Notes for the July 11th, 2008 Meeting

[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

[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