Difference between revisions of "Number of electron-positron pairs/sec out of Al converte"

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#<math>\sigma_{brems}=0.1\ \mbox{photons/electrons/MeV/r.l}</math>
 
#<math>\sigma_{brems}=0.1\ \mbox{photons/electrons/MeV/r.l}</math>
#<math>\mbox{r.l.(Ti)} = 3.59\ cm</math>
+
#<math>\mbox{r.l.(Ti)} = 3.59\ \mbox{cm}</math>
 
#<math>\mbox{radiator}\ \mbox{thickness} = 12.5\ \mu m</math>
 
#<math>\mbox{radiator}\ \mbox{thickness} = 12.5\ \mu m</math>
  

Revision as of 06:14, 29 January 2011

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

Alex factor is 6.85 %
[math]1.64 \cdot 10^{9} \frac{\gamma}{sec} \cdot 6.85\ % = 1.12 \cdot 10^{8} \frac{\gamma}{sec}[/math]

Pair production rate

out of Al converter

[math]\sigma_{pairs} = 0.5\ \frac{barns}{atom}[/math]
[math]l = 3.0\ \mu m[/math] (by varying width we can vary the yield)
[math]N_{Al} = \frac{2.375 \frac{g}{cm^3} \times 6.02 \cdot 10^{23} \frac{atoms}{mol} \times l} {26.98 \frac{g}{mol}} = 1.59 \cdot 10^{23}\ \frac{atoms}{m^2}[/math]
[math]\frac {1.12 \cdot 10^{8} \frac{\gamma}{sec} \times \sigma_{pairs} \times N_{Al}} {f} = 2.96\ \frac{pairs}{pulse} [/math]

through 1 m of air

[math]\sigma_{pairs} = 0.5\ \frac{barns}{atom}[/math]
[math]l = 3.0\ \mu m[/math] (by varying width we can vary the yield)
[math]N_{Al} = \frac{2.375 \frac{g}{cm^3} \times 6.02 \cdot 10^{23} \frac{atoms}{mol} \times l} {26.98 \frac{g}{mol}} = 1.59 \cdot 10^{23}\ \frac{atoms}{m^2}[/math]
[math]\frac {1.12 \cdot 10^{8} \frac{\gamma}{sec} \times \sigma_{pairs} \times N_{Al}} {f} = 2.96\ \frac{pairs}{pulse} [/math]