KIN2 q=19 ° .30 deuterium t min =-0.123 GeV 2 q’=2.78 GeV P’ d =.352 Gev Inner calo 10x9 block R=13.5 cm q g’ =8° t min =-0.33 GeV 2 q’=2.72 GeV P’ d =.579 Gev The Problem to detect the maximum of recoil deutons with reasonable silicon det hout interfere with the accepted experiment PYB
The Problem. KIN2 q=19 ° .30 deuterium. t min =-0.123 GeV 2 q’=2.78 GeV P’ d =.352 Gev. Inner calo 10x9 block. R=13.5 cm q g ’ =8°. t min =-0.33 GeV 2 q’=2.72 GeV P’ d =.579 Gev. How to detect the maximum of recoil deutons with reasonable silicon detector - PowerPoint PPT Presentation
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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
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
PYB
2 cm
16x(5.5x50 mm2) strips
32 strips
Silicon strips: 1mm
100 mm
PYB
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 :
PYB
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