1.Poisson Distribution & Statistics of Radiation

The Geiger counter-scaler is used in this experiment to measure the background radiation due to Cosmic rays, radiation from building materials like granite, radon gas, etc.  Even without a radioactive material sample (like Ba 137 in the radioactive decay experiment) , the background radiation is present.

The Geiger Counter-Scaler

The essential features of the Geiger counter are displayed in the diagram below:

The Geiger tube is the gas filled metal cylinder with center electrode.  The electrode and cylinder are kept at a potential difference V (300 volts for example) with an external high voltage power supply.  The gas in the cylinder is usually nonconducting.  However, when ionizing radiation passes through the transparent mica (sand) window, it ionizing the gas inside the cylinder along the path of the radiation.  If the path connects the electrode and the outer cylinder, the electricity momentarily conducts (somewhat like lightning) and one hears a "click" in the speaker of the Geiger counter.  Quite often the speaker is not included in the apparatus and instead the clicks are counted on an electronic "scaler" or counter.

Resolving Time

After radiation has passed through the Geiger tube, there is a short period of time where the tube is "dead".  If another radiation particle passes through the tube during this time, then it will not be observed.  The tube is dead since there is a characteristic time for the ionized atoms (from the first radiation particle) to recombine into neutral atoms.  The dead time is also sometimes called the resolving time since this is the shortest time (that can be "resolved" or determined) between two radiation events.
	One way to determine the resolving time is to use higher and higher count rate sources.  At some high count rate, increasing the radioactivity of the source will not increase the count rate (even though it should because there is more radiation).  The resolving time then is

\(\(\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)\(Resolving\ Time\  = \ 1\/\(Maximum\ Count\ Rate\)\)\)\)

Setting the High Voltage

The typical laboratory Geiger counter usually has several controls you must set. The first control sets the high voltage.  If the voltage is too low, then the gas does not conduct when the radiation passes through the tube.  If the voltage is too high, then the gas conducts even if the radiation is not present to initiate the process (somewhat like fluorescent lighting).  There is an between range of voltages (the "plateau region") where the Geiger tube operates properly.  A sketch of the voltage versus counts graph appears

The high voltage often has a "course" (or rough) adjustment knob and "fine" (or precise) adjustment knob.

Controls on the Scaler

The other controls of the scaler are as follows:

1. Power on/off switch:  sometimes on back of unit
2. Count switch: initiate the counting interval
3. Stop switch: terminates the counting interval
4. Resent switch: clears the scaler and sets it to zero counts
5. Time switch:  adjusts the length of the counting interval.

Experiment #1:  Repeated measurement of the count rate Background Radiation

Set the "Time" switch for a convenient time interval (e.g. for one minute) and repeatedly measure (N=20 times) the number of counts of the background radiation.  (Other time intervals could be chosen but one minute is not too short or long to be inconvenient.)  Typically you will get a different number of counts each observation you make.  This statistical randomness is a property of the thing measured (the background radiation) and the randomness has nothing to do with the act observation.  Collect the data in the table like the one below.   However, your table should appear in your laboratory notebook.

({{Observation

          Number α

          N =  20, Counts in

          time interval

          (Subscript[n,  α]), \(Deviation\ from

          the\ mean\ 

          \((n\_\(\(\ \)\(α\)\) - 
              n\&_)\)\), \(Square\ of\ the

          deviation

          \((n\_\(\(\ \)\(α\)\) - n\&_)\)\^2\)}, {1, , , }, {2, , , }, {3, , , }, {4, , , }, {5, , , }, {6, , , }, {7, , , }, {8, , , }, {9, , , }, {10, , , }, {11, , , }, {12, , , }, {13, , , }, {14, , , }, {15, , , }, {16, , , }, {17, , , }, {18, , , }, {19, , , }, {20, , , }, {TOTALS, A=                       , B=                          , C=                      }})

A, B, and C are the totals of each column.

For Your Lab Notebook and Report

1.  Calculate the average \(TraditionalForm\`n\&_\)

\(\(\(\ \)\(n\&_\  = \ \(\(A\/N\)\(=\)\(\ \)\)\)\)\)

where N=20 is the number of observations.  More explicitly the average appears

\(\(\(\ \)\(n\&_\  = \ 
      1\/\(\(N\)\(\ \)\)\ \(∑\+\(α = 1\)\%N n\_\(\(\ \)\(α\)\)\)\)\)\)

Also,notice the sum α  is over all N observations which in this case is 20.  (Subscript[n, α])  is the number of counts obtained on the αth observation.

2. Calculate the average of the deviations \(TraditionalForm\`d\&_\) from the mean or average \(TraditionalForm\`n\&_\).

\(\(\(\ \)\(d\&_ = \ \(\(B\/20\)\(=\)\)\)\)\)

More explicitly the average of the deviations \(\(\(\ \)\(d\&_\)\)\)   is given by

\(\(\(\ \ \)\(d\&_\  = \ 
      1\/N\ \(∑\+\(α = 1\)\%N d\_\(\(\ \)\(α\)\)\)\)\)\)

where the αth deviation (Subscript[d,  α])  is defined

\(\(\(\ \)\(d\_\(\(\ \)\(α\)\) = \ 
      n\_\(\(\ \)\(α\)\) - \ n\&_\)\)\)

NOTE:  When using Mathematica to calculate equation (5) use Num=20. for the number of measurements since the symbol N is reserved by Mathematica for use as the "Numerical Approximation" function.

3. Calculate the Sample Standard Deviation S:

S =  (Sqrt[(C/N)])=

It is called the Sample Standard Deviation because it is obtained from the experimental or data sample.  More explicitly the Sample standard deviation (also called the RMS or Root Mean Square deviation)

\(\(\(\ \ \)\(S\  = \ \@\(1\/N\ \(∑\+\(α = 1\)\%N\ d\_α\^2\)\)\)\)\)

4. Calculate the theoretical Statistical Standard Deviation σ:

\(\(\(\ \)\(\(σ\)\(\ \)\(=\)\(\ \)\(\@n\&_\)\(\ \)\)\)\)

This theoretical or Statistical standard deviation σ is obtained after calculating the average \(TraditionalForm\`n\&_\)  from the data.   It is a theoretical consequence of the Poisson distribution that σ=\(TraditionalForm\`\@n\&_\).   So we are assuming the experimental data follows a Poisson distribution when equation (9) is used.  If you take a large number (e.g. N=20) of observations of the experiment, then it is expected that S approaches σ for because then the experimental data distribution approaches the Poisson distribution.  Report the numerical values you obtained for S and σ from the experimental observation of the background radiation.  Are the numerical values of S and σ close to each other?

Experiment #2:  Observation of the Plateau Region in the Geiger Counter or Scaler.

Again set the time interval on the Geiger scaler for something convenient (e.g. one minute).  Observe the number of counts one time at each voltage in the table below, that is applied to the Geiger tube.  Use a  data table like the one indicated below but write this table in your lab notebook.

({{Voltage, 50, 100, 150, 200, 250, 300, 350, 400, 450, 500, 550, 600}, {Counts, \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder]}})

Use Mathematica to graph the counts (vertical axis) versus the voltage (horizontal axis).  You might want  to take more data points (1) at higher voltage to find the breakdown point where counts occur even without radiation and (2) more additional data points at the lower voltage where the tube first turns on.

Experiment #3:  Observations from a β Source

You will be given a source of β radiation. β radiation is another term for moving electrons. Using this source, make 20 observations of the number of counts in one minute.  Repeat the calculations 1-4 above but this time using the β source data.  Again construct a data table in your lab notebook for this purpose and calculate \(TraditionalForm\`n\&_\), S, and σ.   Compare S and σ for the β radiation.  Are they numerically close together?

The Poisson Distribution

The Poisson distribution occurs frequently in experiments involving radioactive isotope decay.  A statistical distribution is a graph of the probability (something happens) versus the thing measured.  For example, P[n] could be the probability that n radiation particles are given off by a sample in time T and the graph of P[n] versus n (where n=0, 1, 2, 3, ... ) is a statistical distribution.  
	The Poison distribution is one of several statistical distributions (the Gauss and Binomial are two other examples) and the Poisson distribution is a special case of the Binomial distribution as we will see elsewhere.  For the present, we take the Poisson distribution as something to be experimentally verified which is something you are going to do in this experiment.
	The Poisson distribution P[n] is given the formula

P[n] =  
      (e^(-μ)) (((μ )^n)/(n!))

where for example, n is the number of counts/time of  background radiation and the time is one minute. n is non-negative integer (n=0, 1, 2, 3, 4,  ...) in this case.  n is a definite number for each observation but n varys for each different observation of the sample.  The parameter μ  appearing in the Poisson distribution is the average determined from the experimental data as we will see below.   n! is n factorial and by definition

n!  =  n×(n - 1)×(n - 
                2)×(n - 
                3)×  ...  3×2×1

While there is a theoretical reason why P[n] should hold for radioactive decay, we choose to have you verify it experimentally.

The Parameter μ-The Average

The average \(TraditionalForm\`n\&_\) for a statistical distribution is by definition (see the appendix for a discussion of equation (11))

\(\(\(\ \)\(n\&_\  = \ ∑\+\(n = 0\)\%∞ 
          n\ P[n]\)\)\)

or using the Poisson distribution equation (10) one obtains

\(\(\(\ \)\(n\&_\  = \ ∑\+\(n = 0\)\%∞ 
          n\ e\^\(-μ\)\ \(\(μ\)\(\ \)\)\^n\/\(n!\)\)\)\)

Notice the n=0 term of the sum vanishes since n*P[n] =0 for n=0 so we may start the summation at n=1.  Since μ is taken as a constant (independent of n) the (e^(-μ)) can be taken outside the sum

\(\(\(\ \)\(n\&_\  = \ 
      e\^\(-μ\)\ \(∑\+\(n = 1\)\%∞ 
            n\ \(\(μ\)\(\ \)\)\^n\/\(n!\)\)\)\)\)

Using the definition of n! in equation (11) one has after cancellation

\(\(\(\ \ \)\(n\&_\  = \ 
      e\^\(-μ\)\ \(∑\+\(n = 1\)\%∞\ \ \(\(μ\)\(\ \)\)\^n\/\(\((n - 1)\)!\)\)\)\)\)

Note as well

\(\(\(\ \ \ \)\(\(\(μ\)\(\ \)\)\^n = \ μ\ \(\(μ\)\(\ \)\)\^\(n - 1\)\)\)\)

so eqt.. (15) can be written

\(n\&_\  = \ μ\ e\^\(-μ\)\ \(∑\+\(n = 1\)\%∞\ \ \(\(μ\)\(\ \)\)\^\(n - 1\)\/\(\((n - 1)\)!\)\)\)

Introducing a new summation index m=n-1 having the range

({{n, 1, ∞}, {m, 0, ∞}})

and equation (17) now appears

\(\(\(\ \)\(n\&_\  = \ μ\ e\^\(-μ\)\ \(∑\+\(m = 0\)\%∞\ \ \(\(μ\)\(\ \)\)\^m\/\(m!\)\)\)\)\)

The summation is just the Taylor series expansion of the exponential function

\(\(\(	\)\(e\^μ\  = \ ∑\+\(m = 0\)\%∞\ \ \(\(μ\)\(\ \)\)\^m\/\(m!\)\)\)\)

so equation (18) can be written

\(\(\(\ \ \)\(n\&_\  = \ μ\ e\^\(-μ\)\ e\^μ\)\)\)

or since the exponentials cancel and

\(\(\(\ \)\(\(n\&_\)\(\ \)\(=\)\(\ \)\(μ\)\(\ \)\)\)\)

This shows that the parameter μ is simply equal to the average \(TraditionalForm\`n\&_\) obtained from the statistical distribution.  If we assume the Poisson distribution is a good representation of the experimental data, then the μ determined from the average  \(TraditionalForm\`n\&_\) of the experimental data.

Standard Deviation for the Poisson Distribution

A calculation of the dispersion (σ^2) is by definition

\(\(\(\ \ \)\(σ\^2\  = \ \ ∑\+\(n = \ 0\)\%∞\(\(\((n - \ n\&_\ )\)\^2\)\(\ \)\(P[
            n]\)\(\ \ \)\)\)\)\)

Using the Poisson distribution equation (10) yields

\(\(\(\ \)\(σ\^2\  = \ \ n\&_\)\)\)

a surprisingly simple result.  See module 5 for a proof of equation (23).  This is the basis of the assertion that you can calculate σ  quite easily in radioactive decay experiments once you have calculated the average.  In particular,

\(\(\(\ \)\(σ\  = \ \@n\&_\)\)\)

σ characterizes the width of the Poisson distribution.  The larger σ is, the wider the distribution.

The Relative Width of the Poisson Distribution

The relative width is defined  σ/\(n\&_\) and is easily calculated

\(\(\(\ \)\(\(\(σ\)\(\ \)\)\/\(\(\ \)\(n\&_\)\) = \ \(\@n\&_\/n\&_ = \ 1\/\@n\&_\)\)\)\)

The average \(n\&_\)  is often quite large (if you have a large number of counts) so the relative width is very small.

Experiment #4:  Another Experiment with Background Radiation

Part #1:  Set the count duration for 10 seconds on the Geiger counter-scaler.  (If your scaler does not have a count timer, then use a stop watch.)  Next count the background radiation 50 times and place you data in the histogram below:

({{   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, {   ,    ,    ,    ,    ,    ,    ,    ,    ,    ,    }, { 0  ,  1  ,  2  ,  3  ,  4  ,  5  ,  6  ,  7  ,  8  ,  9  ,  10  }})

Place an individual x in the nth column of the above table for each occurrence of n counts.  When you are done there should be 50 x's in the above table.  The table above is a graph of the number of occurrences versus the  number of counts.  
	Let  (Subscript[n,  α]) be the number of counts measured on the αth trial (where α goes from 1 to 50).  Calculate the average number of counts \(n\&_\) using

\(\(\(\ \)\(n\&_ = \ \(\(n\_\(\(\ \)\(1\)\) + n\_\(\(\ \)\(2\)\)\  + n\_\(\(\ \)\(3\)\(\ \)\) + n\_\(\(\ \)\(4\)\(\ \)\) + n\_\(\(\ \)\(5\)\(\ \)\) + n\_\(\(\ \)\(6\)\(\ \)\) + \ \ \ \(+n\_\(\(\ \)\(N\)\(\ \)\)\)\)\/N = \ \(∑\+\(α = 1\)\%N\( \(n\)\(\ \)\)\_\(\(α\)\(\ \)\)\)\/N\)\)\)\)

Part #2:  An different way of computing the average number of counts \(n\&_\).  
Group together all the occurrences having the same number of counts  k  and suppose there are (Subscript[(N ), k]) occurrences of k counts.  
	EXAMPLE:   If there are 5 occurrences (5 x's in the table above) of 8 counts then, (Subscript[(N ), k])=5  with  		k=8.   (Subscript[(N ), k]) is the so-called frequency of occurrence.  
Equation (26) can be rearranged in terms of the frequency of occurrence (Subscript[(N ), k]) as

\(\(\(\ \ \)\(n\&_ = \ \(∑\+k k\ \(\(N\)\(\ \)\)\_k\)\/N\)\)\)

Rearrange equation (27) a little more to get

\(\(\(\ \)\(n\&_ = \ ∑\+k\ 
        k \((\(\(N\)\(\ \)\)\_k\/N)\)\)\)\)

The probability of k counts is (Subscript[P, k]) defined

(Subscript[P, k]) = ((Subscript[(N ), k])/N)

so then equation (28) can be written

\(\(\(\ \)\(n\&_ = \ ∑\+k 
          k\ \(\(P\_k\)\_\ \)\)\)\)

Note that equation (30) should yield an equal numerical result for \(\(\(\ \)\(n\&_\)\)\) as equation (26) (since they calculate the same thing)  and this you will show as part of this experiment.

Calculate the probabilities (Subscript[P, k]) using the occurrences (Subscript[(N ), k]) in the histogram above.  Display the probabilities you calculate with a plot of (Subscript[P, k]) versus the number of counts k using Mathematica.

EXAMPLE:  Suppose you have the following data after NTotal=25 observations.

Num[0] = 1;
Num[1] = 6;
Num[2] = 7;
Num[3] = 5;
Num[4] = 3;
Num[5] = 2;
Num[6] = 0;
Num[7] = 1;
Num[8] = 0;
Num[9] = 0;

(Num is used instead of N since N is reserved by Mathematica for the "numerical approximation" function.) One way of computing the average is equation (27)

\(n\&_ = \(∑\+\(k = 0\)\%9 k*Num[k]\)\/25. \)

Another way of computing the average is equation (30).  First we need to define the probability P[k] of having k counts via

NTotal = 25. ;
P[k_] = Num[k]/25. ;

For example, the probability of having 7 counts is

P[7]

The average according to equation (30) is then

nAve = (∑_(k = 0)^9) k*P[k]

which is the same numerical answer as given by equation (27).  You may collect the probability data together using the Mathematica Table function;

data = Table[{k, P[k]}, {k, 0, 9}]

The data is graphed using the Mathematica function ListPlot as below:

DataGraph = ListPlot[data]

The graph may be improved by increasing the dot size with the option PlotStyle->PointSize[{0.02}]. and the option AxesLabel->{"k counts","P[k]"} indicates which axis is which.

DataGraph=ListPlot[data,PlotStyle -> PointSize[0.02],AxesLabel -> {"k counts", "P[k]"} ]

Also a name was assigned to the above graph.  It is assigned the name DataGraph for later use.

Calculate the average number of counts \(TraditionalForm\`n\&_\) using your actual experimental data and equation (30) with the probabilities (Subscript[P, k]) you used to draw the above graph.  Did you get the same numerical value for \(TraditionalForm\`n\&_\) using equation (30) as you did from equation (26)?

Part #3: The value of \(TraditionalForm\`n\&_\) you determined from the experimental data is assigned (according to equation (21)) to the μ appearing in the Poisson distribution equation (10).   Use the value of  μ  (you obtained from \(TraditionalForm\`n\&_\) ) and plot using Mathematica the corresponding Poisson distribution.  Call this graph the TheoryPlot.

EXAMPLE:  For the data above, μ=\(TraditionalForm\`n\&_\)=2.56 and the Poisson probability distribution is given using Mathematica via

μ = 2.56;
PD[k_] := (((E^(-μ))*(μ^k))/(k!))

For example, the Poisson distribution gives the probability of 1 count as

PD[1]

The Poisson distribution is graphed using

PoissonData = Table[{k, PD[k]}, {k, 0, 9}]

TheoryGraph = 
      ListPlot[PoissonData, PlotStyle -> PointSize[0.015]]

5.  Using Show, graph the DataGraph and the TheoryGraph together for comparison.

EXAMPLE:

Show[DataGraph, TheoryGraph]

The smaller dots are the theoretical Poisson distribution.  To the eye there is rough agreement between the experimental data and the Poisson distribution.  We will make a more precise comparison later using the socalled (χ^2)  test.

Hints on Using  Mathematica

Suppose the average is \(TraditionalForm\`n\&_\)=0.8 counts then you may assign the numerical value 0.8 to the symbol μ using

μ = 0.8

The Poisson distribution is a Mathematica function.

P[n_] :=  (E^(-μ)) (((μ )^n)/(n!))

The Mathematica expression for the above function is different in three ways from the usual mathematical expressions:
1. The argument of a Mathematica function is always enclosed in square brackets [...] instead of parenthesis (...)
2. The variable n of the above Mathematica function is followed by a "bar" symbol (see the dash key on the upper right hand corner of your keyboard).  The "bar" is for the "type checking" feature of Mathematica and will NOT be used by us.  None-the-less you must include the bar for the Mathematica function to work properly since otherwise Mathematica will think you mean a subscripted variable.
3. Note the "colon" symbol : before the equal sign which is on the RHS of your keyboard about midway down with the semi-colon symbol. The := allows you to change a parameter (like μ) in the definition of the function.

Also, the symbol "e" for the natural number

N[E]

has a vertical bar in the usual "e" character and e is accessible in the "Basic Input Palette".  Notice that the function N[...] is RESERVED by Mathematica for special use so you cannot have another use for the symbol N.  
	By the way, it is always a good idea to check to see if your user defined function actually works (gives a numerical result) before trying to graph it.  For example,

P[1. ]

indicates the function P[...] works.  So the Poisson distribution for the case μ=0.8 is

Plot[P[n], {n, 0, 10}, GridLines -> Automatic]

The grid lines were put in with the option ,GridLines->Automatic.  The maximum value of the distribution is less than the average 0.8   Another example of a Poisson distribution with a different μ =3 is

μ = 3.0;
P[
          n_] :=  (E^(-μ)) (((μ )^n)/(n!));
Plot[P[n], {n, 0, 10}, 
      AxesLabel -> {"n", "P[n]"}]

Notice we introduced the option AxesLabel->{"n","P[n]"}.  The maximum value of the distribution is about at the average 3.0   Still another example the Poisson distribution with μ=9.0

μ = 9.0;
P[
          n_] :=  (E^(-μ)) (((μ )^n)/(n!));
Plot[P[n], {n, 0, 20}, AxesLabel -> {"n", "P[n]"}, 
      GridLines -> Automatic]

Notice that the maximum value of n has been increased to n=20 in the "iterator" ,{n,0,20} of the Plot function.  Also, the maximum value of the distribution is about at the average 9.0