Notes for the July 11th, 2008 Meeting - Revision history

Oborn at 06:26, 5 February 2009

2009-02-05T06:26:24Z

Oborn at 11:08, 16 July 2008

2008-07-16T11:08:37Z

Oborn at 10:37, 16 July 2008

2008-07-16T10:37:45Z

Oborn at 10:09, 16 July 2008

2008-07-16T10:09:26Z

Oborn at 09:47, 16 July 2008

2008-07-16T09:47:41Z

Oborn at 09:46, 16 July 2008

2008-07-16T09:46:50Z

Oborn at 09:45, 16 July 2008

2008-07-16T09:45:59Z

Oborn at 16:52, 15 July 2008

2008-07-15T16:52:25Z

Oborn at 16:38, 15 July 2008

2008-07-15T16:38:45Z

Oborn at 16:03, 15 July 2008

2008-07-15T16:03:02Z

New page

[[Image:setup07.11.08.jpg|left]]

<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

[[Image:com_and_lab_frame.jpg|left]]

2 p n cos θ = 15 MeV

2 sqrt(pn 2 + m n 2) = 15 MeV = m d

← Older revision		Revision as of 06:26, 5 February 2009
Line 45:		Line 45:

	Beam pulse [[Image:stuff_for_july_11th.jpg]]		Beam pulse [[Image:stuff_for_july_11th.jpg]]
		+
		+	[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

← Older revision		Revision as of 11:08, 16 July 2008
Line 34:		Line 34:

	θ = 85.6° -> Means that since they're moving forward, the angles will be pushed forward to be less than 90°.		θ = 85.6° -> Means that since they're moving forward, the angles will be pushed forward to be less than 90°.
		+
		+
		+
		+	KE <sub>n</sub> <sup>fission</sup> = 1 MeV -> fission neutrons
		+
		+	V = .05c
		+
		+	-> 25 nsec/ft
		+
		+
		+	Beam pulse [[Image:stuff_for_july_11th.jpg]]

@@ Line 27: / Line 27: @@
 <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 = [\frac{15  + 1875.6}{2}]^{2} - 939.6 ^{2} = 104 MeV</math>
 KE = 5.3 MeV

← Older revision		Revision as of 10:09, 16 July 2008
Line 23:		Line 23:
	<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°.

@@ Line 1: / Line 1: @@
-[[Image:setup07.11.08.jpg|left|400 px]]
+[[Image:setup07.11.08.jpg|left|350 px]]
 <math>\frac {m}{e}</math> = we'll have 2.5"
@@ Line 17: / Line 17: @@
-[[Image:com_and_lab_frame.jpg|left|300 px]]
+[[Image:com_and_lab_frame.jpg|left|350 px]]
 <math>2 \cdot p _n cos \theta = 15 MeV</math>

← Older revision		Revision as of 09:46, 16 July 2008
Line 1:		Line 1:
−	[[Image:setup07.11.08.jpg\|left\|~~300~~ px]]	+	[[Image:setup07.11.08.jpg\|left\|400 px]]

	<math>\frac {m}{e}</math> = we'll have 2.5"		<math>\frac {m}{e}</math> = we'll have 2.5"

@@ Line 19: / Line 19: @@
 [[Image:com_and_lab_frame.jpg|left]]
-* p <sub>n</sub> cos θ = 15 MeV
+<math>2 \cdot p _n cos \theta = 15 MeV</math>
-* sqrt(p<sub>n</sub> <sup>2</sup> + m <sub>n</sub> <sup>2</sup>) = 15 MeV = m <sub>d</sub>
+<math>2 \cdot \sqrt{p_n ^2 + m _n ^2} = 15 MeV = m _d