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Homework Help: Radioactive Dating with Potassium Argon

  1. Dec 9, 2007 #1
    [SOLVED] Radioactive Dating with Potassium Argon

    1. The problem statement, all variables and given/known data
    The technique known as potassium-argon dating is used to date old lava flows. The potassium isotope [tex]^{40}{\rm K}[/tex] has a 1.28 billion year half-life and is naturally present at very low levels. [tex]^{40}{\rm K}[/tex] decays by beta emission into [tex]^{40}{\rm Ar}[/tex]. Argon is a gas, and there is no argon in flowing lava because the gas escapes. Once the lava solidifies, any argon produced in the decay of [tex]^{40}{\rm K}[/tex] is trapped inside and cannot escape. A geologist brings you a piece of solidified lava in which you find the [tex]^{40}{\rm Ar}/^{40}{\rm K}[/tex] ratio to be 0.350.

    t = ? [billions of years]

    2. Relevant equations
    Any of these I suppose:
    N = N_0 e^(-t/T)
    T = time constant = 1/r
    r = decay rate = [per seconds]
    (t/2) = half-life = 1.28 billion years
    Beta-plus decay: X becomes Y (A same, Z-1) + e^+1 + energy

    3. The attempt at a solution
    N = given ratio of Ar/K = .350
    N_0 = 1

    ln(1/2) = -(t/2) / T
    T = -t/2 / ln(.5) = 1.846... years
    r = 1 / T = 5.41 * 10^-10 [yr^-1]

    .350 = 1e^(-rt)
    t = ln .350 / -r = 1,938,653,661
    t = 1.94 billion years

    different attempt using
    N = N_0 * (.5)^t/(t/2)
    t/2 = given halflife, N = ratio = .350, N_0 = 1
    t = 1.94 billion years

    I'm guessing I shouldn't be putting in the ratio of Ar to K in for N. But my book only goes into details about Carbon-dating, so I'm not sure where to go from here.

  2. jcsd
  3. Dec 9, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Bit of a tricky question. What you can do however is say that:

    [tex] N_{Ar}/N = 0.35 [/tex] where N is the amount of potassium after decay, and,

    [tex] N_0 = N_{Ar} + N [/tex]

    That should help you if you have some Carbon dating examples.
  4. Dec 9, 2007 #3
    Thanks that worked out flawlessly.
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