Fission

In some cases, a very heavy nucleus, instead of undergoing alpha decay, will spontaneously split in two. Example:

92U238 50Sn129 + 42Mo106 + 30n1 + energy

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 ReactionsIn some cases, we can stimulate the fissioning

of Uranium by hitting it with a neutron:

on1 + 92U238 49In127 + 43Tc110 + 2 0n1 + energy

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 ReactionsThe 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 – CriticalityIn 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 ReactionsFor 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.

ModeratorsThe 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

SafetyNuclear 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:

92U238 50Sn129 + 42Mo106 + 30n1 + energy

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.

SafetyChernobyl 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.

SafetyU.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:

0n1 + 92U238 92U239 decays with a half life of 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

on1 + 90Th232 90Th233 which decays (with 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).

FusionUp 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

1H1 + 1H1 1D2 + +10 + + energy

1H1 + 1D2 1T3 + +10 + + energy

1H1 + 1T3 2He4 + energy

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.

FusionNote: 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.

CosmologyThe 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.

CosmologyIf 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.

CosmologyWhat 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 weak el wkcharge charge ch ch

down -1/3 -1/2 electron -1 -1/2

up +2/3 +1/2 neutrino 0 +1/2

Beta Decay and the Weak Nuclear Force

In Beta- decay, a neutron emits an electron and anti-neutrino 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.

Beta Decay and the Weak Nuclear Force

d(-1/3 el, -1/2 w) υ (0 el, +1/2 w)

d(-1/3 el, -1/2 w) anti-υ (0 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) W- (-1 el, -1 w) υ (0 el, +1/2 w) u(+2/3 el, +1/2 w) anti-υ (0 el, -1/2 w)

W- meson carries electric and weak charge to neutrino

d(-1/3 el, -1/2 w) e- (-1 el, -1/2 w) u(+2/3 el, +1/2 w) u(+2/3 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

d(-1/3 el, -1/2 w) υ (0 el, +1/2 w) u(+2/3 el, +1/2 w) anti-υ (0 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) υ (0 el, +1/2 w) d(-1/3 el, -1/2 w) W+ (+1 el, +1 w) anti-υ (0 el, -1/2 w) u(+2/3 el, +1/2 w)

W+ meson carries electric and weak charge to anti-neutrino

d(-1/3 el, -1/2 w) υ (0 el, +1/2 w) d(-1/3 el, -1/2 w) e+ (+1 el, +1/2 w) u(+2/3 el, +1/2 w)

positron and neutrino are emitted