Radioactive decay and Dating

1. ¹⁴Carbon Dating formula

Cosmic rays provide neutrons that interact 
with Nitrogen (14N) and form 14C 
( 147N + 10n → 
146C + 11P).
146C combine with 168O to form 
146C168O2 
molecules that plants and animals use or beathe.
The ration Ro (original:before death) = 
Number(147C)/Number(126C) in CO2 
molecules in the atmosphere is well known and 
constant equal to 1.2 x 10- 12
Ro = 1.2 x 10- 12

When an organism dies, 147C atoms cease. After a time t, 
in an organism, It remains N(t) = N0 exp[- λt], according 
to the radioactivity decay equation. 
We have then:
R(t) = N(t)/Number(126C)= 
N0 exp[- λt]/Number(126C) = 
R0 exp[- λt], 
since N0 = Number(147C), 
the original.

It follows that:
R(t) /R0 = exp[- λt]. Thus 
		t =(-1/λ)ln(R(t)/R0) 
		= - (t1/2/ln(2))ln(R(t)/R0) 

Where λ is the decay constant; t1/2 is the half-life of 
the element 147C = 5730 years.
R0 is known and R(t) is measured. The value of t is then 
straightforward.

Example: 
If R(t) = 0.6  x 10- 12, then t = (- 5730 /ln(2)). 
ln(0.6/1.2) = 5730 years.

2. Other formulas: Time
dating using lead isotopes

The 238 Uranium decays to 206 Pb according the process:
238U → 206Pb
At the origin, the number of nuclides 238U is  
N(t = 0) = N0
At a later time, we have : N(t, 238U) = N0exp[-λt]. 
At the precise time, 
N(t, 206Pb) = N0 - N0exp[-λt] 
= N0(1 - exp[-λt]) 
is the number of the lead nuclides 206Pb.
The ratio :R(t) =  N(t, 206Pb)/ N(t, 238U) = 
(1 - exp[-λt])/exp[-λt]= exp[λt] - 1
Then the abudance ratio takes the form:
exp[λt] = [N(t, 206Pb)/ N(t, 238U)] + 1
And:
t = ln [N(t, 206Pb)/ N(t, 238U) + 1]/λ  
t=
(t1/2/ln(2))ln[N(t, 206Pb)/N(t,238U) + 1] 

The half-life t1/2 of the 238U 
is 4.47 x 109 years. If  the ratio : 
N(t, 206Pb)/ N(t, 238U) = 
0.5, How old would be the ore?
t = (4.47 x 109/ 0.69) ln(1.5) ≈ 
3 billion years.