9.3 Solutions
527
E = ( p
2
+ m
2 )
1/2
= (900
2
+ 938
2 )
1/2
= 1, 300 MeV
γ = 1, 300/938 = 1.386
β = 0.692
Maximum energy transferred to electron is
E(max) = 2mβ
2
γ
2
= 2 × 0.511 × 0.692
2
× 1.386
2
= 0.94 MeV
9.87 Initially ω 0 = qB /m
ω = qB /mγ = ω 0 /γ
γ = 1 + T /m = 1 + 469/938 = 1.5
f = f 0 /γ = 20/1.5 = 13.33 Mc/s
9.3.13 Synchrotron
9.88 P = 0.3 BR
P = (T
2
+ 2 Tm)
1/2
= (30
2
+ 2 × 30 × 0.938)
1/2
= 30.924 GeV/c
R = p/0.3 B = 30.924/0.3 × 1 = 103.08 m
9.89 p = 0.3 BR
P = (T
2
+ 2Tm)
1/2
= [3
2
+ (2 × 3 × 0.938]
1/2
= 3.825 GeV/c
R = p/0.3 B = 3.825/(0.3 × 1.4) = 9.1 m
9.90 Initially ω 0 = B 0 e/m
Finally ω = 0.95 B 0 e/(m + T )
ω/ω 0 = 0.95m/(m + T )
ω 0 − ω
ω 0
=
T + 0.05 m
T + m
=
313 + 0.05 × 938
313 + 938
= 0.288
Therefore Depth of modulation is 28.8 %
9.91 Radiation loss per revolution
ΔE = (4π e
2
/3R)(E/mc
2 )
4
.1/4 πε 0
= (1.44 × 4π/3R)(E/mc
2 )
4 MeV-fm
Substituting E = 300 MeV, mc
2
= 0.511 MeV
R = 1.0 × 10
15 fm
ΔE = 716 × 10
−6 MeV = 716 eV
9.92 p = qBR
R = p/qB = m 0 γβ c/qB = m 0 c (γ
2
− 1)
1/2
/qB
But γ = (T /m 0 c
2 ) + 1 = n + 1
∴ γ
2
− 1 = n
2
+ 2n
∴ R = (m 0 c/qB )(n
2
+ 2n)
1/2