Number of electron-positron pairs/sec out of Al converte
LINAC parameters used in calculations
1) pulse width 50 ps
2) pulse current 50 A
3) repetition rate 300 Hz
4) energy 44 MeV
Number of electrons/sec on radiator
[math]\displaystyle{ 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 out of radiator
r.l.(Ti) = 3.59 cm
radiator thickness = 12.5 [math]\displaystyle{ \mu m }[/math]
[math]\displaystyle{ \frac{12.5\ \mu m}{3.59\ cm} = 3.48 \cdot 10^{-4} \ r.l. }[/math]
steps together...
[math]\displaystyle{ 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)
Alex factor is 6.85 %
[math]\displaystyle{ 1.64 \cdot 10^{9} \frac{\gamma}{sec} \cdot 6.85\ % = 11.2 \cdot 10^{7} \frac{\gamma}{sec} }[/math]
Number of neutrons/sec
photonuclear cross section for [math]\displaystyle{ ^{238}U(\gamma , F) }[/math] 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, [math]\displaystyle{ ^{238}U }[/math]
[math]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ Y = \frac{\gamma}{sec} \times t \times \sigma \times 2.4 = }[/math]
[math]\displaystyle{ = 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]
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]\displaystyle{ E_{tot} = E_{kin} + E_{rest} = 10\ MeV + 938\ MeV = 948\ MeV }[/math]
[math]\displaystyle{ \gamma = \frac{E_{tot}}{m_p} = \frac{948\ MeV}{938\ MeV} = 1.0107 }[/math]
[math]\displaystyle{ \gamma^2 = \frac{1}{1 - \beta^2} \ \ \ \rightarrow \ \ \ \beta = 0.145\ c }[/math]
take the neutron detector 1 meter away:
[math]\displaystyle{ 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]\displaystyle{ \frac{40\ cm^{2}}{4 \pi\ (100\ cm)^{2}} = 3.2 \cdot 10^{-4} }[/math] <= geometrical acceptance
Yield
the yield per second:
[math]\displaystyle{ 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]\displaystyle{ 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 (25 µm Al converter)
radiation length
r.l.(Al) = 8.89 cm
radiator thickness = 25 [math]\displaystyle{ \mu m }[/math]
[math]\displaystyle{ \frac{25\ \mu m}{8.89\ cm} = 2.81 \cdot 10^{-4} \ r.l. }[/math]
Calibration factor
The only difference from calculations above is:
1) radiation length:
3.48 (for 12.5 µm Ti) / 2.81 (for 50 µm Al) = 1.24
Yield
53.8 neutrons/sec * 1.24 = 66.7 neutrons/sec
0.18 neutrons/pulse * 1.24 = 0.22 neutrons/pulse
Counts Rate for Deuteron (12.5 µm Ti converter)
photonuclear cross section for [math]\displaystyle{ ^2H(\gamma , n) }[/math] 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, [math]\displaystyle{ D_2O }[/math]
take [math]\displaystyle{ D_2O }[/math], liquid (20°C):
[math]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ 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):
relativistic kinematics
An Introduction to Nuclear and Subnuclear Physics. Emilio Segre (1964)
[math]\displaystyle{ \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]\displaystyle{ E^* = \left[(m_1+m_2)^2 + 2T_1m_2\right]^{1/2} }[/math]
[math]\displaystyle{ \overline{\gamma} = \frac{E}{E^*} = \frac{m_1 + m_2 + T_1}{E^*} }[/math]
[math]\displaystyle{ \overline{\beta} = \frac{p}{E} = \frac{p_1}{m_1 + m_2 + T_1} }[/math]
[math]\displaystyle{ E_3^* = \frac{E^{*2} + m_3^2 - m_4^2}{2E*} }[/math]
[math]\displaystyle{ E_4^* = \frac{E^{*2} + m_4^2 - m_3^2}{2E*} }[/math]
[math]\displaystyle{ |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]\displaystyle{ T_{\gamma} }[/math] | [math]\displaystyle{ \Theta_{LAB} }[/math] | [math]\displaystyle{ \Theta_{CM} }[/math] | [math]\displaystyle{ \sigma_{T} }[/math] | [math]\displaystyle{ d \sigma / d \Omega\left(\Theta_{CM}\right) }[/math] | [math]\displaystyle{ \Omega_{Det}=\frac{A}{r^2} }[/math] | [math]\displaystyle{ \frac{d \sigma / d \Omega \times \Omega_{Det}}{\sigma_{T}} }[/math] |
| 20 MeV | [math]\displaystyle{ 90^o }[/math] | [math]\displaystyle{ 94.38^o }[/math] | [math]\displaystyle{ 600\ \mu b }[/math] | [math]\displaystyle{ 63\ \mu b/sr }[/math] | [math]\displaystyle{ 40\cdot 10^{-4}\ sr }[/math] | [math]\displaystyle{ 4.2\cdot 10^{-4} }[/math] |
| 40 MeV | [math]\displaystyle{ 90^o }[/math] | [math]\displaystyle{ 96.06^o }[/math] | [math]\displaystyle{ 350\ \mu b }[/math] | [math]\displaystyle{ 23\ \mu b/sr }[/math] | [math]\displaystyle{ 40\cdot 10^{-4}\ sr }[/math] | [math]\displaystyle{ 2.6\cdot 10^{-4} }[/math] |
geometrical factor
taking average for 20 and 40 MeV photons
geometrical acceptance = [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \frac{3.4\cdot 10^{-4}\ (D)}{3.2\cdot 10^{-4}\ (^{238}U)} = 1.06 }[/math]
total calibration factor is:
[math]\displaystyle{ \frac{1}{130} \times \frac{66}{0.48} \times \frac{1}{2.4} \times \frac{3.4}{3.2} = 0.468 }[/math]
Yield
saying all other factors is the same =>
the yield per second :
[math]\displaystyle{ 53.8\ \frac{neutrons}{sec} \times 0.468 = 25.2\ \frac{neutrons}{sec} }[/math]
the yield per pulse:
[math]\displaystyle{ 23.7\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 0.08\ \frac{neutrons}{pulse} }[/math]
Summary
| converter | target | neutrons/sec | neutrons/pulse |
| 0.5 mil Ti | [math]\displaystyle{ ^{238}U }[/math] | 53.8 | 0.18 |
| 1.0 mil Al | [math]\displaystyle{ ^{238}U }[/math] | 66.7 | 0.22 |
| 0.5 mil Ti | [math]\displaystyle{ D_2O }[/math] | 25.2 | 0.08 |