Intro Nuclear Science v2 - radiochem

advertisement
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!
Download
Related flashcards

Quantum field theory

40 cards

Elementary particles

12 cards

Standard Model

11 cards

Quantum field theory

39 cards

Particle accelerators

13 cards

Create Flashcards