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The nuclear atom

The nuclear atom

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The nuclear atom

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  1. The nuclear atom • More than 2000 years ago, the Greeks suggested that matter was made up of very tiny (small) particles which they called atoms. What is inside the atom? What is the structure of an atom? 19.1 The atomic model

  2. Models of an atom Plum pudding model 19.1 The atomic model

  3. The plum pudding model of the atom by J.J. Thomson, who discovered the electron in 1897, was proposed in 1904. • The atom is composed of electrons, surrounded by a soup of positive charge to balance the electron's negative charge, like negatively-charged "plums" surrounded by positively-charged "pudding". • The model was disproved by the 1909 gold foil experiment, Plum pudding model 19.1 The atomic model

  4. Geiger-Marsden scattering experiment • Experiment: • a particles are made to hit the thin (super thin) gold foil. • Flashes of light will be observed when a particles hit the zinc sulphide screen. 19.1 The atomic model

  5. Geiger-Marsden scattering experiment • We expectall the a particles can pass through the gold foil. Results: Nearly all a particles pass straight through the gold foil. Some α-particles (about 1/8000) were scattered by angles greater than 90∘and very few even rebounded back along original paths. 19.1 The atomic model

  6. Rutherford’s Remark • It was quite the most incredible event that has ever happened to me in my life. • It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. • On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. • It was then that I had the idea of an atom with a minute massive centre, carrying a charge. 19.1 The atomic model

  7. How can the heavy a particles bounce back after hitting the thin gold foil? • a particles – He2+ • Explanation: • All positive charge of the atom and most of the mass were concentrated in a tiny core called nucleus. • The rest of the atom was largely empty space. 19.1 The atomic model

  8. How can the heavy a particle bounce back after hitting the thin gold foil? • Explanation: • Most of a particles passed straight through the empty space of the gold atoms. • Some come close to the nucleus were repelled by a strong electrostatic force, so they were deflected or bounced back. Bounced back deflected 19.1 The atomic model

  9. -particle lose their K.E. on approaching the +ve charged nucleus, being repelled by an electrostatic force. • At P, distance of closest approach K.E. lost = P.E. due to -particle location in electric field of nucleus. • -particle is then ‘reflected’ away from nucleus and finally acquires the same K.E. as it had initially. Collision is elastic. 19.1 The atomic model

  10. Estimated Upper Limit of the Size of a Nucleus • PE = 2Ze2/40r • At P, distance of nearest approach K.E. of ’s, ½mv2 = 2Ze2/40r (P.E.) • hence an estimate of r which is upper limit to size of nucleus. 19.1 The atomic model

  11. 1 Nuclear fission + energy released + energy released When a heavy nucleus splits up, huge amount of energy is released. • Two typical nuclear fission reactions are: The total mass of the nuclear products is slightly less than the total mass of the original particles. Energy is released and can be calculated by using Einstein’s mass-energy relation (E = mc2) 19.1 The atomic model

  12. 3 Nuclear fusion A huge amount of energy is also released when two light nuclei join together to form a heavy nucleus. This process is called nuclear fusion. 19.1 The atomic model

  13. Example of Nuclear Fusion • Deuterium-Tritium Fusion Reaction. The total mass of the nuclear products is slightly less than the total mass of the original particles. 19.1 The atomic model

  14. 3 Nuclear fusion H H He n 2 3 1 4 2 1 1 0 + + + energy Since nuclei carry +ve charges, they repel each other. For fusion to occur, the 2 hydrogen nuclei must approach each other with very high speed.  hydrogen gas of high temperature (108 C)! 19.1 The atomic model

  15. 3 Nuclear fusion Nuclear fusion occurs in the core of the Sun, giving out heat and light. The reaction takes place continuously for billions of years. (Photo credit: US NASA) 19.1 The atomic model

  16. Nuclear Power – controlled fission • The schematic diagram of a nuclear reactor is shown below: 19.1 The atomic model

  17. Fuel • Enriched uranium is used as the fuel (uranium dioxide). • Natural uranium contains only 0.7% uranium-235, which does not undergo fission in these circumstances. • Treatment is required to increase the concentration to 3%. 19.1 The atomic model

  18. + energy released Moderator • The probability that fission caused by a high energy neutron (fast moving neutron) is quite low. • Use materials to slow down neutrons so that the probability of causing a fission is significantly higher. • These neutron slowing down materials are the so called moderators. 19.1 The atomic model

  19. Moderator At rest • These neutron slowing down materials are the so called moderators. u V v m M Before collision After collision By momentum conservation: mu = mv + MV By energy conservation: ½mu2 = ½mv2 + ½MV2 19.1 The atomic model

  20. Moderator V v After collision • The choice of the moderator • The atoms of an ideal moderator should have the same mass as a neutron (M = m). So a neutron colliding elastically with a moderator atom would lose almost all its KE to the moderator atom. • The moderator atoms should not absorb neutrons but should scatter them instead. • In practice, heavy water (D2O i.e. 2H2O) is chosen as the moderator. (Heavy water is different from hard water) 19.1 The atomic model

  21. Control rods • The control rods are made of boron-steel, which absorbs neutrons. • They are raised and lowered to vary the number of neutrons to control the rate of fission. 19.1 The atomic model

  22. 2 Chain reaction When a uranium nucleus splits, 2 or 3 neutrons are emitted. neutron U-235 nucleus splits If these neutrons carry on splitting other uranium nucleus... 19.1 The atomic model

  23. 2 Chain reaction escapes self-sustaining  chain reaction neutron U-235 nucleus splits 19.1 The atomic model

  24. Chain reaction • The fission neutrons enter the moderator and collide with moderator atoms, transferring KE to these atoms. • So the neutrons slow down until the average KE of a neutron is about the same as that of a moderator atom. 19.1 The atomic model

  25. The fission neutrons could be absorbed by the U-238 nuclei without producing further fission. The fission neutron could escape from the isolated block of uranium block without causing further fission. The critical mass of fuel is the minimum mass capable of producing a self-sustaining chain reaction. 19.1 The atomic model

  26. + energy released + energy released • Two typical nuclear fission reactions are: • At the fission of U-235 on the average 2.5 neutrons are released but not all of these cause fission. • multiplication factor (k) 19.1 The atomic model

  27. escapes neutron U-235 nucleus splits 19.1 The atomic model

  28. critical reactor • The number of neutrons in the system is constant, i.e. they cause the same number of fissions in every second. 19.1 The atomic model

  29. k > 1, the system is supercritical • k < 1, the system is subcritical k can be varied by lowering or raising the control rods 19.1 The atomic model

  30. 1 Nuclear power b Potential hazards The nuclear waste remains radioactive for thousands of years.  serious handling and storage problems Nuclear accidents lead to the leakage of radiation.  widespread and long-lasting disasters (Photo credits: BREDL; IAEA) 19.1 The atomic model

  31. 1 Nuclear power c Controlled nuclear fusion Fuel = H-2, H-3 (plentiful in sea water) Waste product = He-4 (inert and non-radioactive) (Photo credit: Princeton Plasma Physics Lab) Cheaper, abundant and safer, but not yet in practice 19.1 The atomic model

  32. 1 Nuclear power d Benefits and disadvantages contentious Solve energy shortage crisis  Create serious social and environmental problems  19.1 The atomic model

  33. 1 Nuclear power d Benefits and disadvantages Benefits Solve energy shortage crisis  No fuel transportation problem  Cheaper than coal/oil for generating power in most cases  Little environment pollution 19.1 The atomic model

  34. 1 Nuclear power d Benefits and disadvantages Disadvantages Accident  serious consequence  Expensive in maintaining safety standards  Unnecessary as alternative energy sources exist  Lead to widespread of nuclear weapons 19.1 The atomic model

  35. END 19.1 The atomic model

  36. Pre-lesson assignment Einstein's mass-energy relation (E = mc2) • According to Einstein’s mass-energy equation E = mc2, what is the amount of energy produced if 1 kg of a certain element completely changes into energy? Solution: Energy released = (1)(3 x 108)2 = 9 x 1016 J 19.1 The atomic model

  37. Q2 • Mass of U-235: 3.9030 x 10-25 kg Ba-144: 2.3899 x 10-25 kg Kr-90: 1.4931 x 10-25 kg Neutron: 1.6749 x 10-27 kg • Calculate the nuclear energy released in the nuclear fission. Solution: • Mass difference = 3.9030 x 10-25 – (2.3899 x 10-25 + 1.4931 x 10-25 + 1.6749 x 10-27) = 3.251 x 10-28 kg • Energy released = (3.251 x 10-28 )(3 x 108)2 = 2.93 x 10-11 J 19.1 The atomic model

  38. Nuclear Energy • unified atomic mass unit • unit of energy: eV • Binding energy 19.1 The atomic model

  39. Atomic mass 1 u = unified atomic mass unit 19.1 The atomic model

  40. The unified atomic mass unit (u) is defined as one twelfth of the mass of the carbon-12 atom. History Question: How to express the unified atomic mass unit (1u) in kg? Solution: Given that: mass of a C-12 atom = 1.9927 x 10-26 kg ∴1u = 1.9927 x 10-26 kg /12 = 1.66 x 10-27 kg 19.1 The atomic model

  41. Example 1 • Given that the atomic mass of a hydrogen atom is 1.00783u. Find the mass of a hydrogen atom. • Solution Mass of a hydrogen atom = 1.00783 x 1.66 x 10-27 = 1.67 x 10-27 kg 19.1 The atomic model

  42. Units of Energy Very large unit of energy: kWh (electric bill) S.I. unit: J Very small unit of energy: eV (Atomic physics) Energy = charge Q x voltage V 19.1 The atomic model

  43. Energy in eV + – 1.6 x 10-19 C By W = QV 1 eV = (1.6 x 10-19)(1) ∴ 1 eV = 1.6 x 10-19 J 1 V 1 eV is the energy (K.E.) gained by an electron when it is accelerated through 1 V. 19.1 The atomic model

  44. 1.6 x 10-19 C + Energy in eV + 1 V – 1 eV is the work done in moving a charge of 1.6 x 10-19 C through 1 V. 19.1 The atomic model

  45. Example 1 page 25 • Show that 1 u of mass is equivalent to 931 MeV by the mass-energy relation. • Solution: The energy equivalent to 1 u of mass = mc2 = (1.66 x 10-27)(2.998 x 108)2 = 1.492 x 10-10 J = (1.492 x 10-10 / 1.602 x 10-19) eV = 931 x 106 eV = 931 MeV 1 u = 1.66 x 10-27 kg 1 eV = 1.602 x 10-19 J 1 u = 931 MeV 19.1 The atomic model

  46. Example 2 Consider the following nuclear fission. Given that: 1u = 931 MeV Mass of neutron: 1.00866 u U-235: 235.044 u Ba-144: 143.923 u Kr-90: 89.9195 u (a) Find the amount of nuclear energy released in nuclear reaction. Solution: Mass difference = (235.044 – 143.923 – 89.9195 – 1.00866)u = 0.19284u Energy released = 0.19284 x 931 MeV = 180 MeV 19.1 The atomic model

  47. (180 MeV) (b) Hence, show that the energy released from 1 kg of U-235 is about 7.4 x 1013 J which is the energy released by burning about 3 x 106 tonnes of coal. Solution: • No of U-235 atom in 1 kg fuel = 1/(235.044 x 1.66 x 10-27) = 2.563 x 1024 • Energy released = (2.563 x 1024) x (180 MeV) = (2.563 x 1024) x (180 x 106) x (1.6 x 10-19) = 7.4 x 1013 J 19.1 The atomic model

  48. Example 3 Consider the following nuclear fusion. Given that: 1u = 1.66 x 10-27 kg = 931 MeV Mass of neutron: 1.00866 u H-2: 2.01355 u H-3: 3.01605 u He-4: 4.0026 u (a) Find the amount of nuclear energy released in the reaction. Solution: Difference in mass = (2.01355 + 3.01605 – 4.0026 – 1.00866)u = 0.01834u Energy released = 0.01834 x 931 MeV = 17.1 MeV 19.1 The atomic model

  49. Hence, show that the energy released from 1 kg of fuel is about 3.3 x 1014 J. Solution: No of H-2 and H-3 in 1 kg fuel = 1/[(2.01355 + 3.01605 ) x 1.66 x 10-27] = 1.1977 x 1026 Total energy released = (1.1977 x 1026) x (2.74 x 10-12) = 3.3 x 1014 J 19.1 The atomic model

  50. Protons & neutrons are collectively called nucleons. A = mass number / Nucleon number Nuclear Energy – nucleus 19.1 The atomic model