Difference between revisions of "Number of electron-positron pairs/sec out of Al converte"

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[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
 
  
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==LINAC parameters used in calculations==
 
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1) pulse width 50 ps <br>
 
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2) pulse current 50 A <br>
 
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3) repetition rate 300 Hz <br>
 
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4) energy 44 MeV <br><br>
 
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==Number of electrons/sec on radiator==
 
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<math> 50\ \frac{Coulomb}{sec} \times \frac{1\cdot e^-}{1.6\cdot 10^{-19}C} \times 50ps \times  300Hz = 4.68 \cdot 10^{12} \frac{e^-}{sec}</math><br><br>
 
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==Number of photons/sec out of radiator==
 
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<math>\sigma_{brems}=0.1 photons/electrons/MeV/r.l</math>
 
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r.l.(Ti) = 3.59 cm
 
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radiator thickness = 12.5 <math>\mu m</math>
 
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<math>\frac{12.5\ \mu m}{3.59\ cm} = 3.48 \cdot 10^{-4} \ r.l.</math><br>
 
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<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}=1.63 \cdot 10^{9} \frac{\gamma}{sec}</math><br><br>
 
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Alex factor is 6.85 %
 
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<math>1.64 \cdot 10^{9} \frac{\gamma}{sec} \cdot 6.85\ % = 1.12 \cdot 10^{8} \frac{\gamma}{sec}</math><br><br>
 
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==Number of electron/positron pairs/sec out of Al converter==
 
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==Worst Case Isotropic Neutrons==
 
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===checking detector distance===
 
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we want:
 
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      the time of flight of neutron >> the pulse width
 
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take the worst case 10 MeV neutron:
 
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<math> E_{tot} = E_{kin} + E_{rest} =  10\ MeV + 938\ MeV = 948\ MeV </math>
 
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<math> \gamma = \frac{E_{tot}}{m_p} = \frac{948\ MeV}{938\ MeV} = 1.0107 </math>
 
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<math> \gamma^2 = \frac{1}{1 - \beta^2} \ \ \  \rightarrow \ \ \ \beta = 0.145\ c</math>
 
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take the neutron detector 1 meter away:
 
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<math> t = \frac{1\ m}{0.145\ c} = \frac{1\ m}{0.145\cdot 3\cdot 10^8\ m/sec} = 23\ ns </math>
 
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      23 ns >> 50 ps  <= time resolution is good
 
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===geometrical factor===
 
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taking real detector 3" x 2"  =>  S is about 40 cm^2
 
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1 meter away
 
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fractional solid angle = <math>\frac{40\ cm^{2}}{4 \pi\ (100\ cm)^{2}} = 3.2 \cdot 10^{-4}</math> <= geometrical acceptance
 
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==Yield==
 
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the yield per second:
 
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<math>1.68 \cdot 10^{5}\ \frac{neutrons}{sec} \times 3.2 \cdot 10^{-4} = 53.8\ \frac{neutrons}{sec} </math><br>
 
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the yield per pulse:
 
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<math> 53.8\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 0.18\ \frac{neutrons}{pulse} </math> <br><br>
 
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  53.8 neutrons/sec  <= this experiment is do able
 
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  0.18 neutrons/pulse <= good for stopping pulse
 
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=Counts Rate for U238 (25 µm Al converter)=
 
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==radiation length==
 
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r.l.(Al) = 8.89 cm
 
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radiator thickness = 25 <math>\mu m</math>
 
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<math>\frac{25\ \mu m}{8.89\ cm} = 2.81 \cdot 10^{-4} \ r.l.</math><br>
 
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==Calibration factor==
 
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The only difference from calculations above is:
 
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1) radiation length:
 
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    3.48 (for 12.5 µm Ti) / 2.81 (for 50 µm Al) = 1.24
 
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==Yield==
 
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  53.8 neutrons/sec * 1.24 = 66.7 neutrons/sec
 
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  0.18 neutrons/pulse * 1.24 = 0.22 neutrons/pulse
 
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=Counts Rate for Deuteron (12.5 µm Ti converter)=
 
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===photonuclear cross section for <math> ^2H(\gamma , n) </math> reaction===
 
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A. De Graeva ''et all.,'' Phys. Rev. '''C45''', 860 (1992):
 
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[[File:photonuc_sigma_deuteron.png]]
 
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in (10,20) MeV region the average cross section, say, is:
 
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    '''1000 μb'''
 
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===target thickness, <math> D_2O </math>===
 
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take <math>D_2O</math>, liquid (20°C):
 
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<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>
 
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<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>
 
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Let's target thickness = 10 cm:
 
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<math>0.66\cdot 10^{23}\ \frac{atoms}{cm^3} \times 10\ cm = 66\cdot 10^{22}\ \frac{atoms}{cm^2}</math>
 
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===angular distribution of neutron===
 
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====P. Rossi ''et all.,'' Phys. Rev. '''C40''', 2412 (1989):====
 
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[[File:sigma_deuteron_20_40MeV.png|600 px]]
 
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[[File:sigma_deuteron_total.png|300 px]]
 
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====relativistic kinematics====
 
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An Introduction to Nuclear and Subnuclear Physics. Emilio Segre (1964)
 
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  <math> \tan(\Theta_i)  =  \frac{1}{\overline{\gamma}} \frac{\sin\Theta_i^*}{\overline{\beta} (E_i^*/p_i^*) + \cos\Theta_i^*}</math>
 
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where
 
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  asterisks are quantities referred to CM<br>
 
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  barred quantities refer to the velocity of the CM
 
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  <math> E^* = \left[(m_1+m_2)^2 + 2T_1m_2\right]^{1/2}</math><br>
 
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  <math> \overline{\gamma} = \frac{E}{E^*} = \frac{m_1 + m_2 + T_1}{E^*}</math><br>
 
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  <math> \overline{\beta} = \frac{p}{E} = \frac{p_1}{m_1 + m_2 + T_1}</math>
 
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  <math> E_3^* = \frac{E^{*2} + m_3^2 - m_4^2}{2E*}</math><br>
 
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  <math> E_4^* = \frac{E^{*2} + m_4^2 - m_3^2}{2E*}</math><br>
 
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  <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>
 
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====calculations====
 
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{| border="1" cellpadding="20" cellspacing="0"
 
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|-
 
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|<math>T_{\gamma}</math>
 
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||<math>\Theta_{LAB}</math>
 
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||<math>\Theta_{CM}</math>
 
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||<math>\sigma_{T}</math>
 
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||<math>d \sigma / d \Omega\left(\Theta_{CM}\right)</math>
 
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||<math>\Omega_{Det}=\frac{A}{r^2}</math>
 
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||<math>\frac{d \sigma / d \Omega \times \Omega_{Det}}{\sigma_{T}}</math>
 
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|-
 
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|20 MeV || <math>90^o</math> || <math>94.38^o</math> || <math>600\ \mu b</math>
 
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|| <math>63\ \mu b/sr</math> || <math>40\cdot 10^{-4}\ sr</math> || <math>4.2\cdot 10^{-4}</math> 
 
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|-
 
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|40 MeV || <math>90^o</math> || <math>96.06^o</math> || <math>350\ \mu b</math>
 
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|| <math>23\ \mu b/sr</math> || <math>40\cdot 10^{-4}\ sr</math> || <math>2.6\cdot 10^{-4}</math> 
 
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|-
 
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|}
 
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====geometrical factor====
 
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taking average for 20 and 40 MeV photons
 
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  geometrical acceptance = <math>\frac{(4.2\cdot 10^{-4} + 2.6\cdot 10^{-4})}{2} = 3.4\cdot 10^{-4}</math>
 
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===Calibration factor===
 
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The only differences from calculations above are:
 
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1) cross section correction:
 
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    1000 μb (D) / 130 mb (238U) = 1/130
 
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2) target thickness correction:
 
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    <math> \frac{66\cdot 10^{22}\ atoms/cm^2\ (D)}{0.48\cdot 10^{23}\ atoms/cm^2\ (^{238}U)} = 66/0.48 </math>
 
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3) neutrons per reaction correction:
 
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    1 neutron (D) / 2.4 neutrons(238U) = 1/2.4
 
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4) geometrical factor correction:
 
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    <math> \frac{3.4\cdot 10^{-4}\ (D)}{3.2\cdot 10^{-4}\ (^{238}U)} = 1.06 </math>
 
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'''total calibration factor is:'''
 
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  <math>\frac{1}{130} \times \frac{66}{0.48} \times \frac{1}{2.4} \times \frac{3.4}{3.2} = 0.468</math>
 
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===Yield===
 
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saying all other factors is the same =>
 
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the yield per second :
 
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<math> 53.8\ \frac{neutrons}{sec} \times 0.468 = 25.2\ \frac{neutrons}{sec} </math><br>
 
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the yield per pulse:
 
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<math> 23.7\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 0.08\ \frac{neutrons}{pulse} </math><br><br>
 
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=Summary=
 
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{| border="1" cellpadding="20" cellspacing="0"
 
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||'''converter'''
 
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||'''target'''
 
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||'''neutrons/sec'''
 
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||'''neutrons/pulse'''
 
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||0.5 mil Ti||<math>^{238}U</math>||53.8||0.18
 
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||1.0 mil Al||<math>^{238}U</math>||66.7||0.22
 
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||0.5 mil Ti||<math>D_2O</math>  ||25.2||0.08
 
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|}
 
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[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
 

Latest revision as of 23:31, 29 January 2011

Retrieved from "https://wiki.iac.isu.edu/index.php?title=Number_of_electron-positron_pairs/sec_out_of_Al_converte&oldid=56088"