Proton detection with the R3B calorimeter, two layer solution IEM-CSIC sept. 2006 report

1. 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

2. outline • Why a two layer solution • Limitations - requirements • “Conclusion”

3. Energy loss of charged particles: Bethe-Bloch equation energy loss detected incident energy (MeV)

4. 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

5. SRIM Simulations: Deposited energy of protons Fit: Gaussian with a constant background

6. SRIM Simulations: protons DE1+s(DE1) DE2+s(DE2) • Material: LaBr3(:Ce) • Thickness: 1mm+20mm • Monte Carlo: SRIM 2003 E D E2 D E1

7. 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%)

8. 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

9. 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)

10. 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

11. SRIM simulations: protons • Materials: LYSO(:Ce) + LaBr3(:Ce) • Thickness: 30mm + 20mm • Monte Carlo: SRIM 2003 D E1 D E2 E

12. 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%)

13. Gamma absorption • Minimum absorption for g ~5MeV • 55% of g absorbed in 30mm LYSO (Prelude)

14. 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.

15. 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.