Question

The half-life of iodine-123 is about 13 hours. You begin with 38 grams of iodine-123.

Answer 1

Half-life, h = 13 hours
Mass initial, a = 38 g
B=base for half-life = 2
(a)
I(t)=a*B^(-t/h)=a*(B^(1/h)^(-t)=a*(B^(-1/13))^t
Substituting values,
I(t)=38*(2^(-1/13))^t
a=38 g
b=2^(-1/13)=0.9480775 (approximately)
=>
I(t)=38*0.9480775^t

(b) I(t)=7
solve
I(t)=7=38*0.948077^t
Take log on both sides and solve for t
t=log(7/38)/log(0.948077)
=31.727 hours.

Check: I(31.727)=38*9.948077^(31.727)=7.000 g  ok