1000 Solved Problems in Modern Physics

9.3 Solutions 531

(b) Power outputW=iV

The current is obtained from (a) and the voltageVcorresponding to final

voltage of 2 GeV would be 2× 109.

ThereforeW= 4 × 10 −^6 × 2 × 109

= 8 ,000 W

9.101 There are five drift tubes in the section but only four gaps. Initial energy is
100 keV. Therefore, the final energy will be

(a)E=100 keV+neV

= 100 + 4 × 100 =500 keV

(b)L=(1/ 2 f)(2eV/m)^1 /^2

∑ 4

1

√

n

= 2 × 501 × 106

(

2 × 1. 6 × 10 −^19 × 105

1. 67 × 10 −^27

) 1 / 2

(1+

√

2 +

√

3 +

√

4)

= 0 .269 m= 26 .9cm

9.102 The longest drift tube will be in the end of the linear accelerator where max-
imum ion energy has been achieved.
L=v/ 2 f
We can do non-relativistic calculations for v as the ion energy is not large
v=(2T/M)^1 /^2 =c(2T/Mc^2 )^1 /^2 =c(2× 80 / 12 ×938)^1 /^2 = 0. 119 c
L= 0. 119 × 3 × 108 / 2 × 25 × 106 = 0 .714 m.

9.3.15 Colliders .........................................

9.103 In the colliding beam experiments, for two colliding beams the reaction rate
is written in terms of the “luminosity”L. The number of interactions per
second is
N=Lσ
whereσis the interaction cross-section in question.
IfN 1 orN 2 =number of particles/bunch in each beam
A=area of cross-section of intersecting beams
N=number of bunches/beam
andf=frequency of revolution
then luminosity is simply given by
L=nN 1 N 2 f/A

9.104 Number of beam electrons/cm^2 /s,N 1 =nf/πr^2
= 6 × 1011 × 2 × 106 /π(0.12)^2 = 2. 65 × 1019
Number of beam positrons/s,N 2 = 6 × 1011
Expected production rate ofμ+μ−pairs/second
=N 1 N 2 σ(e+e−→μ+μ−)
=(2. 65 × 1019 )(6× 1011 )(1. 4 × 10 −^33 )
= 0. 022