10.3 Solutions
565
10.20 The nucleon has T = 1/2 and pion T = 1. The Δ
+ has T = 3/2 and
T 3 = +1/2. Using the C.G.C. for 1 × 1/2 (Table of Chap. 3), we have
3
2
,
1
2
=
2
3
|1, 0
1
2
,
1
2
+
1
3
|1, 1
1
2
, −
1
2
N π;
3
2
,
1
2
=
2
3
Pπ
0
+
1
3
nπ
+
∴
Γ(Δ
+
→ π
0
+ p)
Γ(Δ + → π + + n)
=
2/3
2
1/3
2 = 2
Note that for the charge states Δ
++ , the strong decay is through only one
channel (Δ
++
→ p + π
+ ), so also for Δ
− , viz Δ
−
→ n + π
− .
10.21 X
+
→ n + π
+
→ p + π
0
X
+ can have either T = 3/2 or 1/2. For T = 3/2 the predicted ratio
Γ(X
+
→ n π
+ )/Γ(X
+
→ pπ
0 ) = 1/2 (as in Problem 10.20) which is
in disagreement with the experimental ratio of 36/18 or 2.
If we assume the value T = 1/2 for X
+ the state
1
2
,
1
2
must be orthogonal
to the state
3
2
,
1
2
. Therefore
X
+
N π,
1
2
,
1
2
= −
1
3
pπ
0
+
2
3
nπ
+
upto an overall phase factor.
The branching ratio
∴ Γ(X
+
→ nπ
+ )/Γ(X
+
→ pπ
0 ) = 2/1, which is in agreement with
experimental ratio of 36/18 or 2/1.
10.22 τ =
Γ
=
c
Γc
=
197.3 MeV−fm
158 (MeV)×3×10 8 (m/s)
=
197.3×10
−15 (MeV−m)
474 (MeV)×10 8 (m/s)
= 4 × 10
−24 s
a lifetime which is characteristic of a strong interaction. Therefore, the ρmeson decays via strong interaction.
10.23 We first write down the isospin for a pair of pions and then combine the
resultant with the third pion. (a) Each pion has T = 1, so that the π
+
, π
−
combination has I = 2, 1, 0. When π
0 is combined, possible values are I =
3, 2, 1, 0 (b) Two π
0 ’s give I = 2. When the third π
0 is added total I = 3
or 1.
10.24 From the conservation laws for strong interactions the quantum numbers for
X are B = +1, Q/e = −1, S = −2. It is Xi hyperon (Ξ
− ) and decays
weakly with lifetime τ ∼10
−13 s. Its quark structure is dss; the quantum
numbers for Y are B = +1, Q/e = 0, S = −1. It is a sigma hyperon
(Σ
0 ) which decays electromagnetically with lifetime of the order of 10
−20 s.
Its quark structure is uds.
565
10.20 The nucleon has T = 1/2 and pion T = 1. The Δ
+ has T = 3/2 and
T 3 = +1/2. Using the C.G.C. for 1 × 1/2 (Table of Chap. 3), we have
3
2
,
1
2
=
2
3
|1, 0
1
2
,
1
2
+
1
3
|1, 1
1
2
, −
1
2
N π;
3
2
,
1
2
=
2
3
Pπ
0
+
1
3
nπ
+
∴
Γ(Δ
+
→ π
0
+ p)
Γ(Δ + → π + + n)
=
2/3
2
1/3
2 = 2
Note that for the charge states Δ
++ , the strong decay is through only one
channel (Δ
++
→ p + π
+ ), so also for Δ
− , viz Δ
−
→ n + π
− .
10.21 X
+
→ n + π
+
→ p + π
0
X
+ can have either T = 3/2 or 1/2. For T = 3/2 the predicted ratio
Γ(X
+
→ n π
+ )/Γ(X
+
→ pπ
0 ) = 1/2 (as in Problem 10.20) which is
in disagreement with the experimental ratio of 36/18 or 2.
If we assume the value T = 1/2 for X
+ the state
1
2
,
1
2
must be orthogonal
to the state
3
2
,
1
2
. Therefore
X
+
N π,
1
2
,
1
2
= −
1
3
pπ
0
+
2
3
nπ
+
upto an overall phase factor.
The branching ratio
∴ Γ(X
+
→ nπ
+ )/Γ(X
+
→ pπ
0 ) = 2/1, which is in agreement with
experimental ratio of 36/18 or 2/1.
10.22 τ =
Γ
=
c
Γc
=
197.3 MeV−fm
158 (MeV)×3×10 8 (m/s)
=
197.3×10
−15 (MeV−m)
474 (MeV)×10 8 (m/s)
= 4 × 10
−24 s
a lifetime which is characteristic of a strong interaction. Therefore, the ρmeson decays via strong interaction.
10.23 We first write down the isospin for a pair of pions and then combine the
resultant with the third pion. (a) Each pion has T = 1, so that the π
+
, π
−
combination has I = 2, 1, 0. When π
0 is combined, possible values are I =
3, 2, 1, 0 (b) Two π
0 ’s give I = 2. When the third π
0 is added total I = 3
or 1.
10.24 From the conservation laws for strong interactions the quantum numbers for
X are B = +1, Q/e = −1, S = −2. It is Xi hyperon (Ξ
− ) and decays
weakly with lifetime τ ∼10
−13 s. Its quark structure is dss; the quantum
numbers for Y are B = +1, Q/e = 0, S = −1. It is a sigma hyperon
(Σ
0 ) which decays electromagnetically with lifetime of the order of 10
−20 s.
Its quark structure is uds.