Session I.4.6
Part I Review of Fundamentals
Module 4 Sources of Radiation
Session 6 Basic Reactor Physics Theory
4/2003 Rev 2
IAEA Post Graduate Educational Course
Radiation Protection and Safe Use of Radiation Sources
I.4.6 – slide 1 of 63
Overview
 In this session we will discuss fission and
fusion reactions
 We will also discuss criticality
4/2003 Rev 2
I.4.6 – slide 2 of 63
Fission
fission
fragment
free
neutron
beta
nucleus
alpha
energy
free
neutron
4/2003 Rev 2
gamma
free
neutron
fission
fragment
I.4.6 – slide 3 of 63
Fission
Sample fission reactions:
235U
+ n  141Ba + 92Kr + 3n + 170 MeV
235U
+ n  94Zr + 139La + 3n + 197 MeV
4/2003 Rev 2
I.4.6 – slide 4 of 63
Fission
 Follows neutron capture
 Thermal neutrons fission 233U, 235U, 239Pu
which have odd number of neutrons
 For isotopes with even number of
neutrons, the incident neutron must have
energy above about 1 MeV
4/2003 Rev 2
I.4.6 – slide 5 of 63
Fission
4/2003 Rev 2
I.4.6 – slide 6 of 63
Fission
Neutron Name/Title
Cold Neutrons
Thermal Neutrons
Epithermal Neutrons
Cadmium Neutrons
0 < 0.025
0.025
0.025 < 0.4
0.4 < 0.6
Epicadmium Neutrons
0.6 < 1
Slow Neutrons
1 < 10
Resonance Neutrons
Intermediate Neutrons
Fast Neutrons
Relativistic Neutrons
4/2003 Rev 2
Energy (eV)
10 < 300
300 < 1,000,000
1,000,000 < 20,000,000
>20,000,000
I.4.6 – slide 7 of 63
Fission
4/2003 Rev 2
I.4.6 – slide 8 of 63
fission
products
and
transuranics
from
neutron
capture
4/2003 Rev 2
I.4.6 – slide 9 of 63
Fission
Source of energy released during fission:
 Kinetic energy of fission fragments
 Gamma rays
 Kinetic energy of neutrons emitted
 Prompt
 Delayed
4/2003 Rev 2
I.4.6 – slide 10 of 63
Criticality
 Neutrons ejected during fission equal
neutrons producing more fissions
+ neutrons absorbed
+ neutrons lost from system
 Criticality is constant if balance exists.
Fission rate (power) can be changed by
varying the number of neutrons absorbed
and/or controlling the number lost
4/2003 Rev 2
I.4.6 – slide 11 of 63
Criticality
Multiplication Factor (4 factor formula)
Nf+1
Keff =
Nf
Nf+1 is the number of neutrons produced in
the “f+1” generation by the Nf neutrons of the
previous “f” generation
4/2003 Rev 2
I.4.6 – slide 12 of 63
Criticality
 Sub-Critical (keff < 1) – more neutrons lost
by escape from system and/or non-fission
absorption by impurities or “poisons” than
produced by fission.
 Critical (keff = 1) – one neutron per fission
available to produce another fission
 Super-Critical (keff > 1) – rate of fission
neutron production exceeds rate of loss
4/2003 Rev 2
I.4.6 – slide 13 of 63
Criticality
 Keff depends on the availability of neutrons
with the required energy and the
availability of fissile atoms
 As a result, keff depends on composition,
arrangement and size of fissile material
 If assembly is infinitely large, no neutrons
are lost and Keff = L x k , where L is the
non-leakage probability and K depends
on 4 factors
4/2003 Rev 2
I.4.6 – slide 14 of 63
Criticality
 Let’s follow “n” fission neutrons through
their life cycle
  is mean number of neutrons emitted
per absorption in uranium
 if “n” fission neutrons are captured,
n x  fission neutrons will be produced
4/2003 Rev 2
I.4.6 – slide 15 of 63
Criticality
  is the mean number of neutrons emitted
per fission which depends on the fuel (2.5 for
235U and 3 for 239Pu)
 not every uranium absorption results in
fission (238U can absorb thermal neutrons
without fission)
 mean number of fission neutrons per
absorption is less than 
4/2003 Rev 2
I.4.6 – slide 16 of 63
Sample Calculation
For 100% enrichment of 235U,  = 2.1. What is
 for natural uranium?
f
=
x
a
f = (N5 x f 5)
a = (N5 x a 5) + (N8 x a 8)
 = macroscopic cross section
 = microscopic cross section
 = average number of neutrons per fission
N = atoms per cm3
a = absorption
(5 = 235U)
f = fission
(8 = 238U)
4/2003 Rev 2
I.4.6 – slide 17 of 63
Sample Calculation
(N5 x f 5)
(f 5)
f
=
=
a
(N5 x a 5) + (N8 x a 8)
(a 5) + ( N8 x a 8)
N5
f 5 = 549 barns
a 5 = 650 barns
a 8 = 2.8 barns
 = 2.5 for 235U
For natural U,
N8
= 139
N5
549
=
x 2.5 = 1.32 for natural U
650 + (139 x 2.8)
4/2003 Rev 2
I.4.6 – slide 18 of 63
Criticality
238U
has a small cross section for fission by
fast neutrons (not thermal)
 = 0.29 barns
Fast fission factor is
total number of fission neutrons
=
number of thermal fission neutrons
4/2003 Rev 2
I.4.6 – slide 19 of 63
Criticality
  depends on 3 factors:
 ratio of moderator to fuel
 ratio of inelastic scattering cross
section to fission cross section
 geometrical relationship between fuel
and moderator
 capture of n thermal neutrons will produce
n x  x  fission neutrons
4/2003 Rev 2
I.4.6 – slide 20 of 63
Criticality
 For unmoderated pure U metal,  = 1.29
which is the maximum value
 For homogenous fuel (such as a solution)
 is very close to 1
4/2003 Rev 2
I.4.6 – slide 21 of 63
Criticality
 while fast neutrons are being slowed, they
may be captured by 238U without fission
 resonance for this capture occurs between
5 and 200 eV
 p is the probability that a neutron will
escape this resonance capture
4/2003 Rev 2
I.4.6 – slide 22 of 63
Criticality
 p = resonance escape probability (fraction
of fast, fission produced neutrons that
become thermalized)
 p depends on ratio of moderator to fuel
 for high ratio of moderator to fuel, p  1
 for a low ratio of moderator to fuel, p small
 for pure unmoderated natural U, p = 0
4/2003 Rev 2
I.4.6 – slide 23 of 63
Criticality
 from the original n thermal neutrons we
have n x  x  x p thermal neutrons
 some of the thermal neutrons are absorbed
by non-fuel atoms
 some are absorbed by 235U without fission
(only 84% of thermal neutrons absorbed by
235U cause fission)
4/2003 Rev 2
I.4.6 – slide 24 of 63
Criticality
 f = thermal utilization factor which is the
fraction of all the thermal neutrons which
are absorbed by the fuel (all of the U)
 the total number of new neutrons
produced by the original n thermal
neutrons is n x  x  x p x f
4/2003 Rev 2
I.4.6 – slide 25 of 63
Criticality
aU
f=
aU + aM + ap
aU = macroscopic cross section for U
aM = macroscopic cross section for moderator
ap = macroscopic cross section for other stuff
4/2003 Rev 2
I.4.6 – slide 26 of 63
Criticality
K =
Nf+1
npf
=
= pf
n
Nf
  depends only on the fuel
 , p and f depend on the composition and
arrangement of the fuel
  varies from 1.29 for unmoderated U to almost 1
for homogeneous dispersion of fuel and moderator
 p is about 0.8 to 1 (for pure 235U, p = 1)
4/2003 Rev 2
I.4.6 – slide 27 of 63
Reactivity and Reactor Control
 in a nuclear reactor the factors are combined
to produce a controlled, sustained chain
reaction
 excess reactivity (k) is an increase in the
multiplication factor (k) above 1
k = k - 1
4/2003 Rev 2
I.4.6 – slide 28 of 63
Reactivity and Reactor Control
 For n neutrons in one generation, the
number of additional neutrons in the next
generation is nk
 If the lifetime of a neutron generation is L
sec, the time rate of change of neutrons is
dn
= nk
dt
L
4/2003 Rev 2
I.4.6 – slide 29 of 63
Reactivity and Reactor Control
dn
= nk
dt
L
when integrated from no to n we get
n
=e
no
k
t
L
The reactor period (T) is the time during which
the neutrons (power level) increase by factor of
“e”
4/2003 Rev 2
I.4.6 – slide 30 of 63
Reactivity and Reactor Control
1
k
= T
L
or the reactor period is:
L
T=
k
The mean lifetime of a neutron (birth to
absorption) in pure 235U is about 0.001 sec
4/2003 Rev 2
I.4.6 – slide 31 of 63
Reactivity and Reactor Control
The mean lifetime of a neutron (birth to
absorption) in pure 235U is about 0.001 sec
Assume excess reactivity = 0.1% (k = 0.001)
0.001
T=
= 1 sec
0.001
And the power level would increase by 2.718
each second
4/2003 Rev 2
I.4.6 – slide 32 of 63
Reactivity and Reactor Control
If k is increased to 0.5%
0.001
T=
= 0.2 sec
0.005
The power level increase each second would be
n =e
no
t
T
= e
1
0.2
= 150
This would be hard to control
4/2003 Rev 2
I.4.6 – slide 33 of 63
Reactivity and Reactor Control
The actual mean generation time in a reactor is
much greater than 0.001 because 0.6407% of
fission neutrons are delayed from 0.3 seconds
to 80 seconds. This delay permits the reactor to
be controlled.
4/2003 Rev 2
I.4.6 – slide 34 of 63
Reactivity and Reactor Control
Delayed Neutrons
Group
ni (%)
Ti (sec)
ni x Ti
1
0.0267
0.33
0.009
2
0.0737
0.88
0.065
3
0.2526
3.31
0.836
4
0.1255
8.97
1.125
5
0.1401
32.78
4.592
6
0.0211
80.39
1.688
ni = 0.6407
4/2003 Rev 2
nixTi = 8.315
I.4.6 – slide 35 of 63
Reactivity and Reactor Control
The mean generation time for all fission neutrons is
delayed
prompt
8.315 + (99.359 x 0.001)
niTi
T = n =
= 0.084 sec
100
i
If k  0.006407 the reactor is prompt critical – the
reaction can be sustained by prompt neutrons alone
If k < 0.006407 the reactor is delayed critical – the
delayed neutrons are needed to sustain the reaction
4/2003 Rev 2
I.4.6 – slide 36 of 63
Sample Calculation
What is the reactor period and the increase in power
level in 1 second for a generation time of 0.084 sec for
excess reactivity (k) of 0.1% and 0.5%
For k = 0.001
T = 0.084 = 84 sec
0.001
n =e
no
1
84
= 1.012
For k = 0.005
0.084
T=
= 16.8 sec
0.005
4/2003 Rev 2
n =e
no
1
16.8
= 1.06
I.4.6 – slide 37 of 63
Reactivity and Reactor Control
 Excess reactivity is measured in units of
“dollars” and “cents” (one dollar = 100
cents) and “inhours” (inverse hours)
 One “dollar” worth of reactivity will cause the
reactor to go prompt critical
 One “inhour” is the amount of excess
reactivity which results in a reactor period of
1 hour
4/2003 Rev 2
I.4.6 – slide 38 of 63
Fission
Control of Fission
 Fission typically releases 2-3 neutron
(average 2.5)
 One is needed to sustain the chain reaction
at a steady level of controlled criticality
 The other 1.5 leak from the core region or
are absorbed in non-fission reactions
4/2003 Rev 2
I.4.6 – slide 39 of 63
Fission
Control of Fission
 Boron or cadmium control rods absorb neutrons
 When slightly withdrawn the number of neutrons
available for fission exceeds unity and the power
level increases
 When the power reaches the desired level, the
control rods are returned to the critical position
4/2003 Rev 2
I.4.6 – slide 40 of 63
Fission
Control of Fission
 ability to control the reaction is due to presence of
delayed neutrons
 without delayed neutrons, change in the critical
balance of the chain reaction would lead to a
virtually instantaneous and uncontrollable rise or
fall in the neutron population
 safe design and operation of a reactor sets strict
limits on departures from criticality
4/2003 Rev 2
I.4.6 – slide 41 of 63
Fission
Control of Fission
 fission neutrons initially fast (energy above 1 MeV)
 fission in 235U most readily caused by slow
neutrons (energy about 0.02 eV)
 moderator slows fast neutrons by elastic collisions
 For natural (unenriched) U only graphite and
“heavy” water suitable moderators
 For enriched uranium “light” water may be used
4/2003 Rev 2
I.4.6 – slide 42 of 63
Fission
Control of Fission
 commercial power reactors designed to have
negative temperature and void coefficients
 if temperature too high or excessive boiling occurs
rate of fission is reduced and temperature reduced

238U
absorbs more neutrons as the temperature
rises, pushing neutron balance towards subcritical
 steam within water moderator reduces density
which pushes neutron balance towards subcritical
4/2003 Rev 2
I.4.6 – slide 43 of 63
Fission
Control of Fission
 fuel gradually accumulates fission products and
transuranic elements which increases neutron
absorption (control system has to compensate)
 after about three years, fuel is replaced due to:
 build-up in absorption
 metallurgical changes from constant neutron
bombardment
 burn-up effectively limited to about half of the
fissile material
4/2003 Rev 2
I.4.6 – slide 44 of 63
Fission Summary
4/2003 Rev 2
I.4.6 – slide 45 of 63
Fusion
 In 1920 Arthur Eddington suggested that the energy
of the sun and stars was a product of the fusion of
hydrogen atoms into helium
 In the core of the sun at temperatures of 10-15
million degrees Celsius, hydrogen is converted to
Helium
 Since the 1950's, great progress has been made in
nuclear fusion research however, the only practical
application of fusion technology to date has been the
"hydrogen" or thermonuclear bomb
4/2003 Rev 2
I.4.6 – slide 46 of 63
Fusion
 Fusion has an almost unlimited potential
 The hydrogen isotopes in one gallon of water have
the fusion energy equivalent of 300 gallons of
gasoline
 A fusion power plant would have no greenhouse gas
emissions and would not generate high level
radioactive waste
 Experts predict the world is still at least 50 years and
billions of dollars away from having fusion
generated electricity largely due to the enormous
size and complexity of a fusion reactor
4/2003 Rev 2
I.4.6 – slide 47 of 63
Fusion
 Hydrogen atoms merged to create helium
 Helium mass is slightly less (1%) than the original
mass with the difference being given off as energy
 Rather than using hydrogen atoms, it is easier to
promote fusion by using two isotopes of hydrogen,
deuterium and tritium
 Deuterium is a naturally occurring isotope of
hydrogen which has one extra neutron
 One hydrogen atom in 6700 occurs as deuterium and
can be separated from the rest
4/2003 Rev 2
I.4.6 – slide 48 of 63
Fusion
4/2003 Rev 2
I.4.6 – slide 49 of 63
Fusion
Energy per Nucleon (MeV)
0
-2
Fission
A  56
-4
-6
-8
-10
0
4/2003 Rev 2
Fusion
A  56
50
100
150
200
Atomic Mass Number
250
I.4.6 – slide 50 of 63
Fusion
 Tritium very rare because it is naturally radioactive
and decays quickly
 Tritium can be made by bombarding the naturally
occurring element lithium with neutrons
 Tritium could be created by having a "blanket" made
of lithium surrounding a fusion containment vessel
(this would result in a breeder reactor)
 Fusion can only be accomplished at temperatures
typical of the centre of stars, (100 million degrees
Celsius
4/2003 Rev 2
I.4.6 – slide 51 of 63
Fusion
 The fusion components exist in the form of a
plasma, where atoms are broken down into electrons
and nuclei
 No known solid material could withstand the
temperatures involved in nuclear fusion, so that a
powerful confinement system is required to keep the
plasma away from the walls of the vessel in which it
is contained
4/2003 Rev 2
I.4.6 – slide 52 of 63
Fusion Fuels
 Deuterium can be extracted from water (If all the
world's electricity were provided by fusion,
deuterium would last for millions of years)
 Tritium does not occur naturally and will be
manufactured from lithium within the machine
 Lithium, the lightest metal, is plentiful in the earth's
crust (if all the world's electricity were to be provided
by fusion, known reserves would last for at least
1000 years)
 Even though fusion occurs between Deuterium and
Tritium, the consumables are Deuterium and Lithium
4/2003 Rev 2
I.4.6 – slide 53 of 63
Fusion Fuels
 For example, 10 grams of Deuterium which can be
extracted from 500 litres of water and 15g of Tritium
produced from 30g of Lithium would produce
enough fuel for the lifetime electricity needs of an
average person in an industrialised country
4/2003 Rev 2
I.4.6 – slide 54 of 63
Current Fusion Research
 Fusion research was big news in 1989 when it was
reported that scientists had achieved fusion at room
temperatures with simple equipment
 Unfortunately, the scientists involved could not
prove their claims
 There are currently two methods of confining the hot
plasma which are being studied around the world,
"magnetic confinement" and "inertial confinement"
4/2003 Rev 2
I.4.6 – slide 55 of 63
Current Fusion Research
 Magnetic confinement has the greatest potential and
most of the recent research has been based on the
"TOKOMAK" system (Tokomak is an acronym for the
Russian words "torroidal magnetic chamber“)
 The tokomak system was developed in the former
U.S.S.R. and is under study in the U.S., Japan and
Europe
 A torroidal magnetic chamber is a doughnut shaped
steel structure in which the fusion plasma is
confined by means of powerful coils of
super-conducting material which create a strong
magnetic field
I.4.6 – slide 56 of 63
4/2003 Rev 2
Current Fusion Research
 The other method is inertial confinement (small
amounts of a deuterium-tritium mixture are rapidly
heated to extremely high temperatures with a high
powered laser beam or a beam of charged particles)
 The most advanced test reactors, the Tokomak
Fusion Test Reactor (TFTR) in the U.S. and the Joint
European Torus (JET), use the tokomak design
 To be useful the energy produced must be many
times that required to sustain the reaction
 Even the most optimistic researchers feel that this
will not be achieved until the next century
4/2003 Rev 2
I.4.6 – slide 57 of 63
Fusion Power Plants
 A fusion reactor capable of generating 1000 MW of
electricity would be very large and complex
 While fission reactors can be made small enough to
be used in submarines or satellites, the minimum
size of a fusion reactor would be similar to that of
today's largest commercial nuclear plants
 the difficult part is creating a sustainable fusion
reaction - capturing the energy to generate
electricity is very similar to a fission reactor
 A 1000 MW fusion generator would consume only
150 kg of deuterium and 400 kg of lithium annually
4/2003 Rev 2
I.4.6 – slide 58 of 63
Advantages and
Disadvantages of Fusion
 The fuels required for fusion reactors, deuterium and
lithium, are so abundant that the potential for fusion
is virtually unlimited
 Oil and gas fired power plants as well as nuclear
plants relying on uranium will eventually run into
fuel shortages as these non-renewable resources
are consumed
 Unlike fossil plants, fusion reactors have no
emission of carbon dioxide (contributor to global
warming) or sulphur dioxide (cause of acid rain)
4/2003 Rev 2
I.4.6 – slide 59 of 63
Advantages and
Disadvantages of Fusion
 Barriers to the widespread use of nuclear power
have been public concern over operational safety,
and the disposal of radioactive waste
 Accidents such as Chernobyl are virtually
impossible with a fusion reactor because only a
small amount of fuel is in the reactor at any time
 It is also so extremely difficult to sustain a fusion
reaction (should anything go wrong, the reaction
would invariably stop)
4/2003 Rev 2
I.4.6 – slide 60 of 63
Advantages and
Disadvantages of Fusion
 Long lived highly radioactive wastes are generated
by conventional nuclear plants
 Radioactive wastes generated by a fusion reactor
are the walls of the vessel exposed to neutrons
 Although the quantity of radioactive waste produced
by a fusion reactor might be slightly greater than
that from a conventional nuclear plant, the wastes
would have low levels of short lived radiation,
decaying almost completely within 100 years.
4/2003 Rev 2
I.4.6 – slide 61 of 63
Advantages and
Disadvantages of Fusion
 The major disadvantages of nuclear fusion are the
vast amounts of time and money which will be
required before any electricity is generated by fusion
 Development of other electricity supply technologies
such as photovoltaic cells which convert sunlight
directly into electricity could eliminate the need for
fusion before it is operational
4/2003 Rev 2
I.4.6 – slide 62 of 63
Where to Get More Information
 Cember, H., Introduction to Health Physics, 3rd
Edition, McGraw-Hill, New York (2000)
 Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds.,
Table of Isotopes (8th Edition, 1999 update), Wiley,
New York (1999)
 International Atomic Energy Agency, The Safe Use
of Radiation Sources, Training Course Series No. 6,
IAEA, Vienna (1995)
4/2003 Rev 2
I.4.6 – slide 63 of 63