September 28, 2007 Calorimetry
Location
JLab: Cebaf Center F226
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Agenda
- Presentation of #PE Analysis (Andrei)(Media:bcal_ampl_070928a.pdf)
- Discussion of radiation hardness of lead glass (postpone)
- Update on FCAL standalone MC Simulations (Elizabeth) (time permitting)
- Plans for γp→ηπp toy analysis (Mihajlo/Matt) (time permitting) [8]
Discussion of Npe in BCAL
Using the numbers that we have been working with for some time, which come from your group and George has frequently used for estimates, there are about 8000 photons/MeV in scintillator, 12% scintillator energy fraction, 8% fiber capture fraction in each direction, a 10% quantum efficiency for the tube, and about 40MeV/sector for normal incidence muons (should probably be more like 50MeV, taking into account that the incidence angle in the plane of the block axis and the vertical was not restricted by the cuts) the result is more like 300pe. These inputs are somewhat uncertain, but the number is not going to be off by an order of magnitude even taking into account the attenuation factor. True, the 12% scintillator fraction is measured for shower particles, not muons. These are not different by more than 10%, however.
It assumes that only thing producing deviations between the two ends for a given event is the Poisson statistics of the scintillation light detection. This seems like a reasonable starting assumption, but if I am correct, there is either something seriously wrong with our basic design assumptions listed above, or else something we are not taking into account. To get a number like 22pe/end/sector you need fluctuations that dwarf the expectations based on simple pe statistics. I have some ideas on the latter, but I want to hear if other people agree that there is a problem before launching out on that.
Comments from George
I had made the point during our last meeting at JLab that while the 8000 photons/MeV is the "universally" accepted number of photons created by one MeV of energy deposition in plastic scintillator, this number is the total (integrated) yield over the whole spectrum, say 420-600 nm. However, as we have reported by actual spectroscopic measurements of SciFi's, blue fibers lose essentially all their photons with wavelength below ~ 460 nm and their peak emission - after say one meter - is around 480 nm instead of the source at ~ 425 nm. Therefore, from the 8000 photons/MeV, more than half are lost by absorption and the rest are attenuated by the bulk attenuation of the plastic. My rough spectral integration gives around 40% yield surviving the first few cm's or 3200 photons/Mev but we will get a better number soon since we have the data.
This loss of photons accounts partially for the loss of pe's and it may explain (in principle) the low numbers. Here are some more realistic calculations to argue about:
3200 ph/MeV x 2.0 MeV/cm x 1.9 cm x 0.093 x 0.5 x 0.1 pe/ph x 0.75 = 42 pe's.
The factors are:
1.9 cm = equivalent SciFi thickness in 3.8 cm of read-out cell height
0.093 = fiber capture ratio
0.5 = reduction factor due to bulk attenuation for 2m light propagation
0.1 pe/ph = QE (includes collection efficiency)
0.75 = the transmission efficiency of light guides as it came out of Blake's simulations.
So now we have a difference between 42 and 22 pe's. Some adjustments can be made in the QE because I took the average from Burle curves but collection efficiency was assumed 100% and in reality it's around 85% for an average PMT of the types we used. Also, loses from BCAL to light guide contacts and light guide to PMT further degrade the number of photons reaching the photocathode film.
My conclusion is that the 22 pe's from the Hall B beam tests are lower what we would expect but upon more detailed study the difference is not that much. In fact, some old measurements on plastic scintillators were consistent with 2000 photons/MeV as the effective number of photons reaching the photo-sensors. If this is correct for our SciFi's as well, the difference is essentially eliminated.
The bottom line is that we must extract the actual photon survival yield from the SciFi data we have obtained in Regina. This will determine with great accuracy the photon yield as a fraction of the 8000 ph./MeV that survives the absorption within the first few cm's. All other factors play a role but this number dominates the final number of pe's.
I hope this clarifies somewhat the observations.
