1000 Solved Problems in Modern Physics

(Romina) #1

576 10 Particle Physics – II


Fig. 10.9


Actually the value of Rwill be slightly greater than 2 because of an
increased value ofαs, the running coupling constant.

10.58 E=〈ψ|H|ψ〉


/
〈ψ|ψ〉=

∫∞
0

e−r/a

{

^2
2 μ

[
d^2
dr^2

+

2
r

d
dr

]
+Br

}
e−r/a. 4 πr^2 dr

∫∞
0

e−^2 r/a. 4 πr^2 dr

whereμ=mq

/

2 is the reduced mass

E=

^2

2 μ

1

a^2

+

3

2

Ba (1)
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