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Re: Geiger, not binomial ?



There is every reason to expect your data to fit a binomial
distribution since it is the correct distribution. However, as Mr.
Cleyet correctly points out, the Poisson distribution is more useful
here because of the huge number of nuclei in your sample (on the order
of 10^23) and the virtually zero probability that any one nucleus will
decay during the counting interval. For such a situation (n -->
infinity; p --> 0; pn=mean counts), the binomial distribution reduces
to the Poisson distribution. Mr. Cleyet also correctly points out
that for high counting rates (i.e., large mean number of
counts/counting interval) a simple Gaussian distribution may also be
applicable.

Increasing the counting interval has no bearing on the agreement of
the Poisson and binomial distributions. In fact increasing the
interval to a time comparable to the half-life of the radionuclide
counted will invalidate the assumptions underlying the Poisson
distribution, because, then, the probability of any one nucleus
decaying during a half-life is not "virtually zero". As you can see
in the table below, your data with a 0.5 second counting interval fits
a Poisson distribution very well. I leave it to you and your students
to analyze your data with a Gaussian distribution.

#counts #times c*t c! P(k;mu(c)) Expected #times =
%diff =
(c) (t) (Poisson) P(k;mu(c))*sum(c*t) |Exp-meas|/Exp*100

0 3321 0 1 0.109586319 3294.712675 0.797863968
1 7278 7278 1 0.242300035 7284.750542 0.092666759
2 7990 15980 2 0.267867866 8053.447402 0.7878291
3 5994 17982 6 0.197422434 5935.505472 0.985502046
4 3260 13040 24 0.10912736 3280.914072 0.637446493
5 1439 7195 120 0.048257051 1450.848248 0.816642833
6 557 3342 720 0.017783067 534.6479145 4.180711243
7 163 1141 5040 0.005617017 168.8756272 3.479262981
8 50 400 40320 0.001552433 46.6739037 7.126244092
9 13 117 362880 0.000381388 11.46644399 13.37429465

sum(t) = 30065
sum(c*t) = 66475

mean(c) = mu(c) = 2.211042741

A final note. Remember to keep in mind that your measured count rate
is not the same thing as the decay rate of your sample. The decay
rate is related to the measured count rate by the efficiency of your
detection system (type of detector and detector-sample geometry).

Good luck!

Glenn A. Carlson, P.E.
gcarlson@mail.win.org

Subject: Geiger, not binomial ?
Date: Thu, 23 Mar 2000 00:08:42 -0500
From: Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>

I have a problem with the distribution of counts from a
Geiger counter.It must be a binomial distribution but I
can not fit it with a binomial distribution.

A weak source is placed in front of a Geiger counter
from Vernier. That counter is connected to a computer
via ULI (also from Vernier) and the event counter
software (also Vernier) is used to count the number
of counts in consecutive time intervals of 0.5 seconds.

Typical counts are like 2, 0, 1, 4, 2, 1, 6, 1 etc. After
counting 30,038 times (for nearly 5 hr.) the distribution
of counts is as follows:

0 counts per 0.5 s occurred 3321 times
1 count per 0.5 s occurred 7278 times
2 occurred 7990 times
3 occurred 5994 times
4 occurred 3260 times
5 occurred 1439
6 occurred 557
7 occured 163
8 occurred 50
9 occurred 13

[snip]

The general shape is always reasonable but details
are never satisfactory. Any comments? I expected
a nearly perfect fit because Geiger counters satisfy
all criteria under which the binomial distribution
should be a very good model.

Ludwik Kowalski

----------------------------------------------------------------------

Subject: Re: Geiger, not binomial ?
Date: Wed, 22 Mar 2000 23:28:14 -0800
From: Leigh Palmer <palmer@SFU.CA>

There is no reason to expect this to be a binomial distribution.
One would expect a Poisson distribution. Check that out. A Poisson
distribution would be indistinguishable from a binomial distribution
if you let the counter run for, say, ten seconds instead of a half
second. Try that.

Leigh

----------------------------------------------------------------------

Subject: Re: Geiger, not binomial ?
Date: Thu, 23 Mar 2000 00:11:35 -0800
From: "Bernard G. Cleyet & Nancy Ann Seese" <georgeann@REDSHIFT.COM>

High their!

I've analyzed your data. It's almost too good to be true. I get a
reduced Chi square of ~ 0.6 w/ dF of 9 The probability of doing worse
than this is ~80%. So I conclude your multi-channel scalar is accurate
and you aren't having any spurious counts or dead time problems. The
only problem I had with your data is that you should have had ~ 2.5
times with ten counts, 0.5 times with 11 counts, etc. instead of zero.

I used the Poisson distrib., as that is the std. one to use (the assumed
distribution with which to compare) . You have discovered why in
counting experiments the binomial distrib. is not used! To get a good
fit n must be very large and p very small, typically 1E+20 and 1E-20
!!! This is because the number of trials is the number of radioactive
nuclei and the prob. of success is very low (i.e. the decay).

bc

If you want a reference, write.

----------------------------------------------------------------------

Subject: Re: Geiger, not binomial ?
Date: Thu, 23 Mar 2000 00:47:14 -0800
From: "Bernard G. Cleyet & Nancy Ann Seese" <georgeann@REDSHIFT.COM>

P.s. I forgot to mention that if the average (counting rate) is high
i.e. ~> 50, one may use "the" Gaussian approx. A not unexpected result
as large n also affords "the" Gaussian approx. for the binomial distrib.

bc

P.s. from this one obtains the well known +/- Squrt.(counts).