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NUCLEAR CHEMISTRY Nuclear Particles: Mass Charge Symbol

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Presentation on theme: "NUCLEAR CHEMISTRY Nuclear Particles: Mass Charge Symbol"— Presentation transcript:

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

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

3 Distribution of Stable Nuclides
Protons Neutrons Stable Nuclides % Even Even Even Odd Odd Even Odd Odd Total = 267 100.0%

4 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 R.J. Rusay

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

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

7 Exposure to Radiation

8 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

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

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

11 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

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

13 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 R.J. Rusay

14 The 238U Decay Series

15 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 R.J. Rusay

16 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 R.J. Rusay

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

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

20 Binding Energy per Nucleon of Deuterium
Deuterium has a mass of amu. Hydrogen atom = 1 x amu = amu Neutrons = 1 x amu = amu amu Mass difference = theoretical mass - actual mass = amu amu = amu Calculating the binding energy per nucleon: Binding energy amu x Mev/amu Nucleon nucleons = Mev/nucleon =

21 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 R.J. Rusay

22 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.

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

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26 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

27 The Nuclear Dawn August 6, 1945

28 Nuclear Power Plant

29 Worldwide Nuclear Power Plants

30 Chernobyl, Ukraine April,1986 and April,2001

31 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 R.J. Rusay

32 National Ignition Facility Lawrence Livermore National Laboratory

33 National Ignition Facility Lawrence Livermore National Laboratory
© Copyright R.J. Rusay

34 Hydrogen Burning in Stars and Fusion Reactions
1H + 1H  2H +  Mev 1H + 2H  3He Mev 2H + 2H  3He + 1n Mev 2H + 2H  3H H mev 2H + 3H  4He + 1n Mev Easiest! 2H + 3He  4He + 1H Mev

35 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


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