9.3 Solutions 519
cosθ= 1 /βn (1)
wherenis the refractive index andβcis the particle velocity. The threshold
corresponds toθ= 00 and is given by
β= 1 /n (2)
Kaon and pion of the same energy would have different velocities, the pion
velocity being higher than the kaon velocity. If the medium be chosen such
thatβk< 1 /nbutβπ> 1 /n, then the pion will be counted by the Cerenkov
counter but not the kaon: In practice there can be several modifications of
this principle as by providing coincidence/anticoincidence with a scintillation
counter, and by restricting the emitted light to a prescribed angular interval.
γπ= 1 +(T/mπ)= 1 +(20/ 0 .14)= 143. 857
βπ=[(
1 −(1/γπ^2)] 1 / 2
= 0. 9999758
cosθ= 1 /βπn= 0. 999624
θ= 1. 57 ◦9.59 cosθ= 1 /βn
β= 1 /ncosθ= 1 /(1. 5 ×cos 45◦)= 0. 943
v=βc= 0. 943 × 3 × 108
= 2. 828 × 108 ms−^1
9.60 β=^1 /ncosθ (1)
Alsoβ=cp/E (2)
whereEis the total energy.
Combining (1) and (2)
E=cpn cosθ (3)
The relativistic equation is
E^2 =c^2 p^2 +m^2 c^4 (4)
EliminatingEbetween (3) and (4), we getmc^2 =pc(n^2 cos^2 θ−1)^1 /^29.61θ= 0 .55 radians= 0. 55 × 57. 3 ◦= 31. 515 ◦
β= 1 /ncosθ= 1 / 1. 88 × 0. 55 = 0. 967
γ= 1 /(1−β^2 )^1 /^2 = 3. 925
γ= 1 +(T/m)→m=T/(γ−1)
m= 420 /(3. 925 −1)=107 MeV/c^2
=(107/ 0 .511)me= 209 me
It is a muon
9.62 The number of photonsN(λ)dλradiated per unit path in a wavelength interval
dλcan be shown to be
N(λ)dλ= 2 πα(1− 1 /β^2 n^2 )(1/λ 1 − 1 /λ 2 )
whereα= 1 /137, is the fine structure constant.
Insertingβ= 0. 95 ,n= 1. 33 ,λ 1 = 3500 × 10 −^8 cm andλ 2 = 5500 × 10 −^8 cm
in the above expression, we findN(λ)dλ=178 per cm.