Half life of Oxygen factors?
The half-life of 15 over 8 Oxygen is 122 s.
How long does it take for the number of 15 over 8 Oxygen nuclei in a given sample to decrease by a factor of 5×10^−4?
The half-life of 15 over 8 Oxygen is 122 s.
How long does it take for the number of 15 over 8 Oxygen nuclei in a given sample to decrease by a factor of 5×10^−4?
Karl
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We can be a bit more parsimonious in labeling the nuclide. Just call it "oxygen-15". Since it is oxygen, everyone will know it has eight protons.
Radioactive nuclides decay according to an exponential equation:
A(t) = A(0) * exp(-kt)
where A(t) is the activity at time t (and A(0) is the activity at time zero),
and k is the decay constant.
Rearranging, we get
A(t) / A(0) = exp(-kt).
What we want is to solve for A(t)/A(0) = 10^(-4).
To make things easier, we can do the same thing with the equation
A(t) / A(0) = 2^(t/T)
where T is the half-life.
So we want to know at how many half-lives we raise 2 to in order to yield a value of 5*10^(-4).
2^(-t/T) = 5*10^(-4)
Take logs of both sides.
(-t/T) * log(2) = log(5*10^(-4))
(-t/T) = log(5*10^(-4))/log(2) = -3.301/0.301= -10.97
-t/T = -t/122 = -10.97
t = 10.97 * 122 = 1338 seconds