FYI  Calculating Dates Using Rates of Nuclear Decay
NOTE: the decay of nuclides is independent of factors outside the individual nucleus  it is unaffected by environment. This means it must be a first order process in which the rate depends only on the concentration of unstable nuclei.
For first order decay reactions, whether chemical or nuclear:
 Rate = dN/dt = kN, where k is a rate constant characteristic to the system, and the negative sign tells us the concentration is decreasing.
 This gives rise to a logarithmic decay curve (overhead) with a constant halflife, t_{1/2}.
 First order reactions will give a linear plot with a negative slope when ln [A] is plotted against time.
 ln N = kt + lnN_{o}(Compare to chemical relationship from Chem 109: ln [A] = kt + ln[A]_{o})
 Note that for first order only, the half life, t_{1/2}, is independent of N and therefore of time.
 t_{1/2} = ln 2 / k = 0.693/k
Example: What is the age of a meteorite if it has a ratio of ^{238}_{92}U to ^{206}_{82}Pb of 1:1. t_{1/2, 23892U} = 4.51 x 10^{9} years? Assume the only source of ^{206}Pb is uranium decay.
The first thing we want to do is determine if it has decayed a whole number of halflives, because if it has the solution becomes trivial. If the current ratio of U:Pb is 1:1, then the original amount of uranium, U_{o} = U + Pb = 1 + 1 = 2. Excellent, it has decayed a whole number of half=lives!
The uranium has decayed by one halflife and the meteorite is 4.51 x 10^{9} years old.
Example: What is the age of a rock sample if it has a ^{238}_{92}U to ^{206}_{82}Pb ratio of 0.906?
In this case U/Pb = 0.906, and U = 0.906Pb. For Pb = 1, U_{o} = U + Pb = 1 + 0.906 = 1.906, and we do NOT have a whole number of halflives. What now?
Since r = kN, can say r_{o} = kN_{o}. If we divided both sides of the equation by r_{o} (recall in algebra can do anything we like as long as we do it to both sides) we get:
r/r_{o} = N/N_{o} = kN/kN_{o} = 0.906/1.906 = 0.475
Now recall that the integrated rate law is lnN = kt + lnN_{o}, subtracting initial from final conditions:
lnN  lnN_{o} = kt (kt_{o}) + lnN_{o}  lnN_{o}
assuming t_{o} = 0, dating from initial time 0
ln(N/N_{o}) = kt
and since t_{1/2, 23892U} = 4.51 x 10^{9} years, can substitute into:
t_{1/2} = 0.693/k and solve for k
k = 0.693/(4.51 x 10^{9}y) = 1.540 x 10^{10}y^{1}
Going back and substituting into ln(N/N_{o}) = kt
ln(N/N_{o}) = ln(0.475) = 0.7444 = kt
Finally, rearranging and substituting:
t = (0.7444)/k = 0.7444/(1.540 x 10^{10}y^{1}) = 4.83 x 10^{9} years.
Carbon14 dating: The halflife for carbon14, t_{1/2, 146C = 5.74 x 103 years. It is quite useful for dating human artifacts etc. over the past 50,000 years or so. However, a number of caveats apply to this use.}
 Assume the ^{14}_{6}C levels of living organisms will be in equilibrium with atmospheric ^{14}_{6}C levels since they are constantly exchanging carbon in the form of carbon dioxide.
 The ^{14}_{6}C will then start decaying when the organism dies, since it will no longer exchange carbon dioxide!
 Assume a known initial concentration of ^{14}_{6}C. This seems reasonable since it is produced in the atmosphere at a continuous rate so that it should reach a nice steadystate concentration due to production and decay:
 ^{14}_{7}N + ^{1}_{0}n ^{14}_{6}C + ^{1}_{1}H
 ^{14}_{6}C ^{14}_{7}N + ^{0}_{1}e
 However, it turns out the rate of production turns out to depend of the Solar cycle, since most of the neutrons are from the sun. But we haven't been measuring solar activity for very long, so what to do?
 Look at ^{14}_{6}C levels in objects of known age!
 Bristle cone pine of High Sierra's most well known and most important as Earth's oldest living things.
 If one takes a core through a tree one can count the growth rings to go back in time to a known date and take ^{14}_{6}C samples for each year to create a corrected ^{14}_{6}C level for each year.
 Can also have problems in specific circumstances when a second source of C is incorporated, e.g. snail shells in snails living on ancient limestone.
