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]
  
 
=LINAC parameters used in calculations=
 
=LINAC parameters used in calculations=
1) pulse width 50 ns <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>
+
4) energy 44 MeV <br><br>
 +
 
 +
 
 +
=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><br><br>
 +
 
 +
 
 +
==Number of photons/sec on target==
 +
 
 +
===bremsstrahlung===
 +
 
 +
[[File: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><br>
 +
 
 +
===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>
 +
 
 +
==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><br><br>
 +
 
 +
==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):
 +
 
 +
[[File: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><br><br>
 +
 
 +
==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><br>
 +
 +
the yield per pulse:
 +
 
 +
<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>
 +
 
 +
<math>\frac{12.5\ \mu m}{8.89\ cm} = 1.41 \cdot 10^{-4} \ r.l.</math><br>
 +
 
 +
==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):
 +
 
 +
[[File: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===
  
=number of electrons/sec on radiator=
+
====P. Rossi ''et all.,'' Phys. Rev. '''C40''', 2412 (1989):====
<math> 50ps \times 50A \times 300Hz \times \frac{1\cdot e^-}{1.6\cdot 10^{-19}C} = 0.47 \cdot 10^{13} \frac{e^-}{sec}</math><br><br>
 
  
=number of photons/sec from radiator=
+
[[File:sigma_deuteron_20_40MeV.png|600 px]]
==Bremsstrahlung==
 
from plot above we have about 0.1 photons/electrons/MeV/r.l from 10 MeV to 20 MeV region
 
==Radiation length==
 
<math>_{22}</math>Ti
 
    --> density is 4.506 g/cm^3 <br>
 
    --> radiation length is 16.17 g/cm^2   
 
Assume the radiator thickness is <math>12.5 /mu m</math><br>
 
  
 +
[[File:sigma_deuteron_total.png|300 px]]
  
==  ==
 
<math>10^{-3} \frac{\gamma 's}{(e^-\ MeV\ r.l.)} \times 2 \cdot 10^{-4} r.l. \times 44 MeV \times 0.375 \cdot 10^{12} \frac{e^-}{sec}=0.3 \cdot 10^{7} \frac{\gamma}{sec}</math><br><br>
 
  
=number of neutrons in 1 second=
+
====relativistic kinematics====
  
=geometrical factor, isotropic case=
+
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===
 +
 
 +
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><br>
 +
 
 +
the yield per pulse:
 +
 
 +
<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]

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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