9.3 Solutions 509
En=(
mΣ^2 +mn^2 −mμ^2)
/ 2 mΣ=(1, 1892 + 9392 − 1062 )/ 2 × 1 , 189 =
960 .6MeV
ThereforeTn=En−mn= 960. 6 − 939 = 21 .6MeV9.25 Maximum neutrino momentum in the L-system is obtained when the neutrino
is emitted in the forward direction, that is atθ∗=0 in the rest system of
K-meson (Fig. 9.3).
The energy of the neutrino in the rest frame of K-meson is
Eν∗=
(
mK^2 −mμ^2)
/ 2 mK= 235 .6MeV
Pν∗= 235 .6MeV/c (because mν=0)
Pν=γKpν∗( 1 +βKβν∗cosθ∗)
βν∗=1;θ∗=0;pν∗= 235 .6MeV/c
γK= 1 +TK/mK= 1 + 500 / 494 = 2. 012
βK= 0. 868
pν=(2.012)(235.6)(1+ 0 .868)= 885 .5MeV/cFig. 9.3Decay configuration
9.26 Rest mass energy of the heavy meson,M= 965 × 0. 511 =493 MeV
γ= 1 +T/M= 1 + 50 / 493 = 1. 101
β=(γ^2 −1)^1 /^2 /γ=(1. 1012 −1)^1 /^2 / 1. 101 = 0. 418
Observed mean lifetimeτis related to proper lifetime by
τ=τ 0 γ
τ=d/βc= 1. 7 / 0. 418 × 3 × 108 = 1. 355 × 10 −^8 s= 1. 101 τ 0
Thereforeτ 0 = 1. 23 × 10 −^8 s
9.27 Fraction of pions that will not decay
I/I 0 =exp(−t/τ 0 )=exp(−d/βcγτ 0 )
γ=E/m=(p^2 +m^2 )^1 /^2 /m=[(p^2 /m^2 )+1]^1 /^2 =[(10/ 0 .1396)^2 +1]^1 /^2 =
71. 64
β≈ 1
ThereforeI/I 0 =exp(− 100 / 1 × 3 × 108 × 71. 64 × 2. 6 × 10 −^8 )=exp(− 0 .179)=
0. 836
γμ=((p^2 /m^2 )+1)^1 /^2 =((8^2 / 0. 10572 )+1)^1 /^2 = 75. 69
βμ≈ 1
tμ=dγ/βc= 100 × 75. 69 / 1 × 3 × 108 = 25. 23 × 10 −^6 s
tν=d/c= 100 / 3 × 108 = 0. 333 × 10 −^6 s
ThereforeΔt=tμ−tν=(25. 23 − 0 .33)× 10 −^6 s
= 24. 9 μs
9.28 In the rest system of pion the muon is emitted with kinetic energy
Tμ∗=4 MeV (see Problem 6.54)
Its energy in the LS will be maximum whenθ∗=0 and minimum when
θ∗=180 in the CMS. Applying the relativistic transformation equation