1000 Solved Problems in Modern Physics

9.3 Solutions 529

B= 1 .8T;q= 1. 6 × 10 −^19 C;c= 3 × 108 m/s;mc^2 = 938 × 1. 6 ×

10 −^13 J

Ts=± 2 .5MeV=± 2. 5 × 1. 6 × 10 −^13 J

Substituting the above values in (2) we find

f 1 = 27 .43; f 2 = 27 .51 Mc

(d) The required electrical frequency for 500 MeV protons is

f=Bqc^2 / 2 π(mc^2 +T)(3)

Put (mc^2 +T)=(938+500)× 1. 6 × 10 −^13 J, B= 1 .8T,c= 3 × 108 m/s

andq= 1. 6 × 10 −^19 C, in (3) to obtainf= 17 .94 Mc.

Thus the range of frequency modulation is 27. 51 − 17 .94 Mc

9.96 The particles are constrained to move in a vacuum pipe bent into a torus that
threads a series of electromagnets, providing a field normal to the orbit. The
particles are accelerated once or more per revolution by radio frequency cav-
ities. Both the magnetic field and the R.F. frequency must increase and the
synchronized with the particle velocity as it increases.
The major energy loss is caused by the emission of synchrotron radiation.
The synchrotron radiation loss per turn is
ΔE=(4π/ 3 R)(e^2 / 4 πε 0 )(E/mc^2 )^4
For electron

ΔE=

(

4 π

3

(1.44 MeV.fm)

(10^3 × 1015 fm)

)(

500 × 103

0. 511

) 4

= 5 , 526 × 103 MeV

= 5 ,526 GeV

an energy loss which is an order of magnitude greater than the electron energy

to which the electrons are to be accelerated, which is impossible. On the other

hand for protons the energy loss per turn will be smaller by a factor (1,836)^4

or 1. 1 × 1013

Thus for protonsΔE= 5 , 526 / 1. 1 × 1013 = 5 × 10 −^10 GeV

= 0 .5 eV which is quite small.

As the synchrotron radiation losses for electrons in circular machines are

much beyond tolerable limits, linear accelerators are employed which are

capable of accelerating electrons up to 30–40 GeV.

9.97 Orbital frequencyf= 1 / 10 −^6 = 106 c/s
P= 0. 3 Br
f=qB/ 2 πm
∴ p= 0. 3 × 2 πmfr/q= 0. 3 × 2 π× 1. 67 × 10 −^27 × 106 × 10 /(1. 6 × 10 −^19 )
= 0 .1966 GeV/c
T^2 + 2 mT=p^2
Usingm = 0 .938 GeV/c^2 and solving forT, we findT = 0 .02 GeV or
20 MeV