10.3 Solutions
575
10.51 (a) Δ
− : ddd , Δ
0 : ddu, Δ
+ : uud , Δ
++ : uuu
(b) The fundamental difficulty is that Pauli’s principle is violated. For example consider the spin of Δ
− . All the three d-quarks have to be aligned with
the same j z = −1/2. The same difficulty arises for Δ
++ . The difficulty
is removed by endowing a new intrinsic quantum number (colour) to the
quarks. Thus, the three d-quarks in Δ
− or the three u-quarks in Δ
++ differ
in colour, green, blue, and red.
(c) The higher values of J are accounted for by endowing higher orbital angular momentum values to the quarks. The parity is determined by the value
of (−1)
l , where l is 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 , charge Q/e =
−1/3, mass ∼4.5 GeV/c
2 , and Beauty = −1.
10.54 Figure 10.9a, b are the quark flow diagrams for the decays Φ → K
+ K
−
and ω → π
+
π
0
π
− , respectively. Figure 10.9 c shows that the decay Φ →
π
+
π
0
π
− is suppressed because of unconnected quark lines.
10.55 σ(π N ) = σ(qq)(qqq) = 6 σ (qq)
σ(NN ) = σ (qqq)(qqq) = 9 σ (qq)
where we have assumed σ (qq) = σ (qq)
∴ σ (π N )/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 of π
− (= ud) with
12 C nucleus
(= 18u + 18d), 18u are involved, the cross section being proportional to
18Q
2
U or 18(4/9) or 8. In the annihilation of π
+ (= ud) with
12 C nucleus,
18d are involved, the cross section being proportional to 18Q
2
d or 18(1/9)
or 2.
Therefore the cross section ratio σ(π
− C)/σ(π
+ C) = 8/2 or 4:1
10.57 According to the quark model, the ratio
R =
σ
e
+ e
−
→ hadrons
σ (e + e − → μ + μ − )
=
3
Q
2
i
1
where Q i is 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, and s are +2/3, −1/3, and −1/3
Therefore ΣQ
2
i =
2
3
2 +
−
1
3
2 +
−
1
3
2 =
2
3
and R = 3 × 2/3 = 2
Hence σ
e
+ e
−
→ hadrons
= 3 ×
2
3
× σ
e
+ e
−
→ μ
+
μ
−
= 2 × 20 nb =
40 nb
575
10.51 (a) Δ
− : ddd , Δ
0 : ddu, Δ
+ : uud , Δ
++ : uuu
(b) The fundamental difficulty is that Pauli’s principle is violated. For example consider the spin of Δ
− . All the three d-quarks have to be aligned with
the same j z = −1/2. The same difficulty arises for Δ
++ . The difficulty
is removed by endowing a new intrinsic quantum number (colour) to the
quarks. Thus, the three d-quarks in Δ
− or the three u-quarks in Δ
++ differ
in colour, green, blue, and red.
(c) The higher values of J are accounted for by endowing higher orbital angular momentum values to the quarks. The parity is determined by the value
of (−1)
l , where l is 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 , charge Q/e =
−1/3, mass ∼4.5 GeV/c
2 , and Beauty = −1.
10.54 Figure 10.9a, b are the quark flow diagrams for the decays Φ → K
+ K
−
and ω → π
+
π
0
π
− , respectively. Figure 10.9 c shows that the decay Φ →
π
+
π
0
π
− is suppressed because of unconnected quark lines.
10.55 σ(π N ) = σ(qq)(qqq) = 6 σ (qq)
σ(NN ) = σ (qqq)(qqq) = 9 σ (qq)
where we have assumed σ (qq) = σ (qq)
∴ σ (π N )/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 of π
− (= ud) with
12 C nucleus
(= 18u + 18d), 18u are involved, the cross section being proportional to
18Q
2
U or 18(4/9) or 8. In the annihilation of π
+ (= ud) with
12 C nucleus,
18d are involved, the cross section being proportional to 18Q
2
d or 18(1/9)
or 2.
Therefore the cross section ratio σ(π
− C)/σ(π
+ C) = 8/2 or 4:1
10.57 According to the quark model, the ratio
R =
σ
e
+ e
−
→ hadrons
σ (e + e − → μ + μ − )
=
3
Q
2
i
1
where Q i is 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, and s are +2/3, −1/3, and −1/3
Therefore ΣQ
2
i =
2
3
2 +
−
1
3
2 +
−
1
3
2 =
2
3
and R = 3 × 2/3 = 2
Hence σ
e
+ e
−
→ hadrons
= 3 ×
2
3
× σ
e
+ e
−
→ μ
+
μ
−
= 2 × 20 nb =
40 nb