10.3 Solutions 577
The ground state energy is obtained by minimizingE,∂∂Ea =0;−
^2
μa^3+
3
2
B= 0 →a=(
2 ^2
3 μB)^1 / 3
(2)
Substituting (2) in (1) and puttingμ=mq/
2
E=E 0 = 2. 45
(
B^2 ^2
mq)^1 / 3
10.3.4 Electromagnetic Interactions .........................
10.59
σ=∫dσ
dΩ.dΩ=∫dσ
dΩ. 2 πsinθdθ
=2 πα^2 ^2 c^2
4 E^2 cM∫^1
− 1(1+cos^2 θ)d cosθ=
2 π
4×
1
1372
×
(197.3MeV−fm)^2
(2× 104 MeV)^2[
cosθ+cos^3 θ
3] 1
− 1
= 2. 17 × 10 −^8 fm^2
= 0 .217nB
The Feynman diagram is given in Fig. 10.10Fig. 10.10e+e−→γ
→μ+μ−
10.60 (a)Γ=
τ=
c
τc=
197 .3MeV−fm
(7. 4 × 10 −^20 s)(3× 108 × 1015 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 andK+meson is known to have zero
spin. Pion will be emitted withl=0 as its energy is small. Therefore the
occurrence of the radiative decay would constitute the violation of angular
momentum conservation.