Counts Rate (44 MeV LINAC)

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Counts Rate for U238

LINAC parameters used in calculations

1) pulse width 50 ns
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 from radiator

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]

Collimation factor

Collimation factor is

    4-6 % of total # of photons (Alex, GEANT calculation)

then, incident flux on target is

[math]1.64 \cdot 10^{9} \frac{\gamma}{sec} \cdot 5% = 8.2 \cdot 10^{7} \frac{\gamma}{sec}[/math]

Number of neutrons/sec (yields)

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

in (10,20) MeV region the average cross section 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} = 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]

Target thickness = 1 cm:

[math]0.48\cdot 10^{23}\ \frac{atoms}{cm^3} \times 1\ cm = 0.48\cdot 10^{23}\ \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] = 8.2 \cdot 10^{7} \frac{\gamma}{sec} \times 130\ mb \times 0.48\cdot 10^{23}\ \frac{atoms}{cm^2} \times 2.4 = 1.2 \cdot 10^{6}\ \frac{neutrons}{sec}[/math]

Worst Case Isotropic Neutrons

Let's say we have:

radius detector = 1 cm

1 meter away

fractional solid angle = [math]\frac{\pi * (1 cm)^{2}}{4 \pi (100cm)^{2}} = \frac{1}{4} \cdot 10^{-4}[/math] <= geometrical acceptance

finally we have

[math]1.2 \cdot 10^{6}\ \frac{neutrons}{sec} \times \frac{1}{4} \cdot 10^{-4} = 30\ \frac{neutrons}{sec} [/math]

Therefore, this experiment is really doable.

Counts Rate for Deuteron

photonuclear cross section for [math]^2H(\gamma , n)[/math] reaction=

in (10,20) MeV region the average cross section, say, is:

    1000 mb

calibration factor

The only difference from calculations above is cross sections

yield

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