10.3 Solutions
563
(b) Each photon will carry 67.5 MeV
λ =
1241
67.5 × 10 6 nm = 18.385 × 10
−6 nm = 1.84 × 10
−14 m
10.3.2 Strong Interactions
10.17 (a) τ =
Γ
=
c
Γc
=
197.3 × 10
−15 (MeV − m)
(120MeV)(3 × 10 8 m/s)
= 5.5 × 10
−24 s
γ = 1 +
200
1.236
∼ = 163
β ∼ = 1
d = βcγ τ = 1 × 3 × 10
8
× 163 × 5.5 × 10
−24 m
∼ = 2.7 × 10
−13 m
∼ = 0.0003 nm
a distance which is much less than the resolution obtainable by the available techniques. The best resolution obtained in photographic emulsions
is only 1 μm.
(b) Uncertainty Principle:
Γ.τ ≥
τ ≥
Γ
=
c
Γc
=
197.3 MeV − fm
6.5(MeV) × 3 × 10 8 (m/s)
=
197.3 × 10
−15
19.5 × 10 8
τ ≥ 10
−22 s
10.18 π
+
+ p → π
+
+ p
(1)
π
−
+ p → π
−
+ p
(2)
π
−
+ p → π
0
+ n
(3)
The cross-section is proportional to the square of the matrix element M if
connecting the initial and final states
M if =
ψ f |H | ψ i
where H is the isospin operator, and
σ ∝
M i f
2
As pion has T = 1 and proton T = 1/2, the reactions can proceed either
through I = 1/2 or I = 3/2 channels. Designating the corresponding operators by H 1 and H 3 and the matrix elements for the reactions by M 1 and M 3 ,
we can write
M 1 =
ψ f
1
2
H 1
ψ i
1
2
M 3 =
ψ f
3
2
H 3
ψ i
3
2
As I -spin is conserved there is no operator connecting different isospin
states. Reaction (1) involves a pure state of I = 3/2, I 3 = +3/2.