1000 Solved Problems in Modern Physics

(Tina Meador) #1

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=

∫dσ

. 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. 10.10

Fig. 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.
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