Difference between revisions of "Plastic Scintillator Calculation"

From New IAC Wiki
Jump to navigation Jump to search
Line 1: Line 1:
 
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.
  
Molecules per <math> cm^3 = \frac{grams CH_{2}}{cm^3} * \frac{mol}{gram} * \frac{N[A]}{mol} </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^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>
 
Molecules per <math> cm^2 (K) = \frac{Molecules}{cm^3} * Thickness </math>

Revision as of 01:17, 5 February 2009

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]

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]


Go Back