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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 chain reaction!
Note that a proton, electron, or alpha particle will not easily be able to penetrate
the electron cloud, but a neutron will have no problem with it.
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).
The values at first may seem counter-intuitive, but remember that
the nuclear for is very short-range (more like Velcro). A slower
speed gives the neutron more time to stick.
The slower speeds and higher f 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 (close to the speed
of light = 3 x 108 m/s ) to thermal speeds (½mv2 =
(3/2)kT so v  3 x 103 m/s) 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 re-started,
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 so 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.
Another element found in nature is 90Th232 .
This can be used in a breeder reactor, since
1
232
233 which decays (with a half life
on + 90Th
90Th
of 22.3 minutes and 27 days) to 92U233 which is a fissile
material with a half life of 162,000 years!
BREEDER REACTORS:
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).
BREEDER REACTORS:
However, Pu-239 can absorb a neutron (a = 270
barns) to become Pu-240 which is NOT a fissile
material (which has a half life of 6,580 years)!
Likewise, U233 can absorb a neutron (a = 50 barns)
to become U234 which is NOT a fissile material
(which has a half life of 248,000 years)!
Thus if we leave the fuel in the reactor too long,
some of the Pu-239 or U-233 will burn up and
some will convert to Pu-240 or U-234, and we
will have too low a fraction of Pu-239 or U-233
to make into a bomb.
Military versus Civilian Reactors
To create fissile material for nuclear weapons,
a breeder reactor that can be shut down fairly
often and fuel rods removed often is desired so
the fissile material is not contaminated with the
other isotopes.
Civilian reactors don’t have this problem so
can be designed to run much longer without
shutting down to remove fuel rods. This allows
stronger containment vessels to be used in
civilian reactors.
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 a 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
• Intensity = 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”: 4 H
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.
Cosmology
The sun “burns” hydrogen to make helium.
What happens when the core of 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 the extra heat will 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 no longer be balanced by the
explosive force of fusion and so it will 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. This is called a supernova.
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 a supernova explosion?
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
Basic Forces
Only need four forces:
• Gravitational (holds planets and galaxies together)
– [gravity does not balance out, the other three forces do]
• Electromagnetic
(holds atoms & molecules together)
– [the two nuclear forces balance out at the atomic level]
• Strong nuclear force (hold neutrons & protons together)
– [a new force not seen at the atomic level but needed to hold the nucleus
together, but does not affect electrons]
• Weak nuclear force (involved in  decay)
– [another new force which does affect electrons]
Elementary Particles
Normally, only need a few “elementary” particles
to explain the periodic chart of the elements:
–
–
–
–
proton
neutron
electron (and positron)
neutrino (and anti-neutrino)
Elementary Particles
Are the neutron, proton, electron, neutrino, and
photon all really 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
But when we collide particles that have lots of
kinetic energy, we see all kinds of particles:
p+, n0, e-, 0  -, -, 0, -, -, 0, -,
0, +, -, K+, K-, K0, D+, ...
It does seem that the electron and neutrino are
elementary (we can’t break them into parts), but it
seems the proton and neutron are NOT
elementary! The proton and neutron appear to
be made up of more elementary particles which
have been named quarks.
Elementary Particles
Further, it seems that the e- and 0 do NOT respond
to the same force that holds n0 and p+ together the force we will now call the STRONG
NUCLEAR FORCE.
But electrons and neutrinos are involved in beta
decay, and so we need a fourth force, the WEAK
NUCLEAR FORCE to which neutrons, protons,
electrons and neutrinos all respond.
Elementary Particles
Examples of situations that we need to explain:
- + p+ + Kinetic Energy
 0 + K0
which involves the strong nuclear force,
and
energy + p+
n0 + e+ + 
which involves the weak nuclear force
To explain the many particles, we think that
the proton and neutron are made up of more
basic particles that we call quarks.
Elementary Particles
Since these particles respond to the weak
nuclear force we give them a “weak charge”.
Quarks
Leptons
(don’t)
(respond to strong nuclear force)
electric
charge
weak
charge
down -1/3
up
+2/3
-1/2
+1/2
electric weak
charge charge
electron -1
neutrino 0
-1/2
+1/2
We’ll use d for the down quark, u for the up quark,
e for the electron, and  for the neutrino.
Using this: proton = (u,u,d);
neutron = (u,d,d).
Color charge
Quarks respond to the electric force, the weak nuclear
force, and the strong nuclear force, and so have an
electric charge, a weak charge, and a color charge. The
color charge has three possible values: red, green, and
blue. We haven’t been able to separate a single quark, so
we don’t see single color charges. In fact, there are only
two possible combinations of quarks based on their color
charges: “White” with three quarks (red, green, and blue),
or “black” with two quarks (one color with its anti-color).
The three quark combo is called a baryon,
while the two quark combo is called a meson.
Gluons
Quarks are held together by a force particle called a
“gluon”. It appears that if we try to separate two quarks,
the energy needed to do this is enough to create a (quark
– anti-quark) pair called a meson. For instance:
neutron + energy = (u,d,d) + energy = (u,d,d) + (u,anti-u) which
recombine as (u,u,d) + (d, anti-u) = proton + negative pion (-).
Notice that a neutron (u,d,d) has (+2/3) + (-1/3) + (-1/3) = 0 el chrg
a proton has (u,u,d) has (+2/3) + (+2/3) + (-1/3) = +1 el chrg, and
the negative pion (d, anti-u) has (-1/3) + (-2/3) = -1 electric charge.
Notice that the quarks have non-integer electric charge, but we
have NOT been able to isolate a single quark, and all possible
combinations of quarks have integer electric charges.
Different “flavors”
To explain all the different hadrons in terms of these more elementary particles,
physicists have postulated the existence of particles they call QUARKS (the
name is from a James Joyce novel). The elementary particles found so far are:
LEPTONS
|
QUARKS
name electric weak color
| name
electric weak color
charge charge charge |
charge charge charge
• electron -1 - ½
0
| up
+ 2/3 + ½ r/b/g
neutrinoe 0 + ½
0
| down
- 1/3 - ½
r/b/g
• muon
-1 - ½
0
| charm
+ 2/3 + ½ r/b/g
neutrinom 0 + ½
0
| strange
- 1/3 - ½
r/b/g
• tau
-1 - ½
0
| top (truth) + 2/3 + ½ r/b/g
neutrinot 0 + ½
0
| bottom (beauty) - 1/3 - ½
r/b/g
It appears that the only difference between the three sets is one of mass
(energy). For instance, the muon (-) behaves exactly as an electron (e-) except
the muon has a mass about 200 times that of the electron. The muon is unstable
and will decay into an electron. The different types of elementary particles are
called different FLAVORS.
Hadrons
examples of HADRONS
BARYONS
MESONS
name quarks
name quarks
proton (u,u,d)
0
(u,anti-u)
neutron (u,d,d)
0
(d,anti-d)
+
(u,anti-d)
0
(u,d,s)
(anti-u,d)
Σ(d,d,s)
K+
(u,anti-s)
(s,s,d)
K(anti-u,s)
0
(s,s,u)
K0
(d,anti-s)
(s,s,s)
D+
(c,anti-d)
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 anti-neutrino then leave the
nucleus in the beta decay.
Example
The normal beta decay of a neutron:
n
p+ + - + anti- 
(u,d,d) + ( + anti- )
(u,u,d) + - + anti- 
[(u,d,d) is neutron; energy creates ( + anti- ); (u,u,d) is proton]
Process: one of the d quarks in the neutron and
the  exchange flavors, with the
d quark (-1/3 ec, -1/2 wc) becoming a u quark (+2/3 ec, +1/2 wc)
and the  (0 ec, +1/2 wc) becoming a - (-1 ec, -1/2 wc).
[ec = electric charge, wc = weak charge]
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