PHYSICS TEACHER'S NOTES LESSON
Constructing a Bohr Rutherford Diagram Section 7. Use the Pen tool to complete the following sentences. Then, use the following tools to draw the Bohr Rutherford diagram for oxygen 6. The atomic number for oxygen is and the mass number is. Since the atomic number is, there are protons and electrons. The number of neutrons is found by subtracting the atomic number from the mass number. The number of neutrons is. Check Answer 2
Isotopes Same element, different mass... 3
4
Nuclear Reactions beta_decay carbon4_decay radiation_penetration 5
Radioactive Song 6
Note 3 5 Potassium 4 is the primary Americium 4 Cesium 37 is and a byproduct this type of of radioactive The decay source of Carbon in the is decay nuclear Plutonium 24 is fission the basis and a for is byproduct the used for human commonly body used and in the Section positron element 7.2 Radioactive operation both of nuclear cancer of reactor most Decay treatment smoke operations and the used emission for dating tomography the oldest (PET) Using detectors. irradiation and nuclear the general of food bomb equations to explosions. destroy in the text (pages rocks medical 325 328), on Earth scans. drag (. each 4.5 type billion of radioactive insects, bacteria, decay listed and below viruses. to match the years correct old). decay example. Note 2 4 beta negative decay electron capture gamma decay beta positive decay alpha decay Check Answer 7
half_life 8
Half Life Section 7.3 Dating samples older than half lives is difficult using the carbon 4 (half life = 5,73 years) technique due to the challenge of counting so few radioactive atoms. For the oldest rocks, one radiometric dating method often used is based on the decay of potassium 4 to argon 4 (half life =.3 billion years). If a rock is found and dated to just after the age of Earth (4.5 billion years old), how much of the original potassium 4 remains in the sample? Drag the blue data from the problem into the solution below. ( ) Solution: A = A h % 2 Hint 4.5 t = time =.3 % h = half life ( ) 2 t Discussion A = % Statement: % of the potassium 4 in the rock sample would remain if it existed just after Earth was formed, 4.5 billion years ago. Check Answer 9
carbon_dating
Intro to Nuclear Chemistry (Crash Course) Nuclear Chemistry Fission and Fusion (Crash Course) FISSION FUSION stable_and_unstable_nuclei (chain reaction) controlled_chain_reaction
Section 7.4 Calculating Energy Yield in a Fission Reaction What is the energy yield of the following fission reaction? Use the given mass in the tab. 235 4 U + n Kr + Ba + 3( n) Required: energy released Analysis: E = Δmc 2 Simplify the reaction equation. Write the simplified reaction equation in the box. 235 36 36 4 56 56 U + n Kr + Ba + 3( n) Given Mass Calculate the mass defect. Δm = m U 235 ( + + ) = 235.44u = Δm = Check Answer 2
Section 7.4 Calculating Energy Yield in a Fission Reaction What is the energy yield of the following fission reaction? 235 36 4 56 U + n Kr + Ba + 3( n) Convert the mass defect to kilograms. Hint = Δm =.25 u = Now determine the binding energy. E = Δmc 2 = = = Given Mass Simplied equation Statement: The nuclear fission reaction releases J of energy per reaction. Check Answer 3
Nuclear Fusion Section 7.5 A Tokamak is designed to contain plasma at very high temperatures, holding it in a strong, continuous magnetic field so fusion can occur without having the plasma touch the walls of the reactor and cool back to a gas. The temperature of the plasma in a Tokamak can reach over million degrees; over five times the temperature at the centre of the Sun! Discussion 4
ANSWERS TEACHER'S NOTES LESSON 5
Constructing a Bohr Rutherford Diagram Section 7. Use the Pen tool to complete the following sentences. Then, use the following tools to draw the Bohr Rutherford diagram for oxygen 6. The atomic number for oxygen is 8 and the mass number is. 6 Since the atomic number is, 8 there are 8 protons and 9 electrons. The number of neutrons is found by subtracting the atomic number from the mass number. The number of neutrons is. 8 8p + 8n Back 6
Radioactive Decay Section 7.2 Using the general equations in the text (pages 325 328), drag each type of radioactive decay listed below to match the correct decay example. Note Note 2 beta negative decay electron capture gamma decay beta positive decay alpha decay Note 3 Note 4 Note 5 Back 7
Half Life Section 7.3 Dating samples older than half lives is difficult using the carbon 4 (half life = 5,73 years) technique due to the challenge of counting so few radioactive atoms. For the oldest rocks, one radiometric dating method often used is based on the decay of potassium 4 to argon 4 (half life =.3 billion years). If a rock is found and dated to just after the age of Earth (4.5 billion years old), how much of the original potassium 4 remains in the sample? Drag the blue data from the problem into the solution below. ( ) Solution: A = A h % 2 4.5 =.3 % ( ) 2 t Hint t = time h = half life Discussion A = 9. % Statement: % 9. of the potassium 4 in the rock sample would remain if it existed just after Earth was formed, 4.5 billion years ago. Back 8
Section 7.4 Calculating Energy Yield in a Fission Reaction What is the energy yield of the following fission reaction? Use the given mass in the tab. 235 4 U + n Kr + Ba + 3( n) Required: energy released Analysis: E = Δmc 2 Simplify the reaction equation. Write the simplified reaction equation in the box. 235 36 36 4 56 56 U + n Kr + Ba + 3( n) Given Mass 235 36 4 56 U Kr + Ba + 2( n) Calculate the mass defect. Δm = m U 235 ( + + ) m Kr m Ba 4 2m n = 235.44u [9.897 u + 4.94 u + 2(.9u)] = 235.44 u 234.829 Δm =.25 u Back 9
Section 7.4 Calculating Energy Yield in a Fission Reaction What is the energy yield of the following fission reaction? 235 36 4 56 U + n Kr + Ba + 3( n) Convert the mass defect to kilograms..25 u.25 u (.66 x = 27 kg Hint u ) = 3.237 x 28 Δm =.25 u kg Given Mass Now determine the binding energy. E = Δmc 2 = (3.237 x 28 kg)(3. x 8 m/s) 2 = 2.9 x kg m/s E = 2.9 x J Statement: The nuclear fission reaction releases 2.9 x J of energy per reaction. Back Simplied equation 2
Answers for Discussion Questions: Slide 6 Sample answer: The large number of artificially created radioactive nuclides and the relatively little research that has been carried out on many of these nuclides potentially means that there are many new possibilities for research and discovery, especially in the areas of materials research (e.g., materials with new properties, high strength, etc.) and medical applications (e.g., new treatments for cancer and other diseases). Back 2
Answers for Discussion Questions: Slide 9 Sample answer: Advantages of fusion as a power source include the following: Hydrogen fuel for fusion can be found in water! No fossil fuels are used. The product of fusion, helium, is not a greenhouse gas, not radioactive, and not polluting. There is no risk of the reactor running out of control, so it is very safe. A large amount of power can be produced in a small space. Opportunities for scientists and engineers to use fusion related technologies in other helpful ways. Disadvantages of fusion as a power source include the following: Fusion is difficult and expensive to achieve. Plasma is extremely hard to control and manipulate. Back 22