Nuclear Chemistry Chapter 21. Slide 2 of 24 Review Chapter 3 Z = Atomic Number Atomic Number is...
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Transcript of Nuclear Chemistry Chapter 21. Slide 2 of 24 Review Chapter 3 Z = Atomic Number Atomic Number is...
Nuclear Chemistry
Chapter 21
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 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 of 24
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 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 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 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 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 = mc2
E = energy m = mass c = speed of light = 3.00 x 108 m/s
Now we will find nuclear binding energy in the previous problem.
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 = mc2 & c = 3.00 x 108 m/s
E = (5.0441 x 10-29 kg) (3.00 x 108 m/s)2
E = 4.54 x 10-12 kg * m2/s2
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 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 83 Sig Figs !!!
Mass Defect = 32.26383 – 32.065 = 0.1988336= 0.199 amu Sig Figs !!!
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 = mc2
E = (3.301 632 x 10-28 kg)(3.00 x 108 m/s)2
E = 2.971 468 x 10-11 kg * m2/s2 = 2.97 x 10-11 kg * m2/s2
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
Slide 15 of 24
Slide 16 of 24
Potassium-40
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
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 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 of 24
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 of 24
Slide 23 of 24
Nuclear Weapons
Fission weapons were actually used against Nagasaki and Hiroshima at the end of WW2
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