Plastic Scintillator Calculation

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]

Density of polyvinyl toluene (a common scintillator material) [math] = \frac{1.02grams}{cm^3}[/math]

or is it [math]\rho_{BC408} = \frac{1.032grams}{cm^3}[/math]  H/C = 11/10  [1] (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}{14 g} * \frac{6.022*10^{23}molecules}{mol} = \frac{4.387*10^{22}molecules PVT}{cm^3} [/math]

Molecules per [math] cm^2 (K) = \frac{4.387*10^{22}molecules PVT}{cm^3} * 1cm = \frac{4.387*10^{22}molecules PVT}{cm^2} [/math]

Weighted cross-section [math] (\sigma_w) = (1.030*10^{-26}cm^2 + 9.645*10^{-26}cm^2) + 2(1.716*10^{-27}cm^2 + 2.688*10^{-27}cm^2) = 1.1556*10^{-25}cm^2[/math]

Probability of interaction (%) [math]= 1.1556*10^{-25}cm^2 * \frac{4.387*10^{22}molecules PVT}{cm^2} * 100% = 0.5070%[/math]

Doing the same calculations using the NPOL3 for the material yields a probability of

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