10.3 Solutions
577
The ground state energy is obtained by minimizing E,
∂ E
∂a
= 0;
−
2
μa 3 +
3
2
B = 0 → a =
2
2
3μB
1 / 3
(2)
Substituting (2) in (1) and putting μ = m q
2
E = E 0 = 2.45
B
2
2
m q
1 / 3
10.3.4 Electromagnetic Interactions
10.59
σ =
dσ
dΩ
.dΩ =
dσ
dΩ
.2π sin θ dθ
=
2πα
2
2 c
2
4E
2
cM
1
−1
(1 + cos
2
θ)d cos θ
=
2π
4
×
1
137 2 ×
(197.3 MeV − fm)
2
(2 × 10 4 MeV) 2
cos θ +
cos
3
θ
3
1
−1
= 2.17 × 10
−8 fm
2
= 0.217nB
The Feynman diagram is given in Fig. 10.10
Fig. 10.10 e
+ e
−
→ γ
→ μ
+ μ
−
10.60 (a) Γ =
τ
=
c
τ c
=
197.3 MeV − fm
(7.4 × 10 −20 s)(3 × 10 8 × 10 15 fm/s)
= 8.89 × 10
−3 MeV = 8.89 keV
Note that because of the inverse dependence of decay width on lifetime,
weak decays with relatively long lives (> 10
−3 s) have widths of the order
of a small fraction of eV, while em decays (τ ∼10
−19 –10
−20 s) have widths
of the oder a kV, and strong decays (τ ∼10
−23
− 10
−24 s) have widths of
several MeV.
(b) Photon has spin 1 and pion zero and K
+ meson is known to have zero
spin. Pion will be emitted with l = 0 as its energy is small. Therefore the
occurrence of the radiative decay would constitute the violation of angular
momentum conservation.