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Interaction Between Ionizing Radiation And Matter,
Problems Charged particles
Audun Sanderud
Dep
art
men
t o
f P
hys
ics
Un
iver
sity
of
Os
lo
Problem 8.1 (see fig 8.9)
Solution:
( )
( )( )
2 2
1
Photons:
2 0.14
Charged particels:
2 0
: 1
xp x e e
x
p x
Heavy Charged particles p x
m
m
- -= = =
=
@
@
Solution: Soft: - b>>a: particle passes an atom in a large distance
- Small energy transitions to the atom- The result is excitations (dominant) and ionization;
amount energy transferred range from Emin to a certain energy H
Hard: - b<<a: particle passes trough the atom- Large (but few) energy transactions to single electron- Amount energy transferred range from H to Emax
- Can be seen as an elastic collision between free particles (bonding energy nelectable)
Problem 8.2 (Chap. 8.II.A.B)
Solution:Classic:
Relativistic:
T0=25 MeV
electron: Tmax=To/2=12.5 MeVpositron: Tmax=To=25 MeVproton: Tmax=T’max=55.19 keVα-particle: Tmax=T’max=13.75 keV
( )2 1 2
max 2 2,max 02
1 2
m m1E m v 4 T
2 m m= =
+
( )( )
2' 2max e 2
' e 0max 0 2
2 0 02 20
T 2m c m T1 T 4T 1M 2M c
1 T / M c 1-
üæ ö ïb ÷ ïç ÷= ïç é ù÷ç ï÷ç ï- bè ø ê úÞ = +ý ê úïï ë ûïb = - + ïïþ
1
1.01
1.02
1.03
1.04
1.05
1.06
1.07
0 20 40 60 80 100
T0 (MeV)
Re
l. T
ma
x/C
las.
Tm
ax
Proton
Alfa
Problem 8.5 (Eq. 8.4 and 8.11)
Solution:
T0proton=25 MeV
electron: T0=13.62 keV→ Tmax=6.81 keVpositron: T0=13.62 keV→ Tmax=13.62 keV proton: T0=25.00MeV→ Tmax=55.19 keVα-particle: T0=99.33MeV→ Tmax=55.19 keV
electron proton
e0electron 0pr oton
p
v v
mT T
m
=
ß
=
Problem 8.6 (Eq. 8.11)
Solution: z=+1, M0c2=3*938.26MeV=2815MeV, T=800MeV
→ =0.6274, Tmax=663.4 keV
Copper: H=100eV, I=322eV, Z/A=0.4564
Soft:
Hard:
Total:
( )
( )
2 2 2 22 2s e e
2 2 2C
2 2
dT 2m c z 2m c HZ cm0.150 lngdx A I 1
MeVcm MeVcm0.1778 ln 639.8 0.3936 1.079g g
é ùæ öæ ö ÷bçê ú÷ ÷ç ç÷ = - b÷ç ê úç÷ ÷ç ÷ç çr b ÷è ø - b ÷ê úçè øë û
é ù= - =ë û
( )
2 22 2e maxh
2C
2 2
2m c z TdT Z cm0.150 lngdx A H
MeVcm MeVcm0.1778 ln 6634 0.3936 1.495g g
æ ö é ùæ ö÷ç ÷çê ú÷ = - b÷ç ç÷ ÷çç ÷ ê úç è ør bè ø ë û
é ù= - =ë û
( )
( )
2 2 2 22 2e e
2 2C
2 2
2m c z 2m cdT Z cm2 0.150 lngdx A I 1
MeVcm MeVcm0.3555 ln 2060 0.3936 2.573g g
é ùæ öæ ö ÷bçê ú÷ ÷ç ç÷ = × - b÷ç ê úç÷ ÷ç ÷ç çr b ÷è ø - b ÷ê úçè øë û
é ù= - =ë û
Problem 8.7 (Eq. 8.4, 8.6, 8.7, 8.8 and 8.10)
Solution: a)
b)
( )( ) ( )( )( ) ( )
2 22 2 2
T T
2 2T T
TT
T
T
1 T / M c 1 1 T / M c 1
T / M c T / M c
MT T
M
4 4T T 800 MeV 1067MeV
3 34.0026
T T 1062MeV3.0160
- -
a a
a a
aa
a
a
- + = - + =b
=
=
» = × =
= =
22 2
2TC, C,T
zdT dT MeVcm MeVcm4 2.573 10.29g gdx dx za
a
æ ö æ ö÷ ÷ç ç÷ = ÷ × = × =ç ç÷ ÷ç ç÷ ÷ç çr rè ø è ø
Problem 8.8 (Eq. 8.11 and 8.10)
Solution: Cyclotron:
a) Deuterons:
α-particle:
b) Water: I=75eV, z=2, Z/A=(8+2*1)/(16+2*1)=10/18
( )2 2
22mv zBr 1 1 zF zvB ma v T mv Br
r m 2 2 m= = = Þ = Þ = =
2pD
D p p p D p2p D
mz 1 1 1 1 1.008T T T T , T T 50.05MeV
z m 1 2 2 1 2.014= » = = =
2p
p p p p2p
mz 4 1 4 1.008T T T T , T T 100.7MeV
z m 1 4 1 4.003a
a aa
= » = = =
( )
2
2 11 0.05158,
100.7 / 3753 1
æ ö÷ç ÷b = - ç =÷ç ÷÷ç +è ø2 2
pm c 4 m c 4 938.26MeV=3753.0MeVa » × = ×
( )
( )
2 2 2 22 2e e
2 2C
2 2
2m c z 2m cdT Z cm2 0.150 lngdx A I 1
MeVcm MeVcm13.21 ln 741.1 0.05158 86.61g g
é ùæ öæ ö ÷bçê ú÷ ÷ç ç÷ = × - b÷ç ê úç÷ ÷ç ÷ç çr b ÷è ø - b ÷ê úçè øë û
é ù= - =ë û
Problem 8.9 (next lection and Eq. 8.11 and 8.10)
Solution: Proton: mpc2=938.26MeV, z=1, T=20MeV, β2=0.04131
Lead: I=823eV, Z 82
0.3958A 207.2
= =
( )
( )
2 2 2 22 2e e
2 2C
2 2
2m c z 2m cdT Z cm2 0.150 lngdx A I 1
MeVcm MeVcm2.938 ln 53.51 0.04131 11.57g g
é ùæ öæ ö ÷bçê ú÷ ÷ç ç÷ = × - b÷ç ê úç÷ ÷ç ÷ç çr b ÷è ø - b ÷ê úçè øë û
é ù= - =ë û
Problem 8.10 (Eq. 8.11 and 8.10)
Solution: Electrons: z=1, T=50MeV, =0.9999≈1,
Aluminium: I=166eV, =5.068,
a) electrons:
b) positrons:
Z 130.4818
A 26.98= =
2eT / m c 50 / 0.511 97.85t = = =
( )
( )( )
[ ]
222
22 2C e
2 2
2dT Z zMeVcm0.1535 ln Fgdx A 2 I / m c
MeVcm MeVcm0.07396 29.14 0.1086 5.068 1.788g g
-
é ùæ ö÷æ ö ç t t +ê ú÷ç÷ç ÷÷ = + t - dê úçç ÷÷ çç ÷ ÷ç ê úr bè ø ç ÷÷çè øê úë û
= + - =
( )( )
( )
22
2
/ 8 2 1 ln2 1197-136.3F 1 0.1086
97711- t - t +
t = - b + = =t +
( )( ) ( )
2
2 3
14 10 4F 2ln 2 23 1.386 1.917-0.01168-0-0=-0.5427
12 2 2 2+
ì üï ïb ï ïï ït = - + + + = -í ýï ït + t + t +ï ïï ïî þ
( )
( )( )
[ ]
222
22 2C e
2 2
2dT Z zMeVcm0.1535 ln Fgdx A 2 I / m c
MeVcm MeVcm0.07396 29.14-0.5427 5.068 1.740g g
+
é ùæ ö÷æ ö ç t t +ê ú÷ç÷ç ÷÷ = + t - dê úçç ÷÷ çç ÷ ÷ç ê úr bè ø ç ÷÷çè øê úë û
= - =
Problem 8.11 (Eq. 8.13 and 8.13ab)
Solution:
Electrons: T=50MeV, n=750MeV
a) Al, Z=13:
b) Lead, Pb: Z=82, I=823eV, Z/A=0.3958
R C
dT dT TZ
dx dx n
æ ö æ ö÷ ÷ç ç÷ = ÷ç ç÷ ÷ç ç÷ ÷ç çr rè ø è ø
2 2
R
dT 50 13 MeVcm MeVcm= 1.788 1.550g gdx 750
æ ö ×÷ç ÷ =ç ÷ç ÷çrè ø
Problem 8.12 (Eq. 8.15 and 8.13(b))
( )
( )( )
[ ]
222
22 2C e
2 2
2 2
R
2dT Z zMeVcm0.1535 ln Fgdx A 2 I / m c
MeVcm MeVcm0.06076 25.94 0.1086 3.579 1.365g g
dT 50 82 MeVcm MeVcm1.365 7.462g gdx 750
-
é ùæ ö÷æ ö ç t t +ê ú÷ç÷ç ÷÷ = + t - dê úçç ÷÷ çç ÷ ÷ç ê úr bè ø ç ÷÷çè øê úë û
= + - =
æ ö ×÷ç ÷ = =ç ÷ç ÷çrè ø
Solution: Radiation yield: Y(10MeV)= 0.2265 and Y(7MeV)= 0.1734
Energy of radiation per electron as it slows down to rest is: Ee= Y(T0)T0
Total radiation energy from 10MeV electrons: E10= N Ee =NY(T0)T0
=1015 0.2265 10MeV 1.602 10-19J/eV=362.9J
Total radiation energy from 7MeV electrons: E7= N Ee =NY(T0)T0
=1015 0.1734 7MeV 1.602 10-19J/eV= 194.5J
Radiation energy from 10MeV electrons until 7 MeV electrons:
E10- E7 = 168.4J
Problem 8.13 (Chapter 8.III.G)
Solution: Iron, Z=26Carbon, Z=6:
At 30 MeV proportional to Z0.3:
Problem 8.14 (Eq. 8.23 and text below)
( )1.77
20CSDA C
T 1g / cm
415 670
æ ö÷ç ÷ +ç ÷ç ÷çè ø;
( ) ( )0.3 1.77
2 2CSDA CSDAFe C
3026 11.55 g / cm 1.54g/cm
6 415 670
æ öæ ö ÷ç÷ç ÷  = + =÷ çç ÷÷ çç ÷çè ø è ø;
( )( )
2
0.3
CSDA 1
CSDA 2Z
Z
Z
æ ö ÷ç ÷ç ÷ç ÷ç è ø;
Solution: a) TD=TPmD/mP=2TP=60 MeV
b)
Problem 8.15 (Argument before and after Fig 8.8.The points a.,b.,c.)
( ) ( ) 2DCSDA CSDAFe,D Fe,p
p
m3.085g/cm
m =  =
