# nuclear chemistry chapter 21. slide 2 of 24 review chapter 3  z = atomic number  atomic number...

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• Slide 1
• Nuclear Chemistry Chapter 21
• Slide 2
• Slide 2 of 24 Review Chapter 3 Z = Atomic Number Atomic Number is the number of _______. Mass Number Number of _______ + ________ Average Atomic mass Weighted average of mass numbers of isotopes What is an isotope? Why are electrons not included in the mass number?
• Slide 3
• Slide 3 of 24 Hydrogen Isotopes Protium (99.985%) 1 proton, 0 neutons, 1 electron Deuterium (0.015%) (Heavy Water) __ proton, __ neutron, __ electron Tritium (Rare) (Radioactive) __ proton, __ neutron, __ electron
• Slide 4
• Slide 4 of 24
• Slide 5
• Slide 5 of 24 Mass of an atom? Mass of 1 atom is 2.657 x 10 -23 So use another method Carbon 12 atom weighs 12 Atomic Mass Units Atomic Mass Unit (AMU) Easy to find the mass of an atom: Find mass number or atomic mass + attach AMU as the units Example: Oxygen = 16 amu OR 15.9994 amu
• Slide 6
• Slide 6 of 24 First some vocab Nucleons particles in the nucleus Nuclide another name for an atom Identified by the number of protons + neutrons Nuclear Reaction reaction that affects the nucleus of an atom Transmutation change in proton number Change in the identity of a nucleus Oxygen-16 transmutates via alpha emission to Carbon- 12
• Slide 7
• Slide 7 of 24 Mass Defect When nucleons bind together into a nucleus, they LOSE mass Mass Defect (sum of the masses of the protons + neutrons + electrons) (atomic mass) Proton mass = 1.007 276 amu Neutron mass = 1.008 665 amu Electron mass = 0.000 5486 amu
• Slide 8
• Slide 8 of 24 Find Mass Defect Helium-4 atom (p. 681) Helium atom = 2 protons, 2 neutrons, 2 electrons 2 protons = 2(1.007 276 amu) =2.014 552 2 neutrons = 2(1.008 665 amu) =2.017 330 2 electrons = 2(0.000 5486 amu) =0.001 097 TOTAL: 4.032 979 amu Periodic Table: 4.002 602 Mass Defect = 4.032 979 amu 4.002 602 amu MASS DEFECT = 3.0377 x 10 -2 amu
• Slide 9
• Slide 9 of 24 Nuclear Binding Energy (NBE) Definition The energy released when a nucleus is formed from its nucleons Mass defect can be converted to NBE by Einsteins famous equation: E = mc 2 E = energy m = mass c = speed of light = 3.00 x 10 8 m/s Now we will find nuclear binding energy in the previous problem.
• Slide 10
• Slide 10 of 24 Finding Nuclear Binding Energy Mass defect for Helium-4 = 3.0377 x 10 -2 amu Step 1: Convert units: amu kg Conversion Factor: 1 amu = 1.6605 x 10 -27 kg Calculation: (3.0377 x 10 -2 amu) (1.6605 x 10 -27 kg/amu) Mass = 5.0441 x 10 -29 kg E = mc 2 & c = 3.00 x 10 8 m/s E = (5.0441 x 10 -29 kg) (3.00 x 10 8 m/s) 2 E = 4.54 x 10 -12 kg * m 2 /s 2
• Slide 11
• Slide 11 of 24 Nuclear Binding Energy NBE is also the energy that must be input to break apart the nucleus into its constituent nucleons Since energy is released when a nucleus forms, which is more stable the nucleus or the separated nucleons? Nucleus, since energy is inversely proportional to stability Lower energy = MORE stability
• Slide 12
• Slide 12 of 24 Another Problem Calculate the nuclear binding energy of a Sulfur-32 atom Step 1: Calculate the mass defect 16 protons (16*1.007276) + 16 neutrons (16*1.008665) + 16 electrons (16*0.0005486) = 16.116 416 + 16.138 64 + 0.0087776 = 32.263 833 6 = 32.263 83Sig Figs !!! Mass Defect = 32.26383 32.065 = 0.1988336 = 0.199 amu Sig Figs !!!
• Slide 13
• Slide 13 of 24 Another Problem (Page 2) Step 2: Calculate the NBE Mass in amu = 0.1988336 0.1988336 amu * (1.6605 x 10 -27 kg/amu) Mass in kg = 3.301 632 x 10 -28 kg E = mc 2 E = (3.301 632 x 10 -28 kg)(3.00 x 10 8 m/s) 2 E = 2.971 468 x 10 -11 kg * m 2 /s 2 = 2.97 x 10 -11 kg * m 2 /s 2
• Slide 14
• Slide 14 of 24 Half-Life Half-life time required for of a radioactive material to decay Each radioactive nuclide has its own life Longer life = more stable nuclide After 1 Half-Life = 50% remain 2 Half-Lives = 25% remain 3 Half-Lives = 12.5% remain
• Slide 15
• Slide 15 of 24
• Slide 16
• Slide 16 of 24 Potassium-40
• Slide 17
• Slide 17 of 24 Half-Life = Math Problems Phosphorous-32 has a life of 14.3 days. How many milligrams (mg) remain after 57.2 days, if the sample began with 4.0 mg? 57.2 / 14.3 = 4 Half-Lives 4 Half-Lives = (1/2)(1/2)(1/2)(1/2) of original amount remains 1/16 of the original amount remains 4.0 * (1/16) = 0.25 mg remains
• Slide 18
• Slide 18 of 24 Half-Life Problems (Page 2) Complete problems from Packet Practice Problems which is next to the decay series page. Complete PRACTICE Problems on pp. 689 in textbook
• Slide 19
• Slide 19 of 24 Pp. 693 Bottom Alpha particles cannot go through paper Beta particles can go through paper but not aluminum Gamma particles can go through both, but not lead or concrete
• Slide 20
• Slide 20 of 24
• Slide 21
• Slide 21 of 24 Nuclear Fission Nuclear Fission heavy nucleus splits into more-stable nuclei of intermediate mass Mass will be converted to energy, usually a lot of energy Chain reaction material that begins a reaction is also one of the products so it can begin another reaction Critical Mass minimum amount of nuclide that is required to sustain a chain reaction Nuclear Power Generators use controlled-fission chain reaction to produce energy Also produces unwanted radioactive nuclides Makes fish (and humans) glow!!
• Slide 22
• Slide 22 of 24
• Slide 23
• Slide 23 of 24 Nuclear Weapons Fission weapons were actually used against Nagasaki and Hiroshima at the end of WW2
• Slide 24
• Slide 24 of 24 Nuclear Fusion Low mass nuclei combine to form a heavier, more stable nucleus Immense energy production Source of energy for the Sun and many stars Thermonuclear or H-bombs Fusion of Deuterium + Tritium 100 times power of atomic bombs mile diameter & 320 feet deep This blast contaminated more US residents than any other activity Yucca Flats, NV

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