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Badger
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[SOLVED] Radioactive Dating with Potassium Argon
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]
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
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.
Cheers.
Homework Statement
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]
Homework 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
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.
Cheers.