Difference between revisions of "Notes for the July 11th, 2008 Meeting"

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<math>2 \cdot \sqrt{p_n ^2 + m _n ^2} = 15 MeV = m _d
 
<math>2 \cdot \sqrt{p_n ^2 + m _n ^2} = 15 MeV = m _d
 
</math>
 
</math>
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<math>p _n = [\frac{15 MeV + m _d}{2}]^{2} - m _n ^{2}</math>
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<math>p _n = [\frac{15  + 1875.6}{2}]^{2} - 939.6 ^{2} = 104 MeV</math>
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KE = 5.3 MeV
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β = .1
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θ = 85.6° -> Means that since they're moving forward, the angles will be pushed forward to be less than 90°.
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KE <sub>n</sub> <sup>fission</sup> = 1 MeV -> fission neutrons
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V = .05c
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-> 25 nsec/ft
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Beam pulse  [[Image:stuff_for_july_11th.jpg]]
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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

Setup07.11.08.jpg

[math]\frac {m}{e}[/math] = we'll have 2.5"

[math]\frac {2.5}{x}[/math] = tan(1.95)

radiator distance - collimator

x = 73.4" = 186.5 cm

radiator is 73.4" - 48" upstream of the wall


Something someone needs to do:

Look up which has the highest neutron production energy thresh hold: Cu, Zn, Ni


Com and lab frame.jpg

[math]2 \cdot p _n cos \theta = 15 MeV[/math]

[math]2 \cdot \sqrt{p_n ^2 + m _n ^2} = 15 MeV = m _d [/math]


[math]p _n = [\frac{15 MeV + m _d}{2}]^{2} - m _n ^{2}[/math]

[math]p _n = [\frac{15 + 1875.6}{2}]^{2} - 939.6 ^{2} = 104 MeV[/math]

KE = 5.3 MeV

β = .1

θ = 85.6° -> Means that since they're moving forward, the angles will be pushed forward to be less than 90°.


KE n fission = 1 MeV -> fission neutrons

V = .05c

-> 25 nsec/ft


Beam pulse Stuff for july 11th.jpg

Go Back

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