9.3 Solutions 507
= 6. 02 × 1023 × 7. 9 × 600 × 10 −^27 / 55. 95 = 0 .051 cm−^1
I/I 0 = 1 / 500 =exp(− 0. 051 x)
Take logeon both sides of the equation and solve forx.
We findx=122 cm= 1 .22 m.
Thus a thickness of 1.22 m iron will reduce the pion beam by a factor of- This does not include the reduction due to the decay of pions. Muons have
much less interaction cross-section so that much greater thickness is required.
9.16 From the momentum triangle (Fig. 9.2)
Pν^2 =p 02 +pe^2 − 2 p 0 pecosφ(momentum conservation)
Eν=E 0 −T (energy conservation)
Sincemν= 0 ,pν=Eν
Therefore
Pν^2 =Eν^2 =(E 0 −T)^2 =E 02 +T^2 + 2 mT− 2 E 0 (T^2 + 2 mT)^1 /^2 cosφSimplifying,E 0 (T^2 + 2 mT)^1 /^2 cosφ=T(E 0 −m)≈TE 0 (becauseE 0 >>m)Squaring and simplifying
tanφ≈φ=(2m/T)^1 /^2Fig. 9.2
9.17 (a) In this reaction initially the total strangeness quantum number
Si=Sπ+Sp= 0 + 0 =0, while for the final state
Sf=SK+SΣ=− 1 +(−1)=− 2
ΔS=−2. Therefore, the ruleΔS=0 for strong interactions is violated.
(b) The total isospin for the initial stateI= 0 + 0 =0, while for the final state
I= 0 + 1 =1.
ΔI=1. Therefore, the ruleΔI=0 for strong interactions is violated.9.18 M(^5 HeΛ)=M(Λ)+M(He^4 )−B(Λ)
= 1 , 115. 58 + 3 , 727. 32 − 3. 08 = 4 , 839 .82 MeV/c^2
9.19 LetPiandPfbe the initial and final four-momentum transfer.
Pi=(pi,iEi)
Pf=(pf,iEf)
Q=Pi−Pf=(pi−pf)+i(Ei−Ef)
Q^2 =pi^2 +pf^2 − 2 pi.pf−Ei^2 −Ef^2 + 2 EiEf
=−m^2 −m^2 + 2 EiEf− 2 pipfcosθ
IfEi>>mandEf>>m, thenpi≈Ei, andpf≈Ef.