1000 Solved Problems in Modern Physics

584 10 Particle Physics – II

Meson Quark wave function |ΣaiQi|^2

ρ^0 (uu−dd)/

√

2

[

√^1

2

( 2

3 −

(

−^13

))]^2

=^12

ω^0 (uu+dd)/

√

2

[

√^1

2

( 2

3 −

1

3

)]^2

= 181

Φ^0 ss

( 1

3

) 2

=^19

The expected leptonic widths are in the ratio

Γ(ρ^0 ):Γ(ω^0 ):Γ(φ^0 )=9:1:2

in agreement with the observed ratios.

10.82 For the decay of charmed particles, the selection rules are (i)ΔC=ΔS=
ΔQ=±1, for Cabibbo allowed and (ii)ΔC=ΔQ=± 1 ,ΔS=0, for
Cabibbo suppressed decays
(ΔQapplies to hadrons only). Using these rules (a) is Cabibbo allowed and
(b) is Cabibbo suppressed

10.83 For the charmed particles, the selection rule for the Cabibbo allowed decay
isΔC=ΔS=ΔQ =±1 and for the Cabibbo suppressed decays the
selection rule isΔC=ΔQ=± 1 ,ΔS=0. Using these rules we infer that
the decays are (a) Cabibbo allowed (b) forbidden

10.84 For semileptonic decays the rule isΔS=ΔQ=±1,. In (a) bothQandS
increase by one unit. Hence it is allowed and experimentally observed. (b) is
also a semileptonic decay in which Q decreases by one unit, but S increases
(from−2to−1) by one unit. Therefore, the decay is forbidden. (c) is a
non-leptonic weak decay in whichΔS=+1 and the energy is conserved. It
is allowed and experimentally observed.

10.85 All the decays are semi-leptonic. For Cabibbo allowed decay,
ΔQ=ΔS=±1.
In (a)ΔQ=−1 andΔS=−1(∵Schanges from+ to 0) Therefore it is 1
allowed.
In (b)ΔQ=+1 whileΔS=−1. Therefore, forbidden.
In (c)ΔQ=+1 andΔS=+1(∵S=− 1 →S=0), therefore allowed.
In (d)ΔQ=−1 andΔS=+1, therefore forbidden.

10.86 If lepton number is not absolutely conserved and neutrinos have finite masses,
then mixing may occur between different types of neutrinos (νe,νμ,ντ)In
what follows consider only two types of neutrinosνeandνμ. The neutrino
statesνμandνewhich couple to the muon and electron, respectively could
be linear combinations