# nuclear physics & radioactivity vce physics unit 1 topic 1

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Nuclear Physics & RadioactivityVCE PHYSICSUnit 1

Topic 1

Explain why some atomic nuclei are stable and others are not.Describe the radioactive decay of unstable nuclei in terms of half life.Model radioactive decay as random decay with a particular half life, including mathematical modelling in terms of whole half lives.Apply a simple particle model of the atomic nucleus to the origin of , and radiation, including changes to the number of nucleons.Describe the detection and penetrating properties of , and radiation.Describe the effects of , and , radiation on humans including short- and long-term effects from low and high doses, external and internal sources, including absorbed dose (Gray), dose equivalence (Sieverts), and effective dose (Sieverts)Describe the effects of ionising radiation on living things and the environment.Explain nuclear transformations using decay equations involving , and radiation.Analyse decay series diagrams in terms of type of decay and stability of isotopes.Describe natural and artificial isotopes in terms of source and stability.Describe neutron absorption as one means of production of artificial radioisotopes.Identify sources of bias and error in written and other media related to nuclear physics and radioactivity.Describe the risks for living things and/or the environment associated with the use of nuclear reactions and radioactivity

Unit Outline

1.0 Atomic Structure.Atoms are made up of a nucleus which contains PROTONS and NEUTRONS, surrounded by ELECTRONS, circulating in groups or shells.THE HELIUM ATOMPROTONS have a mass of 1 A.M.U. ( 1 Atomic Mass Unit = 1.67 x 10-27 kg) Each carries a Positive charge of 1.6 x 10-19 Coulomb. NEUTRONS have a mass of 1 A.M.U. and carry NO CHARGE.ELECTRONS have a mass of 1/1840th of an A.M.U. (9.1 x 10-31 kg) Each carries a Negative charge of 1.6 x 10-19 Coulomb.Normal atoms are electrically neutral, thus the number of Protons = the number of Electrons. The number of neutrons varies (from 0 in Hydrogen atoms to a number much greater than the number of protons, eg Uranium atoms have 92 protons and 146 neutrons)

1.1 Atoms and IsotopesA shorthand method of representing the structure of an atom is:AXZ

where, X = the elements chemical symbol A = the MASS NUMBER = total number of Protons + Neutrons in the nucleus, Z = The ATOMIC NUMBER = the number of protons in the nucleus and therefore the number of electrons grouped around the nucleus.For example an atom of Uranium can be represented as:238U92Thus, this atom contains 92 protons, 92 electrons and (238 - 92 = 146) neutrons ISOTOPES are different forms of the same element. They differ because they contain varying numbers of NEUTRONS in their nucleus. Uranium has 4 main isotopes:233U92 92 protons, 92 electrons, 141 neutrons.234U92 92 protons, 92 electrons, 142 neutrons.235U92 92 protons, 92 electrons, 143 neutrons.238U92 92 protons, 92 electrons, 146 neutrons.

Atoms and Isotopes1. Fill in the blank spaces in the table2. Fill in the blank spaces in the table.1939201919243959514895232909014290210841288484214212Po84214Po84

ElementMass Number Atomic Number Number of ProtonsNumber of NeutronsNumber of ElectronsPotassium ( 39K19 )Americium ( 243Am95 ) Thorium ( 232Th90 )

NameNumber of protonsNumber of neutronsMass NumberAtomic NumberSymbolPolonium84126210Po84Polonium84212Polonium13084

1.2 Atomic and Nuclear Energy UnitsIn the large scale world energy is measured in Joules.In the small scale world of individual atomic or nuclear reactions, the Joule is too large a unit, so a smaller unit, the electron volt (eV) is used to quote energy values.By definition 1 electron volt (1 eV) is the energy obtained by 1 electron when passing through a voltage of 1 volt. Attaching metal plates to the terminals of a battery will provide a region where electrons can pass through a voltageAfter crossing between the two charged plates, the electrons energy will have increased by 1 eVIf the voltage between the plates is 1000 V the electrons energy will increase by 1 keVIf the voltage between the plates is 10 million volts, the electrons energy will increase by 10 MeV.An electron carries a charge of 1.6 x 10-19 Coulombs. When passing through a voltage of 1.0 V, its energy will increase by 1.6 x 10-19 J. So 1 eV = 1.6 x 10-19 J

Atomic Energy3. Calculate the energy (in joule) an electron would gain in passing through a potential difference of 6.2 eV.4. In order to raise an electron from one energy level to another within an atom it must absorb all the energy of an incoming photon of energy 1.25 x 10-18 J. How much more energetic will the electron be after the collision ? (Quote your answer in eV)

1 eV = 1.6 x 10-19 Joule. So 6.2 eV = (6.2)(1.6 x 10-19) J = 9.92 x 10-19 J 1.6 x 10-19 J = 1 eV. So 1.25 x 10-18 J = (1.25 x 10-18/1.6 x 10-19 ) eV= 7.8 eV

1.3 Uranium - Mining & EnrichmentUranium ore is mined and processed at the mine site into a greeny-yellow coloured solid material called YELLOWCAKE. Chemically, yellowcake is Uranium Oxide - U3O8This material is packed into 200 L drums and exported to overseas uranium processing plants.The U3O8 is made up of 2 isotopes; 99.3% 238U and 0.7% 235U.It is the 235U which is the desired product. It is this uranium isotope which is FISSIONABLE (able to be split apart) by slow or thermal neutrons (with energies < 5eV)In order to sustain a Nuclear Chain Reaction (see Slide 1.4) in a nuclear reactor or nuclear weapon, the proportion of 235U needs to be increased. This is achieved by the ENRICHMENT process.Nuclear reactor fuel needs to be enriched to about 3% to 4% 235UNuclear weapons fuel needs to be enriched to 90% 235U.Approximately 17 kg of 235U is needed to produce an effective weapon.However, only 4 kg of 90% pure PLUTONIUM (239Pu) is needed to produce an equally effective weapon.

Uranium7. The enrichment processA: Increases the proportion of 234U in the sampleB: Increases the proportion of 235U in the sampleC: Increases the proportion of 238U in the sampleD: Increases the proportion of all the isotopes in the sample5. What is the chemical composition of yellowcake ? U3O86. Naturally occurring Uranium ore containsA: 4 Isotopes of UraniumB: 3 Isotopes of UraniumC: 2 Isotopes of UraniumD: Only a single isotopic form of Uranium

8. Nuclear reactor fuel needs the proportion of 235U in the fuel sample to be at least A: 3% to 5% of the total B: 10% to 12% of the totalC: 25% to 30% of the totalD: 50% to 75% of the sample

1.4 Fissile MaterialsAny nucleus capable of undergoing fission is called a FISSILE MATERIAL.The main fissile materials known are: 233U92, 235U92 and 239Pu94 are more likely to undergo fission by capture of slow (< 5 eV) neutrons.

238U92 and 232Th90 need fast neutrons (> 1 MeV) to undergo fission.

235U92,239Pu94,238U92and 232Th90Fission is defined as the splitting of atomic nuclei233U92,

1.5 Nuclear FissionWhen slow neutron collides head on with a 235U atom, the nucleus undergoes fission . It splits into 2 fission products with atomic numbers approximately half that of the original 235U, PLUS (on average) 2.5 Neutrons PLUS (on average) 160 - 200 MeV of energy.Both Uranium isotopes are capable of being fissioned by neutrons:235U is fissioned by neutrons of all energies with a high probability of fission by low energy (< 5 eV), thermal neutrons. 238U is fissioned by fast neutrons (>1 MeV). It captures neutrons of lesser energy without suffering fission.The products shown here are typical but not unique, many combinations of product nuclei are possible, with Atomic Nos ranging from 34 to 74.Shown on the left is a typical fission process initiated by a neutron with the first target nucleus splitting to release further neutrons.

Fission10. What are the products of the nuclear fission of 235U ?9. Define nuclear fission.Fission is defined as the splitting of atomic nuclei 235U splits into 2 fission products with atomic numbers approximately half that of the original 235U, PLUS (on average) 2.5 Neutrons PLUS (on average) 160 - 200 MeV of energy.

1.6 Mass into EnergyThe typical 235U fission as mentioned on a previous slide is: Adding up the mass of the reactants (measured in a.m.u.s), we get: 1.0087 + 235.0439 = 236.0526 a.m.u.Adding the masses of the products we get: 140.9139 + 91.8973 + 3.0261 = 235.8373 a.m.u.The mass of the products is 0.2153 a.m.u. LESS than the mass of the reactants.This lost mass has been converted to energy, the amount of which can be calculated from the Einsteins famous equation E = mc21 a.m.u. = 1.66 x 10-27 kg. 0.2153 a.m.u. = 3.57 x 10-28 kg. E = (3.57 x 10-28)(3.0 x 108)2= 3.2 x 10-11 JConverting this energy in Joules to energy in eV we get:3.2 x 10-11/1.6 x 10-19 = 2.0 X 108 eV = 200 MeVThus EACH fission of a 235U nucleus releases about 200 MeV of energy, initially as Kinetic Energy of the fragments which is then converted to Heat Energy by collisions with other nuclei.This heat energy is used to create steam to spin a turbine which drives a generator producing electricity.

Mass into Energy11. The energy released in the nuclear fission process arises from the conversion of what to energy ?12. What equation is used to convert mass to energy ? Who formulated this equation ? 13. Show that 0.5 amu is the equivalent of 478 MeV of energy Note: (1 amu = 1.67 x 10-27 kg)In nuclear fission mass is converted into energyE = mc2 , Equation formulated by Albert Einstein 0.5 a.m.u. = (0.5)(1.67 x 10-27) kg = 8.5 x 10-28 kg. E = mc2 = (8.5 x 10-28)(3 x 108)2 = 7.65 x 10-11 JNow 7.65 x 10-11 J = (7.65 x 10-11)/(1.6 x 10-19) eV = 4.78 x 108 eV= 478 MeV

Chapter 2Topics covered:Neutron Moderation.Chain

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