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>
 +
 +
 +
<math>p _n = [\frac{15 MeV + m _d}{2}]^{2} - m _n ^{2}</math>
 +
 +
<math>p _n = [\frac{15 MeV + 1875.6 MeV}{2}]^{2} - 939.6 MeV ^{2}</math>
 +
 +
<math>p _n = 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°.

Revision as of 10:09, 16 July 2008

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 MeV + 1875.6 MeV}{2}]^{2} - 939.6 MeV ^{2}[/math]

[math]p _n = 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°.