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Nuclear Physics IB 12 Nuclide: a particular type of nucleus Nucleon: a proton or a neutron Atomic number (Z) (proton number): number of protons in nucleus Mass number (A) (nucleon number): number of protons + neutrons Neutron number (N): number of neutrons in nucleus (N = A – Z) Isotopes: nuclei with same number of protons but different numbers of neutrons Unified atomic mass unit (u): 1/12th the mass of a carbon-12 nucleus 1 u = 1.661 x 10-27 kg 1 u = 1 g/mol Atomic mass ≈ A * u 1 u = 931.5 MeV/c2 A Z X 56 26 Fe 27 13 Al 12 6 C 14 6 C Atomic Number 26 13 6 6 Mass Number 56 27 12 14 Neutron Number 30 14 6 8 Atomic Mass 56 u 27 u 12 u 14 u Molar Mass 56 g 27 g 12 g 14 g How big are atomic nuclei? 10-15m – 10-14 m How do we know that neutrons exist? By the existence of isotopes How do we know this? Alpha-particle scattering experiments (Geiger-Marsden) How do we know that isotopes exist? By measurements in the mass spectrometer Formulas: ET = ET EK = Ee ½ mv2 = qV ½ mv2 = q(kQ/r) Formulas: v = E/B Note that most nuclei have approximately the same . . .density = 2 x 1017 kg/m3 = 1014 times denser than water r = mv/qB 1 Nuclear Stability IB 12 What interactions exist in the nucleus? 1. Gravitational: (long range) attractive but very weak/negligible 2. Coulomb or Electromagnetic: (long range) repulsive and very strong between protons 3. Strong nuclear force: (short range) attractive and strongest – between any two nucleons 4. Weak nuclear force: (short range) involved in radioactive decay Why are some nuclei stable while others are not? The Coulomb force is a long-range force which means that every proton in the nucleus repels every other proton. The strong nuclear force is an attractive force between any two nucleons (protons and/or neutrons). This force is very strong but is short range (10-15 m) which means it only acts between a nucleon and its nearest neighbors. At this range, it is stronger than the Coulomb repulsion and is what holds the nucleus together. Neutrons in the nucleus play a dual role in keeping it stable. They provide for the strong force of attraction, through the exchange of gluons with their nearest neighbors, and they act to separate protons to reduce the Coulomb repulsion. Each dot in the plot at right represents a stable nuclide and the shape is known as the “band (or valley) of stability.” With few exceptions, the naturally occurring stable nuclei have a number N of neutrons that equals or exceeds the number Z of protons. For small nuclei (Z < 20), number of neutrons tends to equal number of protons (N = Z). As more protons are added, the Coulomb repulsion rises faster than the strong force of attraction since the Coulomb force acts throughout the entire nucleus but the strong force only acts among nearby nucleons. Therefore, more neutrons are needed for each extra proton to keep the nucleus together. Thus, for large nuclei (Z > 20), there are more neutrons than protons (N > Z). After Z = 83 (Bismuth), adding extra neutrons is no longer able to counteract the Coulomb repulsion and the nuclei become unstable and decay in various ways. Nuclei above (to the left of) the band of stability have too many neutrons and tend to decay by alpha or beta-minus (electron) emission, both of which reduce the number of neutrons in the nucleus. Nuclei below (to the right of) the band of stability have too few neutrons and tend to decay by beta- plus (positron) emission which increases the number of neutrons in the nucleus. 2 Binding Energy IB 12 The total mass of a nucleus is always less than the sum of the masses its nucleons. Because mass is another manifestation of energy, another way of saying this is the total energy of the nucleus is less than the combined energy of the separated nucleons. Mass defect (mass deficit) (Δm) Difference between the mass of the nucleus and the sum of the masses of its individual nucleons Nuclear binding energy (ΔE) Formulas: 1. energy released when a nuclide is assembled from its individual mnucleus + Δm = mnucleons components 2 E = ( m )c 2. energy required when nucleus is separated into its individual components Different nuclei have different total binding energies. As a general trend, as the atomic number increases . . . the total binding energy for the nucleus increases. Electric Charge Electric Charge Rest Mass Rest Mass Rest Mass Particle (e) (C) (kg) (u) (MeV/c2) Proton +1 +1.60 x 10-19 1.673 x 10-27 1.007276 938 Neutron 0 0 1.675 x 10-27 1.008665 940 Electron -1 -1.60 x 10-19 9.110 x 10-31 0.000549 0.511 1. The most abundant isotope of helium has a 24He nucleus whose mass is 6.6447 × 10-27 kg. For this nucleus, find the mass defect and the binding energy. helium: 0.050404 x 10-27 kg 4.53636 x 10-12 J 2. Calculate the binding energy and mass defect for 816O whose measured mass is 15.994915 u. Oxygen: 0.132613 u 123.5 MeV 3 Binding Energy per Nucleon IB 12 To see how the nuclear binding energy varies from nucleus to nucleus, it is useful to compare the binding energy for each nucleus on a per-nucleon basis, as shown in the graph below. Your Turn a) This graph is used to compare the energy states of different nuclides and to determine what nuclear reactions are energetically feasible. As binding energy per nucleon increases so does the stability of the nucleus. Higher binding energies represent lower energy states since more energy was released when the nucleus was assembled. b) Binding energy per nucleon increases up to a peak at 2656Fe then decreases, so 2656Fe is the most stable nuclide. Most nuclides have a binding energy per nucleon of about 8 MeV. Lighter nuclei are held less tightly than heavier nuclei. c) Nuclear reactions, both natural (radioactive decay) and artificial/induced (fission, fusion, bombardments) occur if they increase the binding energy per nucleon ratio. Fusion occurs for light nuclei (below 2656Fe) and fission occurs for heavy nuclei (above 2656Fe). d) For both natural and induced nuclear reactions, the total rest mass of the products is less than the total rest mass of the reactants since energy is released in the reaction. Also, the products are in a lower energy state since energy was released in the reaction and so the products have a greater binding energy per nucleon than the reactants. 1. Use the graph above to estimate the total binding energy of an oxygen-16 nucleus. 8 MeV x 16 = 128 MeV Types of Nuclear Reactions 1) Artificial (Induced) Transmutation: A nucleus is bombarded with a nucleon, an alpha particle or another small nucleus, resulting in a nuclide with a different proton number (a different element). 2) Nuclear Fusion: Two light nuclei combine to form a more massive nucleus with the release of energy. 3) Nuclear Fission: A heavy nucleus splits into two smaller nuclei of roughly equal mass with the release of energy. 4) Natural Radioactivity: When an unstable (radioactive) nucleus disintegrates spontaneously, the nucleus emits a particle of small mass and/or a photon. Release of energy in nuclear reactions: m = m + Δm Energy is usually released in the form of kinetic energy for the products. Binding energy per nucleon: greater for product nuclei than for original nuclei since energy is released. 4 Radioactive Decay Reactions IB 12 Alpha Decay Alpha particle: helium nucleus, α, 24He Example reaction: 226 88 Ra 86 Rn 4 He energy 222 2 General equation: A Z X Z 4 Y 2 He energy A 2 4 Where does the kinetic energy come from? Rest mass of nucleons Result: nucleus is in a more stable state with higher binding energy and higher BE per nucleon since it released energy 1. A radium nucleus, initially at rest, decays by the emission of an alpha particle into radon in the reaction described above. The mass of 88226Ra is 226.025402 u and the mass of 86222Rn is 222.017571 u and the mass of the alpha particle is 4.002602 u. Δm = 0.005229 u a) Calculate the energy released in this decay. 4.87 MeV b) Compare the momenta, speeds, and kinetic energies of the two particles produced by this reaction. c) If the kinetic energy of the alpha particle is d) Calculate the recoil speed of the radon nucleus. 4.77 MeV, calculate its speed. Use momentum – no need to change units 5 Beta Decay IB 12 - Beta-minus particle: electron, β , -1 e 0 + Beta-plus particle: positron, β , +10e Consider the following two “mysterious” results of beta decay: a) Observe the before and after picture of beta decay. What’s wrong? b) Inspect the graph of kinetic energy carried away by the beta particles. Why are so few beta particles leaving with the majority of the kinetic energy? Where did this missing kinetic energy go? Conclusion: there is third particle involved with beta decay that carries away some KE and momentum – virtually undetectable Neutrino and anti-neutrino: fundamental particles - no charge – very small mass Beta-minus decay Beta-plus decay Example reaction: Example reaction: 14 6 C 14 N 01 e 0 energy 7 0 12 7 N 12 C 01 e 0 energy 6 0 General equation: General equation: A Z X Z 1 Y 01 e 0 energy A 0 A Z X Z 1 Y 01 e 0 energy A 0 How does this happen? Weak force How does this happen? Weak force 1 0 n 1 p 01 e 0 energy 1 0 1 1 p 1 n 01 e 0 energy 0 0 Gamma Decay Gamma particle: high energy photon, Example reaction: 12 6 C* 12 C energy 6 Before Decay General equation: A Z X * Z X energy A After Decay Where does the photon (energy) come from? Rest mass of nucleons 6 Energy Spectra of Radiation IB 12 The nucleus itself, like the atom as a whole, is a quantum system Importance: discrete energy spectra give evidence with allowed states and discrete energy levels. The nucleus can be for nuclear energy levels in any one of a number of discrete allowed excited states or in its lowest energy relaxed state. When it transitions between a higher energy level and a lower one, it emits energy in the form of alpha, beta, or gamma radiation. When an alpha particle or a gamma photon is emitted from the nucleus, only discrete energies are Gamma Alpha spectra Beta spectra observed. These discrete energy spectra give evidence that a spectra nucleus has energy levels. (However, the spectrum of energies emitted as beta articles is continuous due to its sharing the energy Discrete continuous discrete with a neutrino or antineutrino in any proportion.) Ionizing Radiation Ionizing Radiation – As this radiation passes through materials, it “knocks off” electrons from neutral atoms thereby creating an ion pair: free electrons and a positive ion. This ionizing property allows the radiation to be detected but is also dangerous since it can lead to mutations in biologically important molecules in cells, such as DNA. α β γ Electron or Particle helium nucleus high-energy photon positron Penetration ability low medium high Sheet of paper; a Material needed to 1 mm of few centimeters of 10 cm of lead absorb it aluminum air Path length in air a few cm less than 1 meter effectively infinite Detection of Radiation: the Geiger-Muller tube (Geiger counter) The Geiger counter consists of a gas-filled metal cylinder. The α, β, or γ rays enter the cylinder through a thin window at one end. Gamma rays can also penetrate directly through the metal. A wire electrode runs along the center of the tube and is kept at a high positive voltage (1000-3000 V) relative to the outer cylinder. When a high-energy particle or photon enters the cylinder, it collides with and ionizes a gas molecule. The electron produced from the gas molecule accelerates toward the positive wire, ionizing other molecules in its path. Additional electrons are formed, and an avalanche of electrons rushes toward the wire, leading to a pulse of current through the resistor R. This pulse can be counted or made to produce a "click" in a loudspeaker. The number of counts or clicks is related to the number of disintegrations that produced the particles or photons. Biological Effects of Ionizing Radiation Alpha and beta particles have energies typically measured in MeV. To ionize an atom requires about 10 eV so each particle can potentially ionize 105 atoms before they run out of energy. When radiation ionizes atoms that are part of a living cell, it can affect the ability of the cell to carry out its function or even cause the cell wall to rupture. In minor cases, the effect is similar to a burn. If a large number of cells that are part of a vital organ are affected then this can lead to death. Alternatively, instead of causing the cell to die, the damage done by ionizing radiation might just prevent cells from dividing and reproducing. Or, it could be the cause of the transformation of the cell into a malignant form. If these malignant cells continue to grow then this is called cancer. The amount of harm that radiation can cause is dependent on the number and energy of the particles. When a gamma photon is absorbed, the whole photon is absorbed so one photon can ionize only one atom. However, the emitted electron has so much energy that it can ionize further atoms, leading to damage similar to that caused by alpha and beta particles. On a positive note, rapidly diving cancer cells are very susceptible to the effects of radiation and are more easily killed than normal cells. The controlled use of the radiations associated with radioactivity is of great benefit in the treatment of cancerous tumors. 7 Mathematical Description of Radioactive Decay IB 12 Radioactive decay: 1) Random process: It cannot be predicted when a particular nucleus will decay, only the probability that it will decay. 2) Spontaneous process: It is not affected by external conditions. For example, changing the pressure or temperature of a sample will not affect the decay process. 3) Rate of decay decreases exponentially with time: Any amount of radioactive nuclei will reduce to half its initial amount in a constant time, independent of the initial amount. Half-life (T1/2) Units: time - the time taken for ½ of the radioactive nuclides in a sample to decay = s or hr or d or yr - the time taken for the activity of a sample to decrease to ½ of its initial value N0 = number of radioactive nuclei originally present N = number of radioactive nuclei present at any one time Your Turn Radioactive tritium has a half-life of about 12 years. Complete the graph below. Radioactive X → stable Y + particle Show how amount of daughter Y mirrors X A nuclide X has a half-life of 10 s. On decay the stable nuclide Y is formed. Initially a sample contains only atoms of X. After what time will 87.5% of the atoms in the sample have decayed into nuclide Y? 30 s 8 Activity IB 12 Activity (A) – the number of radioactive disintegrations per unit time (decay rate) Units: decays/time = s-1 or hr-1or d-1 or yr-1 N Formula: A t Standard units: Becquerel (Bq) 1 Bq = 1 decay per second 1. A sample originally contains 8.0 x 10 12 radioactive nuclei and has a half-life of 5.0 seconds. Calculate the activity of the sample and its half-life after: a) 5.0 seconds b) 10. seconds c) 15 seconds 8.0 x 1011 Bq 6.0 x 1011 Bq 4.7 x 1011 Bq 5.0 s 5.0 s 5.0 s 2. Samples of two nuclides X and Y initially contain the same number of radioactive nuclei, but the half-life of nuclide X is greater than the half-life of nuclide Y. Compare the initial activities of the two samples. Activity of sample Y is greater Activity A N The Radioactive Decay Law: The rate at which radioactive nuclei in a sample decay (the activity) is proportional to the number of radioactive nuclei present in the sample at any one time. A N [As the number of radioactive nuclei decreases, so does the average rate of decay (the activity).] Initial Activity A0 N0 The initial activity (A0) is directly related to the number of radioactive nuclei originally present (N0) in the sample. 3. The isotope Francium-224 has a half-life of 20 minutes. A sample of the isotope has an initial activity of 800 disintegrations per second. What is the approximate activity of the sample after 1 hour? N = 1/8 N0 So A = 1/8 A0 A = 100 disintegrations s-1 9 IB 12 Decay constant (λ) Units: – constant of proportionality between the decay rate (activity) and the inverse time number of radioactive nuclei present. = s-1 or hr-1or d-1 or yr-1 - probability of decay of a particular nuclei per unit time. Deriving the Radioactive Decay Law Relating the Decay Constant and Half-life At t T1/ 2 then N N 0 1 1 N 2 1 T A N 2 e 1/ 2 t The decay equation becomes dN N N0e t N 2 eT1/ 2 dt 1 N N eT1/ 2 0 0 2 T ln 2 e 1/ 2 A solution to this equation ln 2T1/ 2 is an exponential function ln 2 0.693 of the form T1/ 2 T1/ 2 N N 0 e t ln 2 0.693 T1/ 2 also t A = Aoe N 0 e t 1. The half-life of a certain radioactive isotope is 2.0 minutes. A particular nucleus of this isotope has not decayed within a time interval of 2.0 minutes. What is the probability of it decaying in: a) the next two minutes b) the next one minute c) the next second 2. A sample of a radioactive isotope X has the same initial activity as a sample of the isotope Y. The sample of X contains twice the number of atoms as the sample of Y. If the half-life of X is TX then the half-life of Y is 0.5 TX 10 IB 12 3. The half-life of a radioactive isotope is 10 days. Calculate the fraction of the sample that will be left after 15 days. 35% 4. The half-life of a radioactive substance is 10 days. Initially, there are 2.00 x 10 26 radioactive nuclei present. a) What is the probability of any one particular nucleus decaying? b) What is the initial activity? c) How many radioactive nuclei are left after 25 days? d) What is the activity of the sample after 25 days? e) How long will it take for the activity to fall to 1.0 x 1024 dy-1? 11 Graphs of Radioactive Decay IB 12 Radioactive nuclei vs. time N N 0 e t Straightening by natural log ln N ln N 0 e t ln(N0) ln N ln N 0 t ln N t ln N 0 y mx b T1/2 = ln 2/(-slope) Slope = -λ Activity of sample vs. time Straightening by natural log Methods of Determining Half-life If the half-life is short, then readings can be taken of activity versus time using a Geiger counter, for example. Then, either 1. A graph of activity versus time would give the exponential shape and several values for the half-life could be read from the graph and averaged. OR 2. A graph of ln (activity) versus time would be linear and the decay constant can be calculated from the slope. If the half-life is long, then the activity will be effectively constant over a period of time. If a way could be found to calculate the number of nuclei present chemically, perhaps using the mass of the sample and Avogadro’s number, then the activity relation or the decay equation could be used to calculate half-life. 12 IB 12 1. Cesium-138 decays into an isotope of barium. Measurements of the activity of a particular sample of cesium-138 were taken and graphed at right. a) Suggest how the data for this graph could have been obtained. Geiger counter b) Use the graph to estimate the half-life of cesium-138. 35 minutes c) Use the graph to estimate the half-life of the barium isotope. Wait until 250/300 minutes when very little cesium left 90 minutes 2. A 2.0 mg sample of carbon-14 is measured to have an activity of 6.5 x 10 10 Bq. a) Use this information to determine the half-life of carbon-14 in years. 5700 years b) A student suggests that the half-life can be determined by taking repeated measurements of the activity and analyzing the data graphically. Use your answer to part (a) to comment on this method of determining the half-life. 3. The radioactive isotope potassium-40 undergoes beta decay to form the isotope calcium-40 with a half-life of 1.3 x 109 yr. A sample of rock contains 10 mg of potassium-40 and 42 mg of calcium-40. a) Determine the age of the rock sample. 3.1 x 109 yr b) What are some assumptions made in this determination of age? No calcium-40 originally present = all potassium-40 No loss of either isotope in intervening years 13 Nuclear Fission IB 12 Nuclear Fission: A heavy nucleus splits into two smaller nuclei of roughly equal mass with the release of energy. Nuclear Fusion: Two light nuclei combine to form a more massive nucleus with the release of energy. Release of energy in nuclear reactions: m = m + Δm Energy is usually released in the form of . . . kinetic energy for the products. Binding energy per nucleon: greater for product nuclei than for original nuclei since energy is released. One Common Fission Reaction 235 92 U 1 n 92 U * X Y neutrons 0 236 There are about 90 different daughter nuclei (X and Y) that can be formed. Here is a typical example: 235 92 U 1 n 141 Ba 92 Kr 31 n 0 56 36 0 1. Estimate the amount of energy released when a uranium nucleus fissions. Use plot above Original: 7.5 MeV per nucleon Daughters: 8.5 MeV Energy difference: 1.0 MeV per nucleon x 235 nucleons = 235 MeV as KE of daughters and neutrons 2. A neutron collides with a nucleus of plutonium and the following fission reaction occurs. Determine the number of neutrons produced and calculate the amount of energy released. 239 94 Pu 1 n 140 Ba 96 Sr 0 56 38 m = m + Δm Masses: 239 94 Pu = 239.052157 u 0.19383 u = 180 MeV 96 38 Sr = 95.921750 u 140 56 Ba = 139.910581 u 1 0 n = 1.008665 u 14 Chain Reactions IB 12 Chain Reaction – neutrons released from one fission reaction go on to initiate further reactions Uncontrolled nuclear fission: nuclear weapons Controlled nuclear fission: nuclear power production 1) some material absorbs excess neutrons before striking nucleus Uncontrolled Chain Reaction 2) leaving only one neutron from each reaction to produce another reaction Critical Mass: if mass of uranium is too small, too many neutrons escape without causing further fission in uranium so the reaction cannot be sustained Controlled Chain Reaction Thermal Neutron: low-energy neutron (≈1eV) that favors fission reactions – energy comparable to gas particles at normal temperatures Naturally Occurring Isotopes of Uranium: 1) Uranium-238: most abundant, 99.3%, very small probability of fissioning when it captures a neutron, not used for fuel, more likely to capture high energy neutron than low energy one 2) Uranium-235: 0.3%, 500 times greater probability of fissioning when captures a neutron but must be a low-energy (thermal) neutron, used for fuel Fuel Enrichment: process of increasing proportion of uranium-235 in a sample of uranium 1) formation of gaseous uranium (uranium hexafluoride) from uranium ores 2) Separated in gas centrifuges by spinning – heavier U-238 moves to outside 3) increases proportion of U-235 to about 3% to be used as fuel in nuclear reactors Advantage: more uranium is available for fission and reaction can be sustained Disadvantage: enriched fuel can be used in the manufacture of nuclear weapons – threat to world peace – 85% = weapons grade 15 Nuclear Reactors IB 12 Most nuclear reactors: thermal fission reactor using uranium-235 as fuel Fuel Rods: enriched solid uranium Moderator: material (water, graphite) used to slow down high-energy neutrons emitted from fission reactions to thermal levels for use in further fission reactions to sustain the chain reaction - slow neutrons by collisions Control Rods: inserted between fuel rods – made of neutron-absorbing cadmium or boron - used to control reactor temperature to prevent overheating – lowered if too many neutrons/reactions and excess thermal neutrons are absorbed Heat Exchanger: hot fluid circulating around fuel rods (primary loop) is fed into tank of water – heat is transferred to water and makes steam – steam expands adiabatically against fan blades of turbines and turns a magnet is a coil of wire to generate electricity 16 Neutron Capture and Plutonium-239 IB 12 Uranium-238 is a non-fissionable isotope but is considered “fertile” 238 92 U 1 n 92 U 93 Np 01 e 0 239 239 Neutron capture and Beta-minus decay 239 93 Np 94 Pu 01 e 239 Beta-minus decay Fission 239 94 Pu 1 n 140 Ba 96 Sr 31 n 0 56 38 0 reaction Advantage: plutonium-239 used as fuel in “breeder reactors” Disadvantage: plutonium-239 used in nuclear weapons Safety Issues and Risks in the Production of Nuclear Power Uranium Mining: open-cast mining: environmental damage, radioactive waste rock (tailings) underground mining: release of radon gas (need ventilation), radioactive rock dangerous for workers, radioactive waste rock (tailings) leaching: solvents pumped underground to dissolve uranium and then pumped back out – contamination of groundwater Thermal Meltdown: overheating and melting of fuel rods – may be caused by malfunction in cooling system or pressure vessel – overheating may cause pressure vessel to burst sending radioactive material and steam into atmosphere (as in Chernobyl, Ukraine 1986) – hot material may melt through floor – “China syndrome” as in Three Mile Island – limited by containment vessel and containment building Nuclear Waste: Low-level waste: radioactive material from mining, enrichment and operation of plant – must be disposed of – left untouched or encased in concrete High-level waste: disposal of spent fuel rods- some isotopes have ½ lives of thousands of years – plutonium 240,000 years 1) stored under water at reactor site for several years to cool of then sealed in steel cylinders, buried underground 2) reprocessed to remove any plutonium and useful uranium, remaining isotopes have shorter ½ lives and long-term storage need is reduced Nuclear Weapons Manufacture: Enrichment technology could be used to make weapons grade uranium (85%) rather than fuel grade (3%) Plutonium is most used isotope in nuclear weapons and can be gotten from reprocessing spent fuel rods 17 IB 12 1. Suppose the average power consumption for a household is 500 W per day. Estimate the amount of uranium-235 that would have to undergo fission to supply the household with electrical energy for a year. Assume that for each fission, 200 MeV is released. 2. A fission reaction taking place in a nuclear power station might be 235 92 U 1 n 141 Ba 92 Kr 31 n 0 56 36 0 Estimate the initial amount of uranium-235 needed to operate a 600 MW reactor for one year assuming 40% efficiency and 200 MeV released for each fission reaction. 18 Nuclear Fusion IB 12 Nuclear Fusion: Two light nuclei combine to form a more massive nucleus with the release of energy. 1. Write the reaction equation for the fusion reaction shown at right. 2 1 H 1 H 2 He 1 n 3 4 0 2. Calculate how much energy is released in this fusion reaction. m = m + Δm 2 1 H (deuterium, 2.0141 u) 3 1 H (tritium, 3.0161 u) 0.0189 u = 17.6 MeV 4 2 He (4.0026 u) neutron (1.0087 u) 3. Calculate the energy released per nucleon and compare this with a fission reaction. 5 nucleons react = 17.6 MeV / 5 = 3.5 MeV Compared with ≈1.0 MeV for fission Important occurrence of fusion: main source of Sun’s energy – fusion of hydrogen to helium Suggested Mechanism: proton-proton cycle 1 1 H 1 H 1 H 1 e 1 2 0 1 1 H 1 H 3 He 2 2 Then either: Or: 1 1 H 3 He 4 He 1 e 2 2 0 3 2 He 3 He 4 He 1 H 1 H 2 2 1 1 19 Fusion Reactor IB 12 Plasma: fuel for reactor – high energy ionized gas (electrons and nuclei are separate) – if energy is high enough (hot enough), nuclei can collide fast enough to overcome Coulomb repulsion and fuse together Magnetic confinement: charged particles are contained via magnetic fields – travel in a circle in a doughnut shaped ring (tokamak) Heating Plasma: accelerate nuclei by means of magnetic fields and forces = high temperatures (high kinetic energies) Problems with current fusion technology: maintaining and confining very high-density and high-temperature plasmas – very difficult to do – uses more energy input than output – not commercially efficient Artificial (Induced) Transmutation Artificial (Induced) Transmutation: A nucleus is bombarded with a nucleon, an alpha particle or another small nucleus, resulting in a nuclide with a different proton number (a different element). Requirement: the bombarding particle must have sufficient kinetic energy to overcome the Coulomb repulsion 1. In 1919, Ernest Rutherford discovered that when nitrogen gas is bombarded with alpha particles, oxygen and protons are produced. Complete the equation for this reaction. 14 7 N 2 He 17 O 1 H 4 8 1 2. Neutron bombardment of lithium can produce the radioactive isotope of hydrogen known as tritium. Complete the reaction. 6 3 Li 1 n 1 H 4 He 0 3 2 NOTE: isotopes produced Importance: artificial isotopes produced are used in medical tests and therapies 20