Difference between revisions of "Notes for the July 11th, 2008 Meeting"
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| − | [[Image:setup07.11.08.jpg|left]] | + | [[Image:setup07.11.08.jpg|left|350 px]] |
<math>\frac {m}{e}</math> = we'll have 2.5" | <math>\frac {m}{e}</math> = we'll have 2.5" | ||
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| − | [[Image:com_and_lab_frame.jpg|left]] | + | [[Image:com_and_lab_frame.jpg|left|350 px]] |
<math>2 \cdot p _n cos \theta = 15 MeV</math> | <math>2 \cdot p _n cos \theta = 15 MeV</math> | ||
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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> | ||
| + | |||
| + | <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°. | ||
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| + | |||
| + | |||
| + | KE <sub>n</sub> <sup>fission</sup> = 1 MeV -> fission neutrons | ||
| + | |||
| + | V = .05c | ||
| + | |||
| + | -> 25 nsec/ft | ||
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| + | |||
| + | Beam pulse [[Image:stuff_for_july_11th.jpg]] | ||
| + | |||
| + | [http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back] | ||
Latest revision as of 06:26, 5 February 2009
= we'll have 2.5"
= 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
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 