Radiation Shielding and Reactor
Criticality
Fall 2012
By Yaohang Li, Ph.D.
Review
• Last Class
– Test of Randomness
– Chi-Square Test
– KS Test
• This Class
– Monte Carlo Application in Nuclear Physics
• Radiation Shielding
• Reactor Criticality
• Simulation of Collisions
– Assignment #3
• Markov Chain Monte Carlo
Monte Carlo Method in Nuclear Physics
• Flux of uncharged particles through a medium
– Uncharged particles
• paths between collisions are straight lines
• do not influence one another
– independence
– allow us to take the behavior of a relatively small sample of
particles to represent the whole
– Randomness
• derive the Monte Carlo methods directly from the physical
processes
Problem Definition
• Particle (Photon or Neutron)
– energy E
– instantaneously at the point r
– traveling in the direction of the unit vector 
• Traveling of the Particle
– At each point of its straight path it has a chance of colliding with an atom of
the medium
• No collision with an atom of the medium
– continue to travel in the same direction  with same energy E
• A probability of cs that the particle will collide with an atom of the
medium
– s: a particle traverses a small length of its straight line
– c: cross section
» depends on the nature of surrounding medium
» energy E
Cross Section
• Determining c
– The medium remains homogeneous within each of a small
number of distinct regions
• over each region, c is a constant
• c change abruptly on passing from one region to the next
– Example
• Uranium rods immersed in water
– c a function of E in the rods
– c another function of E in the water
Collision
• Collision Probability
– cdf of the distance that the particle travels before collision
• Fc(s) = 1 – exp(- c s)
• Three situations of collision
– Absorption
• the particle is absorbed into the medium
– Scatter
• the particle leaves the point of collision in a new direction with a new
energy with probability (Ei)
– fission (only arises when the original particle is a neutron)
• several other neutrons, known as secondary neutrons, leaves the point
of collision with various energies and directions
• Probability of the three situations
– Governed by the physical law
– Known distribution from Monte Carlo point of view
Shielding and Criticality Problems
• The Shielding Problem
– When a thick shield of absorbing material is exposed to radiation (photons), of specified energy and angle of incidence,
what is the intensity and energy-distribution of the radiation
that penetrates the shield?
• The Criticality Problem
– When a pulse of neutron is injected into a reactor assembly,
will it cause a multiplying chain reaction or will it be absorbed,
and in particular, what is the size of the assembly at which the
reaction is just able to sustain itself?
Elementary Approach
• Elementary Approach
– Exact realization of the physical model
• Not very efficient
– Tracking of simulated particles from collision to collision
• Starting with a particle (E, , r)
• Generate a number s with the exponential distribution
– Fc(s) = 1 – exp(- c s)
• If the straight-line path from r to (r+s) does not intersect any boundary
(between regions)
– the particle has a collision
• Otherwise
– proceed as far as the first boundary
– if this is the outer boundary, the particle escapes from the system
• Repeat the procedure
Improvements of the Elementary
Approach
• Problem
– There may be too many or too few particles
– Consider a reactor containing a very fissile component
• Every neutron entering this region may give rise to a very large number coming
out
– Give us more tracks than we have time to follow
• Solution
– “Russian Roulette”
• Pick out one of the particles
– discard it with probability p
– otherwise allow this particle to continue but multiply its weight (initially unity) by
(1-p)-1
• The number of particles is reduced to manageable size
– “Splitting”
• To increase the sample sizes
– a particle of weight w may be replaced by any number k of identical particles of
weights w1, …, wk
» w1+…+wk=w
Special Methods for the Shielding
Problem
• Outstanding feature of the shielding problem
– The proportion of photons that penetrate the shield is very small, say one in
106.
– To estimate an accuracy of 10% require the number of 108 paths.
• Hit or miss
• Quite inefficient
• Solution
– Semi-analytic method
– Allows the same random paths to be used for shields of other thickness
– Simplification
• Only think about three coordinates
– Energy E
– Angle between the direction of motion and the normal to the stab
– Distance z from the incident face of the slab
The Semi-Analytic Method (I)
•A random history
– for a particle which undergoes a suitably large number n of scatterings in the
medium
E , E ,..., En
h  hn   0 1
  0 ,1 ,..., n
•The semi-analytic method
– Pi()
• The probability that a particle has a history hi and also crosses the plane z= 
between its ith and (i+1)th scatterings
– Abbreviation (α is the absorption probability)
• ci=cosi; i=c(Ei); i=[1-(Ei)]c(Ei)
– P0()=exp(- i/c0)
• the probability that the particle passes through z=  before suffering any
scatterings
– Pi+1()
• A particle crosses z=  between its (i+1)th and (i+2)th scatterings
The Semi-Analytic Method (II)
•The semi-analytic method
– the (i+1)th scattering occurred on some a plane z= ’ where 0<’< .
– Compound event
• i: immediately prior to the (i+1)th scattering the particle crossed z= ’
– P(i)=Pi(’)
• ii: the particle suffered the (i+1)th scattering between the planes z= ’ and z=
’+d’
– P(ii)= i d’/|ci|
• iii: after scattering, the particle now travels with energy Ei+1 in direction i+1
– P(iii)= exp(- i+1(- ’) /ci+1)
– Then

Pi 1 ( )   Pi ( ' ) exp{ i 1 (   ' ) / ci 1}
0
 i d '
| ci |
– The probability of penetrating the shield is

E ( Pi (t ))
i 0
Probability of Penetration
• Replace
with
• Approximate unbiased estimator of penetration
probability
• N = 25, 12, 9, 6 is efficient for shields of water, iron, tin,
and lead

E ( Pi (t ))
i 0
N
E ( Pi (t ))
i 0
Neutron Transport
• Transmission of Neutrons
– Bulk matter
• Plate
– thickness t
– infinite in the x and y directions
– z axis is normal to the plate
– Neutron at any point in the plate
• Capture with probability pc
– Proportional to capture cross section
• Scatter with probability ps
– Proportional to scattering cross section
Scattering
•Scattering
– polar angle 
– azimuthal angle 
• we are not interested in how
far the neutron moves in x
or y direction, the value of 
is irrelevant
X
Z
v
v'

Solid Angle
• 2D
– measured by unit angles
(radians)
– full circle subtends 2
• 3D
– measured by unit solid angles
(steradians)
– full sphere subtends 4 
Probability of Scattering
•Scattering equally in all directions
– probability p(,)dd=d/4
•Definition of the Solid Angle
   sin dd
S
– then d = sindd
– we can get p(,) = sin/4
•Probability density for  and 
p( ) 
2
1
p
(

,

)
d


sin 
0
2

p( )   p( ,  )d 
0
1
2
Non-uniform Random Sample
Generation Revisit
•Probability Density p(x)

 p( x)dx  1

•Then
x
P( x) 
 p( x' )dx'  r

– r is a uniform random number
•Inverse Function Method
– use r to represent x
Randomizing the Angles
•
– =2r
–  is uniformly distributed between 0 and 2
•

1
r   sin xdx
20
– Then we can get cos = 1-2r
– cos is uniformly distributed between -1 and +1
Path Length
•Path length
– distance traveled between subsequent scattering events
– obtained from the exponential probability density function
p(l )  (1 /  )e l / 
– l=-lnr
•  is the mean free path
• or the cross section constant c
Neutron Transport Algorithm (1)
• Input parameters
–
–
–
–
thickness of the plate t
capture probability pc
scattering probability ps
mean free path 
• Initial value z=0
Neutron Transport Algorithm (II)
1. Determine if the neutron is captured or scattered. If it is captured,
then add one to the number of captured neutrons, and go to step 5
2. If the neutron is scattered, compute cos by cos = 1-2r and l by
l=-lnr. Change the z coordinate of the neutron by lcos
3. If z<0, add one to the number of reflected neutrons. If z>t, add
one to the number of transmitted neutrons. In either case, skip to
step 5 below.
4. Repeat steps 1-3 until the fate of the neutron has been determined.
5. Repeat steps 1-4 with additional incident neutrons until sufficient
data has been obtained
An Improved Method
• Instead of Considering A Neutron
– Consider a set of neutrons
– ps portion of neutrons are scattered
• All scattered neutrons will move to a new direction
– pc portion of neutrons are captured
• A better convergence rate
Summary
• Nuclear Simulation
– Radiation Shielding
– Reactor Criticality
– Particle Assumption
• Cross Section
• Collision
– Elementary Method
– Improvements for the Elementary Method
• Russian Roulette
• Splitting
– Special methods for the shielding problem
• Semi-Analytic Method
– Neutron Transport Problem
– Nonuniform Distribution Samples
What I want you to do?
• Review Slides
• Review basic probability/statistics concepts
• Work on your Assignment 3