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

Topic 7. 3 Nuclear Reactions, Fission and

Fusion

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.

7.3.1 Describe and give an example of an artificial (induced) transmutation.7.3.2 Construct and complete nuclear equations.

1919 – RutherfordAlpha 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

7.3.1 Describe and give an example of an artificial (induced) transmutation.7.3.2 Construct and complete nuclear equations.

IB DefinitionUnified 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

7.3.3 Define the term unified atomic mass unit.

Lets take a minute to really examine this “unified mass units” 6 x proton masses = 6 x 1.007276 u = 6.043656 u6 x neutron masses = 6 x 1.008665 u = 6.051990 u6 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

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.

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

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.

IB DefinitionMass 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.

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.

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 = 92.2MeV1u

E = 92.2 x 106eV x 1.6 x 10-19J = 1.479 x 10-11J 1eV

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.

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.

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.

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).

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.

First a conversionMass=0.098940u = 0.098940u x 1.661x10-27kg

= 1.643x10-28kg1u

Now the formulaE = mc2

E = 1.643x10-28kg x (3x108m/s)2

E = 1.479x10-11J

Convert from J to eVE = 1.479x10-11J x 1eV = 92.2MeV

1.6 x 10-19J

Fission and fusion

Learn about how a nuclear power plant works.http://science.howstuffworks.com/nuclear-power.htm

7.3.6 Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon.7.3.8 Describe the processes of nuclear fission and nuclear fusion.7.3.9 Apply the graph in 7.3.6 to account for the energy release in the processes of fission and 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

7.3.8 Describe the processes of nuclear fission and nuclear fusion.

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

7.3.8 Describe the processes of nuclear fission and nuclear fusion.

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.

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.

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.

***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

nucleonFission – shallow – much less energy released per nucleon

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.

The graph explains/predicts:Relative stability of elements around atomic number

26Large abundance of iron in universeThe energy released in nuclear reactions.

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.

There are several different possibilities of this sequence. We’ll look at path 1.

7.3.10 State that nuclear fusion is the main source of the Sun’s energy

Here are two sites that go into more detail and explanation.http://burro.astr.cwru.edu/Academics/Astr221/StarPhys/ppc

hain.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/astro/

procyc.htmlThe 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.

7.3.10 State that nuclear fusion is the main source of the Sun’s energy

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

7.3.10 State that nuclear fusion is the main source of the Sun’s energy

Example Problem 3Estimate the energy released in the fission

reaction of Uranium-235 as it transmutes into Barium-141 and Krypton-92

Answer: about 176 MeV

7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions.

Example Problem 3Estimate the energy released in the fission reaction of

Uranium-235 as it transmutes into Barium-141 and Krypton-92

Answer: about 176 MeVSolution: 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 nucleonU-235 = 7.6MeVBa-141 = 8.3MeVKr-8.6 = 8.6MeV

Example Problem 3Estimate the energy released in the fission reaction

of Uranium-235 as it transmutes into Barium-141 and Krypton-92

Answer: about 176 MeVSolution: Calculating the total binding energies:

U-235 = 7.6MeV x 235 = 1786MeV

Ba-141 = 8.3MeV x 141 = 1170.3MeVKr-8.6 = 8.6MeV x 92 = 791.2MeVTotal = 1961.5MeV

Difference = 1961.5MeV – 1786MeV = about 176MeV

Example Problem 4Repeat the previous example using the following

information and find the exact answer.

Atomic mass of U-235 235.043924uAtomic mass of Kr-92 91.926300uAtomic mass of Ba-141 140.914400u

Answer:173 MeV

7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions.

Example Problem 4Repeat the previous example using the following

information and find the exact answer.

Atomic mass of U-235 235.043924uAtomic mass of Kr-92 91.926300uAtomic 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 defectsAtomic mass of U-235 235.043924uConstituent mass U-235 236.958995uMass defect U-235 1.915071u = 1783.9MeV

Atomic mass of Kr-92 91.926300uConstituent mass Kr-92 92.766940uMass defect Kr-92 0.840640u = 783.1MeV

Atomic mass of Ba-141 140.914400uConstituent mass Ba-141 142.174557uMass defect Ba-141 1.260157u =

1173.8MeV

Example Problem 4

Difference in binding energies= (1173.8 + 783.1) – 1783.9 = 173MeV

Example Problem 5Using 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

7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions.

Example Problem 5Using 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 8.6MeV x 88 = 756MeVXenon-1368.4MeV x 136 = 1142MeVTotal = 1898MeV

Example Problem 5Using 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 6Repeat the previous example, using the following

information.

Atomic mass of U-235 235.043924uAtomic mass of Sr-88 87.905618uAtomic mass of Xe-136 135.907210u

Answer: 126.5MeV

7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions.

Example Problem 7Calculate 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 2.014102uAtomic mass of H-3 3.016049uAtomic mass of He-4 4.002602u

Answer: 17.59MeV

7.3.7 Solve problems involving mass defect and binding energy7.3.11 Solve problems involving fission and fusion reactions.