View source for Counts Rate (44 MeV LINAC)

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=LINAC parameters used in calculations=
1) pulse width 50 ps <br>
2) pulse current 50 A <br>
3) repetition rate 300 Hz <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===

====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===

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

|}



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