Difference between revisions of "Counts Rate (44 MeV LINAC)"
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− | [ | + | [https://wiki.iac.isu.edu/index.php/Roman_calculation Go Back] |
− | + | =LINAC parameters used in calculations= | |
− | + | 1) pulse width 50 ps <br> | |
− | |||
− | 1) pulse width 50 | ||
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 14: | Line 15: | ||
− | ==Number of photons/sec | + | ==Number of photons/sec on target== |
===bremsstrahlung=== | ===bremsstrahlung=== | ||
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 | + | <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)== | |
− | == | ||
Collimation factor is | Collimation factor is | ||
− | ''' | + | '''6.85 % of total # of photons''' |
then, incident flux on target is | then, incident flux on target is | ||
− | <math>1.64 \cdot 10^{9} \frac{\gamma}{sec} \cdot | + | <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]] | [[File:phofission_sigma_U238.png]] | ||
Line 61: | 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} = | + | <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 1\ cm = 0.48\cdot 10^{ | + | <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 73: | Line 72: | ||
===steps together...yeild=== | ===steps together...yeild=== | ||
− | <math> Y = \frac{\gamma}{sec} \times t \times \sigma \times 2.4 = </math | + | <math> Y = \frac{\gamma}{sec} \times t \times \sigma \times 2.4 = </math> |
− | <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== | ==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]] | [[File:photonuc_sigma_deuteron.png]] | ||
Line 104: | Line 155: | ||
in (10,20) MeV region the average cross section, say, is: | in (10,20) MeV region the average cross section, say, is: | ||
− | '''1000 mb''' | + | '''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 | ||
− | + | |} | |
− | |||
− | |||
[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
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
Number of photons/sec on target
bremsstrahlung
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
steps together...
Alex factor (GEANT4 calculation)
Collimation factor is
6.85 % of total # of photons
then, incident flux on target is
Number of neutrons/sec
photonuclear cross section for reaction
J. T. Caldwell et all., Phys. Rev. C21, 1215 (1980):
in (10,20) MeV region the average cross section, say, is:
130 mb
target thickness,
Let's target thickness = 1 mm:
neutrons per fission
2.4 neutrons/fission
steps together...yeild
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:
take the neutron detector 1 meter away:
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 =
<= geometrical acceptance
Yield (1/2 mil of Ti and without detector efficiency)
the yield per second:
the yield per pulse:
- 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
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 reaction
A. De Graeva et all., Phys. Rev. C45, 860 (1992):
in (10,20) MeV region the average cross section, say, is:
1000 μb
target thickness,
take
, liquid (20°C):
Let's target thickness = 10 cm:
angular distribution of neutron
P. Rossi et all., Phys. Rev. C40, 2412 (1989):
relativistic kinematics
An Introduction to Nuclear and Subnuclear Physics. Emilio Segre (1964)
where
asterisks are quantities referred to CM
barred quantities refer to the velocity of the CM
calculations
20 MeV | ||||||
40 MeV |
geometrical factor
taking average for 20 and 40 MeV photons
geometrical acceptance =
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:
3) neutrons per reaction correction:
1 neutron (D) / 2.4 neutrons(238U) = 1/2.4
4) geometrical factor correction:
total calibration factor is:
Yield (1/2 mil of Ti and without detector efficiency)
saying all other factors is the same =>
the yield per second :
the yield per pulse:
Summary (counts rate without neutron efficiency for different radiator thickness
converter | target | neutrons/sec | neutrons/pulse |
1/2 mil Ti | 53.8 | 0.18 | |
1/2 mil Al | 21.5 | 0.07 | |
1/2 mil Ti | 25.2 | 0.08 | |
1/2 mil Al | 10.1 | 0.03 |