Difference between revisions of "Positron Yield Estimate"

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In PEPPo: <math>1~nA = 10^{-9}~C/s</math>. So, 1 nC e- makes 1 e+
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In PEPPo: <math>1~nA = 10^{-9}~C/s</math>. So, PEPPo can generate 1 nC e- per second. This 1 nC e- makes 1 e+/s.
  
  
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<math> \frac{1~mA \times 50~ns \times 300}{1s} = \frac{400 50 \times ^{-9} 50~ns \times 300}{1s} =</math>.
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<math> \frac{80~mA \times 50~ns \times 300}{1s} = \frac{4000  \times 10^{-12} \times 300}{1}~A = 1200 \times 10^{-9}~C/s </math>.
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Assuming the same collection efficiency and positrons yield for HRRL, we should see, '''1200 e+/s'''.

Latest revision as of 17:15, 8 June 2012

The annihilation counter observed 1 e+ for every nA of electrons beam.  I was not sure what the compton polarimeter is observing given the large background.
Use the above to predict our rate if we run the HRRL at 80 mA peak current with 50 nsec wide pulses and a rep rate of 300 HZ.


In PEPPo: [math]1~nA = 10^{-9}~C/s[/math]. So, PEPPo can generate 1 nC e- per second. This 1 nC e- makes 1 e+/s.


In HRRL: 80 mA, 50 ns pulse width, 300 Hz reprate:


[math] \frac{80~mA \times 50~ns \times 300}{1s} = \frac{4000 \times 10^{-12} \times 300}{1}~A = 1200 \times 10^{-9}~C/s [/math].

Assuming the same collection efficiency and positrons yield for HRRL, we should see, 1200 e+/s.