1000 Solved Problems in Modern Physics

528 9 Particle Physics – I

9.93 Using the result of Problem 9.92

R=

m 0 c

qB

(n^2 + 2 n)

(^12)

=

m 0 c

0. 0147 q

[(

50

940

) 2

+

(

2 × 50

940

)]^1 /^2

= 22. 48

m 0 c

q

(1)

As the radius of synchrotron does not change, we can use the same relation at

higher energy

R=

(

m 0 c

1. 2 q

)

(N^2 + 2 N)^1 /^2 (2)

Combining (1) and (2) and solving forn, we findN= 26

∴T=Nm 0 c^2 = 26 × 0. 938 = 24 .39 GeV

9.94 Using the results of Problem 9.92

q/m 0 =(c/BR)(n^2 + 2 n)^1 /^2 (1)

Proton:q/m 0 = 1 / 1 = 1

n=T/mp= 1 , 000 / 1 , 000 = 1

Using the above values in (1)

c/BR= 1 /

√

3(2)

Deuteron:q/m 0 = 1 / 2

Therefore^1 / 2 =(n^2 + 2 n)^1 /^2 /

√

3

Solving forn, we findn= 0. 3229

Kinetic energy of deuteron=nmd

= 0. 3229 × 2 ,000 MeV

=646 MeV

(^3) He : q/m 0 = 2 / 3
Therefore 2/ 3 =(1/

√

3)(n^2 +^2 n)^1 /^2

Solving forn, we findn= 0. 527

Therefore Kinetic energy of^3 He=nmHe 3

= 0. 527 × 3 ,000 MeV

= 1 ,583 MeV.

9.95 (a)p= 0 .3BR(GeV/cifBis in Tesla andRin metres)

P=(T^2 + 2 Tmc^2 )^1 /^2 =(0. 52 + 2 × 0. 5 × 0 .938)^1 /^2 = 1 .09 GeV/c

B=18 kG= 1 .8T

R=p/ 0. 3 B= 1. 09 /(0. 3 × 1 .8)= 2 .02 m.

(b) Energy of ions for acceptance

Ts=±(2eV.mc^2 /π)^1 /^2 [(φs−π/2) sinφs+cosφs]^1 /^2 (1)

Substitute in (1), 10 keV= 0 .01 MeV

mc^2 =938 MeV;φs= 300 = 0 .5236 radians

Ts=± 2 .5MeV

(c) The initial frequency

f=Bqc^2 / 2 π(mc^2 +Ts)(2)