Re: keeping score

On Nov 13, 11:57 am, "Dana Tweedy" <reddfr...@xxxxxxxxxxx> wrote:
Skinny Cartman wrote:
On Nov 13, 9:25 am, "richardalanforr...@xxxxxxxxxxxxxx"

snip



And we should believe you rather than the numerous scientists who
have actually *studied* the subject because.....?

I have studied all your science and have the math backbone to do the
calculations myself
idiot.

So, you should be able to show your calulations.  

basic precalculus.
Got an A in it.

http://plus.maths.org/issue14/features/garbett/index.html

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.

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), so our equation for t is


For convenience, we'll now write F for N/N0. Note that that the domain
of F is the interval from zero to 1, which corresponds to the interval
of time from zero to infinity. Plotting t against F with a value of
l=1 gives the graph on the right.

To find <t>, the mean value of time of survival, all we have to do is
find the integral





which is a very tidy result!

Incidentally, our formula for t gives us an easy way of finding the
half-life, the time it takes for half the nuclei in a sample to decay.
The half-life (often denoted t1/2) is just t(1/2) = (1/l) ln(2). The
equivalent thickness for the medium in radiation attenuation is known
as "half-value thickness". Similarly, in a population which grows
exponentially with time there is the concept of "doubling time".

Here isotopes with longer half lives are used, which enables dating of
geological formations and rocks. However, the essential ideas are
analogous. For example, in lava form, molten lead and Uranium-238
(standard isotope) are constantly mixed in a certain ratio of their
natural abundance. Once solidified, the lead is "locked" in place and
since the uranium decays to lead, the lead-to-uranium ratio increases
with time. In this way, some of the oldest rocks have been measured at
approximately 3 billion years.

Which is totally and completely not their age at all period.

.

Relevant Pages

  • Re: keeping score
    ... that for a thin enough slice of matter, the proportion of ... Exactly the same treatment can be applied to radioactive decay. ... Radioactive atoms decay randomly. ...
    (talk.origins)
  • Re: keeping score
    ... that for a thin enough slice of matter, the proportion of ... Exactly the same treatment can be applied to radioactive decay. ... Radioactive atoms decay randomly. ...
    (talk.origins)
  • Re: keeping score
    ... that for a thin enough slice of matter, the proportion of ... Exactly the same treatment can be applied to radioactive decay. ... Radioactive atoms decay randomly. ... that we saw in the case of light attenuation. ...
    (talk.origins)
  • Re: Halflife and Randomness
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  • Re: Re: Halflife and Randomness
    ... enriched this group when s/he wrote: ... will decay or not. ... seconds and we take a unit of that isotope that has 64 atoms. ... for that is that there is no "cause" for radioactive decay. ...
    (talk.origins)