For radioactive decay
The relationship between fraction remaining (f) and the number of half-lives (n) elapsed can be shown by:
n |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
... |
n |
f |
1/2 |
1/4 |
1/8 |
1/16 |
1/32 |
1/64 |
1/128 |
... |
... |
|
1/21 |
1/22 |
1/23 |
1/24 |
1/25 |
1/26 |
1/27 |
... |
1/2n |
Therefore the fraction remaining after time n is: f=1/2n which is equivalent to f=(1/2)n or f=0.5n. This makes the calculation of the age if the fraction remaining is known quite simple. Solving the above relationship for n using the properties of logarithms:
- f=0.5n becomes
- ln f = n·ln 0.5 and
As an example, for a sample that contains 0.06780 of the original C-14:
- solving gives n= 3.88 half lives and
- 3.883 half lives * 5730 yrs/half life = 22,250 yrs.
A more simple example, for a sample containing 0.25 of the original:
- solving gives n= 2 half lives and
- 2 half lives * 5730 yrs/half life = 11460 yrs.