7.3+NUCLEAR+REACTIONS,+FUSION+AND+FISSION

IB Physics > 7 Atomic and Nuclear Physics [|IB Physics Atomic and Nuclear Physics objectives in Word] - link to Google doc [|IB ATOMIC AND NUCLEAR PHYSICS DEFINITIONS AND CONCEPTS] - link to Google doc
 * 1 Measure || 2 Mech || 3 Therm || 4 Waves || 5 Electric || 6 Fields || 7 Atomic || 8 EPCC || 9 MIF || 10 Therm AHL || 11 Wave Phen || 12 EMI || 13 QNP || 14 Digital || OPT || PRAC || REVISE ||
 * 7.1 ATOM || 7.2 RADIOACTIVE DECAY || 7.3 NUCLEAR REACTIONS ||

7.3 NUCLEAR REACTIONS, FUSION AND FISSION
Nuclear reactions useful pictures 7.3.1  Describe and give an example of an artificial (induced) transmutation.
 * NUCLEAR REACTIONS **

ARTIFICIAL TRANSMUTATION: The conversion of one element or isotope into another by an instrument such as a particle accelerator or nuclear reactor.

Rutherford (1919): Nitrogen bombarded with alpha particles transmutes to oxygen. Cockcroft and Walton (1931): Split lithium by bombarding with alpha particles.

7.3.2  Construct and complete nuclear equations. NUCLEAR EQUATIONS: Showing interactions of nuclear particles in which the proton and nucleon numbers balance.
 * ACTIVITY: Write the nuclear equation for the images below. **
 * [[image:nothingnerdy/lithium_fission.jpg width="320" height="149"]] || [[image:http://www.green-planet-solar-energy.com/images/nuclear_fission_good_1a.jpg width="341" height="283" caption="Barium Z=56, N=86"]] ||
 * [[image:http://www.splung.com/nuclear/images/fusion/fusion.jpg]] || [[image:http://www.personal.psu.edu/sab5235/paperimage.gif width="365" height="272" caption="N hit by p, splits to C and n"]] ||

7.3.3  Define the term //unified atomic mass unit//. ATOMIC MASS UNIT: The mass of 1/12 of the nucleus of a carbon-12 isotope. MeV: Mega-electron-volt is a convenient unit for measuring nuclear energies. The energy an electron would have if it were accelerated through 1 000 000 V Students must be familiar with the units MeV c -2 and GeV c -2 for mass. MeVc -2 : Unit of mass calculated from Einstein relation when equivalent energy is measured in MeV.

7.3.4  Apply the Einstein mass–energy equivalence relationship. EINSTEIN MASS-ENERGY EQUIVALENCE: E = mc 2. The mass defect is equivalent to a change in energy which is calculated using this equation.

7.3.5  Define the concepts of //mass defect//, //binding energy// and //binding energy per// //nucleon//. BINDING ENERGY: The amount of work required to pull apart the constituents of a nucleus. A large binding energy implies a stable nucleus. MASS DEFECT: The mass of the particles of the separated nucleus is greater than when they are combined. Mass defect = mass of parts - mass of nucleus BINDING ENERGY PER NUCLEON: A graph is plotted for each isotope which shows which nuclides are the most stable (the most BE/ nucleon). The graph can predict fission and fusion reactions.

Calculate the **Binding Energy** per nucleon of Iron-54 (Z=26) ACTIVITY: Use these nuclear masses to calculate BE per nucleon
 * EXAMPLE **
 * Proton rest mass = 1.00**728** u (change typo in Hamper HL Ed1 p 244)
 * Neutron rest mass = 1.00866 u
 * 1u has mass 931.5 MeVc -2 and has energy 931.5 MeV
 * Step || Calculation ||  ||
 * __ Step A __ : Total mass of ** individual nucleons ** || 26*1.00728 + 28*1.00866 || = 54.4318 u ||
 * __ Step B __ : Total mass of ** nucleus ** || Given as 53.9396 || = 53.9396 u ||
 * __ Step C __ : Find ** mass defect ** || Difference between A and B || = 0.4922 u ||
 * __ Step D __ : Calculate ** binding energy ** || 0.4922*931.5 || = 458.4 MeV ||
 * __ Step E __ : Calculate ** BE per nucleon ** || 458.4/ 54 || = 8.5 MeV per nucleon ||
 * Be-9 has mass = 9.0122 u (Z = 4)
 * N-14 has mass = 14.0031 u (Z = 7)
 * Pb-204 has mass = 203.9730 u (Z = 82)

SOLUTIONS HERE 6.23 MeV/ nucleon 7.22 MeV/ nucleon 7.67 MeV/ nucleon

ACTIVITY: Put the data in Hamper HL p243 in a spreadsheet and use formulae to calculate BE/ nucleon. Result is [|here]. You can cut and paste the data and try to make the formulas yourself.

7.3.6  Draw and annotate a graph showing the variation with nucleon number of the binding energy per nucleon. Students should be familiar with binding energies plotted as positive quantities. Binding energies correspond to losses of mass, so BE is positive when MD is negative.



7.3.7  Solve problems involving mass defect and binding energy.

ENERGY CALCULATIONS
 * 1) Write the formula.
 * 2) Calculate the total masses in u on each side.
 * 3) Calculate the mass difference (if decreasing, energy is emitted; if increasing, energy is required to initiate the reaction).
 * 4) Calculate the energy in MeV

EXAMPLE OF TRANSMUTATION He-4 + O-16 **=====>** F-18 + H-1 + n Mass is gained, so energy was needed to initiate the reaction (a negative mass deficit!!!!). Therefore **energy required** is =931.5 MeV/u * 0.016826 u=
 * **Particle** || **mass/ u** ||
 * He-4 || 4.002603 ||
 * O-16 || 15.994914 ||
 * || ** 19.997517 ** ||
 * F-18 || 17.998403 ||
 * H-1 || 1.007276 ||
 * n || 1.008664 ||
 * || ** 20.014343 ** ||
 * Mass difference || **+0.016826** ||
 * Mass difference || **+0.016826** ||
 * 15.67** **MeV** (which is the energy required to start the reaction)

7.3.8  Describe the processes of nuclear fission and nuclear fusion. NUCLEAR FISSION: A process where a large nucleus splits to make smaller ones. This is permitted when the BE/ nucleon of the products is higher than the initial nucleus. The difference in mass/ energy is released. NUCLEAR CHAIN REACTION: A uranium nucleus can be split by an incident neutron. the fission reaction produces more neutrons. If one of the produced also causes fission, then a chain reaction will follow. If the number of fissions is greater than one, the reaction will go out of control. NUCLEAR FUSION: A process where small nuclei can combine to make a larger nucleus. This is permitted when the BE/ nucleon of the product is higher. The difference in mass/ energy is released. media type="youtube" key="jk6Hm1QoDYY" width="560" height="315" 7.3.9  Apply the graph in 7.3.6 to account for the energy release in the processes of fission and fusion.
 * FUSION AND FISSION **
 * Because the products of a fission or fusion reaction have higher BE per nucleon than the reactants, there must be a mass defect (the products are lighter).
 * This means that some energy is emitted from the reaction equivalent to the change in mass.
 * The energy is in the form of electromagnetic radiation (probably gamma) and KE of the products.

7.3.10  State that nuclear fusion is the main source of the Sun’s energy. NUCLEAR FUSION IN THE SUN: The temperature in the Sun (more than 10 million kelvin) provides enough energy for nuclei to overcome electrostatic repulsion so that they can fuse. Enormous amounts of energy are emitted since the products have more BE per nucleon than the reactants. 7.3.11  Solve problems involving fission and fusion reactions. HOW MUCH ENERGY IS PRODUCED BY FISSION AND FUSION REACTIONS? If the energy is not given, it can be read from the BE/ nucleon graph and multiplied by the nucleon number.

Calculated using BE. There is a gain in BE which means the products are lighter (energy is produced).
 * EXAMPLE OF NUCLEAR FISSION **
 * ** U-236 ====> Ba-141 + Kr-92 ** || **MeV** ||
 * U-236 || 236*7.6 ||
 * || ** 1793.6 ** ||
 * Ba-141 || 141*8.0 ||
 * Kr-92 || 92*8.2 ||
 * || ** 1882.4 ** ||
 * Gain in BE || **88.8** ||
 * || ** 1882.4 ** ||
 * Gain in BE || **88.8** ||
 * Gain in BE || **88.8** ||

Calculated using nuclear masses in u There is a decrease in mass which means energy is produced.
 * EXAMPLE OF NUCLEAR FUSION**
 * ** H-2 + H-3 ====> He-4 + n ** || **u** ||
 * H-2 || 2.013553 ||
 * H-3 || 3.016049 ||
 * || ** 5.029602 ** ||
 * He-4 || 4.002603 ||
 * n || 1.008664 ||
 * || ** 5.011267 ** ||
 * Mass difference || 0.018335 ||
 * || ** 5.011267 ** ||
 * Mass difference || 0.018335 ||
 * Mass difference || 0.018335 ||



CONSTANTS IN ATOMIC AND NUCLEAR PHYSICS FROM IB DATA BOOK