Humboldt State University ® Department of Chemistry

Richard A. Paselk

Evolution and Christianity
Spring 2008 Notes Office: SA560a
Office Hours: T 1000-1150, W 1600-1650, Th 1000-1050, F 1400-1450
other times by appointment
Phone: x 5719
Home: 822-1116

Deep Time and The History of the Earth 1

How do we know ages?


  1. Stellar Life Cycle: Stars burn hydrogen at particular rates depending on how big (bright) they are. Oldest globular clusters have only stars dimmer than the Sun (< 0.7 solar mass) implying ages between 11 By and 18 By (Sun will run out of hydrogen and become a Red Giant at about 10 By).
  2. Hubble Constant: Universe is "expanding." That is all galaxies are going away. If calculate how long it has been since they were in same place, then Universe 12-14 By old by recent calculations (range broader in past, narrowed by increasing quality of measurements).
  3. WMAP Satellite: Measures of microwave background pattern, etc. Complex calculations give an age of 13.7 By ±1%.

Earth and Solar System

  1. Rock layers. Layers in sedimentary rock generally will be seasonal, like tree rings. Again can extend range by looking for overlaps. Determined Earth must be 100's of millions of years old in nineteenth century. (Rock thicknesses for various eras tend to be km to miles thick.)
  2. Glacial Ice. Count layers, like in tree rings (below). In very deep ice-fields (Greenland and Antarctica) have gone back over 200,000 years. Includes info on atmosphere etc.
  3. Radiometric dating. Depends on rate of decay of radioisotopes.


  1. Tree rings. Note that can go past oldest living tree if older wood which overlaps is available (e.g. in old buildings, preserved in bogs, at archeological sites). Limited to about 10,000 y so far.
  2. Radiometric dating. Depends on rate of decay of radioisotopes.

    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 half-life, t1/2.
    • First order reactions will give a linear plot with a negative slope when ln [A] is plotted against time.
      • ln N = -kt + lnNo(Compare to chemical relationship from Chem 109: ln [A] = -kt + ln[A]o)
      • Note that for first order only, the half life, t1/2, is independent of N and therefore of time.
      • t1/2 = ln 2 / k = 0.693/k

    Example: What is the age of a meteorite if it has a ratio of 23892U to 20682Pb of 1:1. t1/2, 23892U = 4.51 x 109 years? Assume the only source of 206Pb is uranium decay.

    The first thing we want to do is determine if it has decayed a whole number of half-lives, because if it has the solution becomes trivial. If the current ratio of U:Pb is 1:1, then the original amount of uranium, Uo = U + Pb = 1 + 1 = 2. Excellent, it has decayed a whole number of half=lives!

    The uranium has decayed by one half-life and the meteorite is 4.51 x 109 years old.

    Example: What is the age of a rock sample if it has a 23892U to 20682Pb ratio of 0.906?

    In this case U/Pb = 0.906, and U = 0.906Pb. For Pb = 1, Uo = U + Pb = 1 + 0.906 = 1.906, and we do NOT have a whole number of half-lives. What now?

    Since r = -kN, can say ro = -kNo. If we divided both sides of the equation by ro (recall in algebra can do anything we like as long as we do it to both sides) we get:

    r/ro = N/No = kN/kNo = 0.906/1.906 = 0.475

    Now recall that the integrated rate law is lnN = -kt + lnNo, subtracting initial from final conditions:

    lnN - lnNo = -kt -(-kto) + lnNo - lnNo

    assuming to = 0, dating from initial time 0

    ln(N/No) = -kt

    and since t1/2, 23892U = 4.51 x 109 years, can substitute into:

    t1/2 = 0.693/k and solve for k

    k = 0.693/(4.51 x 109y) = 1.540 x 10-10y-1

    Going back and substituting into ln(N/No) = -kt

    ln(N/No) = ln(0.475) = -0.7444 = -kt

    Finally, rearranging and substituting:

    t = -(-0.7444)/k = 0.7444/(1.540 x 10-10y-1) = 4.83 x 109 years.

    Carbon-14 dating: The half-life for carbon-14, t1/2, 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 146C levels of living organisms will be in equilibrium with atmospheric 146C levels since they are constantly exchanging carbon in the form of carbon dioxide.
      • The 146C will then start decaying when the organism dies, since it will no longer exchange carbon dioxide!
    • Assume a known initial concentration of 146C. This seems reasonable since it is produced in the atmosphere at a continuous rate so that it should reach a nice steady-state concentration due to production and decay:
      • 147N + 10n 146C + 11H
      • 146C 147N + 0-1e
    • 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 146C 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 146C samples for each year to create a corrected 146C 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.

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© R A Paselk

Last modified 17 January 2008