516 9 Particle Physics – I
Eπ=(
P^2 +mπ^2) 1 / 2
=(1. 192 + 0. 142 )^1 /^2 = 1 .198 GeV
βπ=P/Eπ= 1. 19 / 1. 198 = 0. 9933
tπ=d/βπc= 12 /(0. 9933 × 3 × 108 )= 40. 26 × 10 −^9 s
= 40 .26 ns9.51 Number of photons emitted due to absorption of 1 MeV electron is
15 × 106 / 103 = 15 ,000.
Number of photo – electrons emitted= 15 , 000 / 10 = 1 , 500
The electron multiplication factorM = 310 because the photomultiplier
tube has 10 dynodes and each dynode produces 3 secondary electrons. The
charge collected at the output is
q= 1 , 500 × 310 × 1. 6 × 10 −^19 = 1. 417 × 10 −^11 Coulomb
The pulse height will be
V=q/C= 1. 417 × 10 −^11 / 1. 2 × 10 −^10 = 0 .118 V
9.52 The number of electrons liberated by the phosphor when 5 MeV proton is
stopped
n= 5 × 106 /^100 =^5 ×^104
Allowing for light collection efficiency (η) and conversion efficiency (ε)of
the photocathode, number of electrons released from the cathode
N=nηε= 5 × 104 ×(60/100)(5/100)= 1 , 500
After going through 10 stages the number of electrons reaching the anode
becomes with a gain (G) of 3/stage
NG= 1 , 500 × 310 = 8. 85 × 107
The charge collected at the anode
q=NGe= 8. 85 × 107 × 1. 6 × 10 −^19 = 14. 16 × 10 −^12 Coulomb
The voltage developed
V=q/C= 14. 16 × 10 −^12 / 12 × 10 −^12 = 1 .18 V.
9.53τ=(17+ 0 .5k)μs
For channel 100,τ= 17 + 0. 5 × 100 = 67 μs
True counting rateN=N 0 /(1− 67 × 10 −^6 f)
N 0 /N= 90 / 100 = 1 − 67 × 10 −^6 f
Orf= 1 ,490 s−^1
Similarly for channel 400,f=460 s−^1
9.54 Let the gain/stage beG, so that the net gain due to electron multiplication will
beG^12 (because there are 12 stages).
Number of photons producing electrons from the cathode for each beta particle
absorbed, with 5% photo-cathode efficiency= 550 × 5 / 100 = 27. 5
Number of electrons reaching the anode= 27. 5 ×G^12
Charge collected,q= 27. 5 ×G^12 × 1. 6 × 10 −^19
The voltage developed,V=q/C= 27 ×G^12 × 1. 6 × 10 −^19 / 45 × 10 −^12 =
8 × 10 −^3
Solving forG, we get the gain/stage,G= 2. 567