Counts Rate (44 MeV LINAC)

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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] 50ps \times 50A \times 300Hz \times \frac{1\cdot e^-}{1.6\cdot 10^{-19}C} = 0.47 \cdot 10^{13} \frac{e^-}{sec}[/math]

Number of photons/sec from radiator

bremsstrahlung

plot from Dale

in (10,20) MeV region we have about

    0.1 photons/electrons/MeV/r.l

radiation length

Titanium r.l. is 3.59 cm

take radiator thickness is [math]12.5 \mu m[/math]

[math]3.59 cm/12.5\mu m = 3.48 \cdot 10^{-4} \ r.l.[/math]

steps together...

[math]0.1 \frac{\gamma 's}{(e^-\ MeV\ 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

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

So, 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)

cross section

plot from Berman

From plot above the average cross section for [math]^{238}U[/math] in (10,20) MeV region 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]

Assume the target thickness is 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...

[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

[math]1.2 \cdot 10^{6}\ \frac{neutrons}{sec} \times \frac{1}{4} \cdot 10^{-4} = 300 \frac{neutrons}{sec} = [/math]

Therefore, this experiment is do able.