Topic 7. 3 Nuclear Reactions, Fission and Fusion
These notes were typed in association with Physics for use with the IB Diploma Programme by Michael Dickinson
For further reading and explanation see:
Physics, Tsokos (purple): Ch 6.3
Physics, Giancoli (mountain): Ch 30
7.3.1 Describe and give an example of an artificial (induced) transmutation.
7.3.2 Construct and complete nuclear equations.
 We already stated that natural transmutation is when during alpha and beta decay, the radioactive nuclide
changes into one of a different atomic number or become a different element.
 Artificial transmutation is when the transmutation process is induced by an external means.
 This can happen when high speed particles like neutrons, alpha particles, other nuclides and even
gamma radiation bombards the nucleus with enough force to cause it to split.
 The parent nuclide undergoes transmutation and the two smaller, daughter, nuclides are formed.
 1919 – Rutherford
 Alpha particles into nuclei of nitrogen-14 gas
 Some alpha particles absorbed and proton was emitted.
 Oxygen–17 was formed
 *** see board for equation***
 Lithium-6 bombarded with neutrons will yield an alpha particle and Hydrogen-3
 *** see board for equation***
 The following link will give you so practice writing equations.
 http://tb014.k12.sd.us/Chemistry/Neclear%20Reactions/pr1.html
7.3.3 Define the term unified atomic mass unit.
IB Definition
Unified Mass Unit(u) – is defined as 1/12 of the mass of a neutral carbon-12 atom.
u = 1.661 x 10-27kg = 931.5 MeV/c2




Using this system the mass of carbon-12 is exactly 12.000000u.
The masses of nuclear particles are given in terms of “u”
Take a look at your data book. That is the extra number after “kg”
http://schools.bibb.k12.ga.us/cms/lib01/GA01000598/Centricity/Domain/292/Data%20Book.pdf
7.3.4 Apply the Einstein mass-energy equivalence relationship.
7.3.5 Define the concepts of mass defect, binding energy and binding energy per nucleon.
 Lets take a minute to really examine this “unified mass units”
6 x proton masses = 6 x 1.007276 u = 6.043656 u
6 x neutron masses = 6 x 1.008665 u = 6.051990 u
6 x electron masses = 6 x 0.000549 u = 0.003294 u
Total = 12.098940 u
Mass of Carbon-12 = 12.000000 u
Difference = 0.098949 u
 The difference is known as the mass defect.
 The mass defect occurs because energy is needed to bind the nucleons together in the nucleus.
 This results in a reduction in the energy of the system and a corresponding reduction in the mass of the
bound atom
IB Definition
Mass Defect – the difference between the mass of a combined atom and the sum of the mass of the
constituent parts.
Binding Energy – the energy needed to bind the constituents of an atom together. It corresponds to the
mass defect of the atom.
 The mass defect of the carbon-12 atom corresponds to the binding energy of the atom.
 The energy (MeV), corresponding to a mass defect of 0.09894u can be calculated or converted.

E = 0.09894u x 931.5 MeV = 92.2MeV
1u

E = 92.2 x 106eV x 1.6 s 10-19J = 1.479 x 10-11J
1eV
 This is where Einstein got his ideas.
 He suggest that mass is simply another form of energy.
 The relationship is as follows
IB Formula – E = mc2
Where E is energy, m is mass, and c is the speed of light.
 The equation predicts how much energy corresponds to a particular loss in mass.
 We can use this equation to calculate the binding energy of the carbon-12 atom. (The energy that is needed
to separate a nucleus into its constituents).
 First a conversion
 Mass=0.098940u = 0.098940u x 1.661x10-27kg = 1.643x10-28kg
 1u
 Now the formula
 E = mc2
 E = 1.643x10-28kg x (3x108m/s)2
 E = 1.479x10-11J
 Convert from J to eV
 E = 1.479x10-11J x
1eV
= 92.2MeV
-19
 1.6 x 10 J
Fission and fusion
 Learn about how a nuclear power plant works.
http://science.howstuffworks.com/nuclear-power.htm
7.3.8 Describe the processes of nuclear fission and nuclear fusion.
 Fusion
 The process that powers the sun and the stars.
 Two atoms of hydrogen combine together, or fuse, to form an atom of helium.
 Some of the mass of the hydrogen is converted into energy.
 Easiest fusion reaction is combining deuterium (or “heavy hydrogen) with tritium (or “heavy-heavy
hydrogen”) to make helium and a neutron.
 Both of which are easy to get making it potentially the “perfect” power source
 http://fusioned.gat.com/what_is_fusion.html
 Fission
 Fission is a nuclear reaction in which an atomic nucleus splits, or fissions, into fragments (ie. certain
forms of Uranium)
 Usually two fragments of comparable mass.
 The split releases kinetic energy, as heat, and radiation and neutrons.
 These neutrons can be used to “fission” other Uranium nuclei and start a chain reaction.
 If the mass of the parts is added up, it is less than the original, or starting mass.
 This missing mass, also known as the mass defect, is changed into energy.
 http://reactor.engr.wisc.edu/tour/reactor.htm
7.3.6 Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon.
7.3.9 Apply the graph in 7.3.6 to account for the energy release in the processes of fission and fusion.
IB Definition
Binding Energy per Nucleon – A nuclide’s binding energy is equal to the total binding energy for an atom
divided by the number of constituents nucleons.
 The graph shows several things:
 How the binding energy per nucleon changes with respect to the nuclide’s atomic number, Z.
 The binding energy per nucleon rises sharply from atomic number, (Z=1) until it reaches a maximum at
Z=26 (Iron).
 The trend then falls away more gradually towards the heavier nuclides.
***Remember*** The higher the binding energy between the nucleons, the more stable the nucleus will be.
 Unstable nuclides below atomic number 26(iron) undergo nuclear fusion in an attempt to increase their
stability.
 Unstable nuclides above atomic number 26 undergo nuclear fission.
 The steeper the slope the more energy is releases.
 Fusion – very steep – much more energy released per nucleon
 Fission – shallow – much less energy released per nucleon
 The graph explains/predicts:
 Relative stability of elements around atomic number 26
 Large abundance of iron in universe
 The energy released in nuclear reactions.
7.3.10 State that nuclear fusion is the main source of the Sun’s energy
 There are several different possibilities of this sequence. We’ll look at path 1.
 Here are two sites that go into more detail and explanation.
http://burro.astr.cwru.edu/Academics/Astr221/StarPhys/ppchain.html
http://hyperphysics.phy-astr.gsu.edu/hbase/astro/procyc.html
 The sequence of fusion reactions that occur in the Sun is known as the proton-proton cycle.
 ****See the board for equations or the web sites****
 Part 1:
 Part 2:
 Part 3:
 The first two parts have to happen twice in order to produce enough He–3 for the third part.
 The total energy released in this fusion reaction is:
 (0.42 x 2) + (5.49 x 2) + 12.86 = 24.68 MeV
 The 2 positrons produced in the first part of the sequence also releases 1.02 MeV of energy which has to
be included in the final sum.
 24.68 + (1.02 x 2) = 26.72 MeV
7.3.7 Solve problems involving mass defect and binding energy
7.3.11 Solve problems involving fission and fusion reactions.
Example Problem 3
 Estimate the energy released in the fission reaction of Uranium-235 as it transmutes into Barium-141 and
Krypton-92
 Answer: about 176 MeV
 Solution: The difference between the binding energies of the original Uranium and the new Barium and
Krypton is a good estimate of the total energy released in this reaction.
 Using the graph from 7.3.6 we can find the binding energy per nucleon
 U-235 = 7.6MeV
 Ba-141 = 8.3MeV
 Kr-8.6 = 8.6MeV
 Solution: Calculating the total binding energies:
 U-235 = 7.6MeV x 235 = 1786MeV
 Ba-141 = 8.3MeV x 141 = 1170.3MeV
 Kr-8.6 = 8.6MeV x 92 = 791.2MeV
 Total = 1961.5MeV
 Difference = 1961.5MeV – 1786MeV = about 176MeV
Example Problem 4
 Repeat the previous example using the following information and find the exact answer.
 Atomic mass of U-235 235.043924u
 Atomic mass of Kr-92
91.926300u
 Atomic mass of Ba-141 140.914400u
 Answer:173 MeV
 Solution: The difference between the binding energies of the original Uranium and the new Barium and
Krypton is a good estimate of the total energy released in this reaction.
 Calculate the binding energy using the mass defects
Atomic mass of U-235 235.043924u
Constituent mass U-235 236.958995u
Mass defect U-235
1.915071u = 1783.9MeV
Atomic mass of Kr-92
Constituent mass Kr-92
Mass defect Kr-92
91.926300u
92.766940u
0.840640u = 783.1MeV
Atomic mass of Ba-141 140.914400u
Constituent mass Ba-141 142.174557u
Mass defect Ba-141
1.260157u = 1173.8MeV
Difference in binding energies = (1173.8 + 783.1) – 1783.9 = 173MeV
Example Problem 5
 Using the binding energy per nucleon curve, estimate the energy released in the fission reaction of Uranium235 to Strontium-88 and Xenon-136 shown below. For this calculation, assume that the initial kinetic
energy of the neutron is negligible.
 Answer: about 112MeV
 Solution: The difference between the binding energies of the original Uranium and the new Strontium and
Xenon is a good estimate of the total energy released in this reaction.
 Uranium-235
7.6MeV x 235 = 1786MeV


Strontium-88
Xenon-136
8.6MeV x 88 = 756MeV
8.4MeV x 136 = 1142MeV
Total = 1898MeV
Difference = 1898MeV – 1786MeV = 112MeV
Example Problem 6
 Repeat the previous example, using the following information.
 Atomic mass of U-235
 Atomic mass of Sr-88
 Atomic mass of Xe-136
 Answer: 126.5MeV
235.043924u
87.905618u
135.907210u
Example Problem 7
 Calculate the energy released in the fusion reaction between Deuterium, H-2, and Tritium, H-3, as shown in
the equation below, using the following information.
 Atomic mass of H-2
 Atomic mass of H-3
 Atomic mass of He-4
 Answer: 17.59MeV
2.014102u
3.016049u
4.002602u