5.3.2 Exponential Functions and Models II
Radioactive decay
 over time, radioactive carbon 14 will “decay” into
ordinary (non-radioactive) carbon 12
 e.g, if you start with 16 g. of carbon 14,
 in 5700 years there will remain only 8 grams
o ½ of the original amount
 in another 5700 years, there will remain only 4 grams
o again ½ half of the remaining amount
 the 5700 years is referred to as the half-life of carbon 14
 no matter how much you start with, only half of the
original amount will remain after 5700 years
So we have:
#½-lives 0
amount
16
left
1
2
3
8 = 16(½) 4 = 16(½)(½) 2 = 16(½)2(½)
= 16(½)2
= 16(½)3
let A be the amount left over time:
A = 16(½)(#half lives)
Let’s measure time t in years, not half-lives:
11400 years: 11400/5700 = 2 half lives
divide years by half-life to get #half-lives
t years = t/5700 half-lives
 for C14 we have the amount left after t years:
A(t) = C(½)t/5700 where C is the original amount
 in general, for any radioactive substance with a half-life
of k years, we have
A(t) = C(½)t/k
5.3.2-1
Finding the age of a fossil (optional – use GC)
A fossil contains 5% of the amount of C14 that it contained
when alive. How old is it?
.05C = C(½)t/5700

.05 = (½)t/5700
graph f(x) = (½)t/5700
use TRACE to estimate for what value of t f(x) = .05
that’s your answer!
5.3.2-2