7.3 Solutions 407
(b) Over the whole path total number of ion pairs
=Total energy lost/Ionization energy of each pair= 5. 0 × 106 / 34
= 1. 47 × 105
ForR 1 = 1 .839 cm, we can find the velocity at the middle of the path by the
given formula
R 1 = 1. 839 = 0. 98 × 10 −^27 v 13
orv 1 = 1. 233 × 109. The corresponding energy at the mid-path is
E 1 =mv 12 / 2 =mc^2 v 12 / 2 c^2 =(1/2)× 3728 ×(1. 233 × 109 / 3 × 1010 )^2 =
3 .148 MeV
Energy lost in the first half of the path
E= 5. 000 − 3. 148 = 1 .852 MeV.
Number of ion pairs produced over the first half of the path =
1. 852 × 106 / 34 = 5. 45 × 1047.49vp^2 = 2 E/mp,vd^2 = 2 E/md= 2 E/ 2 mp=v^2 p/ 2
=dE/dx ∝ z^2 /v^2
∴(dE/dx)p
(dE/dx)d=
v^2 d
v^2 p=2(∵z=1 for bothpandd)7.50 (−dE/dx)proton=(−dE/dx)deuteron
As their charges are identical, their velocity must be the same. Therefore,
the ratio of their kinetic energies must be equal to the ratio of their masses.
Ep/Ed=Mp/Md= 1 / 2
7.51 After crossing a radiation length of lead electrons emerge with an aver-
age energy of 2. 7 /e = 2. 7 / 2. 71 = 1 .0 GeV. Thus, the average energy
loss= 1 .7GeV.
7.52 The root mean square multiple scattering angle is approximately given by
(θ^2 )(^12)
=βp(MeV^20 /c)
√
L
LR
For a traversal of a distanceL.ForP=400 MeV/c andL=LR/10, this
angle is 15.8 mr. An electron emitting a photon will be emitted at an angle of
mec^2 /Eγwith the direction of flight. This angle is 0.511/400 or 1.28 mr.
Thus, the angular distribution of Bremsstrahlung photons is determined
mainly by the multiple scattering of electrons.7.3.4 ComptonScattering................................
7.53 (a) LetE=hνbe the energy of the scattered photon andhν/cbe its momen-
tum (Fig. 7.15).
Energy conservation gives
hν 0 =hν+T (1)
whereTis the electron’s kinetic energy.
Balancing momentum along and perpendicular to the direction of
incidence