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SACE Stage 2 Physics

SACE Stage 2 Physics. The Structure of the Nucleus. Composition of the Nucleus.

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SACE Stage 2 Physics

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  1. SACE Stage 2 Physics The Structure of the Nucleus

  2. Composition of the Nucleus NUCLEUS: A nucleus is the positively charged centre of an atom. It accounts for nearly all of the mass of the atom and consists of protons and neutrons referred to as nucleons while they are bound within the nucleus. The plural of nucleus is NUCLEI. Nucleus

  3. Composition of the Nucleus The nucleus of an atom consists of protons and neutrons, which have approximately the same mass. The proton has a positive charge equal in magnitude to that of an electron. The neutron is uncharged.

  4. Composition of the Nucleus ATOMIC NUMBER (Z): The number of protons in a nucleus. An equivalent number is the number of units of positive charge on the nucleus in electron charge units. For example for hydrogen Z = 1 for helium Z = 2 for carbon Z = 6 NEUTRON NUMBER (N): is the number of neutrons in the nucleus. NUCLEON: a word used for both protons and neutrons.

  5. Composition of the Nucleus MASS NUMBER (A): is the total number of nucleons in a nucleus. A = N + Z N = A - Z note that the mass number is an integer. We represent nuclei by writing the atomic number as a subscript and the mass number as a superscript. NUCLIDE: A nuclide is a unique nuclear species. Nuclei of a particular nuclide will have the same number of protons and the same number of neutrons. eg 1H11H26C122He4 represent different nuclides.

  6. Composition of the Nucleus specific examples1H12He46C12 13Al27 means Z =13 ie. 13 protons A = 27 (total number of nucleons) N = A –Z (number of neutrons) = 27 - 13 = 14 neutrons. As Z = 13 for aluminium there are 13 electrons in the neutral atom of aluminium.

  7. p p p p p p p n n n n n The Nucleon Force Nuclei consist of protons and neutrons closely packed in a package (nucleus). The protons in this "package" are all positively charged and so repel each other strongly. STRONG coulombic repulsion of positive charges. Strong because and r is small !  10-15 m

  8. The Nucleon Force Newton’s Law’s suggest that the net force on the protons is zero, there must be another force present to counter act the repulsive force  nuclear force. (1) The nuclear force is very strong. (2) It is very short range. (compare with gravitation and electric interactions which are long range). (3) It acts equally strongly on protons and neutrons. It does not act on electrons. (4) It depends on the spin or angular momentum of particles. (5) There is no mathematical formula that relates the nuclear force to mass and distance as is the case with other force types.

  9. The Nucleon Force Graphical Representation Average distance between nucleons in nuclei Attraction Strength of nuclear force 2 3 8 10-15 m 4 5 6 7 9 10-14 m Distance between two nucleons Repulsion Variation of a nuclear force with a distance (qualitative)

  10. Isotopes ISOTOPES: (First confirmation of existence about 1920) are nuclides which have the same atomic number but different mass numbers. ie. same number of protons but different numbers of neutrons. Example (a) the three isotopes of hydrogen 1H1 hydrogenone proton in the nucleus 1H2 deuterium one proton and one neutron in the nucleus 1H3 tritium one proton and two neutrons in the nucleus [Incidentally tritium is unstable and decays with half life of approx 11 years.]

  11. Isotopes (b) the three isotopes of neon 10Ne20 10Ne2110Ne22 (c) two isotopes of uranium are 92U23592U238

  12. Isotopes Isotopes of an element have the same number of protons and their electron structures are the same. This means they are identical in the way in which they react with other elements eg. hydrogen oxide (H2O - water), deuterium oxide (D2O - sometimes called heavy water) and tritium oxide (T2O) are chemically the same.

  13. n 0 0.9 n=4 -0.9 n=3 -1.5 1.5 3.4 n=2 -3.4 Ionisation energy = 13.6 eV note: an electron in n=2 needs 3.4 eV to remove it from the atom Binding Energy (eV) Actual energy (eV) 13.6 n=1 -13.6 Mass Defect and Binding Energy Binding Energy - This is the amount of energy which must be supplied to an electron (not necessarily in the ground state) so that the electron is completely removed from the atom and placed at rest an infinite distance from the atom.

  14. Mass Defect and Binding Energy THE PARADOX (or puzzle) Consider an electron and proton a long way apart v = 0 EK = 0 long way from proton  potential energy = 0 e- Total energy of system = 0 v = 0 EK = 0 long way from electron  potential energy = 0 p+

  15. e p hydrogen atom Mass Defect and Binding Energy YET when the electron and proton combine energy appears to be lost (13.6 eV). The question is where did the energy lost come from? e 13.6 eV lost

  16. mass of hydrogen atom mass defect converted into 13.6 eV of energy total mass free electron and proton Mass Defect and Binding Energy Einstein postulated that the energy comes from the MASS of the system (not just the electron). He postulated that mass may be converted into energy according to the formula E = mc2 where m is the change in mass and c is the speed of light.

  17. 1H2 + + p n e Mass Defect and Binding Energy Formation of Deuterium nucleus particle mass in kg electron 9.10956 x 10-31 proton 1.672614 x 10-27 neutron 1.674920 x 10-27 3.348444 x 10-27 1H23.34448 x 10-27 mass lost 3.964 x 10-30 E =mc2 = 3.964 x 10-30 x (2.99792 x 108)2 = 3.563 x 10-13 J = 2.22 MeV

  18. Mass Defect and Binding Energy Example – 4He2 has a mass of 6.644 x 10-27kg. Calculate the loss of mass when it forms from its constituent nucleons and the binding energy of the particle.

  19. Mass Defect and Binding Energy 2 protons  2 x 1.673 x 10 -27 kg =3.346 x 10-27 kg 2 nuetrons 2 x 1.675 x 10-27 kg =3.350 x 10-27 kg Total mass  =6.696 x 10-27 kg Mass defect m = 6.696 x 10-27 – 6.644 x 10-27 = 0.052 x 10-27 kg Loss of mass = 5.2 x 10-29 kg.

  20. Mass Defect and Binding Energy Binding Energy Eb = mc2 = 5.2 x 10-29 kg x (3 x 108 ms-1)2 = 4.68 x 1012 J = 2.93 x 107 eV = 29.3 MeV

  21. Mass Defect and Binding Energy • When a nucleus of Fe56 forms a photon of energy 504 MeV is emitted. Calculate, • The frequency of the photon • The binding energy of Fe56 • The mass that is lost when the nucleus forms

  22. Mass Defect and Binding Energy (1) The frequency of the photon

  23. Mass Defect and Binding Energy • The binding energy of Fe56 • The binding energy is equal in magnitude to the energy of the photon. • Thus the binding energy is 8.06 x 10-11J.

  24. Mass Defect and Binding Energy • The mass that is lost when the nucleus forms • Using Einstein’s equation,

  25. Conservation Laws in Nuclear Reactions • In a nuclear reaction the total charge and the total number of nucleons are conserved. • In a nuclear reaction the total mass of the reactants is different from the total mass of the products. • In a nuclear reaction the total energy (including the energy associated with the mass) is conserved. Hence the energy absorbed or released in a reaction can be calculated from the difference in the masses of the products and the reactants. • In a nuclear reaction momentum is conserved

  26. Conservation Laws in Nuclear Reactions Conservation of Charge and Nucleon Number Example – In the following reaction, complete the mass number and charge;

  27. Conservation Laws in Nuclear Reactions Example – In the following reaction, complete the mass number and charge;

  28. Conservation Laws in Nuclear Reactions Conservation of Mass – Energy In all reactions (nuclear & chemical) the total energy remains the same. Nuclear energy comes from a loss of mass. The total energy given out of a nuclear reaction can be determined by; In nuclear reactions, the total mass of the reactants is less than the products.

  29. Conservation Laws in Nuclear Reactions When a proton and neutron join to form the nucleus of deuterium, p + n = 2H1 Total mass of reactants = 3.348 x 10-27 kg Total mass of products = 3.344 x 10-27 kg Mass difference = reactants – products = 4.0 x 10-30kg. The mass difference is converted into energy and is given by E=mc2. In this case = 2.25 MeV (in this case, a gamma ray photon)

  30. Conservation Laws in Nuclear Reactions Conservation of Momentum In a nuclear decay, a single nucleus emits a particle and is transformed into another nucleus. Eg – A uranium nucleus decays emitting an -particle and transforming into a thorium nucleus. Diagrammatically,

  31. Distribution of Kinetic Energy By the law of conservation of momentum, Ratio of velocities are inversely proportional to their masses.

  32. Distribution of Kinetic Energy Ratio of Kinetic Energy, Kinetic energy is in inverse proportion to their masses.

  33. Application: The Production of Radioisotopes Radioactive isotopes may be produced by the absorption of particles into a stable nuclei. Eg – a nuetron is captured by a carbon-12 atom and the carbon-13 atom is produced. Note: the type of element does not change but the isotope created may be radioactive.

  34. Application: The Production of Radioisotopes Energy required to produce Radio Isotopes When isotopes are formed in a nuclear reaction, the products often weigh more than the reactants giving a negative mass defect. Energy must be supplied for the isotope to be created.

  35. Application: The Production of Radioisotopes Consider the following nuclear reaction, Mass of the products is 4 x 10-30 kg more than the reactants. By E = mc2, the energy required is 2.2 MeV.

  36. Application: The Production of Radioisotopes Production of Carbon-14 The Earth is continually bombarded by cosmic radiation that comes from the sun. This radiation mainly consists of electromagnetic radiation and high energy protons. Protons collide with the nitrogen in the air (Nitrogen comprises 79% of the gases in the atmosphere) producing Carbon-14. The following reaction occurs,

  37. Application: The Production of Radioisotopes Production of Nitrogen – 13 Nitrogen-13 is an isotope used in the area of medicine. As it decays quickly, it needs to be made nearby to where it will be used. It is created by bombarding Carbon-12 with high energy deuterium nucleus which are accelerated using a cyclotron. This reaction is endothermic as energy is required for the reaction to take place.

  38. Application: The Production of Radioisotopes This new isotope of nitrogen is not very stable and will decay quickly. Upon decay it will release a positron. This positron will also immediately collide with a negative electron and annihilation energy of 0.51 MeV will be emitted and can be detected outside the body the latest in positron topography imaging devices. This type of radioactive decay in the human body is relatively safe as the isotope has a short half life.

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