Difference between revisions of "Notes from July 2nd, 2008 Meeting"

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<math>10^{-3} \frac{\frac{\gamma 's}{MeV}}{\frac{e^{-}}{radiation lengths}} \times 2 \cdot 10^{-4} radiation lengths \times 15 MeV \times 9.4 \cdot 10^{12} \frac{e^{-}}{sec}=2.8 \cdot 10^{7} \frac{\gamma}{sec}</math>
 
<math>10^{-3} \frac{\frac{\gamma 's}{MeV}}{\frac{e^{-}}{radiation lengths}} \times 2 \cdot 10^{-4} radiation lengths \times 15 MeV \times 9.4 \cdot 10^{12} \frac{e^{-}}{sec}=2.8 \cdot 10^{7} \frac{\gamma}{sec}</math>
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==Number of ɣ + d -> n + p events/sec==
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<math>2.8 \cdot 10^{7} \frac{\gamma}{sec} \times 4 \cdot 10^{-4} = 10^{4}</math>
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==Probability of Photodisintegration Event==
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target thickness in <math>\frac{Dnuclei}{cm^{2}} \times \sigma in cm^{2}</math>
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==Worst Case Isotropic Neutrons==
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Let's say we have:
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radius detector = 1 cm
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1 meter away
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fractional solid angle = <math>\frac{\pi * (1 cm)^{2}}{4 \pi (100cm^{2}} = \frac{1}{4} \times 10^{-4}</math> <= geometrical acceptance
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10° efficient of n° detection
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<math> 10^{4} \frac{photodisintegrations}{sec} \times \frac{1}{4} \cdot 10^{-4} \times 10^{-1} = .025 \frac{events}{sec}</math>
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time for <math>10^{4}</math> events = 100 hours for 1%
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::::24 hours for 2%
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::::6 hours for 4%
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'''Therefore, this experiment is do able.'''
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[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

Latest revision as of 06:26, 5 February 2009

Numbers for rate of Brems intensity spectrum:

[math]\frac{e^{-}}{sec}[/math] = [math]30 \cdot 10^{-3} \frac{Coulomb}{sec} \times \frac{50 \cdot 10^{-9} sec}{10^{-3} sec} \times \frac{1 \cdot e^{-}}{1.6 \cdot 10^{-19}Coulomb} =\gt 9.4 \cdot 10^{12} \frac{e^{-}}{sec} [/math]

[math]10^{-3} \frac{\frac{\gamma 's}{MeV}}{\frac{e^{-}}{radiation lengths}} \times 2 \cdot 10^{-4} radiation lengths \times 15 MeV \times 9.4 \cdot 10^{12} \frac{e^{-}}{sec}=2.8 \cdot 10^{7} \frac{\gamma}{sec}[/math]

Number of ɣ + d -> n + p events/sec

[math]2.8 \cdot 10^{7} \frac{\gamma}{sec} \times 4 \cdot 10^{-4} = 10^{4}[/math]

Probability of Photodisintegration Event

target thickness in [math]\frac{Dnuclei}{cm^{2}} \times \sigma in cm^{2}[/math]

Worst Case Isotropic Neutrons

Let's say we have:

radius detector = 1 cm

1 meter away

fractional solid angle = [math]\frac{\pi * (1 cm)^{2}}{4 \pi (100cm^{2}} = \frac{1}{4} \times 10^{-4}[/math] <= geometrical acceptance

10° efficient of n° detection [math] 10^{4} \frac{photodisintegrations}{sec} \times \frac{1}{4} \cdot 10^{-4} \times 10^{-1} = .025 \frac{events}{sec}[/math]

time for [math]10^{4}[/math] events = 100 hours for 1%

24 hours for 2%
6 hours for 4%


Therefore, this experiment is do able.

Go Back