Anisotropic n's vs. isotropic ones - Revision history

Shaproma at 19:04, 24 May 2012

2012-05-24T19:04:29Z

2011-06-28T18:17:46Z

Let's count isotropic and anisotropic neutrons

2011-06-28T18:17:19Z

Let's count isotropic and anisotropic neutrons

2011-06-28T18:12:53Z

Let's count isotropic and anisotropic neutrons

2011-06-28T18:12:38Z

Let's count isotropic and anisotropic neutrons

2011-06-28T18:12:27Z

Let's count isotropic and anisotropic neutrons

2011-06-28T18:08:25Z

Let's count isotropic and anisotropic neutrons

2011-06-28T17:57:31Z

Let's count isotropic and anisotropic neutrons

2011-06-28T17:56:16Z

Let's count isotropic and anisotropic neutrons

2011-06-28T17:55:50Z

Let's count isotropic and anisotropic neutrons

← Older revision		Revision as of 19:04, 24 May 2012
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−	[https://wiki.iac.isu.edu/index.php/~~HRRL_RS go back~~]	+	[https://wiki.iac.isu.edu/index.php/Roman_calculation Go Back]

← Older revision		Revision as of 18:17, 28 June 2011
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−	The yields above are for 4<math>\pi</math> geometry. Now we need to calculate how many isotropic and anisotropic neutrons comes to ~~detector~~. Because anisotropic neutrons mostly comes at <math>\Theta = 90^o</math> the relative number of anisotropic neutrons at detectors will increase (will be more than 73 %).	+	The yields above are for 4<math>\pi</math> geometry. Now we need to calculate how many isotropic and anisotropic neutrons comes to detectors. Because anisotropic neutrons mostly comes at <math>\Theta = 90^o</math> the relative number of anisotropic neutrons at detectors will increase (will be more than 73 %).

← Older revision		Revision as of 18:17, 28 June 2011
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	*total isotropic neutrons yields is about (35+10)/(35+10+120) = 27 %		*total isotropic neutrons yields is about (35+10)/(35+10+120) = 27 %
	*total anisotropic neutrons yields is about (120)/(35+10+120) = 73 %		*total anisotropic neutrons yields is about (120)/(35+10+120) = 73 %
		+
		+
		+	The yields above are for 4<math>\pi</math> geometry. Now we need to calculate how many isotropic and anisotropic neutrons comes to detector. Because anisotropic neutrons mostly comes at <math>\Theta = 90^o</math> the relative number of anisotropic neutrons at detectors will increase (will be more than 73 %).

← Older revision		Revision as of 18:12, 28 June 2011
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	Using root simple integration procedure I found:		Using root simple integration procedure I found:

−	[[File:Table isotropic vs anisotropic event yields.png \| ~~400px~~]]	+	[[File:Table isotropic vs anisotropic event yields.png \| 300px]]

← Older revision		Revision as of 18:12, 28 June 2011
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	Using root simple integration procedure I found:		Using root simple integration procedure I found:

−	[[File:Table isotropic vs anisotropic event yields.png \| ~~500px~~]]	+	[[File:Table isotropic vs anisotropic event yields.png \| 400px]]

@@ Line 100: / Line 100: @@
 Using root simple integration procedure I found:
-[[File:Anysotropic vs isotropic totaleventsyield.png | 500px]]
+[[File:Table isotropic vs anisotropic event yields.png | 500px]]
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 Say we have total 100 events. Because we have 1n from (<math>\gamma</math>,n) channel, 2n neutron (<math>\gamma</math>,2n) channel and assuming we have 2n from (<math>\gamma</math>,f) channel:
-*total isotropic events yield from (<math>\gamma</math>,n) channel:   35 n's
+*total number of isotropic neutrons from (<math>\gamma</math>,n) channel:   35 n's
-*total isotropic events yield from (<math>\gamma</math>,2n) channel:  10 n's
+*total number of isotropic neutrons from (<math>\gamma</math>,2n) channel:  10 n's
-*total anisotropic events yield from (<math>\gamma</math>,f) channel: 120 n's
+*total number of anisotropic neutrons from (<math>\gamma</math>,f) channel: 120 n's

@@ Line 106: / Line 106: @@
 So we have:
-*total isotropic events yield is about 40%
+*total isotropic events yield from (<math>\gamma</math>,n) channel is about 35 %
-*total anisotropic events yield is about 60%
+*total isotropic events yield from (<math>\gamma</math>,2n) channel is about 5 %
-Say we have total 100 events. Assuming we have 2n per fission we have:
+Say we have total 100 events. Because we have 1n from (<math>\gamma</math>,n) channel, 2n neutron (<math>\gamma</math>,2n) channel and assuming we have 2n from (<math>\gamma</math>,f) channel:
-*total isotropic neutrons yield:     40  (about 25%)
+*total isotropic events yield from (<math>\gamma</math>,n) channel:   35 n's
-*total anisotropic neutrons yield:  120  (about 75%)
+*total isotropic events yield from (<math>\gamma</math>,2n) channel:  10 n's

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 We can calculate the relative number of neutrons from different channels using formulas:
 :<math>N_n = \frac{\int \sigma_n \Phi(E_{\gamma}) dE_{\gamma}}{\int \sigma_{tot} \Phi(E_{\gamma}) dE_{\gamma}}</math>
 :<math>N_{2n} = \frac{\int \sigma_{2n} \Phi(E_{\gamma}) dE_{\gamma}}{\int \sigma_{tot} \Phi(E_{\gamma}) dE_{\gamma}}</math>