Post on 24-Feb-2016
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KIN2 q=19°.30 deuterium
tmin=-0.123 GeV2 q’=2.78 GeV P’d=.352 Gev
Inner calo 10x9 block R=13.5 cm qg’=8°
tmin=-0.33 GeV2 q’=2.72 GeV P’d=.579 Gev
The Problem
How to detect the maximum of recoil deutons with reasonable silicon detectorWithout interfere with the accepted experiment
PYB
Reasonnable silicona+/- 40° angular range ===> close of the Target as possible
1) Move the beam it is possible let said 1cm 2) Detector close to 1cm from the target
2 cm
Silicon detector dimension +/- 2xtan(40°)/sin(18)=+/-5.42cm
which is limit of reasonable
20°
This insure that all the dvcs events produced in the target and with -tmin <t<0.33GeV2
And with fg<90° or fg>270° are emitted in front of the silicon.We favor the Beth-Heither process
PYB
QgD
QgD
TD MEV
-t GeV2
DQ=k-k’
18o
f
PYB
LD2
SILICON
1Stop in LD2
DE=0
2Stop in Si
DE>> MeV
3Stop after Si
DE<10 MeV
T Mev
DE=0
1
2
3
At angle a fixedcos a
DE=0
1
2
3
At T fixed
a
Note: With a Silicon thickness of 1mm (DE)2~5-10MeV . In the silicon of Compton Polarimeter (DE)3~0.3 MeV
PYB
Conclusion: it is possible to do a full f distribution between -80o to+80o f
for the bin [2.6,3.] GeV2 in –t . in this bin DE is >3MeV 10 times bigger that in the Compton Polarimeter
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2 cm
16x(5.5x50 mm2) strips
32 strips
Silicon strips: 1mm
100 mm
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ARSARS
Charge amplifier. + derivation
Using ARS allow:Times coincidence ~ 1-2 nsDE analyze
Vacuum feed-through
Like this of the polarimeter Compton
Only the charge amplifier require a new study
PYB
•Proton from the electro-desintegration of the deuton
e+De+n+p
Are the two dominant backgrounds
•Moller electrons
The proton field was computed by Pavel :
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15 cm LH2 I=10mA
32 stripes : 5,5 x100 mm2
at 2 cm from the beamPYBBut some stop in the deuterium Tp< 21 MeV
2 cm
Tan(a)=d/l=2/16 a=7.12°
Electron Backgroung :
The forward part of the silicon must have difficulties. If it the case 1,2 or n strips can be forgotten
The best will be to use a shorter target 5 or 4 cm a= 18o
PYB
But that will implied to increase beam intensity to 20 µA and this intensity require the use of the raster !!!!!!
Possible Improvement
20 o
Target f=20 mm
An other setup
18o
X
0D=8cm
But an any case the absorption in the LH2 will conduct at the same range in t and f
My feeling:1) It is technically doable* for KIN22) It will be better to used a 4 or 5 cm target !! 3) The range in t and f is limited . • In this range a cross section can be extract • In this range the B-H is dominant
“Le jeu en vaut il la chandelle ?”Knowing that this technique will be and more difficult to be applied at the 12 GeV when Q2 and t will increase.
• mechanical support of the silicon• vacuum feed –through• position of target versus beam• Preamplifier
*
PYB