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Chain Reactions. In some cases, a very heavy nucleus, instead of undergoing alpha decay, will spontaneously split in two. Example: 92 U 238 50 Sn 129 + 42 Mo 106 + 3 0 n 1 + energy

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chain reactions
Chain Reactions

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

92U23850Sn129 + 42Mo106 + 3 0n1 + 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 reactions1
Chain Reactions

In some cases, we can stimulate the fissioning of Uranium by hitting it with a neutron:

on1+ 92U23849In127 + 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 reactions2
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 reactions3
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 reactions4
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
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 probabilities1
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 probabilities2
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 as one of the i’s is the chance that the neutron did not hit anything (escape)!

chain reactions5
Chain Reactions

In order to see what we need, a useful number 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 reactions6
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 Pfis = fis / (fis + absorb + escape)

(we don’t include scattering since this does not remove the neutron).

chain reactions7
Chain Reactions

We can control escapeby 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 absorbby using different materials, called control rods, that are in or near the uranium fuel.

chain reactions8
Chain Reactions

For U-238 using fast neutrons, we have for Pf:

Pf = f / [f + absorb + escape] , and with a large enough mass of uranium, escape = 0, so at best, Pf-U238 = 0.5 / [0.5 + 2.0 + 0] = 0.2, which is less than the required 0.4.

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
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.


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 absorb and scat data


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
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 waste1
Nuclear Waste

In our first example of fissioning, we had:

92U23850Sn129 + 42Mo106 + 3 0n1 + 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.


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.

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 next slide])
breeder reactors
Breeder Reactors:

0n1 + 92U23892U239decays to 94Pu239 which is a fissile material. However, Pu-239 can absorb a neutron to become Pu-240 which is NOT a fissile material!

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 reactors1
Breeder Reactors:

Another element found in nature is 90Th232 .

This can be used in a breeder reactor, since

on1 + 90Th23290Th233which decays to 92U233 which is a fissile material!

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).

relative safety
Relative Safety

Consider second page of nuclear data sheet to compare wastes from different energy sources that provide the same amount of energy.


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).


1H1 + 1H11D2 + +10 +  + energy

1H1 + 1D21T3 + +10 +  + energy

1H1 + 1T32He4 + energy

so we have four hydrogens becoming one helium,with about 24 MeV of energyand two neutrino’sproduced (plus two positrons which will combine with the extra two electrons from the 4 H’s to give energy).


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!


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!


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.


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
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.
the sun1
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 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.


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?


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.


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.


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
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.


Space and Time:

  • time: finite or infinite?
  • space: 3-D and Euclidean?

The Big Bang vs the Steady State

elementary particles
Elementary Particles
  • Normally, only have a few particles:
    • proton
    • neutron
    • electron (and positron)
    • neutrino (and anti-neutrino)
elementary particles1
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 particles2
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 seems that the electron and neutrino are elementary (we can’t break them into parts,), but it seemsthe proton and neutron are NOT elementary!

elementary particles3
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 FORCEto which neutrons, protons, electrons and neutrinos all respond.

elementary particles4
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 involved the weak nuclear force

elementary particles5
Elementary Particles

Need four forces:



strong nuclear force(hold neutrons & protons together)

weak nuclear force (involved in  decay)

elementary particles6
Elementary Particles

Quarks Leptons

(respond to strong nuclear force) (don’t)

electric weak electric weakchargechargechargecharge

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

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


n p+ + - + anti-

(u,d,d) + ( + anti-) [neutron plus energy]

the d quark and the  change 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), which gives the result

(u,u,d) + - + anti-

which is p + - + anti-

[the normal beta decay of a neutron].