Intro Nuclear Science v2 - radiochem

Report
NE 301 - Introduction to Nuclear Science
Spring 2012
Classroom Session 8:
•Radiation
•
•
Interaction with Matter
Non-Charged Radiation
Mass Attenuation Tables and Use
 Absorbed Dose (D), Kerma (K)

Gray (Gy) = 100 rad
 Dose Calculations
•Analysis
of Gamma Information (NAA)
•Chemical
Effects of Nuclear Reactions
Reminder
Load TurningPoint




Reset slides
Load List
Homework #2 due February 9
Next Tuesday February 14 – 1st Demo Session
 MCA
 Gamma Spectroscopy identification of isotopes
 NAA of samples
2
Ionizing Radiation: Electromagnetic Spectrum
Ionizing Radiation
Ionizing
Each radiation have a characteristic , i.e.:
Infrared: Chemical bond vibrations (Raman, IR spectroscopy)
Visible: external electron orbitals, plasmas, surface interactions
UV: chemical bonds, fluorecense, organic compounds (conjugated bonds)
X-rays: internal electron transitions (K-shell)
Gamma-rays: nuclear transitions
Neutrons (@ mK, can be used to test metal lattices for example)
Radiation Interaction with Matter
Five Basic Ways:
1.
2.
3.
4.
5.
Ionization
Kinetic energy transfer
Molecular and atomic excitation
Nuclear reactions
Radiative processes
4
Radiation from Decay Processes
Charged

Directly ionizing (interaction with e-’s)
 β’s, α’s, p+’s, fission fragments, etc.



Coulomb interaction – short range of travel
Fast moving charged particles
It can be completely stopped
R
Uncharged




Indirectly ionizing (low prob. of interaction – more penetrating)
, X-Rays, UV, neutrons
No coulomb interaction – long range of travel
Exponential shielding, it cannot be completely stopped
5
Neutral Interactions
Stochastic (Probabilistic)
With an electron or a nucleus
Can be scattering – elastic or inelastic
Can be absorptive
It is still a collision:

Flux of particles is important
6
Flux or Intensity
Flux is usually for neutrons (n)
Intensity is usually for photons (’s)
Target
Beam
I   n . v
Density of particles in
the beam
Velocity of beam
particles
7
Attenuation of Uncollided Radiation
How do we calculate the change in the flux of
(uncollided) particles as it moves through the slab?
I0
I ( x)
 ( x)
0
x
Uncollided radiation is a simplification. In reality not every
collided photon/neutron is lost and there are buildup factors (Bi)
Attenuation of Uncollided Radiation
Beam with intensity I,
interacting with shield (1-D)
I0
I ( x)
 ( x)
0
x
Change in
 Prod. (i.e. fission/multiplication) - Loss (collisions)
Flux with x
d
 0  N
dx
d
  N dx integrating

Ln    N x  c
 ( x)  0 e  
t
calling  (t=0)=0 and calling N  t
x
9
Microscopic and Macroscopic Cross Sections
Sigma-N =

Linear Attenuation Coefficient or Macroscopic Cross Section ( or )
i   i N   i
Constant of Proportionality or
Microscopic Cross-Section
 Na
A
Notice Different Units:
 is measured in cm-1
 is measured in barns
1 barn = 10-24 cm2
10
A beam of neutrons is normally incident on a slab
20 cm thick. The intensity of neutrons transmitted
through the slab without interactions is found to be
13% of the incident intensity. What is the total
interaction coefficient t for the slab material?
93%
 t x
 ( x)  0e
7%
0%
10
cm
-1
-1
m
cm
1c
-1
0%
0.
1
4.
-1
3.
cm
2.
0.01 cm-1
0.1 cm-1
1 cm-1
10 cm-1
0.
01
1.
11
Log[0.13]
-1
t  
 0.102 cm
20
12
Attenuation of Uncollided Radiation
Beams of particles: with intensity
I0, interacting with shield (1-D)
Point sources: Isotropic source
emitting Sp particles per unit time
I (r )   (r ) 
I0
I ( x)
 ( x)
0
A0   r
e
4 r 2
r
x
I ( x)  I 0e
 t x
 ( x)  0 e  
t
x
A0   r
I (r )   (r ) 
e
2
4 r
13
Related Concepts
Mean Free Path (mfp or

x ):
Average distance a particle travels before an interaction
x=
1
t
Half-thickness (x1/2) of the slab?

Thickness of slab that will decrease uncollided flux by half
x=
Ln 2
t
Similar concepts to
mean-life and half-life
14
It is found that 35% of a beam of neutrons
undergo collisions as they travel across a 50 cm
slab. What is the mfp and x1/2 for the slab?
69
3c
10
00
an
d
80
cm
cm
.. .
0%
an
d
t
13
.8
x1/2 =
Ln 2
0%
11
6
t
an
d
I ( x)  I0e
 t x
3%
20
x=
1
cm
4.
6.
9
3.
an
d
2.
10 and 6.9 cm
20 and 13.8 cm
116 and 80 cm
1000 and 693 cm
10
1.
97%
15
Clicker solution
Log 0.65
In[6]:=
Out[6]=
50
0.00861566
1
In[7]:=
Out[7]=
,
Log 2
116.068, 80.452
16
Photon Interactions -  tables
Photon energies:

10 eV < E < 20 MeV
 IMPORTANT radiation shielding design
For this energy range,
1. Photoelectric Effect
2. Pair Production
3. Compton Scattering
19
Pair Production
Compton Scattering
The Photoelectric Effect
20
Example: Photon Interactions for Pb
Energy
Low
Photoelectric
Effect
Intermediate
Compton
Scattering
High
Pair
Production
: Gammas
22
Problem with Photons
100 mCi  source of 38Cl is placed at the
center of a tank of water 50 cm in
diameter

What is the uncollided -flux at the surface
of the tank?
Problem with Photons
100 mCi  38Cl, water tank 50 cm dia.

What is the uncollided -flux at the surface of
the tank?
I (r )   (r ) 
Sp
4 r 2
e r
r
25
26
Linear Coefficients – Macroscopic Cross Sections
Linear Absorption Coefficient

μt
Linear Scattering Coefficient

μs
Macroscopic Fission Cross-section

Σf, μf for neutrons
t   i  s     f  etc
i
27
Neutrons:
28
29
For homogeneous mixes of any type
mix   i
i
Valid for any cross section type (fission,
total, etc)
Valid for chemical compounds as well
DO NOT add microscopic cross-sections
30
In natural uranium (=19.21 g/cm3), 0.720% of the atoms are 235U,
0.0055% are 234U, and the remainder 238U. From the data in Table C.1.
What is the total linear interaction coefficient (macroscopic cross section)
for a thermal neutron in natural uranium?
t

Nat
U

 t

234
U
.N
 
234
U
t


235
U
.N
 
235
U

t

238
U
.N
 
238
U

atoms
atoms

2.67
e
18
cm3
cm3
atoms
atoms
 0.0072  4.86e22

3.50
e
20
cm3
cm3
atoms
atoms
 0.992745  4.86e22

4.82
e
22
cm3
cm3
N ( 234U )  0.000055  4.86e22
N ( 235U )
N ( 238U )
t

Nat
U

 t

234
U
.N
 
234
U

t

235
U
.N
 
235
U

t

238
U
.N
 
238
U
0.24 cm-1

1024 cm 2
atoms
1024 cm 2
atoms
t NatU  116 b 
 2.67e18

700
b


3.50
e
20
 
1b
cm3
1b
cm3
1024 cm 2
atoms

12.2
b


4.82
e
22
 0.83 cm -1
-1
3
0.0003 cm
1b
cm
Who dominates?
238U:
0.59 cm-1
31
Absorbed Dose, D (Gray, rad)
Energy absorbed per kilogram of matter (J/kg)

Gray:
1 Gy = 1 J/kg
The traditional unit:

Rad:
100 rad = 1 Gy
rad = Radiation Absorbed Man

Dose rate = dose/time
Dose = dose rate  time
Kerma (Approx. dose for neutrons)
Kerma





Kinetic Energy of Radiation absorbed per
unit MAss
For uncharged radiation
Kerma is easier to calculate than dose for
neutrons
Kerma and Dose: same for low energy
Kerma over-estimates dose at high energy
 No account for “Bremsstrahlung” radiation loses.
Calculating Dose Rate and Kerma Rate
D[Gy / s]  1.602 10
10
 en ( E ) 
2
2 1
E[ MeV ] 
[cm / g ] [cm s ]
  
en(E)/ =mass interaction coefficient (table C3)
E = particle energy [MeV]
Notice Difference
 = flux [particles/cm2 s]
K[Gy / s]  1.602 10
10
 tr ( E ) 
2
2 1
E[ MeV ] 
[
cm
/
g
]

[
cm
s ]

  
tr(E)/ =mass interaction coefficient (table C3)
E = particle energy [MeV]
 = flux [particles/cm2 s]
Engineering Equations – PLEASE Watch out for units!

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