327. FissionLearning Objectives
Fission is the opposite of fusion and releases energy only when heavy nuclei are split. As noted in Fusion, energy is released if the products of a nuclear reaction have a greater binding energy per nucleon ($\text{BE}/A$) than the parent nuclei. Figure 2 shows that $\text{BE}/A$ is greater for mediummass nuclei than heavy nuclei, implying that when a heavy nucleus is split, the products have less mass per nucleon, so that mass is destroyed and energy is released in the reaction. The amount of energy per fission reaction can be large, even by nuclear standards. The graph in Figure 2 shows $\text{BE}/A$ to be about 7.6 MeV/nucleon for the heaviest nuclei ($A$ about 240), while $\text{BE}/A$ is about 8.6 MeV/nucleon for nuclei having $A$ about 120. Thus, if a heavy nucleus splits in half, then about 1 MeV per nucleon, or approximately 240 MeV per fission, is released. This is about 10 times the energy per fusion reaction, and about 100 times the energy of the average $\alpha $, $\beta $, or $\gamma $ decay. Example 1: Calculating Energy Released by FissionCalculate the energy released in the following spontaneous fission reaction: $${}^{\text{238}}\text{U}\to {}^{\text{95}}\text{Sr}+{}^{\text{140}}\text{Xe}+3n$$given the atomic masses to be $m({}^{\text{238}}\text{U})=\text{238.050784 u}$, $m({}^{\text{95}}\text{Sr})=\text{94.919388 u}$, $m({}^{\text{140}}\text{Xe})=\text{139.921610 u}$, and $m(n)=\mathrm{1.008665\; u}$. Strategy As always, the energy released is equal to the mass destroyed times ${c}^{2}$ , so we must find the difference in mass between the parent ${}^{\text{238}}\text{U}$ and the fission products. Solution The products have a total mass of $$\begin{array}{lll}{m}_{\text{products}}& =& \text{94.919388 u}+\text{139.921610 u}+3\left(\mathrm{1.008665\; u}\right)\\ & =& \text{}\text{237.866993 u.}\end{array}$$The mass lost is the mass of ${}^{\text{238}}\text{U}$ minus ${m}_{\text{products}}$, or $$\mathrm{\Delta}m=\text{238.050784 u}\text{237.8669933 u}=\mathrm{0.183791\; u},$$so the energy released is $$\begin{array}{lll}E& =& \left(\mathrm{\Delta}m\right){c}^{2}\\ & =& \left(\text{0.183791 u}\right)\frac{\text{931.5 Me}\text{V/}{c}^{2}}{\text{u}}{c}^{2}=\text{171.2 MeV.}\end{array}$$Discussion A number of important things arise in this example. The 171MeV energy released is large, but a little less than the earlier estimated 240 MeV. This is because this fission reaction produces neutrons and does not split the nucleus into two equal parts. Fission of a given nuclide, such as ${}^{\text{238}}\text{U}$ , does not always produce the same products. Fission is a statistical process in which an entire range of products are produced with various probabilities. Most fission produces neutrons, although the number varies with each fission. This is an extremely important aspect of fission, because neutrons can induce more fission, enabling selfsustaining chain reactions. Spontaneous fission can occur, but this is usually not the most common decay mode for a given nuclide. For example, ${}^{\text{238}}\text{U}$
can spontaneously fission, but it decays mostly by $\alpha $ emission. Neutroninduced fission is crucial as seen in
Figure 2. Being chargeless, even lowenergy neutrons can strike a nucleus and be absorbed once they feel the attractive nuclear force. Large nuclei are described by a where ${\text{FF}}_{1}$ and ${\text{FF}}_{2}$ are the two daughter nuclei, called An example of a typical neutroninduced fission reaction is $$n+{}_{\text{92}}^{\text{235}}\text{U}\to {}_{\text{56}}^{\text{142}}\text{Ba}+{}_{\text{36}}^{\text{91}}\text{Kr}+3\mathrm{n.}$$Note that in this equation, the total charge remains the same (is conserved): $\text{92}+0=\text{56}+\text{36}$ . Also, as far as whole numbers are concerned, the mass is constant: $1+\text{235}=\text{142}+\text{91}+3$ . This is not true when we consider the masses out to 6 or 7 significant places, as in the previous example. Not every neutron produced by fission induces fission. Some neutrons escape the fissionable material, while others interact with a nucleus without making it fission. We can enhance the number of fissions produced by neutrons by having a large amount of fissionable material. The minimum amount necessary for selfsustained fission of a given nuclide is called its The reason ${}^{\text{235}}\text{U}$ and ${}^{\text{239}}\text{Pu}$ are easier to fission than ${}^{\text{238}}\text{U}$ is that the nuclear force is more attractive for an even number of neutrons in a nucleus than for an odd number. Consider that ${}_{\text{92}}^{\text{235}}{\text{U}}_{\text{143}}$ has 143 neutrons, and ${}_{\text{94}}^{\text{239}}{\text{P}}_{\text{145}}$ has 145 neutrons, whereas ${}_{\text{92}}^{\text{238}}{\text{U}}_{\text{146}}$ has 146. When a neutron encounters a nucleus with an odd number of neutrons, the nuclear force is more attractive, because the additional neutron will make the number even. About 2MeV more energy is deposited in the resulting nucleus than would be the case if the number of neutrons was already even. This extra energy produces greater deformation, making fission more likely. Thus, ${}^{\text{235}}\text{U}$ and ${}^{\text{239}}\text{Pu}$ are superior fission fuels. The isotope ${}^{\text{235}}\text{U}$ is only 0.72 % of natural uranium, while ${}^{\text{238}}\text{U}$ is 99.27%, and ${}^{\text{239}}\text{Pu}$ does not exist in nature. Australia has the largest deposits of uranium in the world, standing at 28% of the total. This is followed by Kazakhstan and Canada. The US has only 3% of global reserves. Most fission reactors utilize ${}^{\text{235}}\text{U}$ , which is separated from ${}^{\text{238}}\text{U}$ at some expense. This is called enrichment. The most common separation method is gaseous diffusion of uranium hexafluoride (${\text{UF}}_{6}$) through membranes. Since ${}^{\text{235}}\text{U}$ has less mass than ${}^{\text{238}}\text{U}$ , its ${\text{UF}}_{6}$ molecules have higher average velocity at the same temperature and diffuse faster. Another interesting characteristic of ${}^{\text{235}}\text{U}$ is that it preferentially absorbs very slow moving neutrons (with energies a fraction of an eV), whereas fission reactions produce fast neutrons with energies in the order of an MeV. To make a selfsustained fission reactor with ${}^{\text{235}}\text{U}$ , it is thus necessary to slow down (“thermalize”) the neutrons. Water is very effective, since neutrons collide with protons in water molecules and lose energy. Figure 4 shows a schematic of a reactor design, called the pressurized water reactor. Control rods containing nuclides that very strongly absorb neutrons are used to adjust neutron flux. To produce large power, reactors contain hundreds to thousands of critical masses, and the chain reaction easily becomes selfsustaining, a condition called Example 2: Calculating Energy from a Kilogram of Fissionable FuelCalculate the amount of energy produced by the fission of 1.00 kg of ${}^{\text{235}}\text{U}$ , given the average fission reaction of ${}^{\text{235}}\text{U}$ produces 200 MeV. Strategy The total energy produced is the number of ${}^{\text{235}}\text{U}$ atoms times the given energy per ${}^{\text{235}}\text{U}$ fission. We should therefore find the number of ${}^{\text{235}}\text{U}$ atoms in 1.00 kg. Solution The number of ${}^{\text{235}}\text{U}$ atoms in 1.00 kg is Avogadro’s number times the number of moles. One mole of ${}^{\text{235}}\text{U}$ has a mass of 235.04 g; thus, there are $(\text{1000 g})/(\text{235.04 g/mol})=\mathrm{4.25\; mol}$. The number of ${}^{\text{235}}\text{U}$ atoms is therefore, $$\left(\mathrm{4.25\; mol}\right)\left(6.02\times {\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{}^{\text{235}}\text{U/mol}\right)=2\text{.}\text{56}\times {\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}{}^{\text{235}}\text{U}.$$So the total energy released is $$\begin{array}{lll}E& =& \left(2\text{.}\text{56}\times {\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}{}^{\text{235}}\text{U}\right)\left(\frac{\text{200 MeV}}{{}^{\text{235}}\text{U}}\right)\left(\frac{1.60\times {\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}}{\text{MeV}}\right)\\ & =& \text{}\text{8.21}\times {\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}\text{.}\end{array}$$Discussion This is another impressively large amount of energy, equivalent to about 14,000 barrels of crude oil or 600,000 gallons of gasoline. But, it is only onefourth the energy produced by the fusion of a kilogram mixture of deuterium and tritium as seen in Section 326 Example 1. Even though each fission reaction yields about ten times the energy of a fusion reaction, the energy per kilogram of fission fuel is less, because there are far fewer moles per kilogram of the heavy nuclides. Fission fuel is also much more scarce than fusion fuel, and less than 1% of uranium $\text{(the}\phantom{\rule{0.25em}{0ex}}{}^{\text{235}}\text{U}\text{)}$ is readily usable. One nuclide already mentioned is ${}^{\text{239}}\text{Pu}$
, which has a 24,120y halflife and does not exist in nature. Plutonium239 is manufactured from ${}^{\text{238}}\text{U}$
in reactors, and it provides an opportunity to utilize the other 99% of natural uranium as an energy source. The following reaction sequence, called Uranium239 then ${\beta}^{\u2013}$ decays: $${}^{\text{239}}\text{U}\to {}^{\text{239}}\text{Np}+{\beta}^{}+{v}_{e}\text{(}{\text{t}}_{\text{1/2}}=\text{23}\phantom{\rule{0.25em}{0ex}}\text{min)}.$$Neptunium239 also ${\beta}^{\u2013}$ decays: $${}^{\text{239}}\text{Np}\to {}^{\text{239}}\text{Pu}+{\beta}^{}+{v}_{e}\text{(}{\text{t}}_{\text{1/2}}=2\text{.}4\phantom{\rule{0.25em}{0ex}}\text{d}\text{).}$$Plutonium239 builds up in reactor fuel at a rate that depends on the probability of neutron capture by ${}^{\text{238}}\text{U}$
(all reactor fuel contains more ${}^{\text{238}}\text{U}$
than ${}^{\text{235}}\text{U}$
). Reactors designed specifically to make plutonium are called Plutonium239 has advantages over ${}^{\text{235}}\text{U}$ as a reactor fuel — it produces more neutrons per fission on average, and it is easier for a thermal neutron to cause it to fission. It is also chemically different from uranium, so it is inherently easier to separate from uranium ore. This means ${}^{\text{239}}\text{Pu}$ has a particularly small critical mass, an advantage for nuclear weapons. PhET Explorations: Nuclear FissionStart a chain reaction, or introduce nonradioactive isotopes to prevent one. Control energy production in a nuclear reactor! Section Summary
Conceptual QuestionsExercise 1Explain why the fission of heavy nuclei releases energy. Similarly, why is it that energy input is required to fission light nuclei? Exercise 2Explain, in terms of conservation of momentum and energy, why collisions of neutrons with protons will thermalize neutrons better than collisions with oxygen. Exercise 3The ruins of the Chernobyl reactor are enclosed in a huge concrete structure built around it after the accident. Some rain penetrates the building in winter, and radioactivity from the building increases. What does this imply is happening inside? Exercise 4Since the uranium or plutonium nucleus fissions into several fission fragments whose mass distribution covers a wide range of pieces, would you expect more residual radioactivity from fission than fusion? Explain. Exercise 5The core of a nuclear reactor generates a large amount of thermal energy from the decay of fission products, even when the powerproducing fission chain reaction is turned off. Would this residual heat be greatest after the reactor has run for a long time or short time? What if the reactor has been shut down for months? Exercise 6How can a nuclear reactor contain many critical masses and not go supercritical? What methods are used to control the fission in the reactor? Exercise 7Why can heavy nuclei with odd numbers of neutrons be induced to fission with thermal neutrons, whereas those with even numbers of neutrons require more energy input to induce fission? Exercise 8Why is a conventional fission nuclear reactor not able to explode as a bomb? Problem ExercisesExercise 1(a) Calculate the energy released in the neutroninduced fission (similar to the spontaneous fission in Example 1) $$n+{}^{\text{238}}\text{U}\to {}^{\text{96}}\text{Sr}+{}^{\text{140}}\text{Xe}+3\mathrm{n,}$$ given $m({}^{\text{96}}\text{Sr})=\text{95.921750u}$ and $m({}^{\text{140}}\text{Xe})=\text{139.92164}$. (b) This result is about 6 MeV greater than the result for spontaneous fission. Why? (c) Confirm that the total number of nucleons and total charge are conserved in this reaction. Show/Hide Solution Solution(a) 177.1 MeV (b) Because the gain of an external neutron yields about 6 MeV, which is the average $\mathrm{BE/}A$ for heavy nuclei. (c) $A=1+\text{238}=\text{96}+\text{140}+1+1+\mathrm{1,}\phantom{\rule{0.25em}{0ex}}Z=\text{92}=\text{38}+\text{53},\phantom{\rule{0.25em}{0ex}}\text{efn}=0=0$ Exercise 2(a) Calculate the energy released in the neutroninduced fission reaction $$n+{}^{\text{235}}\text{U}\to {}^{\text{92}}\text{Kr}+{}^{\text{142}}\text{Ba}+2\mathrm{n,}$$ given $m({}^{\text{92}}\text{Kr})=\text{91}\text{.}\text{926269 u}$ and $m({}^{\text{142}}\text{Ba})=\text{141}\text{.}\text{916361}\phantom{\rule{0.25em}{0ex}}\text{u}$. (b) Confirm that the total number of nucleons and total charge are conserved in this reaction. Exercise 3(a) Calculate the energy released in the neutroninduced fission reaction $$n+{}^{\text{239}}\text{Pu}\to {}^{\text{96}}\text{Sr}+{}^{\text{140}}\text{Ba}+4n,$$ given $m({}^{\text{96}}\text{Sr})=\text{95}\text{.}\text{921750 u}$ and $m({}^{\text{140}}\text{Ba})=\text{139}\text{.}\text{910581 u}$. (b) Confirm that the total number of nucleons and total charge are conserved in this reaction. Show/Hide Solution Solution(a) 180.6 MeV (b) $A=1+\text{239}=\text{96}+\text{140}+1+1+1+\mathrm{1,}\phantom{\rule{0.25em}{0ex}}Z=\text{94}=\text{38}+\text{56},\phantom{\rule{0.25em}{0ex}}\text{efn}=0=0$ Exercise 4Confirm that each of the reactions listed for plutonium breeding just following Example 2 conserves the total number of nucleons, the total charge, and electron family number. Exercise 5Breeding plutonium produces energy even before any plutonium is fissioned. (The primary purpose of the four nuclear reactors at Chernobyl was breeding plutonium for weapons. Electrical power was a byproduct used by the civilian population.) Calculate the energy produced in each of the reactions listed for plutonium breeding just following Example 2. The pertinent masses are $m({}^{\text{239}}\text{U})=\text{239.054289 u}$, $m({}^{\text{239}}\text{Np})=\text{239.052932 u}$, and $m({}^{\text{239}}\text{Pu})=\text{239.052157 u}$. Show/Hide Solution Solution${}^{\text{238}}\text{U}+n\phantom{\rule{0.25em}{0ex}}\to {}^{\text{239}}\text{U}+\gamma $ 4.81 MeV ${}^{\text{239}}\text{U}\to {}^{\text{239}}\text{Np}+{\beta}^{}+{v}_{e}$ 0.753 MeV ${}^{\text{239}}\text{}\text{Np}\to {}^{\text{239}}\text{}\text{Pu}+{\beta}^{}+{v}_{e}$ 0.211 MeV Exercise 6The naturally occurring radioactive isotope ${}^{\text{232}}\text{Th}$ does not make good fission fuel, because it has an even number of neutrons; however, it can be bred into a suitable fuel (much as ${}^{\text{238}}\text{U}$ is bred into ${}^{\text{239}}\text{P}$). (a) What are $Z$ and $N$ for ${}^{\text{232}}\text{Th}$? (b) Write the reaction equation for neutron captured by ${}^{\text{232}}\text{Th}$ and identify the nuclide ${}^{A}X$ produced in $n+{}^{\text{232}}\text{Th}\to {}^{A}X+\gamma $. (c) The product nucleus ${\beta}^{}$ decays, as does its daughter. Write the decay equations for each, and identify the final nucleus. (d) Confirm that the final nucleus has an odd number of neutrons, making it a better fission fuel. (e) Look up the halflife of the final nucleus to see if it lives long enough to be a useful fuel. Exercise 7The electrical power output of a large nuclear reactor facility is 900 MW. It has a 35.0% efficiency in converting nuclear power to electrical. (a) What is the thermal nuclear power output in megawatts? (b) How many ${}^{\text{235}}\text{U}$ nuclei fission each second, assuming the average fission produces 200 MeV? (c) What mass of ${}^{\text{235}}\text{U}$ is fissioned in one year of fullpower operation? Show/Hide Solution Solution(a) $2\text{.}\text{57}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{MW}$ (b) $8.03\times {\text{10}}^{\text{19}}\phantom{\rule{0.25em}{0ex}}\text{fission/s}$ (c) 991 kg Exercise 8A large power reactor that has been in operation for some months is turned off, but residual activity in the core still produces 150 MW of power. If the average energy per decay of the fission products is 1.00 MeV, what is the core activity in curies?
