520 9 Particle Physics – I
9.63 Number of photons emitted per unit length
N= 2 πα(1− 1 /β^2 n^2 )(1/λ 1 − 1 /λ 2 )(1)Pions:βπ=p/(p^2 +m^2 )^1 /^2 = 15 /(15^2 + 0. 142 )^1 /^2 = 0. 999956
Kaons:βk= 15 /(15^2 + 0. 4942 )^1 /^2 = 0. 999458
If the signal is to be given by pions but not kaons, the condition on the refrac-
tive index is
βπ> 1 /n>βk
0. 999956 > 1 /n> 0. 999458
For a value ofn= 1. 0004 , 1 /n= 0 .9996, the above condition is satisfied.
Inserting
α= 1 / 137 ,βπ= 0. 999956 ,n= 1 .0004,
λ 1 = 4 × 10 −^7 m, andλ 2 = 7 × 10 −^7 m in (1), we findN=35 photons/m.
Therefore to obtain 175 photons a length of 5.0 m is required.9.64 Pions:E=(p^2 +m^2 )^1 /^2 =(900^2 + 1402 )^1 /^2 =911 MeV
βπ=p/Eπ= 900 / 911 = 0. 9879
For threshold,nπ= 1 /β= 1 / 0. 9879 = 1. 012
Protons:E=(900^2 + 9382 )^1 /^2 = 1 ,300 MeV
βp= 900 / 1300 = 0. 6923
For thresholdnp= 1. 444
The material chosen must have the refraction index 1. 012 <n< 1. 444
Forn= 1 .012, Cerenkov light will come off at 0◦with the path. If a higher
index of refraction is chosen, light will come off at wider angle. Thenmust
be less than 1.444, otherwise protons will be counted.
9.3.8 SolidStateDetector................................
9.65 IfAis the area,dthe thickness of depletion layer andKthe dielectric constant
then the capacitance is
C=εAK/d= 8. 8 × 10 −^12 × 1. 5 × 10 −^4 × 10 / 40 × 10 −^6 = 3. 3 × 10 −^10 F
The charge liberated
q= 5 × 106 × 1. 6 × 10 −^19 / 3. 5 = 2. 286 × 10 −^13 Coulomb
Potential developed
V=q/C= 2. 286 × 10 −^13 / 3. 3 × 10 −^10 = 0. 69 × 10 −^3 V
= 0 .69 mV
9.3.9 Emulsions ........................................
9.66 The Range-Energy-Relation must be such that it ensures that the ionization
−dE/dRis a function ofz^2 f(R/M). One such relation isE=Kz2nM^1 −nRn
whereEis in MeV andRin microns,Kandnare empirical constants which