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Statistical Mechanics

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### 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.

``` Web ScientificSentence