Transcript P6S2 - FacStaff Home Page for CBU

Fission
In some cases, a very heavy nucleus, instead
of undergoing alpha decay, will
spontaneously split in two. Example:
238
129 + Mo106 + 3 n1 + energy
U
Sn
92
50
42
0
This fissioning of uranium does not always
result in these two resultant atoms - there is
a whole range of resulting atoms. But it
always gives a few neutrons.
Chain Reactions
In some cases, we can stimulate the fissioning
of Uranium by hitting it with a neutron:
1 + U238
127 + Tc110 + 2 n1 + energy
n
In
o
92
49
43
0
The interesting thing about this process is that
one neutron causes a fission and the emission
of 2 neutrons. If these two neutrons could
causes fissions with each fission releasing
two more neutrons, etc., we could have a
change reaction!
Chain Reactions
Nuclear reactions like the preceding one
happen very quickly, so this process could
conceivably be (and has been) used in a bomb!
The amount of energy coming from one fission
is about 200 MeV, while the amount of
energy coming from converting C + O2 to
CO2 is about 2 eV - a difference of a factor
of 100 million!
Chain Reactions
If this kind of stimulated fission does happen,
then why doesn’t the whole world blow up?
How can we control this process, so that the
bomb blows up where we want it to, or better
yet, can we control this energy source to
provide a steady supply of energy in a power
generating station?
Chain Reactions
The answer lies in looking at all the things
that can happen when a neutron enters a
region where there are uranium atoms
present: the neutron can:
•
cause a fission
•
be absorbed
•
bounce off (be scattered)
•
escape from the area
Nuclear Probabilities
Since nuclear forces do not have the long
range of gravity and electromagnetism, we
can treat the forces as simply do they act or
don’t they. This is like a target: we either
hit the target or we don’t.
The probability of hitting a target depends on
the sizes of the target and the bullet.
Nuclear Probabilities
We can see the size of a normal target by
throwing photons at it (using light) and
watching how the photons bounce off the
target.
With the nucleus, we can’t throw light at it,
but we can throw neutrons at it to see how
big the nucleus is (for different reactions
involving neutrons). This leads to the
relation of probability to “size”.
Nuclear Probabilities
A big nucleus is on the order of 10-14 m, or an
area of 10-28 m2 . This unit of area has its
own name: the barn: 1 barn = 10-28 m2.
Probability of a specific happening is simply:
Pa = a / i) .
where  is the area measured in barns.
Included in the i is the chance that the
neutron did not hit anything (escape)!
Chain Reactions – Criticality
In order to see what we need, there is the
criticality constant, k, defined as:
k = Pf * avg number of neutrons/fission
When k<1, any reaction will die out, since we
have less neutrons coming out than going in
When k=1, we have steady state, the
operating point for a reactor (CRITICAL)
When k>1, we have the start-up for a reactor,
or if continued, the conditions for a bomb.
Chain Reactions
For the fissioning of uranium-238, we have
the average number of neutrons/fission =
2.5. This means, for k=1, we need Pf = .4 .
On the nuclear data sheet, page 2, we have
listed some information about i’s.
Recall that Pf = f / (f + a + esc)
(we don’t include scattering since this does not
remove the neutron).
Chain Reactions
We can control escape by controlling the size
of the uranium fuel: we can ideally have
two sub-critical pieces of uranium and when
we want, put the two sub-critical pieces
together to form a super-critical piece.
We can also control absorption by using
different materials, called control rods, that
are in or near the uranium fuel.
Chain Reactions
For U-238 using fast neutrons, we have for Pf:
Pf = f / [f + absorption + escape] , and with a
large enough mass of uranium, escape = 0 barns,
so at best,
Pf-U238 = 0.5 barns / [0.5 + 2.0 + 0] barns = 0.2,
which is less than the required 0.4, so
U-238 cannot be used for a bomb.
However, U-235 and Pu-239 can be.
(See their values on the data sheet.)
Power Reactors
The cross sections () for neutron reactions
are different for fast neutrons coming
directly from the fission process than they
are for the neutrons after they have
slowed down to normal speeds (called
thermal neutrons). This allows the uranium
in ore (99.3% U-238, 0.7% U-235) to be
used in a reactor for power, although not for
a bomb.
Moderators
The material used to slow the neutrons down
from their originally high speeds to thermal
speeds is called a moderator.
The properties of a moderator should be:
• light (to slow neutrons down better)
• low absorption of neutrons
• cheap
Look at nuclear data sheet for a and scat data
Safety
Nuclear Reactors CANNOT explode as a
nuclear bomb, since the reaction must
involve thermal (slow) neutrons to proceed.
Once the reaction starts to go wild, the heat
generated will destroy the careful geometry
needed to have the neutrons slow down.
Without slow neutrons, the reaction will die
out.
Nuclear Waste
Since the ratio of neutrons/protons is bigger
for U-238 than for the stable isotopes of the
two decay atoms (whichever two they happen to
be), the two decay atoms will have too
many neutrons, and so will be
radioactive. This is the main source of the
radioactive waste associated with nuclear
power. In addition, the high neutron flux will
tend to make everything around the reactor
radioactive as well.
Nuclear Waste
In our first example of fissioning, we had:
238
129 + Mo106 + 3 n1 + energy
U
Sn
92
50
42
0
The stable isotopes of 50Sn are Sn-112,114,
115,116,117,118,119,120,122,124. The stable
isotopes of 42Mo are 92,94,95,96,97,98,100.
In both cases, we have an excess of about 5
neutrons in the decay products which makes
them radioactive.
Safety
Chernobyl accident:
• An experiment was interrupted and restarted, with safety equipment turned off.
• No nuclear explosion, but excess heat
started the carbon moderator on fire, which
then threw up the decay products into the
atmosphere.
• No containment vessel to keep reactor
wastes contained - allowed fueling on run.
Safety
U.S. civilian reactors:
• use water as moderator in civilian reactors,
rather than carbon (carbon moderates better
for creation of Pu-239 in a breeder reactor
[Chernobyl reactor used carbon]).
• use containment vessels (no fueling on the
run - must stop reaction to re-fuel [Chernobyl
reactor allowed fueling on the run, see previous
slide])
Relative Safety
Consider second page of nuclear data sheet to
compare wastes from different energy
sources that provide the same amount of
energy.
BREEDER REACTORS:
1 + U238
n
0
92
239 decays with a half life of
U
92
23.5 minutes and 2.3 days to 94Pu239 which is a
fissile material with a half life of 24,360 years.
However, Pu-239 can absorb a neutron to
become Pu-240 which is NOT a fissile
material (half life of 6,580 years)!
Thus if we leave the fuel in the reactor too
long, some of the Pu-239 will burn up and
some will convert to Pu-240, and we will
have too low a fraction of Pu-239 to make
into a bomb.
BREEDER REACTORS:
Another element found in nature is 90Th232 .
This can be used in a breeder reactor, since
1 + Th232
233 which decays (with
n
Th
o
90
90
a half life of 22.3 minutes and 27 days) to 92U233
which is a fissile material with a half life of
162,000 years!
To run a breeder reactor, we run a normal
reactor and put the Th-232 or the U-238
around the reactor to absorb neutrons. This
otherwise useless material then becomes a
fissile material (a fuel).
Fusion
Up to now, we have been working with
materials that have a binding energy less
than Iron but were on the heavy side of iron,
so they split apart (fission).
Now we investigate the reverse: look at
materials that are lighter than iron so that
they also have a binding energy less than
iron. In this case we will see that they can
combine to release energy (fusion).
Fusion
1 + H1
2 + 0 +  + energy
H
D
1
1
1
+1
1 + D2
3 + 0 +  + energy
H
T
1
1
1
+1
1 + T3
4 + energy
H
He
1
1
2
so we have four hydrogens becoming one
helium, with about 24 MeV of energy and
two neutrino’s produced plus two positrons
which will combine with the extra two
electrons from the 4 H’s to give another 2
MeV’s of energy.
Fusion
Note: no radioactive wastes, although the
reactor will be subject to radiation that will
make the reactor radioactive.
Note: cheap fuel, since we use hydrogen,
and there is plenty of hydrogen in water
(H2O), and it takes only a few eV to break
water apart, but we get several MeV of
energy in the fusion!
Fusion
problem: how do we get the two protons
close enough together so that the nuclear
force overcomes the electrostatic repulsion?
answer: high temperatures and high densities
(which we have in the sun) - but how to
hold this together (the sun uses its gravity) ?
We need a temperature on the order of a
million degrees!
Fusion
One way: inertial confinement
Take pellet with hydrogen in it. Hit it with
laser beams from many high energy lasers,
so that it heats up so quickly that the
hydrogen atoms do not have time to get
away without hitting other hydrogen atoms.
Fusion
Second way: Magnetic confinement
Use magnetic fields to make the hot plasma
(gas that has been ionized) go in a circular
orbit.
Both methods are under development.
The sun
• P/A at earth = 1350 W/m2 ; radius = 93
million miles = 1.49x1011m, so total power of sun,
P = (P/A)*A = (1350 W/m2)*(4)* (1.49x1011m)2
= 3.8x1026 W = 3.8x1026 Joules/sec.
• “burning H into He”: 4H
1 He+24 MeV so
1 gram of H (1 mole) gives 24 MeV*(6x1023/4)*
(1.6x10-13J/MeV)= 5.8x1011J/gram = 5.8x1014J/kg
•
sun must burn 3.8x1026J/sec / 5.8x1014J/kg
= 6.6x1011kg/sec (660 million tons/sec).
The sun
• sun must burn 3.8x1026J/sec / 5.8x1014J/kg =
6.6x1011kg/sec.
• The mass of the sun is 2x1030kg.
• If the sun could burn all of its fuel, it should
then be able to last 2x1030kg/6.6x1011kg/sec =
3x1018sec = 1011 years = 100 billion years.
• Best theory says the sun will run out of fuel in
its core after shining about 10 billion years.
• The amount of mass converted into energy by each fusion is a little less than
1%, so the total mass of the sun will not change much.
Cosmology
The sun “burns” hydrogen to make helium.
What happens when the sun runs out of
hydrogen?
Can the sun “burn” helium to make heavier
elements? YES, but it needs to be hotter,
which it can become by using gravity to
shrink its inside but expand its outside - out
past the orbit of Venus. This will make the
sun a red giant star.
Cosmology
If a star can become hot enough, that is, if it
has enough gravity (enough mass), a star
can burn the lighter elements to make
heavier elements until the atoms reach iron,
since iron has the highest binding energy
per nucleon.
How are these elements re-distributed to the
rest of the universe?
Where do the heavier elements come from?
Cosmology
When the interior of a massive star runs out of
fuel, gravity will win and cause the core to
shrink. Since there is so much mass, there
will be a huge implosion of the core, which
will cause a huge explosion of the surface.
The power output in this supernova explosion
is equivalent to billions of normal stars, but
it lasts only a few weeks.
Cosmology
This huge supernova explosion will blow
material from the star out into the rest of the
universe, enriched in the heavier elements.
It will also have the energy to make the
elements heavier than iron. This is where
we think the heavy elements like gold and
uranium come from.
Cosmology
What happens to the burned out core of the
star that underwent supernova?
It will either end up with all the electrons
being smashed into the nuclei, causing a
neutron star. This star has gravity so strong
that the electrons cannot re-escape.
Or it will end up with even the nuclei being
crushed and so will form a “black hole”.
Black Holes
In a black hole, all the mass of the core of the
star has collapsed past anything that might
hold it into a finite volume.
However, the mass of the core is still there,
and its gravity will decrease with distance
pretty much as normal.
But since the mass has collapsed, we can get
gravities so strong near the star that even
light cannot escape.
Cosmology
Space and Time:
• time: finite or infinite?
• space: 3-D (or more?) and Euclidean (or flat)?
The Big Bang theory vs the Steady State theory
Elementary Particles
Only need four forces:
• Gravitational (holds planets and galaxies together)
• Electromagnetic (holds atoms & molecules together)
• strong nuclear force (hold neutrons & protons together)
• weak nuclear force (involved in  decay)
Elementary Particles
Are the neutron, proton, electron, neutrino,
and photon all elementary particles, or are
any of these able to be broken down into
more fundamental particles?
In beta decay, the neutron decays into
something else, so maybe it has a structure.
When we throw things at it, it does seem to
have a structure! And so does the proton,
but not the electron.
Elementary Particles
Quarks
Leptons
(respond to strong nuclear force)
(don’t)
electric
charge
down -1/3
up
+2/3
weak
charge
-1/2 electron
+1/2 neutrino
el
wk
ch ch
-1 -1/2
0 +1/2
Beta Decay and the Weak
Nuclear Force
In Beta- decay, a neutron emits an electron and antineutrino and becomes a proton. We can see this
occur if we consider the neutron to consist of 2
down quarks and 1 up quark. The extra energy in
the unstable isotope undergoing the decay is
converted into a neutrino / anti-neutrino pair. One
of the down quarks exchanges a force particle that
carries the –1 electric charge and the –1 weak
charge from the down quark (converting it to the
up quark) to the neutrino (converting it to the
electron). This electron and the remaining antineutrino then leave the nucleus in the beta decay.
Beta Decay and the Weak
Nuclear Force
υ (0 el, +1/2 w)
anti-υ (0 el, -1/2 w)
d(-1/3 el, -1/2 w)
d(-1/3 el, -1/2 w)
u(+2/3 el, +1/2 w)
neutrino / anti-neutrino pair created from energy
d(-1/3 el, -1/2 w)
u(+2/3 el, +1/2 w)
u(+2/3 el, +1/2 w)
d(-1/3 el, -1/2 w)
u(+2/3 el, +1/2 w)
u(+2/3 el, +1/2 w)
υ (0 el, +1/2 w)
anti-υ (0 el, -1/2 w)
W- meson carries electric and weak charge to neutrino
W- (-1 el, -1 w)
e- (-1 el, -1/2 w)
anti-υ (0 el, -1/2 w)
electron and anti-neutrino are emitted
Beta Decay and the Weak
Nuclear Force
Beta+ decay occurs in a like fashion, with an
up quark exchanging a W+ meson (force
particle) of +1 electric charge and +1 weak
charge with an anti-neutrino changing the
up quark into a down quark (and hence the
proton into a neutron) and changing the
anti-neutrino into a positron. The positron
and the remaining neutrino then leave the
nucleus.
Beta Decay and the Weak
Nuclear Force
υ (0 el, +1/2 w)
anti-υ (0 el, -1/2 w)
d(-1/3 el, -1/2 w)
u(+2/3 el, +1/2 w)
u(+2/3 el, +1/2 w)
neutrino / anti-neutrino pair created from energy
d(-1/3 el, -1/2 w)
d(-1/3 el, -1/2 w)
u(+2/3 el, +1/2 w)
W+ (+1 el, +1 w)
υ (0 el, +1/2 w)
anti-υ (0 el, -1/2 w)
W+ meson carries electric and weak charge to anti-neutrino
d(-1/3 el, -1/2 w)
d(-1/3 el, -1/2 w)
u(+2/3 el, +1/2 w)
υ (0 el, +1/2 w)
e+ (+1 el, +1/2 w)
positron and neutrino are emitted