526 9 Particle Physics – I
9.82 (a) Resonance condition for protons is
B= 2 πfm/q=(2π× 5 × 106 )(1. 6726 × 10 −^27 )/(1. 6 × 10 −^19 )
= 0 .3284 T
T=(1/2)(Bqr)^2 /m=(0.3284)^2 (1. 6 × 10 −^19 × 0 .762)^2 /(2× 1. 6726 ×
10 −^27 )
= 4. 79 × 10 −^13 J=3MeV
(b) For deuteron, the charge is the same as that of proton but mass is approx-
imately double, the required magnetic field will be that for (a). SoB=
0 .655 T.
The kinetic energy∝B^2 /m, so that it will be (2^2 /2)×3or6MeV
(c) For alpha particle the mass is approximately four times and charge is dou-
ble compared to proton, so that the required magnetic field is twice that for
proton, that isB= 0 .655 T, and kinetic energy will be (2×2)^2 /4, that is,
four times the proton energy or 12 MeV.9.83 The resonance frequency at the beginning
f 0 =B 0 q/ 2 πm= 1. 5 × 1. 6 × 10 −^19 /(2π× 3. 34 × 10 −^27 )= 11. 44 × 106 c/s
= 11 .44 Mc/s
The resonance frequency at the limiting radius is
f=qB/ 2 πm= 1. 43 × 1. 6 × 10 −^19 / 2 π× 3. 34 × 10 −^27
= 10. 91 × 106 c/s= 10 .91 Mc/s
Range of frequency modulation is 11.44–10.91 Mc/s.
Tmax=q^2 B^2 r^2 / 2 m=(1. 6 × 10 −^19 )^2 (1.43)^2 (2.06)^2 / 2 × 3. 34 × 10 −^27
= 3. 3256 × 10 −^11 J
= 3. 3256 × 10 −^11 / 1. 6 × 10 −^13 MeV= 207 .8MeV
9.84 B=ωm/q=(2π× 8 × 106 )(1. 66 × 10 −^27 )/(1. 6 × 10 −^19 )
= 0 .52 T
r=(2T/mω^2 )^1 /^2 =c(2T/mc^2 ω^2 )^1 /^2
= 3 × 108 (2× 5 / 938 × 4 π^2 × 82 × 1012 )^1 /^2
= 0 .616 m
9.85 The cyclotron resonance condition isω=qB/m
For deuterons,ωd= 1 ×Bd/md
For alpha particles,ωα= 2 ×Bα/mα
If the resonance frequency is to remain unalteredωd=ωα
Bd/Bα= 2 ×(2. 014102 / 4 .002603)= 1. 006396
Fractional decrease of magnetic field
(Bα−Bd)/Bd=− 0. 006355
Percentage decrease= 0 .6355%
9.86 The energy of protons extracted from the accelerator is calculated from the
equations
E^2 =p^2 +m^2
p= 0. 3 BR= 0. 3 × 1. 5 × 2 = 0 .9GeV/c
=900 MeV/c.