520
9 Particle Physics – I
9.63 Number of photons emitted per unit length
N = 2πα(1 − 1/β
2 n
2 )(1/λ 1 − 1/λ 2 )
( 1 )
Pions: β π = p/( p
2
+ m
2 )
1/2
= 15/(15
2
+ 0.14
2 )
1/2
= 0.999956
Kaons: β k = 15/(15
2
+ 0.494
2 )
1/2
= 0.999458
If the signal is to be given by pions but not kaons, the condition on the refractive index is
β π > 1/n > β k
0.999956 > 1/n > 0.999458
For a value of n = 1.0004, 1/n = 0.9996, the above condition is satisfied.
Inserting
α = 1/137, β π = 0.999956, n = 1.0004,
λ 1 = 4 × 10
−7 m, and λ 2 = 7 × 10
−7 m in (1), we find N = 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
+ 140
2 )
1/2
= 911 MeV
β π = p/E π = 900/911 = 0.9879
For threshold, n π = 1/β = 1/0.9879 = 1.012
Protons: E = (900
2
+ 938
2 )
1/2
= 1,300 MeV
β p = 900/1300 = 0.6923
For threshold n p = 1.444
The material chosen must have the refraction index 1.012 < n < 1.444
For n = 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. The n must
be less than 1.444, otherwise protons will be counted.
9.3.8 Solid State Detector
9.65 If A is the area, d the thickness of depletion layer and K the 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 × 10
6
× 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/dR is a function of z
2 f(R/M). One such relation is E = K z
2n M
1−n R
n
where E is in MeV and R in microns, K and n are empirical constants which
9 Particle Physics – I
9.63 Number of photons emitted per unit length
N = 2πα(1 − 1/β
2 n
2 )(1/λ 1 − 1/λ 2 )
( 1 )
Pions: β π = p/( p
2
+ m
2 )
1/2
= 15/(15
2
+ 0.14
2 )
1/2
= 0.999956
Kaons: β k = 15/(15
2
+ 0.494
2 )
1/2
= 0.999458
If the signal is to be given by pions but not kaons, the condition on the refractive index is
β π > 1/n > β k
0.999956 > 1/n > 0.999458
For a value of n = 1.0004, 1/n = 0.9996, the above condition is satisfied.
Inserting
α = 1/137, β π = 0.999956, n = 1.0004,
λ 1 = 4 × 10
−7 m, and λ 2 = 7 × 10
−7 m in (1), we find N = 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
+ 140
2 )
1/2
= 911 MeV
β π = p/E π = 900/911 = 0.9879
For threshold, n π = 1/β = 1/0.9879 = 1.012
Protons: E = (900
2
+ 938
2 )
1/2
= 1,300 MeV
β p = 900/1300 = 0.6923
For threshold n p = 1.444
The material chosen must have the refraction index 1.012 < n < 1.444
For n = 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. The n must
be less than 1.444, otherwise protons will be counted.
9.3.8 Solid State Detector
9.65 If A is the area, d the thickness of depletion layer and K the 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 × 10
6
× 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/dR is a function of z
2 f(R/M). One such relation is E = K z
2n M
1−n R
n
where E is in MeV and R in microns, K and n are empirical constants which