Nuclear Energy Basics

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 $r$, as in Eqn.1.

$\begin{equation}F = \dfrac{k q_1 q_2}{r^2} \tag{1} \end{equation}$

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 (Eqn.2), with respect to the state where they are infinitely far apart.

$\begin{equation}U = \dfrac{k q_1 q_2}{r} \tag{2} \end{equation}$

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 atom[note]National Institute of Standards and Technology.  Carbon Dioxide (online).  http://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Units=SI&Mask=1 [2019-09-16].[/note].

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 ago[note]U.S. Geological Survey.  Age of the Earth (online).  http://pubs.usgs.gov/gip/geotime/age.html [2019-09-16].[/note], 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 nucleus[note]K. S. Krane, Introductory Nuclear Physics, (Wiley, New York, 1988).[/note]: 6 × 10^-15 m.

From Eqn.2 we find that the potential energy $U$ is approximately (Eqn.3):

$\begin{equation} U = \dfrac{(9.0\times10^9 \textnormal{Nm}^2/\textnormal{C}^2)(46 \times 1.6 \times 10^{-19} \textnormal{C})^2}{6 \times 10^{-15}\textnormal{m}} = 8\times 10^{-11}\textnormal{J} \tag{3} \end{equation}$

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.

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.

Purpose:

To observe a model of how self-sustaining chemical and fission reactions work and to investigate how concentration affects the reactions.

Introduction:

No doubt you have heard of radioactive materials with “critical mass”, nuclear meltdowns, or gunpowder which either burns quickly or explodes in a bang. These are all flashy examples of different reaction rates.

While material properties are one factor in determining reaction rates (i.e. wood burns slowly while gasoline burns very quickly) there are many external factors that contribute to the rate of chemical and nuclear reactions.

In this experiment we will look at making a model of a reaction that will allow you to vary the parameter of concentration (for chemical reactions) and density (for nuclear reactions).

Challenge:

To compare different rates of reaction based on varying densities of the reactants.

Equipment:

• 50 mouse traps
• 50 ping pong balls
• A large container to contain the reaction with a lid (clear, so that the reaction can be seen as it progresses)
• Video camera (or a still camera capable of taking video, high speed if possible)

Key Concepts:

• Reaction Rates

Skills:

• Data collection and experimental design.
• Graphical analysis and curve fitting.

Method:

1. Prime as many mouse traps you will use in your reaction
2. Place them evenly spaced within the container
3. Place one ping pong ball on each mousetrap
4. Drop a ping pong ball on one of the mouse traps in the container and film the reaction happening
5. Plot, as a function of time, the number of mouse traps that have gone off by watching the video in slow motion
6. Repeat steps 1-5 for various numbers of mouse traps and see how they compare to each other

WARNING: Mouse traps are prone to snapping shut on unwary fingers. Be careful when setting up this experiment.

Questions to Think About:

• Does this model better portray a chemical or nuclear reaction?
• What other things could this reaction model?
• What other factors affect the reaction rate that are both portrayed and not portrayed by this model?
•  What other models can you think of that might display other factors affecting the reaction rates?
• Is there a mathematical relation between the number of mouse traps and the way the reaction progresses?

Suggested assigned time:

• 1 week

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 GW_ey of energy)?

First of all it is necessary to convert 1 GW_ey 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 GW_ey rather than 1 GW_thy we’ll need about 55 tonnes/0.4 = 140 tonnes

(c) Operators of CANDU natural uranium power reactors quote a “fuel burnup” of 180 MW_eh/kg of uranium[note]Rouben, B., CANDU Fuel-Management Course, https://canteach.candu.org/Content%20Library/20043404.pdf [2019-09-16].[/note]. Hence the mass of uranium required for one GW_ey is:

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

The fuel burnup for reactors using enriched uranium is about double this[note]World Nuclear Association.  The Economics of Nuclear Power (online).  http://www.world-nuclear.org/info/inf02.html [2019-09-16].[/note], and so the uranium needed per 1 GW_ey is about 4000 tonnes.

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 years[note] B. Comby, The Solutions for Nuclear Waste.  International Journal of Environmental Studies, Vol. 62, No. 6, December 2005, 725–736 http://www.efn.org.au/NucWaste-Comby.pdf [2019-09-16]. [/note].

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 GW_ey [note]S. Andseta et al., Candu Reactors and Greenhouse Gas Emissions,  Proceedings of the 19th Annual Conference, Canadian Nuclear Society, Toronto, Ontario, Canada, October 18-21, 1998. http://www.computare.org/Support%20documents/Publications/Life%20Cycle.htm [2019-09-16]. [/note]. A GWe coal-fired power station produces around 10 million tonnes of CO₂ each year. 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 year[note]China Labour Bulletin, https://www.marketscreener.com/news/CLB-China-Labour-Bulletin-Coal-mine-accidents-on-the-rise-during-China-s-Work-Safety-Month–28885239 [2019-09-16].[/note], and the enormous environmental and human consequences of our dependence on fossil fuels.

Nuclear reactors generate heat by the splitting into two of uranium (sometimes plutonium) nuclei. This process is called fission. The reaction is self-sustaining because it is initiated by a neutron striking and being absorbed onto a uranium nucleus, and the fission process produces more neutrons, which go on to cause more fission. This is called a chain reaction. The chain reaction is easy to stop. Neutrons can be absorbed harmlessly on many different nuclei (“neutron poisons”, e.g. cadmium), and the introduction of these poisons, in the form of control rods pushed into the reactor core, turn off the chain reaction very quickly.

Now let us consider the fission products. Uranium-235 abs:orbs a neutron, briefly becomes uranium-236, and splits into two “daughters”. Uranium-236 has 92 protons and 144 neutrons. Note there are more than 1.5 times as many neutrons as protons. This is typical of heavy nuclei but not of light and medium-mass nuclei (where the number of neutrons equals that of protons for elements up to around ³²S and ⁴⁰Ca). The reason for the extra neutrons is that there is the enormous electrostatic repulsion between the 92 protons that needs to be balanced by the attractive nuclear force provided by the uncharged neutrons. When the²³⁶U splits, a couple of these excess neutrons escape immediately (and are used to propagate the chain reaction):

n + ²³⁵U → ²³⁶U → ¹¹⁷Pd + ¹¹⁷Pd + 2n $\tag{1}$

It doesn’t often happen that the uranium splits into two identical daughters (palladium, Pd,  is element 92/2 = 46), but we use this example for simplicity. Palladium has several stable isotopes, mass 102 to 110, but mass 117 is not one of them! There are at least 7 too many neutrons. Palladium, with only 46 protons, doesn’t need the same excess of neutrons that uranium does to be stable. As a result, the ¹¹⁷Pa has to undergo at several beta-decays (which convert neutrons into protons in the nucleus with the emission of an electron and a neutrino) to get to stability (tin-117), and in doing so, releases a lot of energy.

¹¹⁷Pd → ¹¹⁷Ag +e^–+ν … etc., then  ¹¹⁷Ag → ¹¹⁷Cd → ¹¹⁷In → ¹¹⁷Sn$\tag{2}$

Have a look at these three plots of nuclei (Figs.1-3), modified from the Interactive Chart of Nuclides[note] Brookhaven National Laboratory Chart of Nuclides, https://www.nndc.bnl.gov/nudat2/ [2019-09-16].[/note] available on the web from  the National Nuclear Data Center in Brookhaven, New York.

Fig.1. All nuclides plotted according to decay mode. Stable = black, electron emission (β^– decay) = pink, positron emission (β⁺decay) or electron capture = blue, alpha-decay = yellow.

The red line connecting ²³⁶U to the origin shows roughly where the fission products can lie. The red fission product line never intersects with the black line of stability, because the black line is curved. Heavy nuclei need more neutrons for stability, proportionally, than do lighter ones.

 Fig.2. Products of neutron-induced ²³⁵U fission.

Real fission is usually asymmetric. This gets the daughter nuclei slightly closer to the line of stability, which releases slightly more energy in the original fission, and is therefore more probable than symmetric fission. But this is a detail that doesn’t affect the basic question here. The fission products are nearly all unstable, and are going to decay their way upwards and to the left, toward the line of stability, by β-decay. How long will this take? See Fig.3:

 Fig. 3. All nuclides colour-coded by half-life, from light pink (<1μs) to green (~1s) to blue (>1y) and black (stable).

The half-life gets longer and longer the closer one gets to stability. Some of the last steps before reaching a stable nucleus take tens of years or more. This is why reactor cores continue to generate heat long after the chain-reaction is shut down.

This article was first written in March 2011 at the time of the Fukushima tsunami and reactor accident[note] https://en.wikipedia.org/wiki/Fukushima_Daiichi_nuclear_disaster [2019-10-03].[/note]. On March 29th, eight days after the tsunami had hit the reactor site and the fission reactions were shut down, each reactor was still generating several MW of heat. To set this in perspective, if one tries to cool this with water, we can easily calculate how much. Raising water from 10C to steam at 100C absorbs (90 K)(4.180 kJ/kg/K)+2260 kJ/kg = 2.6 MJ/kg. That is to say, 2.6 MW will boil off 1 kg of water per second, a tonne in 17 minutes, etc.

Currently Fukushima reactor 1-3 are generating about 14 MW[note] What is decay heat? https://mitnse.com/2011/03/16/what-is-decay-heat/ [2019-09-16].[/note] of heat and the prediction is that this rate will fall very slowly. A year from now the total power will have dropped to 8 MW.

Fig.4. Power generated by Fukushima reactors 1-3 as function of time after the tsunami hit. The data for the first five days are measured, the rest are a prediction from half-life values.