6.3 Solutions 349
Therefore,γ=[1+(p/m)^2 ]^1 /^2 =[1+(5/ 0 .938)^2 ]^1 /^2 = 5. 41
tanθtanφ=tan 82◦tan 2◦ 30 ′= 7. 115 × 0. 04366 = 0. 3106
2 /(γ+1)= 2 /(5. 41 +1)= 0. 3120
Hence the event is consistent with elastic scattering.6.70 Letp,pe,pNbe the momentum of the incident electron, scattered electron
and recoil nucleus, respectively. From the momentum triangle, Fig. 6.9.
pN^2 =pe^2 +p^2 − 2 ppecosθ=EN^2 + 2 ENM (1)
where we have putc= 1
From energy conservation
EN+Ee=E (2)
As
Ee≈pe (3)
E≈P (4)
(2) Can be written as
EN+pe=E (5)
Combiining (1), (3), (4) and (5), we get
EN=E^2 (1−cosθ)/M[1+E/M(1−cosθ)]
Restoringc^2 , we get the desired result.
Fig. 6.9Momentum triangle
6.71 Use the result of Problem 6.53,
T= 2 mc^2 β^2 cos^2 φ/(1−β^2 cos^2 φ)(1)
Putc= 1 ,T=E−m (2)
P=Mβγ=Mβ/(1−β^2 )^1 /^2
whenceβ^2 =P^2 /(P^2 +M^2 )(3)
Use (2) and (3) in (1) and simplify to get the desired result.6.72 The formula for the recoil energy of electron in Compton scattering is
T=(E^2 /mc^2 )(1−cosθ)/[1+α(1−cosθ)]
Here neutrinos are assumed to be massless, so that the same formula which
is based on relativistic kinematics can be used.
The maximum recoil energy will occur when the neutrino is scattered
back, that isθ = 180 ◦. SubstitutingE =2 GeV for the incident neutrino
energy,mc^2 = 0 .511 MeV = 0. 511 × 10 −^3 GeV, andα = E/mc^2 =
2 / 0. 511 × 10 −^3 = 3 ,914, we find the maximum energy transferred to electron
is 1.9997 GeV. The maximum momentum transfer
pmax=(Tmax^2 + 2 mec^2 .Tmax)^1 /^2 = 2 .0437 GeV/c