1000 Solved Problems in Modern Physics

9.3 Solutions 503

9.109 At a collider, a 20 GeV electron beam collides with a 300 GeV proton beam at
a crossing angle of 10◦. Evaluate the total centre of mass energy and calculate
what beam energy would be required in a fixed-target electron machine to
achieve the same total centre-of-mass energy.

9.3 Solutions..................................................

9.3.1 SystemofUnits ...................................

9.1 E=Mc^2 = 1 ×(3× 108 )^2 = 9 × 1016 J
= 9 × 1016 J/(1. 6 × 10 −^10 J/GeV)= 5. 63 × 1026 GeV

9.2 (a)c= 197 .3MeV-fm

In natural units,=c= 1

Therefore 1= 197 .3MeV-fm= 0 .1973 GeV-10−^15 m

Therefore 10−^15 m= 1 / 0 .1973 GeV−^1

or 1 m= 1015 / 0. 1973 = 5. 068 × 1015 GeV−^1

(b) From (a) we have

1m^2 =(5.068)^2 × 1030 GeV−^2 = 25. 6846 × 1030 GeV−^2

Therefore 1 GeV−^2 = 10 −^30 / 25 .6846 m^2 = 0 .389 mb

(c)= 1. 055 × 10 −^34 J-s

1 = 1. 055 × 10 −^34 J-s/ 1. 6 × 10 −^10 J/GeV

Therefore 1 s=(1. 6 × 10 −^10 / 1. 055 × 10 −^34 )GeV−^1 = 1. 5 × 1024 GeV−^1

9.3 (a) In practical unitsλc=/mec

Put=c= 1

In natural unitsλc= 1 /me

(b) In practical units Bohr’s radius of hydrogen atom is

a 0 =ε 0 ^2 /πme^2 =ε 0 c/πmce^2

In natural unitsa 0 = 1 /αme

(c) In the ground state velocity

v=/ma 0 =α

where we have used the results of (b)

Put=1 to findv=αin natural units.

Numerical values:

λc=/mec=(c/mec^2 )MeV.fm/MeV= 197. 3 / 0 .511 fm=385 fm

= 0. 00385 × 10 −^10 m

= 0. 00385 A ̊

a 0 = 4 πε 0 ^2 /e^2 m=(4πε 0 /e^2 )(c)^2 /mc^2

=[1/(1.44 MeV-fm)]×(197.3MeV-fm)^2 / 0 .511 MeV

= 53 ,000 fm= 0. 53 × 10 −^10 m

v=αc=(3× 108 /137) ms−^1

= 2. 19 × 106 ms−^1