Lecture 24' Origin of Elements contd, Nuclear Fission

1
Lecture 24. Origin of Elements (contd), Nuclear
Fission

• Fusion Reactor the Tokamak
• Nuclear Fission
• Fission Reactors

2
Problem (binding energy)
Show that the nucleus 8Be has a positive binding
energy but is unstable against the decay into two
alpha particles.
The binding energy of 8Be
The energy of the decay 8Be ? two alpha
particles
Because the energy of the decay 8Be ? two alpha
particles is positive, 8Be is unstable (an
important factor for the nucleosynthesis in the
Universe).
3
Stellar Nucleosynthesis up to Iron
In stars 12C formation sets the stage for the
entire nucleosynthesis of heavy elements.
T 6108 K and ? 2105 gcm-3
Three helium nuclei would be unlikely to crash
together simultaneously, even in the dense core
of a star. More likely two successive two-body
collisions.
4He 4He ? 8Be
First step
8Be unstable (? 10-16 s)
- unbound by 93 keV (see previous slide)
There is no way through this bottleneck unless Be
and He could stick together especially easily and
quickly. In 1954, Fred Hoyle realized that this
could happen only if there is a resonance in
the C nucleus the total energy of a C nucleus in
this resonance state should match that of the
incoming Be and He nuclei (including their
kinetic energies). Hoyle predicted a resonance in
the C nucleus with an energy of 7.7MeV.
8Be 4He ? 12C
Second step
at excitation energy of 7.7 MeV
In 1957, Cook, Fowler, Lauritsen and Lauritsen at
Kellogg Radiation Laboratory at Caltech
discovered a resonance in 12C with correct
energy (at 7.654 MeV)
Experimental Nuclear Astrophysics was born
4
Problem 3 (binding energy, Final 2008)
(a) (10) How much energy is required to remove a
proton from ?
(b) (15) Calculate the binding energy per nucleon
for and compare with the proton
separation energy. Can you explain the
difference?
(a) The energy required to remove a proton from
(b) The binding energy per nucleon (EB/A)
Can you explain the difference? Iron corresponds
to the most tightly-bound configuration of
neutrons and protons (the largest binding energy
per nucleon). By removing nucleons one-by-one, we
decrease the binding energy per nucleon. Thus,
the average EB/A is smaller than the energy which
is required to remove the first proton.
5
Fusion Reactors
Fusion an appealing source of energy. In stars,
the multi-step process of the transformation of
four H nuclei into one He nucleus takes billion
years (weak interaction is involved)
In practical applications, the fusion reaction
must be a single-step process (we can use
neutrons pre-existing in D and T nuclei).
main advantage of these reactions deuterium is
cheap to extract from seawater
Example
1 kg of a D/T mixture would allow for a number of
fusion reactions N
Generated energy
3.8GW for 24 hours!
6
Challenges
The D (or DT, DHe) plasma must be hot to
initiate the reaction
The Coulomb barrier for deuterium-tritium
reaction
Very challenging engineering problem confinement
of a dense and hot plasma.
However, because of a wide energy distribution of
nuclei tunneling, the temperatures 108K turn
out to be sufficient to initiate the reaction.
For how long must the plasma be confined?
The Lawson criterion the requirement that the
fusion energy release exceeds the losses.
The deuterium-tritium L function (minimum n?E
needed to satisfy the Lawson criterion)
The fusion reaction rate increases rapidly with
temperature until it maximizes and then gradually
drops off. The D-T reaction rate peaks at a lower
temperature (about 70 keV, or 800 million
Kelvins) and at a higher value than other
reactions commonly considered for fusion energy.
At T30keV,
?E - the confinement time
7
Tokamak
A tokamak is a device for creating and storing a
high temperature plasma for energy production.
The plasma cannot be in direct contact with any
solid material surface, thus to confine it the
tokamak device uses strong magnetic fields.

• Several drawbacks commonly attributed to D-T
  fusion power
• It produces substantial amounts of neutrons that
  result in
• induced radioactivity within the reactor
  structure.
• 2. The use of D-T fusion power depends on lithium
  resources, which are less abundant than deuterium
  resources.
• 3. It requires the handling of the radioisotope
  tritium. Similar to hydrogen, tritium is
  difficult to contain and may leak from reactors
  in some quantity. Some estimates suggest that
  this would represent a fairly large environmental
  release of radioactivity.

8
Stability Issues (Stable Stars vs. Unstable Bombs)
Why are the stars stable (in contrast to the
hydrogen bomb)? In stars, the increase of
temperature results in the increase of the
pressure and the subsequent increase of its size
(think the ideal gas law). The density becomes
smaller, and the rate of thermonuclear reactions
decreases. This is the build-in negative feedback.
The negative feedback works well for young stars.
For more dense and old stars, the pressure
increase is not sufficient to produce a
significant increase of volume (the matter in
such stars is not described by gas laws) and
the thermonuclear explosion occurs! This is the
star explosion (supernova carbon-nitrogen
bomb).
The carbon-nitrogen-oxygen cycle (Bethe, 1938)
Sirius A
carbon-nitrogen cycle
Sun
red dwarf
luminosity
proton cycle
T, K
a catalyst
107
109
105
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Nuclear Fission
Endothermic process (absorption of energy)
Exothermic processes (release of energy)
Binding energy per nucleon (EB/A), MeV
Nuclear fission decay process in which an
unstable nucleus splits into two fragments of
comparable mass.
Mass number, A
The energy release (per nucleus) is very large
(1MeV?200nucleons). The problem though is how to
convince a stable nucleus to undergo fission
without paying too much energy. Especially
challenging is to run the fission reaction in a
controllable way.
10
Neutron-Induced Fission
The neutrons do not feel the Coulomb repulsion,
only the nuclear attraction. Therefore nuclear
reactions can be induced by neutrons of
arbitrarily low energies.
1932 discovery of neutrons by James
Chadwick 1932 experiments on neutron
bombardment of uranium and observation of
induced radioactivity in stable elements by
Enrico Fermi (Nobel 1938) 1933 Leo Szilard
proposed nuclear chain reaction 1938 discovery
of neutron-stimulated nuclear fission of 235U
by Otto Hahn (Nobel 1944), Fritz Strassmann,
Lise Meitner, and Otto Frisch 1942 the first
artificial chain reaction (Enrico Fermi) 1945
first nuclear explosion in Alamogordo (New
Mexico, USA)
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Neutron-Induced Fission of 235U
Heavy nuclei (e.g., 238U) undergo fission when it
acquires enough excitation energy (typically a
few MeV or so). A few nuclei, notably 235U, are
sufficiently excited by the mere absorption of a
neutron (even this is just a thermal neutron).
235U absorbs the neutron to become 236U, and
this new nucleus is so unstable that it
explodes into two fragments.
n
n
n
an average 2.5 neutrons per fission
stable
high excitation and strong oscillation
formation of a neck (electrical repulsion pushes
the lobes apart)
Because heavy nuclei have a greater n/p ratio
than the lighter ones, the fragments contain an
excess of neutrons. To reduce this excess, two or
three neutrons are emitted instantly (instant
neutrons), and subsequent beta decays and neutron
emission (delayed neutrons) bring the n/p ratios
in the fragments to stable values.
12
Fission Barrier
Fission occurs if an excitation energy is greater
than the potential barrier that separates the two
configurations (fragments inside the same nucleus
and completely separated fragments) or if there
is an appreciable probability for tunneling
through the potential barrier.
U
barrier
UB
1/r electric potential energy
total energy of fragments
r
r0
range of the nuclear force
Spontaneous fission occurs via a quantum
mechanical tunneling through the fission barrier.
Spontaneous fission is possible only for elements
with A ? 230 and x ? 45.
Ground states spontaneous fission half-lives
for 235U (9.8 ? 2.8) x 1018 y 238Pu (4.70?
0.08) x 1010 y 256Fm 2.86 h 238U (8.2 ? 0.1)
x 1015 y 254Cf 60.7 y 260106Sg 7.2 ms
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Chain Reactions
Because fission reaction produce neutrons, a
self-sustained sequence of fissions is possible.
The threshold for such a chain reaction one
neutron from each fission strikes another 235U
nucleus and initiates another fission.
Neutron multiplication factor f Sub-critical
regime (f lt 1) if too few neutrons initiate
fissions, the reaction will slow down and
stop. Critical regime (f 1) precisely one
neutron per fission causes another fission,
energy is released at a constant rate (nuclear
reactor). Super-critical regime (f gt 1) more
than one neutron per fission causes another
fission, the frequency of fission increases
exponentially, and an explosion occurs (atomic
bomb).
If the reaction will sustain itself, it is said
to be "critical", and the mass of 235U required
to produced the critical condition is said to be
a "critical mass". A critical chain reaction can
be achieved at low concentrations of 235U if the
neutrons from fission are moderated in water to
lower their speed, since the probability for
fission with slow neutrons is greater.
A fission chain reaction produces intermediate
mass fragments which are highly radioactive and
produce further energy by their radioactive
decay. Some of them produce neutrons, called
delayed neutrons, which contribute to the fission
chain reaction.
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How does an A bomb work?
To realize a super-critical regime, we need a
critical mass of the material that undergoes
fission ( 1kg for pure 235U) (the critical mass
depends on the probability of capture of neutrons
by the nuclei that undergo fission, i.e. on the
mean free path of neutrons in the material).
Assuming that each fission produces 2 neutrons,
and both neutrons cause further fission reactions
(an ideal chain reaction), lets find the time T
required to split all 235U nuclei
?10-7 s is the mean life-time of neutrons in
235U A the number of generations
1st gen.
2nd gen.
3d gen.
the number of fission reactions should be equal
to the total of U atoms in 1kg
?
2?
3?
time
of neutrons
2
22
23
The total time of the explosion
- this means that the last generation will have
the number 80
The energy release
15
Nuclear Reactors
For a self-sustained chain reaction, the
multiplication factor f should be 1. What are
the factors that control f ?
1. The probability of absorbtion of a neutron by
235U nuclei is large only for slow neutrons. The
neutrons produced by fission have too much
energy. A moderator should be used to slow them
down. An effective moderator should contain
nuclei whose mass is close to that of neutrons.
Hydrogen would be a good moderator, but it
absorbs neutrons to form deuterium. Deuterium
does not absorb neutrons, it is used as a
moderator in the form of heavy water. Another
common moderator graphite.
2. Neutrons can produce reactions other than
further fission. For example, 238U can absorb
neutrons to form 239U. Naturally occurring
uranium contains 99.3 238U and only 0.7 of
fissionable 235U - enrichment is required to
increase the percentage of 235U. A reactor that
uses highly enriched uranium can use ordinary
water (instead of heavy water) as a moderator.
3. Neutrons can escape from the reaction zone
(the mean free path ? the size of the zone). Thus
the mass of the nuclear fuel must be sufficiently
large for a self-sustained chain reaction to take
place (critical mass). The value of the critical
mass depends on the fuel and the moderator
(typically, a few kg).
To maintain critical regime (f 1), the reactors
should have a negative feedback. They are
equipped with movable control rods (usually made
of cadmium or boron) whose function is to absorb
neutrons (if the rods malfunction Chernobyl !).
The time constant of the feedback loop can be
reasonably long due to an existence of delayed
neutrons (1 of the total amount of neutrons)
emitted by neutron-rich fission fragments having
lifetimes on the order of seconds (even thermal
neutrons move with v 2km/s, so without delayed
neutrons, the feedback should operate on the time
scale 0.1m/2000m/s 10-4 s !)
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Nuclear Reactors (contd)
1W ? 1011 fissions per second. 1MW per day ? one
gram of fuel. This means that one gram of waste
products is produced per megawatt per day, which
includes 0.5 grams of 239Pu.
Pressurized Water Reactor
Boiling Water Reactor
Liquid-Metal Fast-Breeder Reactor
BWR the most common in case of a leak
the water can become
radioactive PWR the moderating and turbine
waters are separated
more expensive LMFBR liquid sodium is used as
moderator and heat transfer
medium
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Problem (radioactive decay)

• At present the naturally occurring uranium
  contains 99.28 of 238U and 0.72 of 235U.
• What was the percentage of 238U and 235U at the
  moment of the Earth formation (the Earths age is
  4?109y)? The half-life of 238U is 4.5?109y, the
  half-life of 235U is 7.1?108y.
• The increased percentage of 235U probably allowed
  natural nuclear reactors to occur. Explain why
  such a reaction could not occur today.

(a)
(b) 16.3 is significantly greater than the
concentration of 235U in the enriched uranium
fuel used in nuclear reactors (3.5). The
current concentration (0.72) is insufficient to
produce a self-sustained chain reaction.