proton detection with the r3b calorimeter two layer solution iem csic sept 2006 report
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CONSEJO SUPERIOR DE INVESTIGACIONES CIENTÍFICAS. MINISTERIO DE EDUCACIÓN Y CIENCIA. Proton detection with the R3B calorimeter, two layer solution IEM-CSIC sept. 2006 report. O. Tengblad, M. Turrión Nieves, C. Pascual Izarra, A. Maira Vidal. outline. Why a two layer solution

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proton detection with the r3b calorimeter two layer solution iem csic sept 2006 report

CONSEJO SUPERIOR

DE INVESTIGACIONES

CIENTÍFICAS

MINISTERIO

DE EDUCACIÓN

Y CIENCIA

Proton detection with the R3B calorimeter,

two layer solution

IEM-CSIC sept. 2006 report

O. Tengblad, M. Turrión Nieves, C. Pascual Izarra, A. Maira Vidal

outline
outline
  • Why a two layer solution
  • Limitations - requirements
  • “Conclusion”
energy loss of charged particles bethe bloch equation
Energy loss of charged particles: Bethe-Bloch equation

energy loss detected

incident energy (MeV)

proposed scenario
Proposed scenario
  • Two layers detector:
  • Simplification

Ë=f(D E1 )+ g(D E2)

E

D E1

D E2

the estimated final energy is proportional to the energy deposited in each layer

srim simulations deposited energy of protons
SRIM Simulations: Deposited energy of protons

Fit: Gaussian with a constant background

srim simulations protons
SRIM Simulations: protons

DE1+s(DE1) DE2+s(DE2)

  • Material: LaBr3(:Ce)
  • Thickness: 1mm+20mm
  • Monte Carlo: SRIM 2003

E

D E2

D E1

energy resolution
Energy resolution

E DE1+D(sE1) DE2+s(DE2)

sE?

protons of 200MeV

deposit an energy of:

1mm LaBr3= 1.49±0.23 MeV

20mm LaBr3= 31.32±1.13 MeV

200±50MeV

(sE/E=25%)

200±10MeV

(sE/E=5%)

first conclusions
First Conclusions
  • If not fully stopped, two DE-detectors are required
  • A too thin detector gives bad estimation of the energy leading to bad resolution first detector should be thick in order to totally absorb protons up to rather high energy
  • Second detector placed to solve the ambiguity on the signal
  • The gammas will deposit most of the energy aroundthe first hit, which we want to be the first detector, why this crystal should have a good Eg resolution.
  • Two detectors of different materials with a unique PM or APD? Optically compatible
detector spectral response matching

Emission

Absorption

Detector spectral response matching
  • Emission and absorption spectra do not overlap » emitted light is not re-absorbed
  • Emission spectra shifted to lower energies
  • LYSO:
  • lexcitation [nm] =262, 293, 357
  • Max. lemission [nm] =398, 435

Hautefeuille et al. J. of Crystal Growth (in press)

emission spectra
Emission spectra

NaI(Tl)

CsI(Tl)

BGO

Max. lemission [nm] Decay time[ns]

CsI(Tl) 550 1000

BGO 478 300

CsI pure 315 16

LYSO (Ce) 420 45-60

CsI(Na) 420 630

NaI(Tl) 400 230

LaBr3 (Ce) 380 16

LaCl3 (Ce) 350 28

LYSO

LaBr3

srim simulations protons1
SRIM simulations: protons
  • Materials: LYSO(:Ce) + LaBr3(:Ce)
  • Thickness: 30mm + 20mm
  • Monte Carlo: SRIM 2003

D E1

D E2

E

energy resolution1
Energy resolution

DE1+D(sE1) + DE2+s(DE2) E + sE?

protons of 200MeV

deposit an energy of:

30mm LYSO= 67.44±1.77 MeV

20mm LaBr3= 43.50±3.11MeV

200±7MeV

(sE/E=3.5%)

200±10MeV

(sE/E=5%)

gamma absorption
Gamma absorption
  • Minimum absorption for g ~5MeV
  • 55% of g absorbed in 30mm LYSO (Prelude)
second conclusion
Second Conclusion
  • Protons
    • Two detectors are required to detect 300 MeV protons
    • The energy of the incident protons can be estimated with an error of ~3-4% with the LYSO + LaBr3(Ce) solution
  • Gammas
    • Most of the energy of the gammas is deposited around the first hit, why this should happen in the first layer!
    • 55% of g are absorbed in 30mm LYSO when the energy of the incident gammas is 5MeV
    • >55% of g are absorbed for E≠5MeV in 30mm LYSO
    • The rest will be absorbed in the second layer
    • If first gamma detected in second layer; event discarded
    • However, the gamma resolution in LYSO is about 6%
    • If this g-resolution is good enough one would choose LSO + LYSO as the resolution of LaBr is too good to be place as second layer.
final conclusion
Final Conclusion
  • To obtain the optimum situation both for protons and gammas;
    • First crystal layer relatively thick and of a material with excellent gamma resolution,

 LaBr3(Ce) of 30 mm l= 380nm decaytime= 16ns

    • Second crystal layer of a material emitting at shorter wavelength and with a decay constant different in order to separate the signals and that the second detector is transparent to the first.

 LaCl3(Ce) of 150 mml= 350nm decaytime= 25ns

This will detect protons up to 280 MeV with an

proton energy resolution of the order of 2%.

One could, however, live with a much shorter LaCl3(Ce) or one could combine the LaBr3(Ce) with pure CsI as a cheaper solution.

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