Number of electron-positron pairs/sec out of Al converte

From New IAC Wiki
Jump to navigation Jump to search

Go Back

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

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