Plastic Scintillator Calculation

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

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

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

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

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

All cross sections listed here are pair production cross-sections

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

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

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

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

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

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

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

or is it [math]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ (\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]\displaystyle{ = 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]\displaystyle{ 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]\displaystyle{ cm^3 }[/math]. Therefore there is a quicker way to calculate the probability of interaction which is listed below.

Probability of interaction [math]\displaystyle{ = (\frac{NumCarbonAtoms}{cm^3} * MF#H(\sigma_{elec}C + \sigma_{nucleus}C)) + (\frac{NumCarbonAtoms}{cm^3}*MF#H(\sigma_{elec}H + \sigma_{nucleus}H }[/math]

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