Notes from July 2nd, 2008 Meeting: Difference between revisions

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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>
==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.'''
[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]\displaystyle{ \frac{e^{-}}{sec} }[/math] = [math]\displaystyle{ 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} => 9.4 \cdot 10^{12} \frac{e^{-}}{sec} }[/math]

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

Probability of Photodisintegration Event

target thickness in [math]\displaystyle{ \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]\displaystyle{ \frac{\pi * (1 cm)^{2}}{4 \pi (100cm^{2}} = \frac{1}{4} \times 10^{-4} }[/math] <= geometrical acceptance

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

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

24 hours for 2%
6 hours for 4%


Therefore, this experiment is do able.

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