Interaction Between Ionizing Radiation And Matter, Problems Photons Audun Sanderud Department of...
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Transcript of Interaction Between Ionizing Radiation And Matter, Problems Photons Audun Sanderud Department of...
Interaction Between Ionizing Radiation And Matter,
Problems Photons
Audun Sanderud
Dep
art
men
t o
f P
hys
ics
Un
iver
sity
of
Os
lo
Problem 7.1• Is the mass Compton attenuation/energy transfer
coefficient larger in carbon or lead?
Solution: ←Independent of Z
Carbon: NAZ/A=0.49954NA Lead: NAZ/A=0.39575NA
NA=6.0022 1023
( )20 cm electrone Zs µ%
( )2cm ga A
e
N Z
A
ss
r= ×
( )2cm ga tr a T
h
s sr r n
=
NAZ/A as function of Z
0
1E+23
2E+23
3E+23
4E+23
5E+23
6E+23
7E+23
0 20 40 60 80Atomnumber Z
Problem 7.2• Why is Rayleigh scattering not plotted in Fig. 7.16a,b,
although quite significant in Fig. 7.13a,b?
Solution:( ) ( )
( )
2
2
2
cm g
cm0 g
a R
a R tr
Z
h
sr n
sr
µ
=
%
Problem 7.3• On the basis of the K-N theory, what is the ratio of the
Compton interaction cross section per atom for lead and carbon?
Solution: ( )( )
20
2
cm electron
cm atom
82
6
e
a e
Pb Pb e
C C e
Z
Z
Z
Z
s
s s
s ss s
µ
= ×
ß×
= =×
%
Problem 7.4• Calculate the energy of the Compton-scattered photon
at = 0, 45 , 90 and 180 for h = 50 keV, 500 keV and 5 MeV.
Solution:( )21 1 cos
e
hh
hm c
nn
nq
¢=æ ö÷ç ÷+ -ç ÷ç ÷çè ø
h\ 0 45 90 180
50 50 48.60 45.54 41.82 keV
500 500 388.6 252.7 169.1 keV
5000 5000 1293 463.6 243.1 keV
h´:
Problem 7.5• What are the corresponding energies and angels of
the recoiling electrons for the cases in problem 4?
Solution:2
, cot 1 tan2e
hT h h
m c
n qn n j
æ ö æö÷ç ÷ç¢ ÷= - = +ç ÷ç÷ ÷ç ç÷ç è øè ø
h\ 0 45 90 180
50 0 1.393 4.456 8.183 keV
500 0 111.4 247.3 330.9 keV
5000 0 3707 4536 4757 keV
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 45 90 135 180scattering angle q
T/h
n50 keV
500keV
5000kev
T:
h\ 0 45 90 180
50 90 65.55 42.33 0 keV
500 90 50.67 26.81 0 keV
5000 90 12.62 5.298 0 keV
:
Problem 7.6• Calculate for 1-MeV photons the total K-N cross
section from Eq.(7.15), and derive the Compton mass attenuation coefficient for copper in cm2/g and m2/kg.
Solution:( ) ( ) ( )
( )2
0 22
2 13 25 20
23
2 1 ln 1 2 ln 1 21 1 3 2
1 2 2 1 2
/ 1.96, 2.818 10 , 2.112 10 /
6.0022 10 /,
e
e e
a A CuAe
Cu
r
h m c r cm cm electron
N ZN Z amu g
A A
a a aa as p
a a a a a
a n s
ss
r
- -
ì üï ïé ù+ + ++ +ï ïï ïê ú= - + -í ýê úï ï+ +ï ïë ûï ïî þ= = = × = ×
×= × = 23
23 25 2
2 2 2 3 2
292.748 10 /
63.55
2.748 10 / 2.112 10 /
0.05804 / 0.05804 (10 ) /10 0.005804 /
a
electronelectron g
amu
electron g cm electron
cm g m kg m kg
sr
-
- -
×= ×
æ ö÷ç ÷= × × ×ç ÷ç ÷çè ø
= = × =
Problem 7.7• What is the maximum energy, what is the average
energy, of the Compton recoil electrons generated by 20-keV and 20-MeV -rays?
Solution: ( )max min 2
2
max,20
max,20
20
20
11 1 cos(180)2
1.452
19.75
Eq.(7.20) page 134
0.7210
14.53
e
e
keV
MeV
keV
MeV
h hT h h h
m chhm c
T keV
T MeV
Based on
T keV
T MeV
n nn n n
nn
¢= - = - =æ ö÷ç +÷+ -ç ÷ç ÷çè ø
=
=
=
=
Problem 7.8• Calculate the energy of a photoelectron ejected from
the K-shell in tin by a 40-keV photon. Calculate tr/you may estimate from fig. 7.15.
Solution: Tin: K-edge 29.20keV T=hv-Tb=(40-29.20)keV=10.80keV
( ) ( )( )
2
1
2 2
Sn,40keV Sn,40keV
1 cm g
Tin Z= 50 and from Fig 7.15: 0, 19
40 19 0=18.9cm /g 18.9cm /g
40
K K K K L L La tr a
L L L L L b K K KL
a a tr
h P Y h P P Y h
h
P Y h P Y E P Y h keV
n n nt tr r n
n n
t tr r
æ ö- - - ÷ç ÷= ç ÷ç ÷çè ø
» » =
æ ö æ ö æ - -÷ ÷ç ç ç÷ Þ ÷ =ç ç ç÷ ÷ çç ç÷ ÷ç ç èè ø è ø29.92cm /g
ö÷=÷÷ø
Problem 7.9• What is the average energy of the charged particles
resulting from pair production in (a) the nuclear field (b) the electron field, for photons of h = 2 and 20 MeV?
Solution: a)
b)
22
min
22
min
2: 2
2
2: 4
3
ee
ee
h m cpair h m c T
h m ctrip h m c T
nn
nn
-= =
-= =
pair trip
pair trip
h = 2MeV : T = 0.489 MeV, T = 0
h = 20MeV : T = 9.49 MeV, T = 6.33 MeV
n
n
Problem 7.10• A narrow beam containing 1020 photons at 6 MeV
impinges perpendicularly on a layer of lead 12 mm thick, having a density 11.3 g/cm3. How many interactions of each type (photoelectric, Compton, pair, Rayleigh) occur in the lead?Solution:Number of interactions ∆Ntot =N0(1-e-∆x µ)
∆N´tot =N0∆x µ = N0∆x (p.el+C+pair +R)
=∆N´p.el+∆N´C+∆N´pair+∆N´R >∆Ntot
∆Np.el =∆N´p.el·∆Ntot/ ∆N´tot
Photoelectric Compton Pair Rayleigh
9.999E+17 1.792E+19 2.541E+19 5.375E+16
Problem 7.10• A narrow beam containing 1020 photons at 6 MeV
impinges perpendicularly on a layer of lead 12 mm thick, having a density 11.3 g/cm3. How many interactions of each type (photoelectric, Compton, pair, Rayleigh) occur in the lead?Solution:Number of interactions ∆Ntot =N0(1-e-∆x µ)
∆N´tot =N0∆x µ = N0∆x (p.el+C+pair +R)
=∆N´p.el+∆N´C+∆N´pair+∆N´R >∆Ntot
∆Np.el =∆N´p.el·∆Ntot/ ∆N´tot
Photoelectric Compton Pair Rayleigh
9.999E+17 1.792E+19 2.541E+19 5.375E+16