CHAPTER 2
2-1
RADIATION INTERACTION WITH MATTER
Beta Ray Interaction with Matter
The swift electrons (e+ or e-) are produced from the following mechanisms:
 -decay;
e+ or e- in continuous energy spectrum
 pair production;
e+ or e- in continuous energy spectrum
 internal conversion;
e-, mono-energetic orbital electron (X-rays or Auger
electrons accompany)
 cathode rays, synchrotron, betatron, etc.;
 Compton scattering;
e-, mono-energetic
e-, in continuous energy spectrum
 interaction of heavy ionizing particles with matter;
e-, -rays (avalanche of
electrons)
 muon decay;
-  e- +  +  (neutrino + antineutrino)
+  e+ +  + 
where,
(+  + + :
 = 2.15 sec
 = 0.02 sec)
 = mean time = 1/.
The beta-ray reactions with matter are summarized as follows:
 collision with atomic electrons: ionization, excitation (dominates in few MeV
region),
 collision with nucleus: bremsstrahlung (in high energy region, especially in high Z
material, ex. Ec of Pb =7.8 MeV), and
 annihilation: positron + negatron.
1) ionization and excitation:
The linear energy loss (MeV/cm) due to ionization and excitation by Bethe (relativistic
due to very small mass of electron) is given by:
dE 2q 4 NZ (3x10 9 ) 4

dx Em  2 (1.6 x10 6 ) 2
2


  Em Ek  

 2
ln  2
2 


  I (1   ) 

1
MeV/cm
(2-1.1)
where,
q = electron charge, 1.6x10-19 C
N = # of absorber atoms per cm3
Z = absorber atomic number (NZ of STP air = 3.88x1020 e/cm3)
STP (standard temperature and pressure) conditions in air:
T = 0oC,
P = 1 atm,
air= 1.293x10-3 g/cm3
Em = electron mass energy (= 0.511 MeV)
Ek = kinetic energy of beta particles, MeV
 = v/c
I = mean (geometric) ionization and excitation potential of absorbing atom, MeV
I  1.35x10-5 Z (I = 8.6x10-5 for air); Check Table 5.1 of Cember for details.
The specific ionization (S.I.) is defined by:
S .I . 
dE
w
dx
ion pairs/cm
(2-1.2)
where,
w = mean energy expenditure per ion pair. w = 34 eV/i.p. for air (independent of
incident  energy). The mean energy expenditure per ion pair is summarized in Table
5.1 of Cember as follows:
Table 5.1 Average Energy Lost by a Beta Particle in the Production of an Ion Pair
Gas
Ionization Potential (eV)
Mean Energy Expenditure per Ion
Name
Pair (eV)
H2
13.6
36.6
hydrogen
He
24.5
41.5
helium
N2
14.5
34.6
nitrogen
O2
13.6
30.8
oxygen
Ne
21.5
36.2
neon
A
15.7
26.2
argon
Kr
14.0
24.3
krypton
2
Xe
12.1
Air
21.9
xenon
33.7
air
CO2
14.4
32.9
carbon dioxide
CH4
14.5
27.3
methane
C2H2
11.6
25.7
acetylene
C2H4
12.2
26.3
ethylene
C2H6
12.8
24.6
ethane
Example 2-1-1:
Specific ionization: Find S.I. of 0.1 MeV beta particle through STP
dry air.
Solution:


1
Ek  mo c 2 
 1
 1  2





1
0.1 MeV = 0.511 MeV 
 1
 1 2



dE 2 (1.6 x10 19 ) 4 (3.88 x10 20 )(3x109 ) 4

dx
(0.511)(0.3010)(1.6 x10 6 ) 2
 2 = 0.3010
  (0.511)(0.1)(0.3010) 

 0.3010
ln 

5 2
  (8.6 x10 ) (1  0.3010) 

= 4.75x10-3 MeV/cm
Hence,
S.I. = (4750 eV/cm)/(34 eV/i.p.) = 140 i.p./cm
The similar terms of ionization and excitation loss are expressed as follows:
 LET (linear energy transfer)
keV/
 mass stopping power:
S
dE

dx
MeV/g/cm2
(2-1.3)
 relative mass stopping power: (~ 1)
m 
S medium
S air
(2-1.4)
3
Example 2-1-2:
Relative Mass Stopping Power: Find the relative mass stopping power
of graphite for 0.1 MeV beta particle.
Solution:
graphite = 2.25 g/cm3
NZ = (6.02x1023 atoms/mole)(2.25 g/cm3)(6 e/atom)/(12 g/mole)
= 6.77x1023 electrons/cm3
I = (1.35x10-5)(6) = 8.1x10-5
MeV

dE 2 (1.6 x10 19 ) 4 (6.77 x10 23 )(3x109 ) 4   (0.511)(0.1)(0.3010) 

 0.3010
ln 

6 2
5 2
dx
(0.511)(0.3010)(1.6 x10 )
  (8.1x10 ) (1  0.3010) 

= 8.67 MeV/cm
Sgraphite = 8.67/2.25 = 3.85 MeV/g/cm2
Sair
= (4.75X10-3)/ (1.293X10-3) = 3.67 MeV/g/cm2
 m = 3.85/3.67 = 1.05
2) bremsstrahlung (radiative energy loss):
“Bremsstrahlung” is the radiative loss of energy which appears as the continuous
electromagnetic radiation produced by inelastic scattering of beta particle with nuclei.
Bremsstrahlung becomes significant with high energy beta particles in high Z materials.
In particular, we should be very careful about using most common Pb (lead: Z=82)
shield since shielding of high energy beta using lead creates an accompanying problem
of secondary radiation of electromagnetic radiation. Extra shielding should be
considered for photons due to bremsstrahlung.
The fraction of the incident beta energy converted into bremsstrahlung photon is
proportional to beta particle energy and atomic number of shield medium, which is
given by:
4
f  3.5x104 ZEmax
(2-1.5)
where,
Z = absorber atomic number
Emax = maximum energy of beta particle, MeV
Example 2-1-3:
Shielding of bremsstrahlung:
Find the photon energy flux (MeV/cm2
sec) from a source of 1 Ci of P-32 which is surrounded by Pb shield.
Solution:
P-32  S-32 + -
( E m ax = 1.71 MeV)
T1/2 = 14.3 d
a) Find the energy fraction converted to photon in the Pb shield:
f = (3.5X10-4)(82)(1.71) = 0.049
b) Find the photon energy flux:
E m ax = 1.71 MeV or E avg  1/3 E m ax
I  E 
fEavg
4r 2
S
1
(0.049) (1.71MeV /  )
3

(3.7 x1010  / s e c )
2
4x10
= 8.22x105 MeV/cm2 sec
2)
range of beta rays:
The range is defined as the flight distance of beta particle in a medium. The range of
beta particle (in “mass thickness”, x (cm)(mg/cm3)= mg/cm2 ) for any material is
R(mg / cm 2 )  412 E 1.2650.0954ln E
:0.01 E  2.5 MeV
5
R(mg / cm 2 )  530E  106
: E  2.5 MeV
(2-1.6)
where, E = Emax (maximum beta energy in MeV).
Example 2-1-4:
Range of beta particles: Find the range of P-32 beta particle in air
and aluminum.
Solution:
Since Emax of P-32 beta particle is 1.71 MeV, the first equation of Eq.2-1.6 is used.
R(mg / cm 2 )  412(1.71)1.2650.0954ln1.71 = 760 mg/cm2
R (cm) in air = 760 mg/cm2 / 1.293 mg/cm3 = 611 cm = 6.11 m
R (cm) in Al = 760 mg/cm2 / 2700 mg/cm3 = 0.293 cm
We notice that P-32 beta particles ( Emax = 1.71 MeV) can be stopped by a thin
aluminum shield (0.293 cm) while can travel quite a long distance in air (6.11 m) on
average.
6
2-2
Alpha Ray Interaction with Matter
The heavy charged particles are summarized in the table as follows:
names
-meson (muon)
-meson (pion)
symbols
mass
spin
charge
+
207 m0
1/2
+e
-

207 m0
1/2
-e
+
273 m0
0
+e
273 m0
0
-e
1 amu(1836 m0)
1/2
+e
2 amu
1
+e
3 amu
1/2
+e
1
proton
p, H
deuteron
d, H2
3
triton
t, H
helium-3
He3
3 amu
1/2
+2e
-ray
, He4
4 amu
0
+2e
note: electron mass = m0 = ~10-27 gram
The heavy charged particle reactions with matter are as follows:
 inelastic collision with atomic electrons: ionization, excitation
 inelastic collision with nucleus: bremsstrahlung
 very high energy interaction: nuclear reaction (400-500 MeV), star reaction (BeV)
1) linear energy loss:
The linear energy loss due to ionization and excitation (non-relativistic) is:
 v2  v2 
dE 4z 2 q 4 NZ (3 x10 9 ) 4  2 Mv 2

ln

ln

1  2   2 
dx
I
Mv 2 x1.6 x10 6 
 c  c 
MeV/cm
(2-2.1)
where,
q = electron charge, 1.6x10-19 C
N = # of absorber atoms per cm3
z = atomic # of the ionizing particle (z=2 for -particle)
Z = absorber atomic number
M = rest mass of ionizing particle in gram (M=6.60x10-24 g for -particle)
7
v = velocity of ionizing particle, cm/sec
c = light velocity, 3x1010 cm/sec
I = mean (geometric) ionization and excitation potential of absorbing atom, ergs
I  2.16x10-11 Z (I = 1.38x10-10 for air)
2) -particle ranges:
In the STP air, the -particle ranges are given by:
R(cm) = 0.56 E (MeV)
; E < 4 MeV
R(cm) = 1.24 E (MeV) - 2.62
; 4 < E < 8 MeV
(2-2.2)
The ranges of other heavy charged particles in air are:
proton;
RH(E) = R(4E) - 0.2 cm
deuteron;
Rd(E) = 2 RH(1/2 E)
helium-3;
RHe-3(3E) = 3/4 R(4E)
(2-2.3)
We can see that the range of heavy charged particles are proportional to A/z2 for the
same speed of particles, i.e.,
R(cm)  A/z2
for the same speed of particles
(2-2.4)
where,
A = mass number of ionizing particle (A=4 for )
z = atomic number of ionizing particle (z=2 for )
ex.
(A/z2)H / (A/z2) = 1
(A/z2)d / (A/z2) H = 2
(A/z2)He-3 / (A/z2) = 3/4
The ranges in different absorber media are:
R1  2

R2  1
A1
A2
; Bragg-Kleeman Rule
(2-2.5)
8
where,  = density of absorber medium, g/cm3.
Range of Alpha Particles: Find the thickness of Al foil to stop Po210-
Example 2-2-1:
 (E=5.3 MeV).
Solution:
Al = 2.7 g/cm3
A Al = 27
air = 1.293x10-3 g/cm3
A air = 14.5
From Eq.2-2.2, Rair = (1.24) (5.3) - 2.62 = 3.95 cm;
From Eq.2-2.5, R Al 
2
1
A1
A2
Rair 
range in air
1.293x10 3 27
3.95 = 2.58x10-3 cm
2.7
14.5
Hence, heavy charged particles can be stopped by a thin aluminum foil, and only travel a
short distance even in air.
3) secondary electrons:
The heavy charged particles produces electrons by excitation and ionization. These
electrons are called primary electrons, and these primary electrons also produce
secondary electrons due to succeeding excitation and ionization. Hence, the energy
required for the production of an ion-pair, W (eV/ion-pair) is almost independent of
incident particle energy, which can be observed in the following table.
W (eV/ion-pair)
Gas
H
He
N
O
Ne
A
air
Po-
35.0
30.2
36.2
32.3
29.0
27.7
34.8
350MeV-p
35.3
29.9
33.6
31.5
28.6
25.5
33.3
We sometimes used the term called -ray, which is defined as the swiftly ejected
electrons and heavy charged particles due to primary electrons.
9
2-3
Electromagnetic Radiation Interaction with Matter
The electromagnetic radiations could be categorized by origin as follows:
 X-ray (electromagnetic radiation produced by atomic transition)
 -ray (electromagnetic radiation produced by nuclear transition)
And by energy shape, it is classified as:
 discrete:
-
nuclear transition (-ray)
-
atomic transition (characteristic X-rays)
-
annihilation
0.511 MeV -ray (e) & MeV -ray (e++e-)
 continuous:
-
bremsstrahlung (inelastic scattering of charged particles with
-
absorber nucleus)
-
Compton scattered photons (elastic scattering of photons with
-
orbital electrons)
The reactions of electromagnetic radiation with matter are:
 photo-electric effect
 Compton effect
 pair production
 neutron production (photo-neutron)
1) photo-electric effect:
The photo-electric effect of electromagnetic radiation with matter is the total absorption
of photon by absorber atoms and subsequent ejection of orbital electron. Hence, the
ejected photon carries the kinetic energy of incident photon subtracted by the binding
energy of the ejected orbital electron.
10
h
atom (total absorption)
e- (photo electron)
T = h - Be
where, T = kinetic energy of ejected electron
Be= binding energy of ejected electron
h = incident photon energy.
2) Compton effect:
The Compton effect of electromagnetic radiation with matter is the elastic scattering of
photon with atomic orbital electron.
T = ho - h’
e- (Compton electron)
E=ho


E’= h’
(scattered photon)
Energy conservation gives,
h 0  h   T
(2-3.1)
In the parallel to incident particle direction, momentum conservation gives,
h 0 h 

cos   p cos 
c
c
(2-3.2)
In the normal to incident particle direction, momentum conservation gives,
0
h 
sin   p sin 
c
(2-3.3)
11
The relativistic relation gives,
pc  T (T  2m0c2

Let,
(2-3.4)
h 0
m0 c 2
(2-3.5)
Eliminating  and T,
c
c
h

    0 
(1  cos  ) ;
 0
m0 c 2
“Compton shift”
(2-3.6)
The scattered photon energy (scattering angle dependent) is given by:
E, 
E
(2-3.7)
E
1
(1  cos  )
mo c 2
The energy of Compton electron is:
T  h 0
 (1  cos  )
1   (1  cos  )
(2-3.8)
The angular differential X-section (called “Klein-Nishina formula”) is:
d ro2   ,
 
d 2   o



2
 o  ,

 , 
 sin 2  
  o

cm2/electron/steradian
(2-3.9)
where, ro = e2/moc2 = 2.818x10-13 cm: “classical radius of electron”
And the angular differential scattering X-section is:
d s   ,  d
 
d   o  d
(2-3.10)
12
Example 2-3-1:
After Compton scattering, find the followings.
(1) 90o scattered energy of 1 MeV photon (from Eq.2-3.7)
1
E, 
1
1
(1  cos 90 o )
0.511
= 0.388 MeV
(2) total angular differential X-section
(from Eq.2-3.9)
d (2.818 x10 13 ) 2  0.338   1
0.338



 sin 2 90 o 

 
d
2
1
 1   0.338

2
= 1.04x10-26 cm2/electron/steradian
(3) scattering angular differential X-section (from Eq.2-3.10)
d s  0.388 
 26

(1.04 x10 )
d  1 
= 0.352x10-26 cm2/electron/steradian
3) pair production:
The creation of (e+ + e-) pair due to photon interaction with the electric field formed
around atomic nucleus is called “pair production.”
T+
e+ (positron)
ho
electric field
e- (electron)
T-
T+ + T- = h - 2 moc2 (2 x 0.511 MeV)
4) photo-neutron production:
13
The photo-neutron production is a threshold reaction. The threshold energies of photoneutron production for various media are as follows:
nuclide
H-2(,n)
Li-6(,n)
Li-6(,n+p)
Li-7(,n)
Be-9(,n)
C-13(,n)
Eth(MeV)
2.225
5.67
3.698
7.251
1.665
4.946
14
2-4
Attenuation of Photons
1) linear attenuation coefficient of photon:
The photon reaction rate (# of reactions/sec.cm3) with matter is proportional to the
intensity of photon flux (# of photons/ sec.cm2), i.e.,
F
reactions/sec.cm3
where,  = photon flux, # of photons/ sec.cm2
Hence,
F   F =  
(2-4.1)
where,
 = “total linear attenuation coefficient” (proportionality constant), cm-1
=++
(2-4.2)
 = linear attenuation coefficient by photo-electric effect
 = linear attenuation coefficient by pair production
 = linear attenuation coefficient by Compton effect
2) linear energy absorption coefficient of photon:
e =  +  + e =  - s ( =e + s)
(2-4.3)
where, e = total linear energy absorption coefficient, cm-1
e = linear energy absorption coefficient by Compton effect
s = linear scattering coefficient by Compton effect
The linear coefficients are sometimes given by:
/
cm2/g
- energy absorption mass attenuation coefficient; e/
cm2/g
- mass attenuation coefficient;
15
For the mixture of absorbers, it is given by



 w1 1  w2 2  

1
2
(2-4.4)
where, w1, w1, ... = weight fractions.
Total Mass Attenuation Coefficients (/ in cm2/g)
Mat.
Photon Energy, MeV
0.1
0.15
0.2
0.3
0.5
0.8
1.0
1.5
2
3
5
8
10
C
.149
.134
.122
.106
.0870
.0707
.0636
.0518
.0444
.0356
.0270
.0213
.0194
Al
.161
.134
.120
.103
.0840
.0683
.0614
.0500
.0432
.0353
.0282
.0241
.0229
Fe
.344
.183
.138
.106
.0828
.0664
.0595
.0485
.0424
.0361
.0313
.0295
.0294
Pb
5.29
1.84
.896
.356
.145
.0836
.0684
.0512
.0457
.0421
.0426
.0459
.0489
Air
.151
.134
.123
.106
.0868
.0706
.0655
.0517
.0445
.0357
.0274
.0220
.0202
H2O
.167
.149
.136
.118
.0966
.0786
.0706
.0575
.0493
.0396
.0301
.0240
.0219
Conc.
.169
.139
.124
.107
.0870
.0707
.0635
.0517
.0445
.0363
.0287
.0243
.0229
Tiss.
.163
.144
.132
.115
.0936
.0761
.1683
.0556
.0478
.0384
.0292
.0233
.0212
Energy Absorption Mass Attenuation Coefficients (e/ in cm2/g)
Photon Energy, MeV
Mate.
0.1
0.15
0.2
0.3
0.5
0.8
1.0
1.5
2
3
5
8
10
C
.0215
.0246
.0267
.0288
.0297
.0289
.0280
.0256
.0237
.0209
.0177
.0153
.0145
Al
.0373
.0283
.0275
.0283
.0286
.0278
.0270
.0248
.0232
.0212
.0192
.0183
.0182
Fe
.225
.0810
.0489
.0340
.0294
.0274
.0261
.0242
.0231
.0224
.0227
.0239
.0250
Pb
5.193
1.753
.821
.294
.0994
.0505
.0402
.0306
.0293
.0305
.0352
.0412
.0450
Air
.0233
.0251
.0268
.0288
.0297
.0289
.0280
.0256
.0238
.0211
.0181
.0160
.0153
H2O
.0253
.0278
.0300
.0321
.0330
.0321
.0311
.0285
.0264
.0233
.0198
.0173
.0165
Conc.
.0416
.0300
.0289
.0294
.0296
.0287
.0278
.0256
.0239
.0216
.0194
.0180
.0177
Tiss.
.0271
.0282
.0293
.0312
.0320
.0311
.0300
.0276
.0256
.0220
.0192
.0168
.0160
3) uncollided flux;
The uncollided flux is the “flux penetrated a material without any interactions.” The
characteristics of the uncollided flux is that it can be presented analytically.
Let’s consider a mono-directional photon source as described in the figure as follows:
16
Absorber material
o
(x)
#/cm2.sec
dA
x=0 x’ x’+dx’
x=x
In order to keep the particle conservation within the small thickness dx’, the reduction
rate of particles while traveling dx’ equals the particle interaction rate in dV, i.e.
(x’)dA - (x’+dx’)dA = (x’)dV
where, dV = dA dx’
d(x’)  (x’+dx’) - (x’) = -(x’)dx’
d(x’)/(x’) = -dx’
(x) = o exp(-x):
“uncollided flux”
analytically presentable (2-4.5)
where,
o = (x=0), and
x = absorber thickness, cm.
For an isotropic source,
So(#/sec)
r
the uncollided flux at r away from the source is given by:
17
 (r )  S 0
exp(  r )
4r 2
(2-4.6)
4) buidup factor;
uncollided flux
source
detector
scattered flux
There are two contributions to the detector: uncollided and scattered fluxes. Hence, the
observed response by the detector is given as the addition of response due to uncollided
flux and response due to scattered flux. Here, the responses could be flux, dose, energy
absorption, heat, etc.
The incorporation of contribution by scattered flux is usually done by “buildup
factors.” Let’s define “buildup factor, B”:
B
observed response
response due to uncollided flux
 1
then,
response due to scattered flux
response due to uncollided flux
1
observed response = B x (response due to uncollided flux).
We better note that B is a function of: effects of interest (energy, flux, dose, etc.),
source-shield-detector geometry, r (absorber material thickness = relaxation length)
and E (beam energy). It is quite customary to use the exposure buildup factor (Br) for
dose or flux calculations, and the energy buildup factor (Ba, or Be) for heat generation in
shield. As an example, for selected absorber materials, the dose buildup factors for a
18
point source in infinite media are as given in the following table.
Dose Buildup Factor B(r) for a Point Source
Material
H2O
Al
Fe
Pb
Photon Energy, MeV
r
0.5
1.0
2.0
3.0
4.0
6.0
8.0
10.0
1
2.52
2.13
1.83
1.69
1.58
1.46
1.38
1.33
2
5.14
3.71
2.77
2.42
2.17
1.91
1.74
1.63
4
14.3
7.68
4.88
3.91
3.34
2.76
2.40
2.19
7
38.8
16.2
8.46
6.23
5.13
3.99
3.34
2.97
10
77.6
27.1
12.4
8.63
6.94
5.18
4.25
3.72
15
178
50.4
19.5
12.8
9.97
7.09
5.66
4.90
20
334
82.2
27.7
17.0
12.9
8.85
6.95
5.98
1
2.37
2.02
1.75
1.64
1.53
1.42
1.34
1.28
2
4.24
3.31
2.61
2.32
2.08
1.85
1.68
1.55
4
9.47
6.57
4.62
3.78
3.22
2.70
2.37
2.12
7
21.5
13.1
8.05
6.14
5.01
4.06
3.45
3.01
10
38.9
21.2
11.9
8.65
6.88
5.49
4.58
3.96
15
80.0
37.9
18.7
13.0
10.1
7.97
6.56
5.63
20
141
58.5
26.3
17.7
13.4
10.4
8.52
7.32
1
1.98
1.87
1.76
1.55
1.45
1.34
1.27
1.20
2
3.09
2.89
2.43
2.15
1.94
1.72
1.56
1.42
4
5.98
5.39
4.13
3.51
3.03
2.58
2.23
1.95
7
11.7
10.2
7.25
5.85
4.91
4.14
3.49
2.99
10
19.2
16.2
10.9
8.51
7.11
6.02
5.06
4.35
15
35.4
28.3
17.6
13.5
11.2
9.89
8.50
7.54
20
55.6
42.7
25.1
19.1
16.0
14.7
13.0
12.4
1
1.24
1.37
1.39
1.34
1.27
1.18
1.14
1.11
2
1.42
1.69
1.76
1.68
1.56
1.40
1.30
1.23
4
1.69
2.26
2.51
2.43
2.25
1.97
1.74
1.58
7
2.00
3.02
3.66
3.75
3.61
3.34
2.89
2.52
10
2.27
3.74
4.84
5.30
5.44
5.69
5.07
4.34
15
2.65
4.81
6.87
8.44
9.80
13.8
14.1
12.5
20
2.73
5.86
9.00
12.3
16.3
32.7
44.6
39.2
19
Hence, the observed exposure rate at r from a point source is

x(r )  Br ( r , E )
S o  r
e C
4r 2
(2-4.7)
where,
Br(r,E) = exposure buildup factor with absorber thickness of r for the incident energy
E (usually given for an infinite medium for a point source)
So = photon point source with energy E, #/sec or MeV/sec
r = distance between source and detector, cm
 = total linear attenuation coefficient, cm-1
C = conversion factor from flux to exposure.
20
Example 2-4-1:
Find the observed flux of 1 MeV -rays (So = 1010 photons/sec) with
10 cm Pb shield.
Solution:
a) uncollided flux
u(r=10 cm) = So exp(-r) / 4r2
= (1010 photons/sec) exp(-[(0.0684 cm2/g)(11.34 g/cm3)][10 cm]) / 4(10 cm)2
= 3.6x103 photons/sec.cm2
b) observed flux (1 MeV equivalent)
(r=10 cm) = Br(r=7.71,E=1 MeV) u
= (3.1)(3.6x103) = 1.12x104 photons/sec.cm2
21
2-5
Neutron Interactions with Matter
The sources of neutron are:
 fission neutrons;
U-235 + n  f.p. + 2.5 n (Eav = 2 MeV)
 spontaneous fission; Cf-252 (2.34x1012 n/sec.g, Eav=2.35 MeV, T1/2=2.65 y)
 delayed neutrons:
Br-87(54.5s), I-137(24.4s), Br-88(16.3s), etc.
 activation neutrons;
Be-9 (d,n) B-10, Be-9 (,n) C-12
 photo-neutrons;
Be-9 (,n) B-8, d (,n) p
The neutron interactions with matter are with atomic nucleus: scattering and absorption:
 = s + a.
1) scattering (s)
The energy transfer fraction to the target nucleus during collision is
f  1
E
 1  e 
Eo
(2-5.1)
where,

= average logarithmic energy decrement
Eo = incident neutron energy
E
= average scattered energy after collision
material
H
C

1
0.159
  1
U
0.00838
Fe
0.920
0.0357
 ln 
1
 A 1

 A  1
(2-5.2)
 
n
 
H2O

i 1
n
si

i 1
N i i
si
Ni
22
2) absorption (a)
Typical neutron absorption reactions are presented in the following table.
reaction
H-1(n,)H-2
X-section
N-14(n,p)C-14
0.33 b
B-10(n,)Li-7
4.01x103 b
1.70 b
Cd-113(n,)Cd-114
2.1x104 b
The average thermal absorption X-section is given by the Westcott formula as follows:
  To 
1/ 2
a 
   a ( Eo ) g a (T )
2 T 
(2-5.3)
where, To = 293 K and Eo = 0.0253 eV (V =2200 m/sec).
23
Example 2-5-1:
Find the activity of Cr-51 after 1 week thermal neutron activation of
Cr-50 under thermal neutron flux of 1x1011 n/cm2.sec. The facility operates at the
temperature of 20oC.
Solution:
Cr-50 natural abundance = 4.31%
n (v=2200 m/sec)= 15.2 b, considering 1/v-absorber,
a 
  To 

1/ 2
 a ( Eo ) = (0.886)(15.2 b) = 13.5 b
   a ( Eo ) g a (T ) 
2 T 
2
chromium molecular weight = 52.01 g
Cr foil mass = 9.46x10-5 g
thermal neutron flux;  = 1x1011 n/cm2.sec
half-life of Cr-51 = 27.8 days
dN
  n    N
dt
dN
 N   n 
dt
N (t )  e t  et  n dt  ce t 
N (t  0)  0 
 n 

c
 n 

p= et
 ce t
c
 n 

 N = n (1 - e-t)
n = Ncr n 
= [(9.46x10-5 g)(0.0431)/52.01 g](6.02x1023) (13.5x10-24) (1011)
= 6.36x104 Bq (saturated activity, i.e., at t = )
For 1 week activation, t= 7days,
A = N
= n (1 - e-t)
= (6.36x104 Bq) (1 - e-(0.693 /27.8)(7))
= 1.018x104 Bq
= 0.275 Ci
24