1000 Solved Problems in Modern Physics

(Romina) #1

10.3 Solutions 569


Assuming that the decay proceeds via strong interaction, the parity
PB=PωPπ=(−1)(−1)(−1)^0 =+ 1
Jω= 1 + 0 =1 (becausel=0)
I= 0 + 1 = 1
For theB+meson,Jp= 1 +andI= 1
In case of weak decay, the spin would still be 1 but it would not be mean-
ingful to talk aboutIorP.

10.33 The nucleon hasT = 1 /2 and pionT =1. TheΔ^0 hasT = 3 /2 and
T 3 =− 1 /2. Using Clebsch – Gordon coefficients (C.G.C) for 1× 1 / 2
(Table 3.3 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


Δ^0 π^0 n π− p

The ratio of the amplitudes for the decaysΔ^0 →π−pandΔ^0 →π^0 n
is given by the ratio of the corresponding C.G.C. and the ratio of the cross
sections by the squares of the C.G.C. Thus, the branching ratio would be
(√
2
3

) 2 /(√

1

3

) 2

= 2

10.34

Fig. 10.4S−I 3 plot for
pseudoscalar meson octet


10.35

Fig. 10.5S−I 3 plot for
vector meson nonet

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