NUCLEAR CHEMISTRY Nuclear Particles: Mass Charge Symbol PROTON 1 amu +1 H+, H, p NEUTRON 1 amu 0 n

Number of Stable Nuclides Elements 48 through 54 Atomic Number of Element Number (Z) Nuclides Cd 48 8 In 49 2 Sn 50 10 Sb 51 2 Te 52 8 I 53 1 Xe 54 9 Magic numbers are 2, 8, 20, 28, 50, or 82 protons or neutrons. Even numbers of protons and neutrons are more stable than odd.

Distribution of Stable Nuclides Protons Neutrons Stable Nuclides % Even Even 157 58.8 Even Odd 53 19.9 Odd Even 50 18.7 Odd Odd 7 2.6 Total = 267 100.0%

Nuclear Decay / Radioactivity Unstable nuclei “decay” i.e. they lose particles which lead to other elements and isotopes. The elements and isotopes produced may also be unstable and go through further decay. Nuclear decay reactions conserve mass & energy. © Copyright 1994-2002 R.J. Rusay

Nuclear Particles emitted from unstable nucleii Emitted Particles: Mass Charge Symbol alpha particle 4 amu +2 beta particle very small -1 gamma very very small 0 positron very small +1 © Copyright 1994-2002 R.J. Rusay

Nuclear Penetrating Power alpha particle: low beta particle: moderate gamma: high X-rays? Water © Copyright 1994-2002 R.J. Rusay

Exposure to Radiation

NUCLEAR STABILITY Patterns of Radioactive Decay Alpha decay () –heavy isotopes Beta decay () –neutron rich isotopes Positron emission ()–proton rich isotopes Electron capture–proton rich isotopes X-rays: Gamma-ray emission (-most all unstable isotopes Spontaneous fission–very heavy isotopes

Alpha Decay–Heavy Elements 238U  234Th +  + e t1/2 = 4.48 x 10 9 years 210Po  206Pb +  + e t1/2 = 138 days 256Rf  252No +  + e t1/2 = 7 ms For Cobalt 60 predict the alpha decay product:

Beta Decay–Electron Emission n  p+ +  + Energy 14C  14N +  + Energy t1/2= 5730 years 90Sr  90Y +  + Energy t1/2= 30 years 247Am  247Cm +  + Energy t1/2= 22 min 131I  131Xe +  + Energy t1/2 = 8 days

Electron Capture–Positron Emission p+ + e-  n + Energy = Electron capture 51Cr + e-  51V + Energy t1/2 = 28 days p+  n + e+ + Energy = Positron emission 7Be  7Li +  + Energy t1/2 = 53 days ? + e-  177Ir + Energy 144Gd  ? +  + Energy

Nuclear Decay Predict the Particle or element: Thallium 206 decays to Lead 206. What particle is emitted? Cesium 137 goes through Beta decay. What element is produced and what is its mass? Thorium 230 decays to Radium 226. What particle is emitted? © Copyright 1994-2002 R.J. Rusay

Nuclear Decay Series If the nuclei produced from radioactive decay are unstable, they continue to decay until a stable isotope results. An example is Radium which produces Lead © Copyright 1994-2002 R.J. Rusay

The 238U Decay Series

Nuclear Decay Measurement Decay can be measured with an instrument that detects “disintegrations”. The rate can be obtained by “counting” them over a period of time. e.g. disintegrations / second (dps) The rate of decay follows first order kinetics The rate law produces the following equation for the half life: © Copyright 1994-2002 R.J. Rusay

Nuclear Reactions The mass of the visible universe is 73% H2 and 25% He. The remaining 2%, “heavy” elements, have atomic masses >4. The “heavy” elements are formed at very high temperatures (T>106 oC) by FUSION, i.e. nuclei combining to form new elements. There is an upper limit to the production of heavy nuclei at A=92, Uranium. Heavy nuclei split to lighter ones by FISSION © Copyright 1994-2002 R.J. Rusay

NUCLEAR ENERGY EINSTEIN’S EQUATION FOR THE CONVERSION OF MASS INTO ENERGY E = mc2 m = mass (kg) c = Speed of light c = 2.998 x 108 m/s

Mass  Energy Electron volt (ev) The energy an electron acquires when it moves through a potential difference of one volt: 1 ev = 1.602 x 10-19J Binding energies are commonly expressed in units of megaelectron volts (Mev) 1 Mev = 106 ev = 1.602 x 10 -13J A particularly useful factor converts a given mass defect in atomic mass units to its energy equivalent in electron volts: 1 amu = 931.5 x 106 ev = 931.5 Mev

Binding Energy per Nucleon of Deuterium Deuterium has a mass of 2.01410178 amu. Hydrogen atom = 1 x 1.007825 amu = 1.007825 amu Neutrons = 1 x 1.008665 amu = 1.008665 amu 2.016490 amu Mass difference = theoretical mass - actual mass = 2.016490 amu - 2.01410178 amu = 0.002388 amu Calculating the binding energy per nucleon: Binding energy -0.002388 amu x 931.5 Mev/amu Nucleon 2 nucleons = -1.1123 Mev/nucleon =

Nuclear Reactions Fission and Fusion reactions are highly exothermic (1 Mev / nucleon). This is 10 6 times larger than “chemical” reactions which are about 1 ev / atom. Nuclear fission was first used in a chain reaction: © Copyright 1994-2002 R.J. Rusay

Nuclear Reactions: _________________________ Fission: Fusion: Fission and Fusion reactions are highly exothermic (1 Mev / nucleon). This is 10 6 times larger than “chemical” reactions which are about 1 ev / atom.

Nuclear Reactions Nuclear fission was first used in a chain reaction:

Nuclear Reactions / Fission The Fission Chain Reaction proceeds geometrically: 1 neutron  3  9  27  81 etc. 1 Mole of U-235 (about 1/2 lb) produces 2 x 1010 kJ which is equivalent to the combustion of 800 tons of Coal! Commercial nuclear reactors use fission to produce electricity....Fission [“atomic”] bombs were used in the destruction of Hiroshima and Nagasaki, Japan, in August 1945. © Copyright R.J.Rusay 1994-2002

The Nuclear Dawn August 6, 1945

Nuclear Power Plant

Worldwide Nuclear Power Plants

Chernobyl, Ukraine April,1986 and April,2001

Nuclear Reactions / Fusion Fusion has been described as the chemistry of the sun and stars. It too has been used in weapons. It has not yet found a peaceful commercial application The application has great promise in producing relatively “clean” abundant energy through the combination of Hydrogen isotopes particularly from 2H, deuterium and 3H, tritium: (NIF/National Ignition Facility, LLNL) © Copyright 1994-2002 R.J. Rusay

National Ignition Facility Lawrence Livermore National Laboratory

National Ignition Facility Lawrence Livermore National Laboratory © Copyright 1994-2002 R.J. Rusay

Hydrogen Burning in Stars and Fusion Reactions 1H + 1H  2H +  + 1.4 Mev 1H + 2H  3He + 5.5 Mev 2H + 2H  3He + 1n + 3.3 Mev 2H + 2H  3H + 1H + 4.0 mev 2H + 3H  4He + 1n + 17.6 Mev Easiest! 2H + 3He  4He + 1H + 18.3 Mev

Helium “Burning” Reactions in Stars 12C + 4He  16O 16O + 4He  20Ne 20Ne + 4He  24Mg 24Mg + 4He  28Si 28Si + 4He  32S 32S + 4He  36Ar 36Ar + 4He  40Ca