = 4.25 minutes

Answer 2

128/2 = (1/2)^6
So, now you want
25.5/6 = 4.25

For a half-life of k minutes, your function is A = P*(1/2)^(t/k)