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Re: [GlueX] Radiation lengths in Hall B beam line



Title:
Thanks Elton. It will be considered in the simulations.
Cheers, Rafael.

Elton Smith wrote:
Hi All,

The one item missing is the veto counter (5 mm thick) in front of the
calorimeter. This would be a 0.5/42cm = 0.012 rad lengths. However, we
should read this out into an ADC for use as a software veto.

Cheers, Elton.




Elton Smith
Jefferson Lab
elton@jlab.org
(757) 269-7625

On Tue, 1 Aug 2006, Rafael Hakobyan wrote:

  
Dear Daniel,

Thanks a lot for your detailed estimate. As I mentioned this morning,
the simulation showed such pair
production that hits the BCAL, and the rate of those cases is very
small. Of course, in addition to helium bags,
hydrogen target and two Aluminium foils (71 micrometer), in my
simulation, I considered the existens of BPM (scintillator of 4mm
thickness, I took the one that has been used in PRIMEX) and aluminized
mylar windows as well, but of course the latters (BPM and Al-mylar) will
not change the result of your estimate, i.e. will not be larger than 2%,
because your estimate showed that the fraction of photons producing
pairs is well under 2%.
Thanks again for your very useful estimate.
With kind regards,
Rafael

Daniel Sober wrote:

    
Here are some numbers for radiation lengths in the beam line:

2 x 71 micrometer Al foils:  142E-6m/0.089m = 0.0016 RL
18 cm H2 target filled with H2 at STP:  18cm/731000cm = 0.00002 RL
(ignore)
25 m He at STP:  (2500cm*0.1786E-3 g/cm^3)/(94.3 g/cm^2) = 0.0047 RL

So, if there is vacuum up to the target and helium afterwards, the
total is 0.0016+0.0047=0.0063 RL
If there is helium both before and after the target, the total is
0.0016+2*0.0047=0.0110 RL

Either way, the fraction of photons producing pairs is well under 2%,
so it should not be a problem.
Other processes will be much less probable; for example the total
hadronic photproduction probability on 25 m of He (0.44 g/cm^2) at the
gamma+p cross section maximum (~ 0.6 mb at 300 MeV) is about
0.6E-27 cm^2/atom * 0.44 g/cm^2 * 6.022E23 atom/mole / (4 g/mole) * 4
nucleons/atom  =   0.00015
(more than an order of magnitude lower than pair production).

Most of  the pairs will result in both e+ and e- hitting the BCAL, so
the energy sum will be the same as for a photon, but with a larger
spatial distribution (the characteristic angle, like that of
bremsstrahlung, is about m/E, which for 200 MeV gives a separation of
about 6 cm at the BCAL if the pair is produced 25 m upstream, i.e. in
the target region.)

Dan
      
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