Nuclear Energy Basics

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What is the physical basis of nuclear power? Is nuclear power good or bad for the environment, safe or dangerous for us?

Big Ideas: 
  • The fundamental source of most usable energy lies in the arrangement of electrical charge inside the atom. The atomic nucleus is much smaller than the whole atom, so the electrostatic potential energies are enormous.

Introductory video produced by Alex Hass and Van Ly.

A starting point for comparison: Chemical Energy

All the energy transfer in chemical reactions comes from or goes into the rearrangement of electrons in the atom. The amount of energy involved is therefore controlled by Coulomb’s Law relating the magnitude of the repulsive force F between two charges to the amounts of charge $ q_1 $ and $ q_2 $, and the distance between them $ r $.

$ F = \dfrac{k q_1 q_2}{r^2} $                      1)

Here $ k $ is the universal constant, 9.0 × 109 Nm 2/C2 in SI units. If $ q_1 $ and $ q_2 $ are of opposite sign, the force is attractive. From this expression we can find the potential energy $ U $ of two charges a distance $ r $ apart, with respect to the state where they are infinitely far apart.

$ U = \dfrac{k q_1 q_2}{r} $                           2)

If $ q_1 $ and $ q_2 $ are of opposite sign, this potential energy is negative; it takes positive work to pull to the two charges apart. This much is standard textbook work.

Consider a hydrogen atom, consisting of a proton and an electron in its lowest orbital state. Both particles have a charge of magnitude 1.6 × 10-19 C, and the mean distance between them is the so-called Bohr radius 5.3 × 10-11 m. The potential energy of this state can be found from the above formula, and is -4.35 × 10-18 J. Because the electron also has kinetic energy, it only takes half of 4.35 × 10-18 J to rip it out of the hydrogen atom. Most chemical reactions involve electronic changes far less dramatic than ionizing a hydrogen atom, and a typical energy is of order 10-19 J.

For example burning carbon in oxygen produces 393.5 kJ/mol or 6.54 × 10-19 J per atom1.

Energy in the Atomic Nucleus

Now consider what happens when a uranium nucleus splits in half, or, as we say, undergoes fission.

The first thing we have to appreciate is that atomic nuclei are made up of a collection of protons and neutrons (collectively called nucleons) held together by the strong nuclear force. This force is very different from the Coulomb force in that it is only attractive and much stronger, but only over short distances (just a few nucleon radii, after which it disappears to zero), and is altogether too complicated to be described by a simple formula like Coulomb’s Law. Hence, atomic nuclei are a fine balance between the attractive strong nuclear force and the electrostatic repulsion between the positively-charged protons. When nuclei become too big, the short-range nuclear force can no longer hold everything together against the electrostatic force. This is why the heaviest stable nucleus is 209Bi, with 83 protons and 126 neutrons. Heavier common nuclei like 238U are stable enough (half-life 4.46 Gyr) that a large fraction is left over from the formation of the solar system 4.5 or 4.6 Gyr ago2, but heavier nuclei than this only last for a few thousand years, days, or mere fractions of a second. 

Food for thought: why don’t we see nuclei made up of only neutrons, that experience only the attractive strong force and not the replusive electrostatic force? 

Now consider a nucleus that is teetering on the edge of stability, 235U (half-life 700 Myr, 0.7% of natural uranium, which is mostly 238U). If we add one more neutron, it becomes just too big to hold together, and splits in two. This process is called fission and it releases a large amount of energy in the form of the kinetic energy of the two mutually repelling fragments. This energy we can estimate from Coulomb’s Law. For now, assume the two fragments have equal numbers of protons (this is usually not the case, but good for our rough estimate). Uranium is element 92, so that means 46 protons each. To estimate the mean distance between the two fragments, lets take it to be the radius of the original nucleus3: 6 × 10-15 m.


From equation 2 we find that the potential energy is approximately:

$ U = \dfrac{(9.0\times10^9 \textnormal{Nm}^2/\textnormal{C}^2)(46 \times 1.6 \times 10^{-19} \textnormal{C})^2}{6 \times 10^{-15}\textnormal{m}} = 8\times 10^{-11}\textnormal{J} $

This is an over-estimate by a factor of about 2.5, (a) because the nucleus has to deform considerably before it breaks, and so the effective distance between the two fission fragments is much greater than the original radius, and (b) because real nuclei seldom split into two equal parts (so $ q_1q_2 $ is less than if it did); but we're in the right ball-park. The bottom line is that the energy released in the fission of one uranium nucleus is of the order of 100 000 000 times more than that released in burning a carbon atom in air (a few times 10-11 J compared to a few times 10-19 J). And the result of a fission event, however nasty the fission products (and they do tend to be highly radioactive) does not involve putting yet another carbon atom into the atmosphere. Thus fission as a potential source of useful energy is worth looking at very closely.

The Chain Reaction

How do we keep uranium nuclei fissioning to provide a reliable source of power? It turns out that every time a uranium nucleus splits, it releases two or three free neutrons. If the lump of uranium is large enough, these neutrons can cause another nucleus to fission, which causes more neutrons to be released, etc. etc. Plainly if one is doubling the energy output for each generation of neutrons, and the time for a neutron to find its target nucleus in a lump of uranium is very short (these neutrons travel at around 107 m/s) and the neutrons are not being lost to other reactions or the outside world, then we have a problem on our hands. These conditions occur in a few kg of fairly pure 235U (238U absorbs neutrons harmlessly). This is a bomb. 

On the other hand it is possible to slow the neutrons down before they find another uranium nucleus and yet still keep the reaction going. This slowing occurs in a moderator, a material that allows neutrons to rattle around and lose their kinetic energy without absorbing them. Common materials used are water (heavy water is better than ordinary "light" water) and carbon. In addition, if one ensures that by losing enough neutrons to absorption or escape each uranium fission causes precisely one other fission, nor more or less, then the system calms down and produces power at a constant, controllable rate.  This is a reactor.

Some reactors can work with natural uranium if the moderator is good enough (i.e. absorbs very few neutrons). The Canadian CANDUs are like this: they use heavy water as a moderator. Most reactors use slightly enriched uranium (~3% 235U) and ordinary water as a moderator.

The choice is basically between isotopically enriching the uranium or isotopically enriching water. Both processes are expensive, in energy and money.


How much natural uranium is needed to produce a GW of electricity (the output of a typical big power station) for one year (i.e. a GWey of energy)?

First of all it is necessary to convert 1 GWey to joules: (109 W)(3600 s/h)(24 h/d)(365 d/y) = 3.15 ×1016 J

Each 235U atom produces 3.2 ×10-11 J of energy.

In principle (3.15 ×1016 J)/(3.2 ×10-11 J per atom) = 9.84 ×1026 atoms can produce the required energy.

Each atom has a mass of (235)(1.66 ×10-27 kg) = 3.90 ×10-25 kg, so the total mass of 235U is 384 kg.

However, this is not the whole story because (a) natural uranium is only 0.7% 235U by mass, (b) the thermal energy produced by fission is only convertible to electricity with at most 40% efficiency, and (c) reactors can only "burn" a small fraction of the 235U fuel before the build-up of neutron-absorbing products reduces the neutron population below the level of a self-sustaining chain reaction. 

Let us take these points in turn:

(a) The amount of natural uranium containing 384 kg of 235U is 384 kg/0.007 = 54 900 kg ≈ 55 tonnes.

(b) If we want 1 GWey rather than 1 GWthy we'll need about 55 tonnes/0.4 = 140 tonnes

(c) Operators of CANDU natural uranium power reactors quote a "fuel burnup" of 180 MWeh/kg of uranium4. Hence the mass of uranium required for one GWey is:

(3.15 ×1016 J/GWey)/(3.6 ×109 J/MWeh)/(180 MWeh/kg) = 8.8 ×106 kg = 8800 tonnes.

The fuel burnup for reactors using enriched uranium is about double this5, and so the uranium needed per 1 GWey is about 4000 tonnes.  

Nuclear Waste and the Environment 

How much nuclear waste does each 1000 tonnes of spent uranium produce? As only a tiny fraction of the original mass ends up as energy (via E=mc2), the answer is pretty much 1000 tonnes. This material contains many different isotopes of widely varying radioactivity and chemical toxicity. Short-lived isotopes (minutes, hours, days) decay away very quickly. Very long-lived isotopes (thousand, millions of years) have low radioactivity. The most dangerous isotopes are those that are biologically active (they mimic common atoms in the human body) and have half-lives of the same order as the human lifespan,  e.g. 90Sr (29 years, replaces the calcium in our bones). The total radioactivity of all reactor waste products will decay to the level of the original uranium ore in a few thousand years6.

Greenhouse gas emissions from nuclear power are not zero, as is sometimes assumed. Mining, refining, transport and construction all contribute CO2 to the atmosphere. The total CO2 production is estimated to be about 100 000 tonnes per GWey 7. A GWe coal-fired power station produces around 10 million tonnes of CO2 each year8. Unless we take active and expensive steps to remove it from the atmosphere, most of the CO2 will stay there forcing the climate forever; this is a useful fact to bear in mind when worrying about the long-term storage of nuclear waste.

In addition, any discussion of the danger of nuclear power should be set against the approximately 6000 miners who die in China's coal mines each year9, and the unfolding disaster in the Gulf of Mexico. 

Take Home Experiment: 
Nuclear Fission Take-Home Experiment


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