10.3 Solutions
579
10.64 σ ∝
1
q 2 + m 2
2
At low energies q → 0
σ
σ 0
=
80
100
=
m
2
30 2 + m 2
2
m = 87 GeV/c
2
10.65 Number of W
+
→ e
+
ν e =
σ ( pp
−
→ W
+ )
σ ( pp − → anything)
× 10
9
=
1.8 × 10
−9
70 × 10 −3 × 10
9
= 26
10.66 The decay D
+ (= cd) → K 0 (= sd) + μ
+
+ ν μ involves 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
+
+ ν μ
the 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) the Q - values in
the decays are different. For Σ
−
→ n + e
−
+ ν e , Q 1 = 257 MeV, while for
Σ
−
→ Λ + e
−
+ ν e , 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
ω will be proportional to sin
2
θ c , where θ c is the Cabibbo angle. The second
one involves |ΔS| = 1, so that ω is proportional to cos
2
θ c . In all
R =
ω
Σ
−
→ n + e
−
+ ν e
ω (Σ − → Λ + e − + ν e )
=
sin
2
θ c
cos 2 θ c
Q 1
Q 2
5
= 0.0533
257
81
5
= 17.14
where we have used θ c = 13
◦ . The experimental value for R is 17.8
10.68 The leptonic decays of τ
− are
τ
−
→ e
−
+ ν e + ν τ
(1)
τ
−
→ μ
−
+ ν μ + ν τ
(2)
which from lepton – quark symmetry have equal probability. In addition, the
possible hadronic decays are of the form τ
−
→ ν τ + X
where X can be du or sc so that X may have negative charge. However the
latter possibility is ruled out because m τ < m s + m c . We are then left with
only one possibility for X so that the hadronic decay will be
τ
−
→ ν τ + du
(3)