1000 Solved Problems in Modern Physics

9.3 Solutions 527

E=(p^2 +m^2 )^1 /^2 =(900^2 + 9382 )^1 /^2

= 1 ,300 MeV

γ= 1 , 300 / 938 = 1. 386

β= 0. 692

Maximum energy transferred to electron is

E(max)= 2 mβ^2 γ^2 = 2 × 0. 511 × 0. 6922 × 1. 3862

= 0 .94 MeV

9.87 Initiallyω 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 .3B= 30. 924 / 0. 3 × 1 = 103 .08 m

9.89 p= 0. 3 BR
P=(T^2 + 2 Tm)^1 /^2 =[3^2 +(2× 3 × 0 .938]^1 /^2 = 3 .825 GeV/c
R=p/ 0 .3B= 3. 825 /(0. 3 × 1 .4)= 9 .1m

9.90 Initiallyω 0 =B 0 e/m
Finallyω= 0. 95 B 0 e/(m+T)
ω/ω 0 = 0. 95 m/(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=(4πe^2 / 3 R)(E/mc^2 )^4. 1 / 4 πε 0
=(1. 44 × 4 π/ 3 R)(E/mc^2 )^4 MeV-fm
SubstitutingE=300 MeV,mc^2 = 0 .511 MeV
R= 1. 0 × 1015 fm
ΔE= 716 × 10 −^6 MeV=716 eV

9.92 p=qBR
R=p/qB=m 0 γβc/qB=m 0 c(γ^2 −1)^1 /^2 /qB
Butγ=(T/m 0 c^2 )+ 1 =n+ 1
∴γ^2 − 1 =n^2 + 2 n
∴R=(m 0 c/qB)(n^2 + 2 n)^1 /^2