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Introduction• Nuclear power stations tap into the energy storedinside an atom’s nucleus.• Huge amounts of energy can be released.• But where does it come from and how can we get atit?Learning Outcomes:• At the end of this lesson you will be able to:1. Explain Mass Defect2. Calculate the energy stored by an atomic nucleus2
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Missing mass – Mass Defect• A helium nucleus is made up of 2 protons and 2neutrons.• When the mass of two protons + mass of 2 neutronsis compared with mass of helium nucleus – somethingvery odd is seen.• The mass of the nucleus is less than the total mass ofthe individual particles that it contains• True for all nuclei containing more than one nucleon.• Missing mass is known as mass difference or massdefect.• Masses involved are very small – measured in atomicmass units u, where:• 1u = 1.6605 x 10-27kg3
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Mass defectMass(atomic mass unit)Proton 1.00728 uNeutron 1.00867 uHelium nucleus 4.00151 uTable 1:Atomic massunits.Example 1.Using the data in table 1, calculate the mass defect for ahelium nucleus in atomic mass units and in kilograms.Mass of 2p + 2n = (2 x 1.00728 u) + (2 x 1.00867 u)= 4.03190 uMass defect = 4.03190 u – 4.00151 u = 0.03039 uMass defect in kg = 0.03039 x 1.6605 x 10-27kg= 5.046 x 10-29kg4
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Mass defect• A mass of 5.046 x 10-27kg may not sound a lot – but onatomic scale it is significant.• The mass defect is about 3% of the mass of a proton(or 55 electrons)• It is important to measure nuclear masses precisely –masses are quoted to 6 significant figures.Mass and energy equivalence• Splitting the nucleus into individual nucleons results inincrease of total mass.• Where does this extra mass come from?• Splitting the nucleus is very difficult – nucleons heldtogether by a very strong but short-range nuclearforces.• Overcoming these forces requires energy5
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Mass defect• What happens to this energy?• Energy disappears into the system and mass iscreated!• Goes against the conventional conservation laws.6Albert EinsteinIt was Albert Einstein whosuggested that mass and energyare equivalent.He linked mass and energy in hisfamous equation: E = m . c 2E is the energy equivalent in joulesof a mass m in kilograms, c is thevelocity of light (3.00 x 108m s-1)
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7Example 2:Calculate the energy equivalent of the mass defectcalculated in Example 1.E = m c2= 5.046 x 10-29kg x (3.00 x 108m s-1)2= 4.54 x 10-12J (3 s.f.)This is the energy needed to separate the helium nucleusinto individual protons and neutrons.42He + 4.54 x 10-12J → 2p + 2nenergyIn any system, the total amount of mass and energy isconserved.
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Energy equivalent of 1 u• Einstein’s equation uses mass in kg and energy in J.• In nuclear physics we are more likely to be working inatomic mass units (u) and electron-volts (eV)• Need to work out energy equivalent (in eV) of 1 u:• Using Einstein’s equation with a precise value for thevelocity of light:• E = m c2= 1.6605 x 10-27kg x (2.9979 x 108m s-1)2= 1.4924 x 10-10J• But 1 eV = 1.6022 x 10-19J• Hence E = 1.4924 x 10-10J / 1.6022 x 10-19= 9.315 x 108eV = 931.5 x 106eV• 1 u = 931.5 MeV8
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9Example 3:A carbon nucleus has a mass of 11.9967 uHow much energy, in MeV, would ne needed to split itinto its 6 protons and 6 neutrons?(mp = 1.00728 u, mn = 1.00867 u)Mass of 6p + 6 n = (6 x 1.00728 u) + (6 x 1.00867 u)= 12.0957 uMass defect = 12.0957 u – 11.9967 u = 0.0990 uEnergy equivalent = 0.0990 x 931.5 MeV= 92.2 MeV needed to separate the 12 nucleons.Question:Calculate the mass defect in u and kg for the followingnuclei:(a)Lithium (73Li), nuclear mass = 7.014353 u(b)Silicon (2814Si), nuclear mass = 27.96924 u
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Binding energy• Energy needed to separate a nucleus into individualnucleons is its binding energy.• Also the energy equivalent of the mass defect –found from E = m c2• Binding energy indicates the stability of the nucleus.• Total binding energy is linked to the size of thenucleus.• The more nucleons there are, the greater the energyneeded to separate them all out.• More useful comparison is the binding energy pernucleon.• Average energy needed to remove each nucleon fromthe nucleus.10
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11Example 4:Use the figures calculated in example 3 to find thebinding energy per nucleon for a carbon-12 nucleus.Carbon-12 contains 12 nucleons (6p + 6n)Total binding energy for carbon-12 = 92.2 MeVBinding energy per nucleon = 92.2 MeV / 12= 7.68 MeVQuestionFor each of the two nuclei in question on slide 9, calculate:(a)Total binding energy in eV(b)The binding energy per nucleon in eV
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Binding energy12Fig 1: Graph of binding energy per nucleon (eV) againstnucleon number (A)
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Binding energy• Notice that binding energy per nucleon is typicallyaround 8 MeV• The most stable nucleus has the highest bindingenergy per nucleon.• This is 8.79 MeV for 5626FeRadioactive Decay and Binding Energy• Unstable nucleus emits radiation and becomes morestable.• Daughter nucleus always has a higher binding energyper nucleon than the parent.• Energy is given out when a nucleus decays• Where does it come from?13
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Radioactive Decay and Binding Energy• Total mass of products is less than mass of parentnucleus.• Mass difference is released as energy – look at thefollowing examples:Example 5: α-decay• Thorium-228 decays by α-emission:22890Th → 22488Ra + 42α• Mass of thorium-228 nucleus = 227.97929 u• Mass of radium-224 nucleus + α-particle= 223.97189 u + 4.00151 u = 227.97240 uMass difference = 227.97929 u – 227.97340 u= 0.00589 u = 5.49 MeV (as 1 u = 931.5 MeV)14
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Radioactive Decay and Binding Energy• The surplus energy appears mostly as K.E. of the α-particle.• Radium nucleus also recoils slightly (momentum isconserved).Example 6: β-decay• Aluminium-29 decays by β-emission152913Al → 2914Si + 0-1β + 00ν-Mass of Si-29 nucleus + β-particle + antineutrino= 28.96880 u + 0.000549 u + 0 = 28.969349 uMass of aluminium-29 nucleus = 28.97330 uMass difference = 28.97330 u – 28.960349 u = 0.003951 u= 3.68 MeV (as 1 u = 931.5 MeV)
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Radioactive Decay and Binding Energy• Some of this energy is carried away by a γ-ray.• Rest becomes KE of the decay products.Transmutation and energy• Transmutation is conversion of one element toanother• Radioactive decay is a spontaneous transmutationwith a release of energy.• Some elements can be made by firing very fastmoving charged particles at a stable nucleus –artificial transmutation. (energy must be supplied)• First achieved by Rutherford, Marsden and Chadwickin 1919.16
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Transmutation and energy• Converted nitrogen into oxygen by bombarding it withα-particles.147N + 42α → 178O + 11H• Here mass of products is greater.• Energy must be supplied to make this reaction happenand balance the nuclear equation• Energy comes from K.E. of the α-particle. Modern useof artificial transmutation is the production ofmedical radioisotopes.Questions1. The decay of radium into radon is22688Ra → 22286Rn + 42αcalculate the energy released in eV 17
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Questions182. A possible reaction for the fission of uranium is:23592U + 10n → 14657La + 8735Br + 310nCalculate the energy released in the reaction in MeV.3. Calculate the minimum energy needed in MeV to makethe following reaction happen:6028Ni + 42α → 6330Zn + 10nNuclear masses: radium-226 = 225.9771 uradon-222 = 221.9703 uuranium-235 = 234.9934 ulanthanum-146 = 145.8684 ubromine-87 = 86.9153 unickel-60 = 59.9153 uzinc-63 = 62.9205 u
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