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Re: Geiger (a challenge)



According to Bob Sciamanda:

fm(t) = (t^m)*r^(m+1)*exp(-rt)/m!

where fm(t) is the probability of recording m counts in time t
and r is the rate of decay (dN/dt).

I have experimental data on fm(t) for t=0.2 seconds. The
histogram of the m distribution (for t=0.2 s) was constructed
on the basis of 2676 measurements. It shows that m=0
happened 123 times, m=1 happened 467 times, etc.

The experimental distribution of arrivals (see below) showed
that the mean waiting time was 0.325 s. Thus r=1/0.325=3.08
counts per second.

fm(t) =(t^m)*r^(m+1)*exp(-rt)/m!

f0(0.2)= 1 * 3.08 *exp(-3.08*0.2)=1.13

Opps, the probability should never exceed
1.00. Perhaps r should be the number of
counts per t=0.2s not per 1s. Let me try
again with r=3.08/5=0.616

f0(0.2)= 1 * 0.616 *exp(-0.616*0.2)=0.545

In other words, the formula predicts that
m=0 should be observed 0.544*2676=1457 times.
This is much larger than what was actually
observed (1457>>123). I will not compare
experimental data with other predictions
before Bob corrects me. What am doing wrong?
Perhaps he (or somebody else) will make
numerical predictions. Then I will be able
to compare them directly with experimental
data.

Ludwik Kowalski, refering to the next experiment (with
exactly the same geometry) wrote this morning:

... I took 3000 numbers from the second column and
constructed the histogram. Here are the results:

(0.00 to 0.20 s) dt=0.1 Bin #1, --> 1202
(0.20 to 0.40 s) dt=0.3 Bin #2, --> 726
(0.40 to 0.60 s) dt=0.5 Bin #3, --> 403
(0.60 to 0.80 s) dt=0.7 Bin #4, --> 266
(0.80 to 1.00 s) dt=0.9 Bin #5, --> 154
(1.00 to 1.20 s) dt=1.1 Bin #6, --> 100
(1.20 to 1.40 s) dt=1.3 Bin #7, --> 56
(1.40 to 1.60 s) dt=1.5 Bin #8, --> 34
(1.60 to 1.80 s) dt=1.7 Bin #9, --> 19
(1.80 to 2.00 s) dt=1.9 Bin #10, --> 11

... Numbers of occurrences (last column) were converted
into approximate probabilities (dividing each by 3000).
Thus the probability of dt=0.1 was 0.400, the probability
of dt=0.3 was 0.242, etc. The average waiting time (the
inter-arrival time) turned out to be 0.325 s, the standard
deviation 0.155 s.