Difference between revisions of "Counts Rate (44 MeV LINAC)"

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−
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
+
[https://wiki.iac.isu.edu/index.php/Roman_calculation Go Back]
  
−
=Counts Rate for U238=
+
=LINAC parameters used in calculations=
−
 
 
−
==LINAC parameters used in calculations==
 
 
1) pulse width 50 ps <br>
 
1) pulse width 50 ps <br>
 
2) pulse current 50 A <br>
 
2) pulse current 50 A <br>
 
3) repetition rate 300 Hz <br>
 
3) repetition rate 300 Hz <br>
 
4) energy 44 MeV <br><br>
 
4) energy 44 MeV <br><br>
 +
 +
 +
=Counts Rate for U238 (1/2 mil of Ti radiadot)=
  
 
==Number of electrons/sec on radiator==
 
==Number of electrons/sec on radiator==
Line 26: Line 27:
 
===radiation length===
 
===radiation length===
  
−
r.l.(Ti) = 3.59 cm
+
r.l.(Ti) = 3.59 cm
 
    
 
    
−
radiator thickness = 12.5 <math>\mu m</math>
+
radiator thickness = 12.5 <math>\mu m</math>
  
−
<math>12.5\mu m/3.59 cm = 3.48 \cdot 10^{-4} \ r.l.</math><br>
+
<math>\frac{12.5\ \mu m}{3.59\ cm} = 3.48 \cdot 10^{-4} \ r.l.</math><br>
  
 
===steps together...===
 
===steps together...===
−
<math>0.1\ \frac{\gamma 's}{(e^- \cdot MeV \cdot r.l.)} \times 3.48 \cdot 10^{-4} r.l. \times 10\ MeV \times 0.47 \cdot 10^{13} \frac{e^-}{sec}=1.64 \cdot 10^{9} \frac{\gamma}{sec}</math><br><br>
+
<math>0.1\ \frac{\gamma 's}{(e^- \cdot MeV \cdot r.l.)} \times 3.48 \cdot 10^{-4} r.l. \times 10\ MeV \times 0.47 \cdot 10^{13} \frac{e^-}{sec}=1.64 \cdot 10^{9} \frac{\gamma}{sec}</math><br><br>
  
 
==Alex factor (GEANT4 calculation)==
 
==Alex factor (GEANT4 calculation)==
Line 43: Line 44:
 
then, incident flux on target is
 
then, incident flux on target is
  
−
<math>1.64 \cdot 10^{9} \frac{\gamma}{sec} \cdot 6.85\ % = 11.2 \cdot 10^{7} \frac{\gamma}{sec}</math><br><br>
+
<math>1.64 \cdot 10^{9} \frac{\gamma}{sec} \cdot 6.85\ % = 1.12 \cdot 10^{8} \frac{\gamma}{sec}</math><br><br>
  
−
==Number of neutrons/sec (yields)==
+
==Number of neutrons/sec==
  
−
===photonuclear cross section for <math>^{238}U(\gamma , n)</math> reaction===
+
===photonuclear cross section for <math>^{238}U(\gamma , F)</math> reaction===
  
−
From the paper "Giant resonance for the actinide nuclei: Photoneutron and photofission cross sections for 235U, 236U, 238U, and 232Th", J. T. Caldwell and E. J. Dowdy, B. L. Berman, R. A. Alvarez, and P. Meyer. Physical Review C, (21), 1215, April 1980:
+
J. T. Caldwell ''et all.,'' Phys. Rev. '''C21''', 1215 (1980):
  
 
[[File:phofission_sigma_U238.png]]
 
[[File:phofission_sigma_U238.png]]
Line 59: Line 60:
 
===target thickness, <math>^{238}U</math>===
 
===target thickness, <math>^{238}U</math>===
  
−
<math>\frac{19.1\ g/cm^3}{238.02\ g/mol} = 0.08\ \frac{mol}{cm^3} \times \frac{6.02\cdot 10^{23}\ atoms}{mol} = 0.48\cdot 10^{23}\ \frac{atoms}{cm^3}</math>
+
<math>\frac{19.1\ g/cm^3}{238.02\ g/mol} = 0.08\ \frac{mol}{cm^3} \times \frac{6.02\cdot 10^{23}\ atoms}{mol} = 0.48\cdot 10^{23}\ \frac{atoms}{cm^3}</math>
  
 
Let's target thickness = 1 mm:
 
Let's target thickness = 1 mm:
  
−
<math>0.48\cdot 10^{23}\ \frac{atoms}{cm^3} \times 0.1\ cm = 0.48\cdot 10^{22}\ \frac{atoms}{cm^2}</math>
+
<math>0.48\cdot 10^{23}\ \frac{atoms}{cm^3} \times 0.1\ cm = 0.48\cdot 10^{22}\ \frac{atoms}{cm^2}</math>
  
 
===neutrons per fission===
 
===neutrons per fission===
Line 71: Line 72:
 
===steps together...yeild===
 
===steps together...yeild===
  
−
<math> Y = \frac{\gamma}{sec} \times t \times \sigma \times 2.4 = </math><br>
+
<math> Y = \frac{\gamma}{sec} \times t \times \sigma \times 2.4 = </math>
  
−
<math> = 11.2 \cdot 10^{7} \frac{\gamma}{sec} \times 130\ mb \times 0.48\cdot 10^{22}\ \frac{atoms}{cm^2} \times 2.4 = 1.68 \cdot 10^{5}\ \frac{neutrons}{sec}</math><br><br>
+
<math> = 1.12 \cdot 10^{8} \frac{\gamma}{sec} \times 130\ mb \times 0.48\cdot 10^{22}\ \frac{atoms}{cm^2} \times 2.4 = 1.68 \cdot 10^{5}\   \frac{neutrons}{sec}</math><br><br>
  
 
==Worst Case Isotropic Neutrons==
 
==Worst Case Isotropic Neutrons==
Line 85: Line 86:
 
take the worst case 10 MeV neutron:
 
take the worst case 10 MeV neutron:
  
−
<math> E_{tot} = E_{kin} + E_{rest} =  10\ MeV + 938\ MeV = 948\ MeV </math>
+
<math> E_{tot} = E_{kin} + E_{rest} =  10\ MeV + 938\ MeV = 948\ MeV </math>
  
−
<math> \gamma = \frac{E_{tot}}{m_p} = \frac{948\ MeV}{938\ MeV} = 1.0107 </math>
+
<math> \gamma = \frac{E_{tot}}{m_p} = \frac{948\ MeV}{938\ MeV} = 1.0107 </math>
  
−
<math> \gamma^2 = \frac{1}{1 - \beta^2} \ \ \  \rightarrow \ \ \ \beta = 0.145\ c</math>
+
<math> \gamma^2 = \frac{1}{1 - \beta^2} \ \ \  \rightarrow \ \ \ \beta = 0.145\ c</math>
  
 
take the neutron detector 1 meter away:
 
take the neutron detector 1 meter away:
  
−
<math> t = \frac{1\ m}{0.145\ c} = \frac{1\ m}{0.145\cdot 3\cdot 10^8\ m/sec} = 23\ ns </math>
+
<math> t = \frac{1\ m}{0.145\ c} = \frac{1\ m}{0.145\cdot 3\cdot 10^8\ m/sec} = 23\ ns </math>
−
 
 
  
 
       23 ns >> 50 ps  <= time resolution is good
 
       23 ns >> 50 ps  <= time resolution is good
Line 102: Line 102:
 
taking real detector 3" x 2"  =>  S is about 40 cm^2
 
taking real detector 3" x 2"  =>  S is about 40 cm^2
  
−
1 meter away
+
1 meter away
  
−
fractional solid angle = <math>\frac{40\ cm^{2}}{4 \pi\ (100\ cm)^{2}} = 3.2 \cdot 10^{-4}</math> <= geometrical acceptance
+
fractional solid angle = <math>\frac{40\ cm^{2}}{4 \pi\ (100\ cm)^{2}} = 3.2 \cdot 10^{-4}</math> <= geometrical acceptance
  
−
==Yield==
+
==Yield (1/2 mil of Ti and without detector efficiency)==
  
 
the yield per second:
 
the yield per second:
  
−
<math>1.68 \cdot 10^{5}\ \frac{neutrons}{sec} \times 3.2 \cdot 10^{-4} = 54\ \frac{neutrons}{sec} </math><br>
+
<math>1.68 \cdot 10^{5}\ \frac{neutrons}{sec} \times 3.2 \cdot 10^{-4} = 53.8\ \frac{neutrons}{sec} </math><br>
 
   
 
   
 
the yield per pulse:
 
the yield per pulse:
  
−
<math> 54\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 0.2\ \frac{neutrons}{pulse} </math> <br><br>
+
<math> 53.8\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 0.18\ \frac{neutrons}{pulse} </math> <br><br>
 +
 
 +
:'''53.8 neutrons/sec  <= this experiment is do able'''
 +
 
 +
:'''0.18 neutrons/pulse  <= good for stopping pulse'''
 +
 
 +
=Counts Rate for U238 (1/2 mil of Al converter)=
 +
 
 +
==radiation length==
 +
 
 +
r.l.(Al) = 8.89 cm
 +
 
 +
radiator thickness = 12.5 <math>\mu m</math>
  
−
  538 neutrons/sec <= this experiment is do able
+
  <math>\frac{12.5\ \mu m}{8.89\ cm} = 1.41 \cdot 10^{-4} \ r.l.</math><br>
  
−
  1.8 neutrons/pulse <= is not good for stopping pulse
+
==Calibration factor==
  
−
<br><br>
+
The only difference from calculations above is:
  
−
=Counts Rate for Deuteron=
+
1) radiation length:
  
−
===Photonuclear cross section for <math> ^2H(\gamma , n) </math> reaction====
+
    1.41 (1/2 mil Al) / 3.48 (1/2 mil Ti) = 0.40
  
−
From the paper "Absolute total cross sections for deuteron photodisintegration between 7 and 19 MeV", A. De Graeva and other. Physical Review C, (45), 860, February 1992:
+
==Yield (1/2 mil of Al and without detector efficiency)==
 +
 
 +
 
 +
:'''53.8 neutrons/sec * 0.40 = 21.5 neutrons/sec '''
 +
 
 +
:'''0.18 neutrons/pulse * 0.40 = 0.07 neutrons/pulse '''
 +
 
 +
=Counts Rate for Deuteron (1/2 mil of Ti converter)=
 +
 
 +
===photonuclear cross section for <math> ^2H(\gamma , n) </math> reaction===
 +
 
 +
A. De Graeva ''et all.,'' Phys. Rev. '''C45''', 860 (1992):
  
 
[[File:photonuc_sigma_deuteron.png]]
 
[[File:photonuc_sigma_deuteron.png]]
Line 138: Line 161:
 
take <math>D_2O</math>, liquid (20°C):
 
take <math>D_2O</math>, liquid (20°C):
  
−
<math> \frac{1.1056\ g/mL}{20.04\ g/mol} = 0.055\ \frac{mol}{cm^3} \times \frac{6.02\cdot 10^{23}\ molecules}{mol} = 0.33\cdot 10^{23}\ \frac{molecules}{cm^3} </math>
+
<math> \frac{1.1056\ g/mL}{20.04\ g/mol} = 0.055\ \frac{mol}{cm^3} \times \frac{6.02\cdot 10^{23}\ molecules}{mol} = 0.33\cdot 10^{23}\ \frac{molecules}{cm^3} </math>
  
−
<math> 0.33\cdot 10^{23}\ \frac{molecules}{cm^3} \times \frac{2\ deuterons}{molecule} = 0.66\cdot 10^{23}\ \frac{deuterons}{cm^3} </math>
+
<math> 0.33\cdot 10^{23}\ \frac{molecules}{cm^3} \times \frac{2\ deuterons}{molecule} = 0.66\cdot 10^{23}\ \frac{deuterons}{cm^3} </math>
  
−
Target thickness = 1 cm:
+
Let's target thickness = 10 cm:
  
−
<math>0.66\cdot 10^{23}\ \frac{atoms}{cm^3} \times 1\ cm = 0.66\cdot 10^{23}\ \frac{atoms}{cm^2}</math>
+
<math>0.66\cdot 10^{23}\ \frac{atoms}{cm^3} \times 10\ cm = 66\cdot 10^{22}\ \frac{atoms}{cm^2}</math>
 +
 
 +
 
 +
===angular distribution of neutron===
 +
 
 +
====P. Rossi ''et all.,'' Phys. Rev. '''C40''', 2412 (1989):====
 +
 
 +
[[File:sigma_deuteron_20_40MeV.png|600 px]]
 +
 
 +
[[File:sigma_deuteron_total.png|300 px]]
 +
 
 +
 
 +
====relativistic kinematics====
 +
 
 +
An Introduction to Nuclear and Subnuclear Physics. Emilio Segre (1964)
 +
 
 +
  <math> \tan(\Theta_i)  =  \frac{1}{\overline{\gamma}} \frac{\sin\Theta_i^*}{\overline{\beta} (E_i^*/p_i^*) + \cos\Theta_i^*}</math>
 +
 
 +
where
 +
 
 +
  asterisks are quantities referred to CM<br>
 +
  barred quantities refer to the velocity of the CM
 +
 
 +
 
 +
  <math> E^* = \left[(m_1+m_2)^2 + 2T_1m_2\right]^{1/2}</math><br>
 +
  <math> \overline{\gamma} = \frac{E}{E^*} = \frac{m_1 + m_2 + T_1}{E^*}</math><br>
 +
  <math> \overline{\beta} = \frac{p}{E} = \frac{p_1}{m_1 + m_2 + T_1}</math>
 +
 
 +
 
 +
  <math> E_3^* = \frac{E^{*2} + m_3^2 - m_4^2}{2E*}</math><br>
 +
  <math> E_4^* = \frac{E^{*2} + m_4^2 - m_3^2}{2E*}</math><br>
 +
  <math> |p_3^*| = |p_4^*| = \left( E_3^{*2} - m_3^2 \right)^{1/2} = \left( E_4^{*2} - m_4^2 \right)^{1/2}</math>
 +
 
 +
====calculations====
 +
 
 +
{| border="1" cellpadding="20" cellspacing="0"
 +
|-
 +
|<math>T_{\gamma}</math>
 +
||<math>\Theta_{LAB}</math>
 +
||<math>\Theta_{CM}</math>
 +
||<math>\sigma_{T}</math>
 +
||<math>d \sigma / d \Omega\left(\Theta_{CM}\right)</math>
 +
||<math>\Omega_{Det}=\frac{A}{r^2}</math>
 +
||<math>\frac{d \sigma / d \Omega \times \Omega_{Det}}{\sigma_{T}}</math>
 +
|-
 +
|20 MeV || <math>90^o</math> || <math>94.38^o</math> || <math>600\ \mu b</math>
 +
|| <math>63\ \mu b/sr</math> || <math>40\cdot 10^{-4}\ sr</math> || <math>4.2\cdot 10^{-4}</math> 
 +
|-
 +
|40 MeV || <math>90^o</math> || <math>96.06^o</math> || <math>350\ \mu b</math>
 +
|| <math>23\ \mu b/sr</math> || <math>40\cdot 10^{-4}\ sr</math> || <math>2.6\cdot 10^{-4}</math> 
 +
|-
 +
|}
 +
 
 +
====geometrical factor====
 +
 
 +
taking average for 20 and 40 MeV photons
 +
 
 +
  geometrical acceptance = <math>\frac{(4.2\cdot 10^{-4} + 2.6\cdot 10^{-4})}{2} = 3.4\cdot 10^{-4}</math>
  
 
===Calibration factor===
 
===Calibration factor===
  
−
The only difference from calculations above is:
+
The only differences from calculations above are:
  
−
1. cross section:
+
1) cross section correction:
  
 
     1000 μb (D) / 130 mb (238U) = 1/130
 
     1000 μb (D) / 130 mb (238U) = 1/130
  
−
2. target thickness:
+
2) target thickness correction:
  
−
     <math> \frac{0.66\cdot 10^{23}\ atoms/cm^2\ (D)}{0.48\cdot 10^{23}\ atoms/cm^2\ (238U)} = 0.66/0.44 </math>
+
     <math> \frac{66\cdot 10^{22}\ atoms/cm^2\ (D)}{0.48\cdot 10^{23}\ atoms/cm^2\ (^{238}U)} = 66/0.48 </math>
  
−
3. neutrons per reaction:
+
3) neutrons per reaction correction:
  
 
     1 neutron (D) / 2.4 neutrons(238U) = 1/2.4
 
     1 neutron (D) / 2.4 neutrons(238U) = 1/2.4
  
−
total calibration factor is:
+
4) geometrical factor correction:
  
−
  <math>\frac{1}{130} \times \frac{0.66}{0.44} \times \frac{1}{2.4} = 4.8 \cdot 10^{-3}</math>
+
    <math> \frac{3.4\cdot 10^{-4}\ (D)}{3.2\cdot 10^{-4}\ (^{238}U)} = 1.06 </math>
  
−
===Yield===
+
'''total calibration factor is:'''
 +
 
 +
  <math>\frac{1}{130} \times \frac{66}{0.48} \times \frac{1}{2.4} \times \frac{3.4}{3.2} = 0.468</math>
 +
 
 +
===Yield (1/2 mil of Ti and without detector efficiency)===
  
 
saying all other factors is the same =>
 
saying all other factors is the same =>
Line 172: Line 256:
 
the yield per second :
 
the yield per second :
  
−
<math> 538\ \frac{neutrons}{sec} \times 4.8 \cdot 10^{-3} = 2.6\ \frac{neutrons}{sec} </math><br>
+
<math> 53.8\ \frac{neutrons}{sec} \times 0.468 = 25.2\ \frac{neutrons}{sec} </math><br>
  
 
the yield per pulse:
 
the yield per pulse:
  
−
<math> 2.6\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 8.6 \cdot 10^{-3}\frac{neutrons}{pulse} </math><br><br>
+
<math> 25.2\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 0.08\ \frac{neutrons}{pulse} </math><br><br>
 +
 
 +
=Summary (counts rate without neutron efficiency for different radiator thickness=
 +
 
 +
{| border="1" cellpadding="20" cellspacing="0"
 +
 
 +
||'''converter'''
 +
||'''target'''
 +
||'''neutrons/sec'''
 +
||'''neutrons/pulse'''
 +
|-
 +
||1/2 mil Ti||<math>^{238}U</math>||53.8||0.18
 +
|-
 +
||1/2 mil Al||<math>^{238}U</math>||21.5||0.07
 +
|-
 +
||1/2 mil Ti||<math>D_2O</math>  ||25.2||0.08
 +
|-
 +
||1/2 mil Al||<math>D_2O</math>  ||10.1||0.03
 +
 
 +
|}
 +
 
  
  
 
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
 
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

Latest revision as of 19:03, 24 May 2012

Go Back

LINAC parameters used in calculations

1) pulse width 50 ps
2) pulse current 50 A
3) repetition rate 300 Hz
4) energy 44 MeV


Counts Rate for U238 (1/2 mil of Ti radiadot)

Number of electrons/sec on radiator

[math] 50\ \frac{Coulomb}{sec} \times \frac{1\cdot e^-}{1.6\cdot 10^{-19}C} \times 50ps \times 300Hz = 0.47 \cdot 10^{13} \frac{e^-}{sec}[/math]


Number of photons/sec on target

bremsstrahlung

Bremss44MeV.png

in (10,20) MeV region we have about 

    0.1 photons/electrons/MeV/r.l

radiation length

r.l.(Ti) = 3.59 cm
  
radiator thickness = 12.5 [math]\mu m[/math]

[math]\frac{12.5\ \mu m}{3.59\ cm} = 3.48 \cdot 10^{-4} \ r.l.[/math]

steps together...

[math]0.1\ \frac{\gamma 's}{(e^- \cdot MeV \cdot r.l.)} \times 3.48 \cdot 10^{-4} r.l. \times 10\ MeV \times 0.47 \cdot 10^{13} \frac{e^-}{sec}=1.64 \cdot 10^{9} \frac{\gamma}{sec}[/math]

Alex factor (GEANT4 calculation)

Collimation factor is 

    6.85 % of total # of photons

then, incident flux on target is 

[math]1.64 \cdot 10^{9} \frac{\gamma}{sec} \cdot 6.85\ % = 1.12 \cdot 10^{8} \frac{\gamma}{sec}[/math]

Number of neutrons/sec

photonuclear cross section for [math]^{238}U(\gamma , F)[/math] reaction

J. T. Caldwell et all., Phys. Rev. C21, 1215 (1980):

Phofission sigma U238.png

in (10,20) MeV region the average cross section, say, is: 

    130 mb

target thickness, [math]^{238}U[/math]

[math]\frac{19.1\ g/cm^3}{238.02\ g/mol} = 0.08\ \frac{mol}{cm^3} \times \frac{6.02\cdot 10^{23}\ atoms}{mol} = 0.48\cdot 10^{23}\ \frac{atoms}{cm^3}[/math]

Let's target thickness = 1 mm: 

[math]0.48\cdot 10^{23}\ \frac{atoms}{cm^3} \times 0.1\ cm = 0.48\cdot 10^{22}\ \frac{atoms}{cm^2}[/math]

neutrons per fission

   2.4 neutrons/fission

steps together...yeild

[math] Y = \frac{\gamma}{sec} \times t \times \sigma \times 2.4 = [/math]

[math] = 1.12 \cdot 10^{8} \frac{\gamma}{sec} \times 130\ mb \times 0.48\cdot 10^{22}\ \frac{atoms}{cm^2} \times 2.4 = 1.68 \cdot 10^{5}\   \frac{neutrons}{sec}[/math]

Worst Case Isotropic Neutrons

checking detector distance

we want: 

     the time of flight of neutron >> the pulse width

take the worst case 10 MeV neutron: 

[math] E_{tot} = E_{kin} + E_{rest} =  10\ MeV + 938\ MeV = 948\ MeV [/math]

[math] \gamma = \frac{E_{tot}}{m_p} = \frac{948\ MeV}{938\ MeV} = 1.0107 [/math]

[math] \gamma^2 = \frac{1}{1 - \beta^2} \ \ \  \rightarrow \ \ \ \beta = 0.145\ c[/math]

take the neutron detector 1 meter away: 

[math] t = \frac{1\ m}{0.145\ c} = \frac{1\ m}{0.145\cdot 3\cdot 10^8\ m/sec} = 23\ ns [/math]

     23 ns >> 50 ps  <= time resolution is good

geometrical factor

taking real detector 3" x 2" => S is about 40 cm^2 

1 meter away

fractional solid angle = [math]\frac{40\ cm^{2}}{4 \pi\ (100\ cm)^{2}} = 3.2 \cdot 10^{-4}[/math] <= geometrical acceptance

Yield (1/2 mil of Ti and without detector efficiency)

the yield per second: 

[math]1.68 \cdot 10^{5}\ \frac{neutrons}{sec} \times 3.2 \cdot 10^{-4} = 53.8\ \frac{neutrons}{sec} [/math]

the yield per pulse: 

[math] 53.8\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 0.18\ \frac{neutrons}{pulse} [/math]
53.8 neutrons/sec <= this experiment is do able
0.18 neutrons/pulse <= good for stopping pulse

Counts Rate for U238 (1/2 mil of Al converter)

radiation length

r.l.(Al) = 8.89 cm
  
radiator thickness = 12.5 [math]\mu m[/math]

[math]\frac{12.5\ \mu m}{8.89\ cm} = 1.41 \cdot 10^{-4} \ r.l.[/math]

Calibration factor

The only difference from calculations above is:

1) radiation length: 

    1.41 (1/2 mil Al) / 3.48 (1/2 mil Ti) = 0.40

Yield (1/2 mil of Al and without detector efficiency)

53.8 neutrons/sec * 0.40 = 21.5 neutrons/sec
0.18 neutrons/pulse * 0.40 = 0.07 neutrons/pulse

Counts Rate for Deuteron (1/2 mil of Ti converter)

photonuclear cross section for [math] ^2H(\gamma , n) [/math] reaction

A. De Graeva et all., Phys. Rev. C45, 860 (1992):

Photonuc sigma deuteron.png

in (10,20) MeV region the average cross section, say, is: 

    1000 μb

target thickness, [math] D_2O [/math]

take [math]D_2O[/math], liquid (20°C): 

[math] \frac{1.1056\ g/mL}{20.04\ g/mol} = 0.055\ \frac{mol}{cm^3} \times \frac{6.02\cdot 10^{23}\ molecules}{mol} = 0.33\cdot 10^{23}\ \frac{molecules}{cm^3} [/math]

[math] 0.33\cdot 10^{23}\ \frac{molecules}{cm^3} \times \frac{2\ deuterons}{molecule} = 0.66\cdot 10^{23}\ \frac{deuterons}{cm^3} [/math]

Let's target thickness = 10 cm: 

[math]0.66\cdot 10^{23}\ \frac{atoms}{cm^3} \times 10\ cm = 66\cdot 10^{22}\ \frac{atoms}{cm^2}[/math]


angular distribution of neutron

P. Rossi et all., Phys. Rev. C40, 2412 (1989):

Sigma deuteron 20 40MeV.png

Sigma deuteron total.png


relativistic kinematics

An Introduction to Nuclear and Subnuclear Physics. Emilio Segre (1964) 

  [math] \tan(\Theta_i)  =   \frac{1}{\overline{\gamma}} \frac{\sin\Theta_i^*}{\overline{\beta} (E_i^*/p_i^*) + \cos\Theta_i^*}[/math]

where 

  asterisks are quantities referred to CM

  barred quantities refer to the velocity of the CM



  [math] E^* = \left[(m_1+m_2)^2 + 2T_1m_2\right]^{1/2}[/math]

  [math] \overline{\gamma} = \frac{E}{E^*} = \frac{m_1 + m_2 + T_1}{E^*}[/math]

  [math] \overline{\beta} = \frac{p}{E} = \frac{p_1}{m_1 + m_2 + T_1}[/math]



  [math] E_3^* = \frac{E^{*2} + m_3^2 - m_4^2}{2E*}[/math]

  [math] E_4^* = \frac{E^{*2} + m_4^2 - m_3^2}{2E*}[/math]

  [math] |p_3^*| = |p_4^*| = \left( E_3^{*2} - m_3^2 \right)^{1/2} = \left( E_4^{*2} - m_4^2 \right)^{1/2}[/math]

calculations

[math]T_{\gamma}[/math] [math]\Theta_{LAB}[/math] [math]\Theta_{CM}[/math] [math]\sigma_{T}[/math] [math]d \sigma / d \Omega\left(\Theta_{CM}\right)[/math] [math]\Omega_{Det}=\frac{A}{r^2}[/math] [math]\frac{d \sigma / d \Omega \times \Omega_{Det}}{\sigma_{T}}[/math]
20 MeV [math]90^o[/math] [math]94.38^o[/math] [math]600\ \mu b[/math] [math]63\ \mu b/sr[/math] [math]40\cdot 10^{-4}\ sr[/math] [math]4.2\cdot 10^{-4}[/math]
40 MeV [math]90^o[/math] [math]96.06^o[/math] [math]350\ \mu b[/math] [math]23\ \mu b/sr[/math] [math]40\cdot 10^{-4}\ sr[/math] [math]2.6\cdot 10^{-4}[/math]

geometrical factor

taking average for 20 and 40 MeV photons 

 geometrical acceptance = [math]\frac{(4.2\cdot 10^{-4} + 2.6\cdot 10^{-4})}{2} = 3.4\cdot 10^{-4}[/math]

Calibration factor

The only differences from calculations above are:

1) cross section correction: 

    1000 μb (D) / 130 mb (238U) = 1/130

2) target thickness correction: 

   [math] \frac{66\cdot 10^{22}\ atoms/cm^2\ (D)}{0.48\cdot 10^{23}\ atoms/cm^2\ (^{238}U)} = 66/0.48 [/math]

3) neutrons per reaction correction: 

   1 neutron (D) / 2.4 neutrons(238U) = 1/2.4

4) geometrical factor correction: 

   [math] \frac{3.4\cdot 10^{-4}\ (D)}{3.2\cdot 10^{-4}\ (^{238}U)} = 1.06 [/math]

total calibration factor is: 

  [math]\frac{1}{130} \times \frac{66}{0.48} \times \frac{1}{2.4} \times \frac{3.4}{3.2} = 0.468[/math]

Yield (1/2 mil of Ti and without detector efficiency)

saying all other factors is the same =>

the yield per second : 

[math] 53.8\ \frac{neutrons}{sec} \times 0.468 = 25.2\ \frac{neutrons}{sec} [/math]

the yield per pulse: 

[math] 25.2\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 0.08\ \frac{neutrons}{pulse} [/math]

Summary (counts rate without neutron efficiency for different radiator thickness

converter target neutrons/sec neutrons/pulse
1/2 mil Ti [math]^{238}U[/math] 53.8 0.18
1/2 mil Al [math]^{238}U[/math] 21.5 0.07
1/2 mil Ti [math]D_2O[/math] 25.2 0.08
1/2 mil Al [math]D_2O[/math] 10.1 0.03


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