Interaction Between Ionizing Radiation And Matter, Problems Photons Audun Sanderud Department of Physics University of Oslo.

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