Question

Q5: Colunlate t aupltd sipnal plse epeted froe NOICT-PMT uder follouinp Thursday, 9 May 2019 6:11 PM aru mstauces A2 Hev loss Ratatan eutgy DE NoI(T deuay tiu NT) Ladpop AE3eV fferiog Sciuhlaon Lpht collection f 0 Pisto cathad Qitum Efoa.y e- 20% fMT gan Anode capoetaut And load firet anode faecy Ery ox

Answer 1

Aniutr! Radiuhom (n(1y loss . AE = 1.2 Mey = 1.210'eV NaI(T) decuy Hme, t = 230A105 bundyap, AEBu 3ey NuI 12. O.12 eflvtien cy, E = 10 v ht collectin Photviuthude Buuntum e fhcincy, En = 20/. - 20 - 0.20 PMT jain, Ga 105 Anude Cuputitun ce Ca leo pf Anode loud Rc (Anoce loud ) x (Anudt Cupuitunee ) Nru 1pF RCa (l05л) x (160 PF ) lo-5 Sec R C NUI (T1) : 130 x 10-sec 4s hayht Pulse t. the muximum Since Rc C No. 5 Photms ,m -^E.EsclinhiuH m E Radrattm cncny loss (1.2у105 ev). (о.л2) (3ev/Photm) L4 ma ч.8х10Ч Photms

Locffcunt ftcny ) X (photr Carhode tt)x Ve= Chaye o elecm I.62x,01C = (4.PX1D")x (0.70)x (0.20)x (0.80) X (lo5) X (1.62x10") 8.11 Xlo"C (3.11 x10c) Muximum pulse heght= (1010 F) C O.811 V