12.2
- 2. Rutherford’s Experiment In 1909, Rutherford investigated high speed alpha particles that passed through thin, metal foils. The experiments were actually carried out by two young physicists named Hans Geiger and Ernest Marsden.
- 3. Rutherford’s Experiment Alpha particles are double charged helium ions; two protons and neutrons. Emitted spontaneously from heavy radioactive atoms such as uranium and plutonium.
- 4. Rutherford’s Experiment The beam of alpha particles were directed at a metal foil,detector picks up the emerging particles on the other side. This would give an indication of the nature of the atom.
- 5. Rutherford’s Experiment q a- source lead collimators vacuum gold foil ZnS crystal microscope
- 6. Rutherford’s Experiment A lead block holds the alpha particle source and emitted through a fine hole in the lead. It is collimated into a beam by aligning holes in a series of lead plates. The beam is then aimed at a gold foil. Gold is used as it can be beaten into a thin foil, about 1 m (8000 atoms) thick.
- 7. Rutherford’s Experiment The particles are detected by using a microscope fitted with, a zinc sulfide crystal. If an alpha particle hits the crystal, a flash can be seen and counted.
- 8. Rutherford’s Experiment Rutherford made predictions as to what he would find. He said there would be very few particles would suffer deflections those which did would be deviated, only a few degrees. Theory suggested that only 1% would be deflected more than 3o.
- 9. Rutherford’s Experiment Rutherford also determined there was almost no way there could be any backscattering ie. angles greater than 90o. It could only happen with a direct collision with an electron and only 1 particle in 103500 would take this path.
- 10. Rutherford’s Experiment The calculations were on the basis of previous model of the atom by Thomson. Rutherford predicted that the deflections would be due to electrostatic attraction.
- 11. Rutherford’s Experiment The actual results came as a great surprise everyone was expecting Thomson’s model was correct. One particle in 10 000 was backscattered compared to the theory of 1 in 103500.
- 12. Rutherford’s Atom He suggested that all the positive matter was concentrated in a very small sphere, at the centre of the atom. As electrons have very little mass all the mass must be in this sphere, called the nucleus. Around the nucleus is essentially empty space.
- 13. Rutherford’s Atom This explains the backscattering found in the experiment. The alpha particle would strongly be repelled when it is directed at the nucleus, due to electrostatic repulsion. Those particles that do not hit the nucleus; would pass through, with little or no deflection.
- 15. The kinetic energy of the alpha particle is EK This kinetic energy is converted into potential energy as the alpha particle overcomes the coulombic force as it approaches the nucleus All the kinetic energy = potential energy when the approach is the closest
- 16. The potential energy near a point charge is Ep = Q1 Q2 / 40 r EK = Ep = Ze 2e/ 40 d EK = Ze2/ 20 d Calculate the distance of closest approach for an alpha particle of 5MeV approaching a gold nucleus
- 17. Nuclear Physics Modern scattering techniques show that the nucleus is not hard like a billiard ball but has a ‘fuzzy’ surface. This is due to a variation in the density of the nuclear material in the outside layer of the nucleus.
- 18. Nuclear Physics The mass of an atom is one of the characteristic properties that can give an insight into the structure of an atom. One device that can be used to determine the mass is a mass spectrometer. One particular type is the Bainbridge Mass Spectrometer.
- 19. Nuclear Energy Levels Remember how e/m radiation is given off when electrons that have been excited, drop back down to lower energy levels. The photons given off have specific energies equal to the difference in energy levels.
- 20. Nuclear Energy Levels Previously, we have studied a, , and radiation Often, radiation has accompanied the emission of a or particles. The emission of photons is evidence that the nucleus has energy levels
- 21. Nuclear Energy Levels Radium decays to Radon at different energy levels. a-particles are ejected at certain discrete velocities (energies). The energy depends on which level the Radium decays to in the Radon. HeRnRa 4 2 224 86 226 88
- 22. Nuclear Energy Levels A B C D Nucleus Shells Radon 222 Excited State 3 Radon 222 Ground State a A aB a C a D Energy above the ground state of Radon
- 23. Nuclear Energy Levels In the above diagram, 226Ra decays giving off an aB particle that has a specific K (aB) when it decays to Rn in the 2nd excited state. The 222Rn then might move to the ground state giving off a photon of energy in the MeV range called a GAMMA PHOTON ()
- 24. Radioactive Decay We will look at the following areas: Beta Decay The Neutrino Half-life Decay constant
- 25. BETA DECAY Nuclei that have an imbalance of protons or neutrons can be unstable and also undergo radioactive decay. The process involves the change of a proton into a neutron or more commonly a neutron into a proton with the ejection of an electron from the nucleus. This decay is called beta decay, and the electron is referred to as a beta particle.
- 26. BETA+ DECAY BETA+ DECAY (too many protons)- When a nucleus has to increase its neutron number to become more stable, a proton can spontaneously change into a neutron. B+
- 27. BETA+ DECAY An electron (positively charged) is ejected with a neutrino. The positive electron is called a positron and is an example of antimatter. The atomic number is reduced by one but the mass number is unaffected.
- 28. BETA+ DECAY On the line stability on the graph, any atom below the line would decay this way. +eY+X 0 1+1 A Z A Z B+
- 29. BETA+ DECAY In the nucleus, the reaction is: An example of this is: +e0 1+ 1 0 1 1 np +e+CN 0 1+ 13 6 13 7
- 30. BETA+ DECAY Notice that both mass and charge are conserved. A ‘positron’, a positively charged electron (the same mass as an electron) is ejected. The positron is an example of antimatter (“opposite of”). This is known as ‘proton decay”. +e+CN 0 1+ 13 6 13 7
- 31. BETA+ DECAY The positron is known as the B+ particle. A Neutrino (v) is also released. Note a new element is formed. There are no natural positron emitters since positron half- lives are very small. Note- as the 13N might decay into a metastable form of 13C, the 13C could then drop down to a more stable state, giving off a GAMMA RAY.
- 32. NEUTRINOS AND ANTINEUTRINOS Beta particles are emitted with a range of energies up to a maximum of a few MeV. It seemed strange that the electrons with the maximum kinetic energy carried away all the available energy, yet those with less than he maximum kinetic energy appeared to have energy missing.
- 33. NEUTRINOS AND ANTINEUTRINOS This did not obey the law of conservation of energy. Other experiments with momentum confirmed that linear momentum was not conserved. 7N14 6C14 e- Speed and direction of the electron if momentum was conserved. e- Actual speed and direction of the electron.
- 34. NEUTRINOS AND ANTINEUTRINOS In 1934 Enrico Fermi developed the theory of beta decay and that the conservation laws did hold because there was a particle that had yet to be detected carrying the lost energy and momentum. He called this particle a neutrino (Italian for ‘little neutral one’). The antimatter of the neutrino ( ) is the antineutrino ( ). _
- 36. NEUTRINOS AND ANTINEUTRINOS Using the conservation laws, he postulated the properties for the neutrino. Neutrinos are uncharged. This is because charge is already conserved. A neutron decays into a proton and an electron. Neutrinos have zero rest mass but carry energy and momentum. The conservation laws would not hold otherwise.
- 37. NEUTRINOS AND ANTINEUTRINOS Neutrinos react very weakly with matter. It took 25 years to detect them and there are millions of neutrinos that pass through the Earth from the sun as if the Earth was not there. This is because they have no real mass or charge. Neutrinos travel at the speed of light. As they have no mass but have energy, they must travel at the maximum speed possible - the speed of light.
- 38. NEUTRINOS AND ANTINEUTRINOS The neutrino was accepted readily as it solved awkward problems but was not discovered until 1956. It is given the symbol (the Greek letter nu) and has zero atomic number and mass number.
- 39. Radioactive Decay Radioactive decay is a completely random process. No one can predict when a particular parent nucleus will decay into its daughter. Statistics, however, allow us to predict the behaviour of large samples of radioactive isotopes.
- 40. Radioactive Decay We can define a constant for the decay of a particular isotope, which is called the half-life. This is defined as the time it takes for the activity of the isotope to fall to half of its previous value.
- 41. Radioactive Decay From a nuclear point of view, the half-life of a radioisotope is the time it takes half of the atoms of that isotope in a given sample to decay. The unit for activity, Becquerel (Bq), is the number of decays per second.
- 42. Radioactive Decay An example would be the half-life of tritium (3H), which is 12.5 years. For a 100g sample, there will be half left (50g) after 12.5 years. After 25 years, one quarter (25g) will be left and after 37.5 years there will be one eighth (12.5g) and so on.
- 43. Radioactive Decay The decay curve is exponential.The only difference from one sample to another is the value for the half-life.
- 44. Radioactive Decay Below is a decay curve for 14C. Determine the half-life for 14C.
- 45. Radioactive Decay The half-life does not indicate when a particular atom will decay but how many atoms will decay in a large sample. Because of this, there will always be a ‘bumpy’ decay for small samples.
- 46. Radioactive Decay If a sample contains N radioactive nuclei, we can express the statistical nature of the decay rate (-dN/dt) is proportional to N:
- 47. Radioactive Decay in which , the disintegration or decay constant, has a characteristic value for every radionuclide. This equation integrates to: No is the number of radioactive nuclei in a sample at t = 0 and N is the number remaining at any subsequent time t. N dt dN t oeNN You have to derive this
- 48. Radioactive Decay -dN/dt = N Collect like terms dN/N = -dt Integrate ln N = -t + c But c = ln N0 So, ln N = -t + ln N0 ln N - ln N0 = -t N/ N0 = e-t
- 49. Radioactive Decay Solving for t½ yields, that is when N = N0/2 t1/2 = 0.693 t1/2 = 0.693/ 2ln 2 1 t or 2 1 2ln t
- 50. Radioactive Decay The half-life of an isotope can be determined by graphing the activity of a radioactive sample,over a period of time
- 51. Radioactive Decay The graph of activity vs time can be graphed in other ways As the normal graph is exponential it does not lead to a straight line graph Semi logarithmic graph paper can solve this problem
- 53. Radioactive Decay If we take the natural log of N = Noe-t we get: ln N = ln No -t The slope of the line will determine the decay constant
- 54. Radioactive Decay The accuracy in determining the half-life depends on the number of disintegrations that occur per unit time. Measuring the number of disintegrations for very long or short half-life isotopes could cause errors.
- 55. Radioactive Decay For very long half-life isotopes i.e. millions of years Only a small number of events will take place over the period of a year Specific activity is used Activity of sample is measured against a calibrated standard
- 56. Radioactive Decay Standard is produced by reputable organisations i.e. International Atomic Energy Agency Calibrated standard measures the accuracy of the detector making sure it is accurate Specific activity and atomic mass of isotope is then used to calculate the half- life
- 57. Radioactive Decay With very short half-life isotopes, the isotope may disintegrate entirely before it is measured.Time is therefore of the essence As most of these isotopes are artificial. Produce them in or near the detector This eliminates or reduces the transfer time problem
- 58. EXAMPLE 1 (a)Radium-226 has a half-life of 1622 years. A sample contains 25g of this radium isotope. How much will be left after 3244 years? (b)How many half-lives will it take before the activity of the sample falls to below 1% of its initial activity? How many years is this?
- 59. EXAMPLE 1 SOLUTION (a) 3244 years is 2 half lives (2 x 1622) N= No(1/2)n = 25 x (1/2)2 = 25 x (1/4) =6.25
- 60. EXAMPLE 1 SOLUTION (b) The activity of a radioactive sample is directly proportional to the number of remaining atoms of the isotope. After t1/2, the activity falls to ½ the initial activity. After 2 t1/2, the activity is ¼. It is not till 7 half-lives have elapsed that the activity is 1/128th of the initial activity. So, 7 x 1622 = 11354 years
- 61. EXAMPLE 2 A Geiger counter is placed near a source of short lifetime radioactive atoms, and the detection count for 30-second intervals is determined. Plot the data on a graph, and use it to find the half-life of the isotope.
- 62. EXAMPLE 2 Interval Count 1. 12456 2. 7804 3. 5150 4. 3034 5. 2193 6. 1278 7. 730
- 63. EXAMPLE 2 SOLUTION The data are plotted on a graph with the point placed at the end of the time interval since the count reaches this value after the full 30 seconds.
- 64. EXAMPLE 2 SOLUTION A line of best fit is drawn through the points, and the time is determined for a count rate of 12 000 in 30 seconds. Then the time is determined for a count rate of 6000, and 3000.
- 65. EXAMPLE 2 SOLUTION t(12 000) = 30s t( 6000) = 72s, so t1/2 (1) = 42s t( 3000) = 120s, so t1/2 (2) = 48s The time difference should have be the half- life of the sample.
- 66. EXAMPLE 2 SOLUTION Since we have two values, an average is taken. t s1 2 42 48 2 45/
- 67. methods Short half life Long half life