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
Nuclear Reactions
 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
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/Ce
ntricity/Domain/292/Data%20Book.pdf
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 x 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
1.6 x 10-19J
= 92.2MeV
Fission and fusion
Learn about how a nuclear power plant works.
http://science.howstuffworks.com/nuclear-power.htm
 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
 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.
 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.ht
ml
 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
Example Problem 3
 Estimate the energy released in the fission reaction of Uranium235 as it transmutes into Barium-141 and Krypton-92
 Answer: about 176 MeV
Example Problem 3
 Estimate the energy released in the fission reaction of Uranium235 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
Example Problem 3
 Estimate the energy released in the fission reaction of Uranium235 as it transmutes into Barium-141 and Krypton-92
 Answer: about 176 MeV
 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
Atomic mass of Kr-92
Atomic mass of Ba-141
 Answer:173 MeV
235.043924u
91.926300u
140.914400u
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
 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
Example Problem 4
 Calculate the binding energy using the mass defects
Atomic mass of U-235
Constituent mass U-235
Mass defect U-235
235.043924u
236.958995u
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
Constituent mass Ba-141
Mass defect Ba-141
140.914400u
142.174557u
1.260157u = 1173.8MeV
Example Problem 4
 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 Uranium-235 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
Example Problem 5
 Using the binding energy per nucleon curve, calculate the energy
released in the fission reaction of Uranium-235 to Strontium-88
and Xenon-136 shown below. For this calculation, assume that
the initial kinetic energy of the neutron is negligible.
 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
Example Problem 5
 Using the binding energy per nucleon curve, calculate the energy
released in the fission reaction of Uranium-235 to Strontium-88
and Xenon-136 shown below. For this calculation, assume that
the initial kinetic energy of the neutron is negligible.
 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.
 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