Counts Rate (44 MeV LINAC)

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Counts Rate for U238 (0.5 mil Ti converter)

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] 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

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]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\ % = 11.2 \cdot 10^{7} \frac{\gamma}{sec}[/math]

Number of neutrons/sec

photonuclear cross section for [math]^{238}U(\gamma , n)[/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]^{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] = 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] 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

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 Deuteron (0.5 mil Ti converter)

photonuclear cross section for [math] ^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] 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]

Calibration factor

The only difference from calculations above is:

1) cross section:

    1000 μb (D) / 130 mb (238U) = 1/130

2) target thickness:

   [math] \frac{66\cdot 10^{22}\ atoms/cm^2\ (D)}{0.48\cdot 10^{23}\ atoms/cm^2\ (238U)} = 66/0.48 [/math]

3) neutrons per reaction:

   1 neutron (D) / 2.4 neutrons(238U) = 1/2.4

4) geometrical factor:

   assume for now is 1

total calibration factor is:

  [math]\frac{1}{130} \times \frac{66}{0.48} \times \frac{1}{2.4} \times 1= 0.481[/math]

Yield

saying all other factors is the same =>

the yield per second :

[math] 53.8\ \frac{neutrons}{sec} \times 0.441 = 23.7\ \frac{neutrons}{sec} [/math]

the yield per pulse:

[math] 23.7\ \frac{neutrons}{sec} \times \frac{1\ sec}{300\ pulses} = 0.079\ \frac{neutrons}{pulse} [/math]

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