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Nuclear Chemistry Chapter 21. Slide 2 of 24 Review Chapter 3  Z = Atomic Number  Atomic Number is the number of _______.  Mass Number  Number of _______.

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Presentation on theme: "Nuclear Chemistry Chapter 21. Slide 2 of 24 Review Chapter 3  Z = Atomic Number  Atomic Number is the number of _______.  Mass Number  Number of _______."— Presentation transcript:

1 Nuclear Chemistry Chapter 21

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?

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

4 Slide 4 of 24

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

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

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

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

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 Einstein’s 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.

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

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

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 !!!

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

14 Slide 14 of 24 Half-Life  Half-life – time required for ½ of a radioactive material to decay  Each radioactive nuclide has its own ½ life  Longer ½ life = more stable nuclide  After 1 Half-Life = 50% remain  2 Half-Lives = 25% remain  3 Half-Lives = 12.5% remain

15 Slide 15 of 24

16 Slide 16 of 24 Potassium-40

17 Slide 17 of 24 Half-Life = Math Problems  Phosphorous-32 has a ½ 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

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

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

20 Slide 20 of 24

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!!

22 Slide 22 of 24

23 Slide 23 of 24 Nuclear Weapons  Fission weapons were actually used against Nagasaki and Hiroshima at the end of WW2

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 ¼ mile diameter & 320 feet deep  This blast contaminated more US residents than any other activity  Yucca Flats, NV


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