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This powerpoint relates to Topic 2 Section 4 of the SACE in South Australia

This powerpoint relates to Topic 2 Section 4 of the SACE in South Australia

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The structure of the nucleus 07 The structure of the nucleus 07 Presentation Transcript

  • THE STRUCTURE OF THE NUCLEUS 12 SACE PHYSICS-STAGE 2 SECTION 4 TOPIC 2 PRINCE ALFRED COLLEGE
  • NUCLEAR TERMS
      • A specific nucleus can be exactly identified using the following notation: z X A or Z X A .
      • X is the symbol of the element.
      • Z the atomic number (number of protons).
      • A the mass number (number of protons and neutrons). The term ‘nucleon’ refers to the protons or neutrons in the nucleus.
  • ISOTOPES
    • As early as 1911, it became clear that elements could include atoms of different masses.
    • They were named isotopes (from the Greek, isos meaning ‘same’ and topos meaning ‘place’) since, although different, they appear at the same place on the periodic table.
  • ISOTOPES
    • Remember, if you change the number of protons, you change the element.
    • If you change the number of neutrons, you have the same element but a different isotope.
  • ISOTOPES
    • Isotopes can be shown to exist by the study of the element carbon. They have 6 electrons and 6 protons and so are electrically neutral.
    • This accounts for its chemical properties. The number of neutrons can change without altering the chemical properties.
  • ISOTOPES
    • Most carbon nuclei are unstable, only 6 C 12 and 6 C 13 are stable.
    • All the rest will decay naturally to more stable arrangements.
    • Unstable isotopes are referred to as radioisotopes as they are radioactive.
    • We will discuss radioactivity shortly.
  • ISOTOPES
    • For all the possible arrangements of protons and neutrons that form nuclei, only about 300 are stable.
    • All the others are radioactive and will spontaneously breakdown over time.
    • Most isotopes are unstable (radioisotopes).
  • ISOTOPES
    • Take care not to mix up the words ‘isotope’ and ‘ion’.
    • An ion has an unequal number of protons and electrons. It is electrically charged. THIS CHANGES THE CHEMISTRY OF THE ATOM.
    • An isotope has a unequal number of protons and neutrons. This DOES NOT CHANGE THE CHEMISTRY OF THE ATOM.
  • THE NUCLEAR FORCE
    • Most atoms have more than one proton in the nucleus.
    • This means that there will be a Coulombic repulsive force.
    • Why does the nucleus remain stable and not fall apart?
    • This is because there is an attractive force between nucleons called the nuclear force. There are a number characteristics that can be deduced.
  • THE NUCLEAR FORCE
      •  The force must be very strong to overcome electrostatic repulsion. It is 1000 times stronger than the electric force and 10 38 times stronger than gravitational attraction.
      •  The force is independent of charge. This means the force is the same whether it acts between two protons, two neutrons or a proton and a neutron.
  • THE NUCLEAR FORCE
      •  The force acts over a very short range. Within the nucleus, the force acts between a nucleon and its very nearest neighbours. The range of the force is only about 1 x 10 -15 m or about the diameter of a proton.
      • The electric force is different in that it acts between all charged pairs and over any distance.
  • THE NUCLEAR FORCE
    • This explains why there are only a certain number of elements in the universe.
    • As the size of the nucleus increases by adding nucleons, the attractive force acts between neighbours but the repulsive force acts between all protons.
  • THE NUCLEAR FORCE
    • At some stage, the repulsive force becomes so great that the element becomes unstable.
    • This occurs at Z = 83 (Z = atomic number). This means that any element with an atomic number greater than 83 is unstable and is radioactive.
  • MASS DEFECT
    • The simplest bound nuclear system is the nucleus of deuterium ( 1 H 2 ) that is often called heavy hydrogen. Accurate measurements of the mass of the nucleus have found it to be 3.34374 x 10 -27 kg.
  • MASS DEFECT
    • Using the accurate measurements of the nucleons, there appears to be a discrepancy in the mass.
    • m proton = 1.67268 x 10 -27 kg
    • m neutron = 1.67499 x 10 -27 kg
    • total mass of 1 H 2 = 3.34767 x 10 -27 kg
    • actual mass of 1 H 2 = 3.34374 x 10 -27 kg
    • mass difference = 0.00393 x 10 -27 kg
  • MASS DEFECT
    • This loss appears to come from the process that fuses the proton and neutron together.
    • It is 4 times as great as the mass of an electron and too great to be explained as experimental error.
  • MASS DEFECT
    • In every nucleus there is some missing mass. The correct name for this is ‘mass defect’  m and the calculation requires the rest mass of each particle.
  • MASS DEFECT
    • Einstein’s special theory of relativity requires that mass increases dramatically at speeds close to the speed of light.
    • The increase in mass prohibits any object with mass reaching the speed of light so the value of the mass of an object at rest is used to avoid confusion.
  • MASS DEFECT
    • The mass defect for a nucleus (  m) is defined as the difference between the rest mass of the atomic nucleus and the sum of the rest masses of its individual nucleons in an unbound state.
  • BINDING ENERGY
    • Where does this mass go? Einstein’s mass-energy relationship E =  mc 2 gives us the answer.
    • The lost mass is converted into either kinetic energy or electromagnetic radiation (as gamma rays).
    • Einstein audio on his equation
  • BINDING ENERGY
    • The law of the conservation of energy should be modified to become the law of conservation of mass-energy.
    • If a nucleus loses mass, energy is released by the system.
    • Binding Energy
    • If a nucleus gains mass, energy is required for the nuclear reaction to occur.
  • BINDING ENERGY
    • A nucleus is considered to be more stable than individual neutrons or protons.
    • Energy is always released when something reverts to its most stable state.
    • As an example, remember that electrons give off energy when they drop down to their innermost shells (their most stable energy state).
  • BINDING ENERGY
    • When separate protons or neutrons join together to form a new nucleus, energy will be released.
    • The amount of energy released in a deuterium nucleus ( 1 H 2 ) due to the mass defect is:
  • BINDING ENERGY
    • E =  mc 2 =3.93 x 10 -30 x (3.0 x 10 8 ) 2
            • = 3.54 x 10 -13 J
            • = 2.2 x 10 6 eV
            • = 2.2 MeV
    • This is equal to 2.2 X 10 6 eV !!!!
    • Compare this to 1-10 eV when an electron falls to a lower energy level.
  • BINDING ENERGY
    • This can be written as a formula…
    • You can see that the combining of a proton and a neutron makes a deuterium nucleus and gives off 2.2 MeV of energy in the form of a gamma ray.
    • This is an EXOTHERMIC REACTION.
  • BINDING ENERGY
    • To reverse the process, the energy from the mass defect must be added to the nucleus.
    • This is usually in the form of a gamma ray or kinetic energy from a particle. This energy is called the binding energy .
    • This would be an ENDOTHERMIC REACTION.
  • BINDING ENERGY
    • The binding energy is the energy equivalent to the mass defect when nucleons bind together to form a nucleus. E b =  mc 2
  • BINDING ENERGY
    • The typical value for binding energy of nucleons is 1 to 10 MeV and 1 to 10 eV for electrons.
    • It is for this reason that mass defect for electrons are not detected - they are too small (approx. 10 -35 kg).
  • EXAMPLE 1
      • (a) Given that the rest mass of a proton is 1.67268 x 10 -27 kg and the rest mass of a neutron is 1.67499 x 10 -27 kg, determine the mass defect for the helium isotope ( 2 He 4 ) whose rest mass is 6.64489 x 10 -27 kg.
      • (b) What is the binding energy for the helium nucleus in MeV?
  • EXAMPLE 1
      • (c) Compare the answer in (b) to the binding energies of electrons bound to the atom.
      • (d) What is the maximum wavelength gamma ray that can be used to disintegrate a helium nucleus into its component protons and neutrons?
  • EXAMPLE 1
      • m proton = 1.67268 x 10 -27 kg
      • m neutron = 1.67499 x 10 -27 kg
      • m helium = 6.64489 x 10 -27 kg
  • EXAMPLE 1 SOLUTION
      • (a)  m = Z x m p + N x m a - m He
      • = (2 x 1.67268 + 2 x 1.67499 -6.64489) x 10 -27
      • = (6.69534 - 6.64489) x 10 -27
      • = 5.045 x 10 -29 kg
  • EXAMPLE 1 SOLUTION
      • (b) E b =  mc 2
      • =5.05 x 10 -29 x (3.0 x 10 8 ) 2
      • =4.54 x 10 -12 J
      • = 2.83 x 10 7 eV
      • = 28.3 MeV
  • EXAMPLE 1 SOLUTION
      • (c) Typical binding energies for the electron in the hydrogen atom were between 1 and 10 eV. This nuclear binding energy is of the order of 10 MeV. This is a factor of a million times as great.
  • EXAMPLE 1 SOLUTION
      • (d) In order to disintegrate the nucleus, the binding energy must be supplied by the gamma ray.
      • E photon = hf =
      • The maximum wavelength will correspond to the minimum energy photon, ie. the photon with the energy equal to the binding energy of the nucleus.
  • EXAMPLE 1 SOLUTION = 4.38 x 10 -14 m
  • EXAMPLE 1 SOLUTION
    • Chemical reactions involving the transition of electrons will produce enough energy to produce a UV (100nm) or a visible photon (400-700nm).
    • Nuclear reactions involving the formation of a nucleus will produce enough energy to produce a gamma ray photon (.00004nm).
  • BINDING ENERGY PER NUCLEON
    • As mass number increases for stable nuclei, so does the binding energy.
    • By determining the average binding energy per nucleon (E b /A), the average energies for various nuclei can be compared directly.
  • BINDING ENERGY PER NUCLEON
    • In a previous example, we found the total binding energy for the helium nucleus to be 28.3 MeV.
    • This means the average binding energy is 28.3/4 = 7.1 MeV. This means on average, 7.1 MeV of energy is needed to remove one nucleon.
  • BINDING ENERGY PER NUCLEON
    • In the graph below, note the shape.
    • The value of E b /A rises rapidly with increasing A and then flattens off to an almost constant value.
    • Almost all of the nuclei have values between 7.3 and 8.8 MeV.
  • BINDING ENERGY PER NUCLEON
  • BINDING ENERGY PER NUCLEON
    • Careful inspection shows that the maximum value occurs at 26 Fe 56 and slowly decreases.
    • This shows that iron has the most stable nucleus of all nuclei in the graph (takes the most energy to pry away a nucleon).
    • The curve can also be used to explain the production of energy from nuclear reactions that are covered below.
  • BINDING ENERGY PER NUCLEON
    • If the graph is drawn for the light elements, peaks can be seen.
  • BINDING ENERGY PER NUCLEON
    • The peaks occur for nuclei whose mass number is divisible by four: He-4, Be-8, C-12, O-16, Ne-20 and Mg-24.
    • This indicates how stable alpha particles ( 2 He 4 ) are. The peaks also indicate the existence of shells within the nucleus that are similar to the shells of the electrons in orbit.
  • EXAMPLE 2
    • Using the graph shown below, explain why the nucleus 26 Fe 56 is more stable than the nucleus 1 H 2 .
  • EXAMPLE 2 SOLUTION
    • By inspection, 26 Fe 56 has a higher binding energy per nucleon than 1 H 2 . This means that there is a greater mass defect for the iron nucleus when compared to the heavy hydrogen nucleus. So, to remove a nucleon from each nucleus, more mass needs to be replaced for iron. This will require more energy, so the iron nucleus is more stable.
  • PARTICLE MASSES
    • Energy can be converted over time from one form to another such as heat, light, kinetic etc.
    • Einstein showed that mass could be converted to energy and vice -versa using E = mc 2 .
  • PARTICLE MASSES
    • Mass particles can be expressed in terms of their equivalent energies.
    • A proton has energy of 1.50 x 10 -10 J or 939 MeV.
    • The ‘mass’ of an electron is 0.511 MeV.
  • PARTICLE MASSES
    • This energy is useful in calculations involving collisions between particles.
    • The initial kinetic energies of the particles and their masses is simply the total initial energy of the system.
  • CONSERVATION LAWS IN NUCLEAR REACTIONS
    • In all interactions in nature, certain quantities are always conserved such as charge.
    • Example:
    • 2 He + 7 N  8 O + 1 H
    • There are 9 positive charges on each side of the equation.
  • CONSERVATION LAWS IN NUCLEAR REACTIONS
      •  This leads to the conservation of atomic number as the atomic number refers to the number of protons.
      •  Mass number is also conserved. The individual nucleons may be converted from one type to another but the total number will remain constant.
      • Example:
      • He 4 + N 14  O 17 + H 1
  • CONSERVATION LAWS IN NUCLEAR REACTIONS
      •  Linear and angular momentum are also conserved as they are isolated systems.
  • CONSERVATION LAWS IN NUCLEAR REACTIONS
      •  The total amount of mass and energy is conserved. It may be converted from one form to another, i.e. mass to energy by E =  mc 2 .
      • The mass of the elements in a nuclear reaction is always greater than the products. This has already been covered earlier.
  • EXOTHERMIC NUCLEAR REACTIONS
    • EXOTHERMIC NUCLEAR REACTIONS - give off energy.
    • 1 H 2 = 3.3445 x 10 -27 kg
    • 7 N 14 = 2.3252 x 10 -26 kg
    • 6 C 12 = 1.9926 x 10 -26 kg
    • 2 He 4 = 6.644 x 10 -27 kg
  • EXOTHERMIC NUCLEAR REACTIONS
    • 2.65965 x 10 -26 kg  2.657 x 10 -26 kg
    • Lost 2.65 x 10 -29 kg of mass
  • EXOTHERMIC NUCLEAR REACTIONS
    • E = mc 2
    • E = (2.65 x 10 -29 )(3 x 10 8 ) 2
    • E = 2.385 x 10 -12 J
    • E = 1.49 x 10 7 eV
    • E = 14.9 MeV (Exothermic – energy given out)
  • ENDOTHERMIC NUCLEAR REACTION
    • ENDOTHERMIC NUCLEAR REACTION- Incoming particles must have enough K to produce a nuclear reaction.
    • 2 He 4 = 6.644 x 10 -27 kg
    • 7 N 14 = 2.3252 x 10 -26 kg
    • 8 O 17 = 2.8227 x 10 -26 kg
    • 1 H 1 = 1.673 x 10 -27 kg
  • ENDOTHERMIC NUCLEAR REACTION
    • 2.9896 x 10 -26 kg  2.99 x 10 -26 kg
    • Gain of 4.0 x 10 -30 kg of mass
    • E = mc 2
    • E = 3.6 x 10 -13 J
    • E = 2.25 x 10 6 eV
    • E = 2.25 MeV of Kinetic Energy required by the incoming particles to produce the nuclear reaction.
  • EXAMPLE 3
    • Complete the following nuclear reaction:
  • EXAMPLE 3 SOLUTION
  • Conservation of Momentum
    • In all interactions, the total momentum before the interaction is equal to the total momentum after the interaction.
    • This is also true for nuclear interactions.
  • Conservation of Momentum
    • If a particle radioactively decays, its initial momentum before it decays is zero. This means the total momentum after the decay is also zero. As momentum is a vector quantity, we must consider both magnitude and direction.
  • Conservation of Momentum
    • Thorium 90 Th 230 decays to an isotope of Radium 88 Ra 226
    • And also emits a helium nucleus 2 He 4 . The formula for this reaction is:
    • Notice that the mass number and the atomic number is conserved.
  • Conservation of Momentum
    • Consider the direction that the products will move off at after the decay.
    • The only way that momentum can be conserved is for them to move off in opposite directions.
    • This is because the total final momentum must equal zero.
  • Conservation of Momentum
    • As the mass of the two particles is very different, so will the velocities. This is because the total momentum must equal zero ( p = m v ).
  • Conservation of Momentum
    • To study the momentum in more detail, as the initial momentum is zero, we can say:
    • m Ra v f Ra = m He v f He
  • Conservation of Momentum
    • This indicates that the helium nucleus will move away with a velocity 56.5 times that of the radium nucleus.
    • This is of course an approximation as mass has been lost in this reaction due to E = mc 2 .
  • Conservation of Momentum
    • Originally, kinetic energy is also equal to zero.
    • After the decay, the kinetic is a value much more than zero.
    • This is because mass has been converted to energy.
  • Conservation of Momentum
    • The ratio of kinetic energies are shown below:
  • Conservation of Momentum
    • As a general rule, the particle with the smaller mass has the greater kinetic energy.
    • If a particle, such as an electron is ejected, it will have a much higher kinetic energy, as its mass is much smaller.
  • Conservation of Momentum
    • The above process is called alpha decay as the Helium nucleus, being of smaller mass, will have the greatest kinetic energy and move off at the highest speed.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • An atom can be changed from one type to another by the absorption or emission of a photon.
    • A nucleus can also be changed from one type to another by the absorption or emission of a proton or neutron.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • The emission of protons or neutrons occurs naturally in radioactive material or in the bombardment of atmospheric gases by high-energy particles from space.
    • Artificially, radioisotopes are produced in one of two ways:
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    •  Nuclear Reactor
    • Nuclear Fission produces many radioisotopes in small quantities.
    • Stable isotopes are introduced and bombarding them with the many neutrons that are a part of the nuclear reactions.
    • The unstable nucleus can absorb the neutron and form a radioactive isotope of the same element or eject a proton and form a radioisotope of a different element.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    •  Cyclotron
      • or other particle accelerator
    • Charged particles such as protons or deuterons
      • heavy hydrogen (deuterium) nucleus
    • are accelerated in the cyclotron and directed towards a stable nucleus.
    • As protons are fired at stable nuclei, only isotopes of different elements are formed.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • Neutrons
      • produced from a fission reactor
        • Why not a cyclotron?
      • can react with 32 S
      • To form a useful therapeutic radionuclide medical isotope
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • Used in the treatment of:
      • polycythemia vera
        • excess red blood cells
      • chronic myelocytic leukaemia,
      • chronic lymphocytic leukaemia,
      • certain ovarian and prostate carcinomas,
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • Also
      • palliation of metastatic skeletal disease,
      • and treatment of metastatic intrapleural and intraperitoneal effusions.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • Is a beta emitter
    • Equation?
    • Described as a Radiopharmaceutical
  • Radiopharmaceutical
    • A molecule that consists of a radioisotope tracer attached to a pharmaceutical.
    • After entering the body, the radio-labelled pharmaceutical will accumulate in a specific organ or tumour tissue.
  • Radiopharmaceutical
    • The radioisotope attached to the targeting pharmaceutical
    • undergoes decay and produces
    • specific amounts of radiation
    • used to diagnose or treat human diseases and injuries.
  • Radiopharmaceutical
    • The amount of radiopharmaceutical administered is carefully selected to ensure each patient’s safety.
    • Radioisotopes are an essential part of radiopharmaceuticals.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • When protons are accelerated in a cyclotron at oxygen, a radioactive isotope of fluorine is produced along with a neutron.
    • Notice that the atomic numbers and the atomic masses balance.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • Using a table of masses, we can calculate the energy released.
    • Mass of 1 H 1 atom = 0.167353 x 10 -26 kg
    • Mass of 8 O 18 atom = 2.9896606 x 10 -26 kg
    • Total mass reactants =3.1570136 x 10 -26 kg
    • Mass of 9 F 18 atom = 2.9899558 x 10 -26 kg
    • Mass of 0 n 1 atom = 1.665 x 10 -27 kg
    • Total mass products =3.1564558 x 10 -26 kg
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    •  m =
    • 3.1570136 x 10 -26 -3.1564558x 10 -26
    •  m = 0.0005578 x 10 -26 kg
    • E=  mc 2 =
    • 5.578 x 10 -30 x (2.998 x 10 8 ) 2
    • E = 5.0135 x 10 -13 J
    • E = 3.133 MeV
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • We are not making mass
    • Why is energy required for reaction to occur?
    • Conservation of momentum and energy
    • How much energy is required?
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • Minimum energy will produce fluorine and a neutron but they will be at rest.
    • As the proton was moving to begin with, this energy will not allow for the conservation of momentum.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • Through a derivation outside this course, the minimum kinetic energy required for a particle striking a stationary nucleus is:
    •  
    • K min = (1 + m / M )  E 
    •  
    • Where m is the mass of the incoming particle and M is the mass of the stationary target.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • Therefore the binding energy required for the nuclear reaction of the Hydrogen and Oxygen atom to occur will be…
    • = 3.308 MeV
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • This amount of energy (3.308 MeV) is higher than previously calculated ( 3.133 MeV) as some additional energy is required to get the process started.
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • 15 O is produced by firing a deuteron at 14 N
    • Equations is?
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • Determine binding energy of reactants and products.
    • Determine mass defect
    • Determine energy
  • APP - PRODUCTION OF MEDICAL RADIOISOTOPES
    • 18 F and 15 O are used in PET scans,
    • Will study this in more detail in the next topic
  • FORMATION OF C 14 IN THE ATMOSPHERE
    • As the neutrons collide with other nuclei, they transfer energy without reacting, eventually slowing down sufficiently to be absorbed.
    • The molecules then become incorporated into the molecules of CO 2 and are absorbed into the food chain. This forms the basis for carbon dating.
  • NITROGEN 13 PRODUCTION IN MEDICINE
    • This is an example of how a radioactive isotope can be produced naturally.
    • It forms the basis for Positron Emission Tomography (PET), which shows the body’s anatomy or structure. PET images can be used to show aspects of body metabolism or function.
  • NITROGEN 13 PRODUCTION IN MEDICINE
    • 7 N 13 is produced in hospitals using cyclotrons. One way it is produced is shown below:
    • The energy released is negative (-0.281 MeV) in this reaction implying that the cyclotron must accelerate the deuterium ( 1 H 2 ) nucleus.
  • NITROGEN 13 PRODUCTION IN MEDICINE
    • The other method discussed previously is the method is preferred by hospitals.
    • Although more energy is required, there is a greater abundance of 1 H 1 than deuterium.