10.3 Solutions 57510.51 (a)Δ−:ddd,Δ^0 :ddu,Δ+:uud,Δ++:uuu
(b) The fundamental difficulty is that Pauli’s principle is violated. For exam-
ple consider the spin ofΔ−. All the three d-quarks have to be aligned with
the samejz=− 1 /2. The same difficulty arises forΔ++. The difficulty
is removed by endowing a new intrinsic quantum number (colour) to the
quarks. Thus, the three d-quarks inΔ−or the three u-quarks inΔ++differ
in colour, green, blue, and red.
(c) The higher values ofJare accounted for by endowing higher orbital angu-
lar momentum values to the quarks. The parity is determined by the value
of (−1)l, wherelis the orbital angular momentum quantum number.
10.52 (a)Σ+:uus,Σ−:dds,n:udd,p:uud
(b)K+:us,K−:ds,π+:ud,π−:ud
10.53 (a)Δ++,Δ−,Ω−
(b) The quantum numbers of the b-quark are spin=^1 / 2 , chargeQ/e =
− 1 /3, mass∼ 4 .5GeV/c^2 , and Beauty=−1.
10.54 Figure 10.9a, b are the quark flow diagrams for the decaysΦ→ K+K−
andω→π+π^0 π−, respectively. Figure 10.9 c shows that the decayΦ→
π+π^0 π−is suppressed because of unconnected quark lines.
10.55σ(πN)=σ(qq)(qqq)= 6 σ(qq)
σ(NN)=σ(qqq)(qqq)= 9 σ(qq)
where we have assumedσ(qq)=σ(qq)
∴σ(πN)/σ(NN)= 6 / 9 = 2 / 3
which is in agreement with the ratio 25/38.10.56 The cross section for this electromagnetic process is proportional to the
square of the quark charge. In the annihilation ofπ−(=ud) with^12 C nucleus
(= 18 u+ 18 d), 18 uare involved, the cross section being proportional to
18 Q^2 Uor 18(4/9) or 8. In the annihilation ofπ+(=ud) with^12 C nucleus,
18 dare involved, the cross section being proportional to 18Q^2 dor 18(1/9)
or 2.
Therefore the cross section ratioσ(π−C)/σ(π+C)= 8 /2or4:1
10.57 According to the quark model, the ratio
R=
σ(
e+e−→hadrons)
σ(e+e−→μ+μ−)=
3
∑
Q^2 i
1whereQiis the charge of the quark and the factor 3 arises due to the three
colours in which the quarks can appear. Now the charges of the three quarks,
u, d, andsare+ 2 / 3 ,− 1 /3, and− 1 / 3
ThereforeΣQ^2 i=( 2
3) 2
+
(
−^13
) 2
+
(
−^13
) 2
=^23
andR= 3 × 2 / 3 = 2
Henceσ(
e+e−→hadrons)
= 3 ×^23 ×σ(
e+e−→μ+μ−)
= 2 ×20 nb=
40 nb