[Phys-L] Re: Survey Meters

[edmiston at BLUFFTON.EDU (Michael Edmiston)]

Actually I think the simple-decay observed half-life for 214Pb and 214Bi
starting at equilibrium is about 37 minutes if you fit a single
exponential to generated data that represents a time period of a 5 or 6
hours.  If you only look at early data the observed half-life is longer
than that, and if you look only at later data the observed half-life is
shorter than that.  I am not sure of the calculation John M used, but
the calculation Brian W. showed is faulty.

First, some data.  Let's get the decay constants and equilibrium values
for 214Pb and 214Bi

214Pb half-life is 26.8 min = 1608 s.  Decay constant is ln(2)/t2 =
4.311E-4.

214 Bi half-life is 19.8 min = 1194 s.  Decay constant is ln(2)/t2 =
5.805E-4

When these are in equilibrium with radon and other radon daughters,
their decay rates are equal.  That means the numbers of atoms of 214Pb
and 214Bi present are in the inverse ratio of the decay constants, or
the direct ratio of their half-lifes.  This ratio of N(Bi) to N(Pb) is
0.7388, or the inverse of this is the equilibrium value of 214Pb is
1.354 times the equilibrium value of 214Bi.  If we start with 10,000
214Bi atoms then at that time there are 13,540 214Pb atoms.

Now, in the first second, the number of 214Pb atoms that decay is
4.311E-4*N(Pb) which is rate(Pb) and the number of 214Bi atoms that
decay is 5.805E-4*N(Bi).  At the end of one second N(Pb) becomes (what
it was ) minus rate(Pb); N(Bi) becomes (what it was) minus rate(Bi) plus
rate(Pb); and the number of events observed (if we could observe 100% of
the decays) would be rate(Pb) plus rate(Bi).

Then we just keep iterating these steps to generate our data.

Oh... I just received John M's message linking to his spreadsheet.  He
has done it mostly correctly.

There is a further complication that John did not consider.  You won't
observe the 214Pb and 214Bi decays equally well.  Assuming we are
measuring the gamma radiation, there are two problems.  (1) a gamma is
not emitted in every decay, and the percentage of time a gamma is
emitted is different for 214Pb and 214Bi. (2) The 214Pb gammas are lower
energy than those from 214Bi.  This means our detector is less efficient
at detecting the 214Bi gammas.

Thus, when calculating the observed count rate, you add a fraction of
the 214Pb decays to a fraction of the 214Bi decays (rather than simply
adding the two rates) and the fractions of each are not the same.  That
changes the "observed counts versus time" curve.

It is possible to model this quite well, but it is complicated, and you
need to know the detector efficiency difference for the two nuclides.


Michael D. Edmiston, Ph.D.
Professor of Physics and Chemistry
Bluffton University
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu

John M. wrote

> > The result of the two beta decays from initially stable populations
> > of Pb-214 and Bi-214 produces radiation with an approximately 40
> > minute half life.  I constructed a simple spread sheet to model the
> > decay sequence and it produced a graph of beta production versus time
> > that very nicely matches the Buda figure.

Brian W. wrote

> If instead of using a spreadsheet, I synthesize the effect
> of a double decay cascade like this,
> where the first species decays at one third the rate of
>  the daughter species,......
>
>   'exponential decay of double series
>     count = 1000
>     for a = 1 to 1000
>     count = count*0.99
>     count2 = count* 0.97
>     print a; " " ;count; " "; count2
>     next a
>     end
>
> ... a single exponential of this form:
> count2 = 970* exp(-0.001005 *a)
>
> leads to a fit with exceptionally good statistics
> anovar: F = 9.6E20, SE estimate = 1.4E-7
> using just a single exponential equation.
>
> Therefore, I expect I am misunderstanding some prior comments
> about the possibility of pulling details of a double decay out of an
> exponential time series. I cannot do this.
> Or was my synthetic dataset faulty?