556
10 Particle Physics – II
10.91 The neutral kaons K
0 and K 0 are the charge conjugate of each other and are
distinguished by their strangeness. However, they decay similarly and mixing
can occur by a virtual process like K
0
⇔ π
+
+ π
−
⇔ K 0 . Starting with a
pure beam of K
0 ’s at t = 0 obtain the intensity of K
0 ’s and K 0 ’s at time t,
in terms of the mean lifetimes τ L (0.9 × 10
−10 s) and τ s (0.5 × 10
−7 s) for the
components K L and K s , long lived and short lived respectively.
10.92 Suppose one starts with a pure beam of K
0 ’s which traverses in vacuum
for a time of the order of 100 K s mean lives so that all the K s - component
has decayed and one is left with K L only. If now the K L traverses a carbon
screen some of the K S states are regenerated. Explain this phenomenon of
regeneration.
10.2.6 Electro-Weak Interactions
10.93 The observation of the process ν μ e
−
→ ν μ e
− , signifies the presence of a
neutral current interaction. Similarly, why does the process ν e e
−
→ ν e e
− ,
not indicate the presence of such an interaction?
10.94 From the data on the partial and full decay width of Z
0 boson show that the
number of neutrino generations is 3 only.
Γ z (total) = 2.534 GeV, Γ(z
0
→ hadrons) = 1.797 GeV,
Γ(z
0
→l
+ l
− ) = 0.084 GeV.Theoretical value for Γ(z
0
→ν l ν l ) = 0.166 GeV.
10.95 (a) What are the experimental signatures and with what detectors would one
measure (a) W → eν and W → μν (b) Z
0
→ e
+ e
− and Z
0
→ μ
+
μ
−
(b) The weak force is due to W , Z exchange, mass ∼ = 100 GeV. Give the
range in meters.
[University of London 2000]
10.96 Using the results of electro-weak theory of Salam and Weinberg, calculate
the masses of W and Z bosons. Take the fine-structure constant α = 1/128
and Fermi’s constant G F /
3 c
3
= 1.166 × 10
−5 GeV
−2 and Weinberg angle
θ W = 28.17
◦
10.2.7 Feynman Diagrams
10.97 Explain which force is responsible for the following particle interactions
and draw a Feynman diagram for each:
[University of Wales, Aberystwyth 2004]
(i) τ
+
→ μ
+
+ ν μ + ν τ
(ii) K
−
+ p → Ω
−
+K
+
+K
0
(K
−
= su; K
+
= us; K
0
= ds; Ω
−
= sss)
(iii) D 0 → K
+
+ π
− (D 0 = uc; K
+
= us; π
−
= du)
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