10.3 Solutions 579
10.64σ∝
1
[
q^2 +m^2] 2
At low energiesq→ 0σ
σ 0=
80
100
=
[
m^2
302 +m^2] 2
m=87 GeV/c^210.65 Number ofW+→e+νe=
σ(pp−→W+)
σ(pp−→anything)× 109
=
1. 8 × 10 −^9
70 × 10 −^3
× 109 = 26
10.66 The decayD+(=cd)→K^0 (=sd)+μ++νμinvolves the transformation
of a c-quark to an s-quark which is Cabibbo favored. In the case of the decay
D+(=cd)→π^0 (=uu−dd
√
2)+μ++νμthe transformation of a c-quark to a u-quark is Cabibbo suppressed. This
explains why the former process is more likely than the latter.10.67 The difference in the decay rates is due to two reasons (i) theQ- values in
the decays are different. ForΣ−→n+e−+νe,Q 1 =257 MeV, while for
Σ−→Λ+e−+νe,Q 2 =81 MeV, so that by Sargent’s law, the decay rate
(ω) will be proportional to Q^5. (ii) The first decay involves|ΔS|=1, so that
ωwill be proportional to sin^2 θc, whereθcis the Cabibbo angle. The second
one involves|ΔS|=1, so thatωis proportional to cos^2 θc.Inall
R=
ω(
Σ−→n+e−+νe)
ω(Σ−→Λ+e−+νe)=
sin^2 θc
cos^2 θc(
Q 1
Q 2
) 5
= 0. 0533
(
257
81
) 5
= 17. 14
where we have usedθc= 13 ◦. The experimental value forRis 17.810.68 The leptonic decays ofτ−are
τ−→e−+νe+ντ (1)
τ−→μ−+νμ+ντ (2)which from lepton – quark symmetry have equal probability. In addition, the
possible hadronic decays are of the formτ−→ντ+X
whereXcan beduorscso thatXmay have negative charge. However the
latter possibility is ruled out becausemτ<ms+mc. We are then left with
only one possibility for X so that the hadronic decay will be
τ−→ντ+du (3)