9.3 Solutions 525
Furtherp=qBR
T=p^2 / 2 m=q^2 B^2 R^2 / 2 m
Maximum radius
R(max)=(2mTmax)^1 /^2 /qB
=(2× 1. 67 × 10 −^27 × 6 × 1. 6 × 10 −^13 )^1 /^2 /(1. 6 × 10 −^19 B)
= 0. 354 /Bm9.80 The resonance condition is
ω= 2 π f=qB/m
B= 2 πfm/q
For deuterons,Bd= 2 π× 1. 2 × 107 × 2. 01 × 1. 66 × 10 −^27 / 1. 6 × 10 −^19
= 1 .5715 T
For alpha particles,Bα= 2 π× 1. 2 × 107 × 4. 00 × 1. 66 × 10 −^27 / 2 × 1. 6 × 10 −^19
= 1 .5637 T
Deuterons: At ejection kinetic energy
Ef=(qBR)^2 / 2 m=(1. 6 × 10 −^19 × 1. 5715 × 0 .3)^2 /(2× 2. 01 × 1. 66 × 10 −^27 )J
= 0. 085 × 10 −^11 J
= 5 .3MeV
IfNis the number of orbits,ttotal time, andT 0 the time period then the
mean energy increment per orbit isE/Nand the average time for each orbit
(supposed to be constant)
T 0 = 2 π/ω 0 = 1 /f 0
N=t/T 0 =tf 0
ThereforeEf/N=Ef/tf 0 =2 eV (the factor 2 is introduced because there
are two gaps)
t=Ef/(2eV)f 0 = 5. 3 × 106 /(2× 5 × 104 )× 1. 2 × 107 = 4. 42 × 10 −^6 s
= 4. 42 μs
α-particles: At ejectionEf=(qBR)^2 /2m
=(2× 1. 6 × 10 −^19 × 1. 5637 × 0 .3)^2 /(2× 4. 0 × 1. 66 × 10 −^27 )
= 0. 1697 ×10 J
= 10 .6MeV
Total timet=Ef/(2×2eV)f 0
= 10. 6 × 106 /(4× 5 × 104 × 1. 2 × 107 )
= 4. 42 × 10 −^6 s= 4. 42 μs
9.81 Cyclotron resonance condition is
ω=qB/m
Because of relativistic increase of mass the resonance condition would be
ω′=qB′/mγ
Ifω′=ω, thenB′=Bγ
Fractional increase of magnetic flux density required is
(B′−B)/B=ΔB/B=γ− 1
For protons of 20 MeV.γ=(T/m)+ 1 =(20/940)+ 1 = 1. 0213
Therefore percentage increase of B is (γ−1)× 100
=(1. 0213 −1)× 100 = 2. 13