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Module 4

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If you add up a certain number of neutrons and protons and ... e.g. Helium from the fusion of Deuterium and Tritium. Original mass, 2.0141u 3.0160u = 5.0301u ... – PowerPoint PPT presentation

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Title: Module 4


1
Module 4
  • Nuclear applications

2
Fission and Fusion
  • Fission large unstable nuclei split apart e.g.
    Uranium-235
  • Induced by bombarding with neutrons (e.g in
    reactors)
  • During fission energy is released
  • Fusion light nuclei release energy when they
    combine - in fusion reactions

3
Mass Defect
  • If you add up a certain number of neutrons and
    protons and put them together in a nucleus you
    end up with less mass than you started with!

The missing mass is known as the Mass
Defect
4
Atomic (or unified) Mass Unit, u
  • Definition of the a.m.u.
  • 1u 1/12 of the mass of a Carbon-12 atom
  • 1u 1.66043 x 10-27 kg

5
proton mass 1.0073 u neutron mass 1.0087
u Mass Defect Example Uranium-235 has 92 protons
and 143 neutrons You may therefore expect its
mass to be 92 x 1.0073u 143 x 1.0087u
236.92u However its mass is actually 235.04u The
missing mass or mass defect 236.92u 235.04u
1.88u This principle applies to all
nuclei Where does the mass go??
6
Einsteins theory of Special Relativity (1905)
showed that - Mass and Energy are
Equivalent The total energy of a particle of
rest mass m0 and velocity v is given by
,where m is the total relativistic mass,
In Newtonian mechanics kinetic energy is given by
½ mv2 For vltltc the relativity equation can be
shown to approximate E m0c2 ½ m0v2 , the
first term is the rest energy that only depends
on the objects mass, the second term is the KE
N.B. If v is large KE is given by the full
equation, i.e. (m-mo)c2 So in relativity if the
energy of a body changes by E its mass will
change by E/c2
7
Relativity can explain the mass defect
e.g. Uranium-235 has a mass defect of 1.88 u By E
mc2 the energy equivalent of this mass is
(1.88 x 1.66043x10-27) x (3 x 108)2 2.8 x
10-10 J
This is the energy that would have to be put in
to split the U-235 nucleus into its constituents
it is the energy required to bind the nucleus
together binding energy. i.e. Zmp Nmn
mnuc mb.e.
8
Binding Energy
  • Binding Energy mass defect x c2
  • This energy is so small that it is usually
    written down using a new unit of energy
  • The electronvolt or eV
  • 1eV is the energy that would be gained by 1
    electron if accelerated through a potential
    difference of 1 Volt
  • 1eV 1.6 x 10-19 J
  • For U-235 the binding energy is 1.749x109eV

9
For a given nucleus the Binding energy / atomic
mass number Binding energy per nucleon How many
eVs are equivalent to 1 u ? 1u 1.66043x10-27
kg By Emc2 the energy equivalent of this mass
1.66043x10-27 x c2 J 1.491x10-10 J In eV this
energy 1.491x10-10 / 1.6x10-19 eV 9.31x108 eV
931 MeV If we know the mass defect for a
particular reaction in us we can work out the
energy released in MeV
10
Nuclear Stability
  • The higher the binding energy per nucleon the
    more stable a nucleus will be
  • For U-235 the binding energy is 1.749x109 / 235
    7.44 MeV per nucleon
  • This value is different for every nucleus and has
    a maximum for Iron-56

11
The Binding Energy Curve
12
A nucleus will become more stable if a reaction
brings it closer to the iron peak Nuclei heavier
than iron do this by breaking up into smaller
nuclei via Fission Nuclei lighter than iron do
this by joining together via Fusion reactions
13
Energy from Fusion e.g. Helium from the fusion of
Deuterium and Tritium  
Original mass, 2.0141u 3.0160u 5.0301u After
reaction, 4.0026u 1.0087u 5.0113u Mass defect
5.0301u 5.0113u 0.0188u
3.1216x10-29kg Energy release mc2 2.81x10-12J
(approx. 18MeV)
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