Difference between revisions of "Anisotropic n's vs. isotropic ones"

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[https://wiki.iac.isu.edu/index.php/HRRL_RS go back]
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[https://wiki.iac.isu.edu/index.php/Roman_calculation Go Back]
  
  
  
==Winhold and Halpern, Phys.Rev. 103 4, 990 (1956)==
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==Motivation: there are a lot of source of isotropic neutrons===
  
:'''The observation were consistent with the assumption that'''
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1. Winhold and Halpern, Phys.Rev. 103 4, 990 (1956).
:'''anisotropic fission is due solely to photons with-in about'''
 
:'''3 MeV of the fission threshold".'''
 
  
 +
:The observation were consistent with the assumption that
 +
:anisotropic fission is due solely to photons with-in about
 +
:3 MeV of the fission threshold".
  
:'''The photons in the giant resonance region were found'''
+
 
:'''to produce essentially isotropic fission'''
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:The photons in the giant resonance region were found
 +
:to produce essentially isotropic fission
  
  
 
[[File:Winhold fig.4.png | 400px]]
 
[[File:Winhold fig.4.png | 400px]]
 +
 +
 +
2. A lot of other papers.
 +
 +
3. Isotropic neutrons from (<math>\gamma</math>,n) channel and (<math>\gamma</math>,2n) channel
  
 
==Anisotropic n's vs. isotropic ones==
 
==Anisotropic n's vs. isotropic ones==
  
Say, we have only anisotropic neutrons
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Say, we have only anisotropic neutrons with 1.25 calculated ratio:
 +
 
 +
:<math>N(90^o) = 125</math>
 +
:<math>N(0^o) = 100</math>
 +
 
 +
The calculated asymmetry would be:
  
:<math>\Theta=90^o,\ \phi = 0^o:\ 100\ n's</math>
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:<math>A = \frac{N(90) - N(0)}{N(90) + N(0)} = \frac{25}{225} = 11.1%</math>
:<math>\Theta=90^o,\ \phi = 90^o:\ 125\ n's</math>
 
  
So the calculated asymmetry would be:
 
  
:<math>A = \frac{125}{100} = 1.25</math>
 
  
  
 +
Now, say, we have extra 100 isotropic neutrons for each detector:
  
 +
:<math>N(90^o) = 100</math>
 +
:<math>N(0^o) = 100</math>
  
Now, say, we have extra 200 isotropic neutrons
+
The total number of neutrons become:
  
:<math>\Theta=90^o,\ \phi = 0^o:\ 100\ n's</math>
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:<math>N(90^o) = 100 +125 = 225</math>
:<math>\Theta=90^o,\ \phi = 90^o:\ 100\ n's</math>
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:<math>N(0^o) = 100 + 100 = 200</math>
  
 
So the calculated asymmetry would be:
 
So the calculated asymmetry would be:
  
:<math>A = \frac{225}{200} = 1.12</math>
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:<math>A = \frac{N(90) - N(0)}{N(90) + N(0)} = \frac{25}{425} = 5.9%</math>
  
  
  
That really reduce the measured asymmetry. If isotropic neutron's are much more than anisotropic ones we would not be able probably to see any asymmetry. Need to count isotropic and anisotropic neutrons.
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So if we would have about the same number of isotropic and anisotropic neutrons that would be reduced the expectation values of asymmetry almost in two times!!!.
 +
 
 +
 
 +
'''We need to count isotropic and anisotropic neutrons to estimate the effect of washing out the anisotropy by isotropic neutrons.'''
  
  
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Source of isotropic neutrons:
 
Source of isotropic neutrons:
 
*(<math>\gamma</math>,f) channel due to photons out of 3 MeV of the fission threshold
 
*(<math>\gamma</math>,f) channel due to photons out of 3 MeV of the fission threshold
*(<math>\gamma</math>,n) channel? Need to check the angular distribution of neutrons.
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*(<math>\gamma</math>,n) channel.
*(<math>\gamma</math>,2n) channel? Need to check the angular distribution of neutrons.
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*(<math>\gamma</math>,2n) channel.
  
 
==U Photo-fission Cross Section==
 
==U Photo-fission Cross Section==
  
[[File:U235 sigma 01.png | 1000 px]]
 
  
[[File:U238 sigma 01.png | 1000 px]]
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[[File:U235.sigma.01.png | 1000px]]
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 +
[[File:U235.sigmaflux.01.png | 1000px]]
 +
 
 +
From the plot above we can expect significant number of isotropic neutrons from (<math>\gamma</math>,n) and (<math>\gamma</math>,2n) channels. So the effect of washing out the anisotropy by isotropic neutrons should be significant and should be estimated.
 +
 
 +
==Let's count isotropic and anisotropic neutrons==
 +
 
 +
Assuming conservatively:
 +
 
 +
 
 +
Source of anisotropic neutrons:
 +
*(<math>\gamma</math>,f) channel for all incident gammas. In reality the  number of anisotropic neutrons will reduce with increasing the energy of gammas.
 +
 
 +
Source of isotropic neutrons:
 +
*(<math>\gamma</math>,n) channel
 +
*(<math>\gamma</math>,2n) channel
 +
 
 +
 
 +
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>
 +
 
 +
:<math>N_f = \frac{\int \sigma_f \Phi(E_{\gamma}) dE_{\gamma}}{\int \sigma_{tot} \Phi(E_{\gamma}) dE_{\gamma}}</math>
 +
 
 +
 
 +
Using root simple integration procedure I found:
 +
 
 +
[[File:Table isotropic vs anisotropic event yields.png | 300px]]
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 +
 
 +
 
 +
So we have:
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 +
*total isotropic events yield from (<math>\gamma</math>,n) channel is about 35 %
 +
*total isotropic events yield from (<math>\gamma</math>,2n) channel is about 5 %
 +
*total anisotropic events yield from (<math>\gamma</math>,f) channel is about 60 %
 +
 
 +
 
 +
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 number of isotropic neutrons from (<math>\gamma</math>,n) channel:  35 n's
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*total number of isotropic neutrons from (<math>\gamma</math>,2n) channel:  10 n's
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*total number of anisotropic neutrons from (<math>\gamma</math>,f) channel: 120 n's
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 +
 
 +
So we have:
 +
 
 +
*total isotropic neutrons yields is about    (35+10)/(35+10+120) = 27 %
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*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 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 %).

Latest revision as of 19:04, 24 May 2012

Go Back


Motivation: there are a lot of source of isotropic neutrons=

1. Winhold and Halpern, Phys.Rev. 103 4, 990 (1956).

The observation were consistent with the assumption that
anisotropic fission is due solely to photons with-in about
3 MeV of the fission threshold".


The photons in the giant resonance region were found
to produce essentially isotropic fission


Winhold fig.4.png


2. A lot of other papers.

3. Isotropic neutrons from ([math]\gamma[/math],n) channel and ([math]\gamma[/math],2n) channel

Anisotropic n's vs. isotropic ones

Say, we have only anisotropic neutrons with 1.25 calculated ratio:

[math]N(90^o) = 125[/math]
[math]N(0^o) = 100[/math]

The calculated asymmetry would be:

[math]A = \frac{N(90) - N(0)}{N(90) + N(0)} = \frac{25}{225} = 11.1%[/math]



Now, say, we have extra 100 isotropic neutrons for each detector:

[math]N(90^o) = 100[/math]
[math]N(0^o) = 100[/math]

The total number of neutrons become:

[math]N(90^o) = 100 +125 = 225[/math]
[math]N(0^o) = 100 + 100 = 200[/math]

So the calculated asymmetry would be:

[math]A = \frac{N(90) - N(0)}{N(90) + N(0)} = \frac{25}{425} = 5.9%[/math]


So if we would have about the same number of isotropic and anisotropic neutrons that would be reduced the expectation values of asymmetry almost in two times!!!.


We need to count isotropic and anisotropic neutrons to estimate the effect of washing out the anisotropy by isotropic neutrons.


Source of anisotropic neutrons:

  • ([math]\gamma[/math],f) channel due to photons with-in about 3 MeV of the fission threshold

Source of isotropic neutrons:

  • ([math]\gamma[/math],f) channel due to photons out of 3 MeV of the fission threshold
  • ([math]\gamma[/math],n) channel.
  • ([math]\gamma[/math],2n) channel.

U Photo-fission Cross Section

U235.sigma.01.png

U235.sigmaflux.01.png

From the plot above we can expect significant number of isotropic neutrons from ([math]\gamma[/math],n) and ([math]\gamma[/math],2n) channels. So the effect of washing out the anisotropy by isotropic neutrons should be significant and should be estimated.

Let's count isotropic and anisotropic neutrons

Assuming conservatively:


Source of anisotropic neutrons:

  • ([math]\gamma[/math],f) channel for all incident gammas. In reality the number of anisotropic neutrons will reduce with increasing the energy of gammas.

Source of isotropic neutrons:

  • ([math]\gamma[/math],n) channel
  • ([math]\gamma[/math],2n) channel


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]
[math]N_f = \frac{\int \sigma_f \Phi(E_{\gamma}) dE_{\gamma}}{\int \sigma_{tot} \Phi(E_{\gamma}) dE_{\gamma}}[/math]


Using root simple integration procedure I found:

Table isotropic vs anisotropic event yields.png


So we have:

  • total isotropic events yield from ([math]\gamma[/math],n) channel is about 35 %
  • total isotropic events yield from ([math]\gamma[/math],2n) channel is about 5 %
  • total anisotropic events yield from ([math]\gamma[/math],f) channel is about 60 %


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 number of isotropic neutrons from ([math]\gamma[/math],n) channel: 35 n's
  • total number of isotropic neutrons from ([math]\gamma[/math],2n) channel: 10 n's
  • total number of anisotropic neutrons from ([math]\gamma[/math],f) channel: 120 n's


So we have:

  • total isotropic neutrons yields is about (35+10)/(35+10+120) = 27 %
  • 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 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 %).