Pair Production Rate Calculation

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LINAC parameters used in calculations

1) pulse width 50 ps
2) pulse current 50 A
3) repetition rate 300 Hz
4) energy 44 MeV

Number of electrons/sec on radiator

[math] 50\ \frac{C}{sec} \times \frac{1\cdot e^-}{1.6\cdot 10^{-19}\ C} \times 50\ \mbox{ps} \times 300\ \mbox{Hz} = 0.47 \cdot 10^{13}\ \frac{e^-}{sec}[/math]

Number of photons/sec out of radiator

1/2 mil of Ti

  1. [math]\sigma_{brems}=0.1\ \mbox{photons/electrons/MeV/r.l}[/math]
  2. [math]\mbox{r.l.(Ti)} = 3.59\ \mbox{cm}[/math]
  3. [math]\mbox{radiator}\ \mbox{thickness} = 12.5\ \mu m[/math]


[math]\frac{12.5\ \mu m}{3.59\ cm} = 3.48 \cdot 10^{-4}\ r.l.[/math]
[math]0.47 \cdot 10^{13}\ \frac{e^-}{sec} \times 0.1\ \frac{\gamma 's}{(e^- \cdot MeV \cdot r.l.)} \times 3.48 \cdot 10^{-4}\ r.l. \times 10\ MeV =1.63 \cdot 10^{9} \frac{\gamma}{sec}[/math]

Alex factor is 6.85 %

[math]1.63 \cdot 10^{9} \frac{\gamma}{sec} \cdot 6.85\ % = 1.12 \cdot 10^{8} \frac{\gamma}{sec}[/math]

1/2 mil of Al

  1. [math]\sigma_{brems}=0.1\ \mbox{photons/electrons/MeV/r.l}[/math]
  2. [math]\mbox{r.l.(Al)} = 8.89\ \mbox{cm}[/math]
  3. [math]\mbox{radiator}\ \mbox{thickness} = 12.5\ \mu m[/math]


[math]\frac{12.5\ \mu m}{8.89\ cm} = 1.41 \cdot 10^{-4}\ r.l.[/math]
[math]0.47 \cdot 10^{13}\ \frac{e^-}{sec} \times 0.1\ \frac{\gamma 's}{(e^- \cdot MeV \cdot r.l.)} \times 1.41 \cdot 10^{-4}\ r.l. \times 10\ MeV =0.66 \cdot 10^{9} \frac{\gamma}{sec}[/math]

Alex factor is 6.85 %

[math]0.66 \cdot 10^{9} \frac{\gamma}{sec} \cdot 6.85\ % = 0.45 \cdot 10^{8} \frac{\gamma}{sec}[/math]

Conversion factor from Ti to Al

All my following calculation for pair production rate are based on (1/2) mil of Ti radiator. If we want to recalculate for (1/2) mil of Al converter we need to use the conversion factor:

[math]\frac{0.45}{1.12} = 0.40 [/math]

Appendix

Bremss44MeV.png Al 44MeV.png

in (10,20) MeV region we have about 0.1 photons/electrons/MeV/r.l both for Ti and Al radiators

Pair production rate

out of Al converter

[math]\sigma_{pairs} = 0.5\ \frac{\mbox{barns}}{\mbox{atom}}[/math]
[math]l = 3.0\ \mu m[/math] (by varying width we can vary the yield)
[math]N_{Al} = \frac{2.70\ \frac{g}{cm^3} \times 6.02 \cdot 10^{23}\ \frac{atoms}{mol} \times 3.0\ \mu m} {26.98\ \frac{g}{mol}} = 1.81 \cdot 10^{23}\ \frac{\mbox{atoms}}{m^2}[/math]
[math]\frac {1.12 \cdot 10^{8}\ \frac{\gamma}{sec} \times \sigma_{pairs} \times N_{Al}} {f} = 3.38\ \frac{\mbox{pairs}}{\mbox{pulse}} [/math]

through 1 m of air

Assume air consists entirely from Nitrogen:

[math]\sigma_{pairs}\ (\mbox{Nitrogen}) = 0.15\ \frac{\mbox{barns}}{\mbox{atom}}[/math] 
[math]l = 1.0\ \mbox{m}[/math]
[math]N_{\mbox{Nitrogen}} = \frac{0.00125\ \frac{g}{cm^3} \times 6.02 \cdot 10^{23}\ \frac{atoms}{mol} \times 1.0\ \mbox{m}} {14.01\ \frac{g}{mol}} = 5.37 \cdot 10^{25}\ \frac{\mbox{atoms}}{m^2}[/math]
[math]\frac {1.12 \cdot 10^{8}\ \frac{\gamma}{sec} \times \sigma_{\mbox{pairs}} \times N_{\mbox{Nitrogen}}} {f} = 300\ \frac{\mbox{pairs}}{\mbox{pulse}} [/math]

1 m of air vs. [math]l = 3.0\ \mu m[/math] of Al converter

[math]\frac{300\ \frac{\mbox{pairs}}{\mbox{pulse}}} {3.38\ \frac{\mbox{pairs}}{\mbox{pulse}}} = 88.8\ \mbox{times!}[/math]

Appendix

pair production cross sections in an Al target

Ref. Geant4 and Theoretical Pair Production Cross Sections for 1 MeV - 100 GeV photons in Aluminum. Vakho Makarashvili, December 18, 2007


Pair production Al.png

pair production cross sections in a Nitrogen

Ref. Photon Cross Section, Attenuation Coefficients, and Energy Absorption Coefficients From 10 keV to 100 Gev. J.H.Hubbell. Center for Radiation Research.National Bureau of Standards. Washington, D.C. 20234

Pair prodiction nitrogen.png


Here I just plotted the table above for pair production cross section in (0, 40) MeV energy region


Sigma nitrogen fitted m2.png



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