Prove that radioactive decay law is exponencial
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Radioactive Decay
In the previous article, we saw that light attenuation obeys an exponential law. To show this, we needed to make one critical assumption: that for a thin enough slice of matter, the proportion of light getting through the slice was proportional to the thickness of the slice.
Exactly the same treatment can be applied to radioactive decay. However, now the "thin slice" is an interval of time, and the dependent variable is the number of radioactive atoms present, N(t).
Radioactive atoms decay randomly. If we have a sample of atoms, and we consider a time interval short enough that the population of atoms hasn't changed significantly through decay, then the proportion of atoms decaying in our short time interval will be proportional to the length of the interval. We end up with a solution known as the "Law of Radioactive Decay", which mathematically is merely the same solution that we saw in the case of light attenuation. We get an expression for the number of atoms remaining, N, as a proportion of the number of atoms N0 at time 0, in terms of time, t:
N/N0 = e-lt,
where the quantity l, known as the "radioactive decay constant", depends on the particular radioactive substance.
Again, we find a "chance" process being described by an exponential decay law. We can easily find an expression for the chance that a radioactive atom will "survive" (be an original element atom) to at least a time t. The steps are the same as in the case of photon survival.
Mean lifetime of a Radioactive Atom
On average, how much time will pass before a radioactive atom decays?
This question can be answered using a little bit of calculus. Suppose that we invert our function for N/N0 in terms of t, to get an expression for t as a function of N/N0. Once we have an expression for t, a "definite integral" will give us the mean value of t (this is how "mean value" is defined).
From the equation above, taking logarithms of both sides we see that lt = -ln(N/N0) = ln(N0/N),
that's much I know
In the previous article, we saw that light attenuation obeys an exponential law. To show this, we needed to make one critical assumption: that for a thin enough slice of matter, the proportion of light getting through the slice was proportional to the thickness of the slice.
Exactly the same treatment can be applied to radioactive decay. However, now the "thin slice" is an interval of time, and the dependent variable is the number of radioactive atoms present, N(t).
Radioactive atoms decay randomly. If we have a sample of atoms, and we consider a time interval short enough that the population of atoms hasn't changed significantly through decay, then the proportion of atoms decaying in our short time interval will be proportional to the length of the interval. We end up with a solution known as the "Law of Radioactive Decay", which mathematically is merely the same solution that we saw in the case of light attenuation. We get an expression for the number of atoms remaining, N, as a proportion of the number of atoms N0 at time 0, in terms of time, t:
N/N0 = e-lt,
where the quantity l, known as the "radioactive decay constant", depends on the particular radioactive substance.
Again, we find a "chance" process being described by an exponential decay law. We can easily find an expression for the chance that a radioactive atom will "survive" (be an original element atom) to at least a time t. The steps are the same as in the case of photon survival.
Mean lifetime of a Radioactive Atom
On average, how much time will pass before a radioactive atom decays?
This question can be answered using a little bit of calculus. Suppose that we invert our function for N/N0 in terms of t, to get an expression for t as a function of N/N0. Once we have an expression for t, a "definite integral" will give us the mean value of t (this is how "mean value" is defined).
From the equation above, taking logarithms of both sides we see that lt = -ln(N/N0) = ln(N0/N),
that's much I know
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