Difference between revisions of "Plastic Scintillator Calculation"

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Below is the calculations done to determine the probability of pair production depending on thickness of the scintillator.
 
Below is the calculations done to determine the probability of pair production depending on thickness of the scintillator.
  
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NOTE: For the equation below <math> M_f(Atomic Symbol)</math> is just the number of that atom present from the molecular formula.
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Probability of interaction per cm thickness<math> = ((\frac{NumCarbonAtoms}{cm^3} *(\sigma_{elec}C + \sigma_{nucleus}C)) + (\frac{NumHydrogenAtoms}{cm^3} *(\sigma_{elec}H + \sigma_{nucleus}H)))*100%</math>
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Using this method of calculation for Rexon RP 200 yields a probability <math> = \frac{0.5234%}{cm}</math>
  
Probability of interaction <math>= ((\frac{NumCarbonAtoms}{cm^3} * M_fH(\sigma_{elec}C + \sigma_{nucleus}C)) + (\frac{NumCarbonAtoms}{cm^3} * M_fH(\sigma_{elec}H + \sigma_{nucleus}H)))*100%</math>
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Using this method for Bicron BC 408 yields a probabiltiy <math> = \frac{0.5303%}{cm} </math>
  
  
 
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Latest revision as of 18:11, 5 February 2009

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Below is the calculations done to determine the probability of pair production depending on thickness of the scintillator.

Molecules per [math] cm^3 = \frac{grams CH_{2}}{cm^3} * \frac{mol}{gram} * {N[A]}[/math] (NOTE: [math] \frac{gram}{cm^3} [/math] is just the density of the scintillator material and N[A] is Avogadro's number)

Molecules per [math] cm^2 (K) = \frac{Molecules}{cm^3} * Thickness [/math]

Weighted cross-section [math] (\sigma_w) = (\sigma_{elec}C + \sigma_{nucleus}C) + 2(\sigma_{elec}H + \sigma_{nucleus}H)[/math]

Probability of interaction (%) [math]= \sigma_w * K * 100%[/math]


All cross sections listed here are pair production cross-sections

For carbon [math]\sigma_{nucleus} = 9.645*10^{-2} barns[/math] or [math]9.645*10^{-26}cm^2[/math]

For carbon [math]\sigma_{elec} = 1.030*10^{-2} barns[/math] or [math]1.030*10^{-26}cm^2[/math]

For hydrogen [math]\sigma_{nucleus} = 2.688*10^{-3} barns[/math] or [math]2.688*10^{-27}cm^2[/math]

For hydrogen [math]\sigma_{elec} = 1.716*10^{-3} barns[/math] or [math]1.716*10^{-27}cm^2[/math]

Avogadro's number [math] = \frac{6.022*10^{23}molecules}{mol}[/math]

Molecular formula for PVT [math] = C_{10}H_{11} [/math]

Density of polyvinyl toluene (a common scintillator material) [math] = \frac{1.02grams}{cm^3}[/math] (NOTE: this value is from Rexon RP 200 [1])

or is it [math]\rho_{BC408} = \frac{1.032grams}{cm^3}[/math]  H/C = 11/10  [2] (TF)

For the sample calculation the thickness will be set to 1 cm just to get probability per cm


So entering all the numbers into the 4 initial equations gives the following answers:

Molecules per [math] cm^3 = \frac{1.02g PVT}{cm^3} * \frac{1 mol}{131 g} * \frac{6.022*10^{23}molecules}{mol} = \frac{4.689*10^{21}molecules PVT}{cm^3} [/math]

Molecules per [math] cm^2 (K) = \frac{4.689*10^{21}molecules PVT}{cm^3} * 1cm = \frac{4.689*10^{21}molecules PVT}{cm^2} [/math]

Weighted cross-section [math] (\sigma_w) = (10*(1.030*10^{-26}cm^2 + 9.645*10^{-26}cm^2)) + (11*(1.716*10^{-27}cm^2 + 2.688*10^{-27}cm^2)) = 1.116*10^{-24}cm^2[/math]

Probability of interaction (%) [math]= 1.116*10^{-24}cm^2 * \frac{4.689*10^{21}molecules PVT}{cm^2} * 100% = 0.5233%[/math]

Doing the same calculations using the Bicron BC 408 PVT with anthracene [3] for the material yields a probability of [math]0.5294%[/math]

A different way to calculate probability of interaction

I checked out a few of the physics material supply sites and most of them list with their products the amounts of each individual atom per [math] cm^3[/math]. Therefore there is a quicker way to calculate the probability of interaction which is listed below.


Probability of interaction per cm thickness[math] = ((\frac{NumCarbonAtoms}{cm^3} *(\sigma_{elec}C + \sigma_{nucleus}C)) + (\frac{NumHydrogenAtoms}{cm^3} *(\sigma_{elec}H + \sigma_{nucleus}H)))*100%[/math]

Using this method of calculation for Rexon RP 200 yields a probability [math] = \frac{0.5234%}{cm}[/math]

Using this method for Bicron BC 408 yields a probabiltiy [math] = \frac{0.5303%}{cm} [/math]


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