User:Vsmith/Dating calc

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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
n = \frac{\ln \left({f}\right)}{\ln \left({0.5}\right)}

As an example, for a sample that contains 0.06780 of the original C-14:

n = \frac{\ln \left({0.06780}\right)}{\ln \left({0.5}\right)} 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:

n = \frac{\ln \left({0.25}\right)}{\ln \left({0.5}\right)} solving gives n= 2 half lives and
2 half lives * 5730 yrs/half life = 11460 yrs.