Nuclear Physics
The nucleus
Radioactivity
Applications
Particle accelerators
© The scientific sentence. 2010

Radioactive decay and Dating
1. ^{14}Carbon Dating formula
Cosmic rays provide neutrons that interact
with Nitrogen (^{14}N) and form ^{14}C
( ^{14}_{7}N + ^{1}_{0}n →
^{14}_{6}C + ^{1}_{1}P).
^{14}_{6}C combine with ^{16}_{8}O to form
^{14}_{6}C^{16}_{8}O_{2}
molecules that plants and animals use or beathe.
The ration R_{o} (original:before death) =
Number(^{14}_{7}C)/Number(^{12}_{6}C) in CO_{2}
molecules in the atmosphere is well known and
constant equal to 1.2 x 10^{ 12}
R_{o} = 1.2 x 10^{ 12}
When an organism dies, ^{14}_{7}C atoms cease. After a time t,
in an organism, It remains N(t) = N_{0} exp[ λt], according
to the radioactivity decay equation.
We have then:
R(t) = N(t)/Number(^{12}_{6}C)=
N_{0} exp[ λt]/Number(^{12}_{6}C) =
R_{0} exp[ λt],
since N_{0} = Number(^{14}_{7}C),
the original.
It follows that:
R(t) /R_{0} = exp[ λt]. Thus
t =(1/λ)ln(R(t)/R_{0})
=  (t_{1/2}/ln(2))ln(R(t)/R_{0})
Where λ is the decay constant; t_{1/2} is the halflife of
the element ^{14}_{7}C = 5730 years.
R_{0} 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 isotopesThe 238 Uranium decays to 206 P_{b} according the process:
^{238}U → ^{206}P_{b}
At the origin, the number of nuclides ^{238}U is
N(t = 0) = N_{0}
At a later time, we have : N(t, ^{238}U) = N_{0}exp[λt].
At the precise time,
N(t, ^{206}P_{b}) = N_{0}  N_{0}exp[λt]
= N_{0}(1  exp[λt])
is the number of the lead nuclides ^{206}P_{b}.
The ratio :R(t) = N(t, ^{206}P_{b})/ N(t, ^{238}U) =
(1  exp[λt])/exp[λt]= exp[λt]  1
Then the abudance ratio takes the form:
exp[λt] = [N(t, ^{206}P_{b})/ N(t, ^{238}U)] + 1
And:
t = ln [N(t, ^{206}P_{b})/ N(t, ^{238}U) + 1]/λ
t=
(t_{1/2}/ln(2))ln[N(t, ^{206}P_{b})/N(t,^{238}U) + 1]
The halflife t_{1/2} of the ^{238}U
is 4.47 x 10^{9} years. If the ratio :
N(t, ^{206}P_{b})/ N(t, ^{238}U) =
0.5, How old would be the ore?
t = (4.47 x 10^{9}/ 0.69) ln(1.5) ≈
3 billion years.

