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

From New IAC Wiki
Jump to navigation Jump to search
 
(5 intermediate revisions by the same user not shown)
Line 13: Line 13:
  
 
target thickness in <math>\frac{Dnuclei}{cm^{2}} \times \sigma in cm^{2}</math>
 
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]\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