7.3 Solutions 415
Fig. 7.17
hν/c≤p++p− (2)
Or (hν)^2 ≤(cp++cp−)^2
By virtue of (1)
(E++E−)^2 ≤c^2 p+^2 +c^2 p−^2 + 2 c^2 p+p−
OrE+^2 +E−^2 + 2 E+E−≤E+^2 −m^2 c^4 +E−^2 −m^2 c^4 +2[(E+^2 −m^2 c^4 )
(E−^2 −m^2 c^4 )]^1 /^2
E+E−≤[(E+^2 −m^2 c^4 )(E−^2 −m^2 c^4 )]^1 /^2 −mc^2
Now the left hand side of the inequality is greater than the value of the rad-
ical on the right hand side. It must be still greater than the right hand side. We
thus get into an absurdity, which has resulted from the assumption that both
momentum and energy are simultaneously conserved in this process. Thus an
electron–positron pair cannot be produced by an isolated photon.7.3.7 CerenkovRadiation ................................
7.78 Pμ=150 MeV/c
Eμ=(p^2 μ+m^2 μ)^1 /^2 =(150^2 + 1062 )^1 /^2 = 183 .67 MeV
βμ=Pμ/Eμ= 150 / 183. 67 = 0. 817
nμ= 1 /βμ= 1. 224
Pπ=150 MeV/c
Eπ=(P^2 π+m^2 π)^1 /^2 =(150^2 + 1402 )^1 /^2 = 205. 18
βπ=Pπ/Eπ= 150 / 205. 18 = 0. 731
nπ= 1 /βπ= 1. 368
Therefore the range of the index of refraction of the material over which the
muons above give Cerenkov light is 1. 224 − 1 .368.
7.79 cosθ= 1 /β μ
β= 1 /μcosθ= 1 /(1. 8 ×cos 11◦)= 0. 566
γ=(1−β^2 )−^1 /^2 = 1. 213
Kinetic energy of protonT=(γ−1)mpc^2
=(1. 213 −1)× 938. 3
=200 MeV