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Quantum and Nuclear Physics The Photoelectric effect Waves or Particles? The Photoelectric effect How are the electrons released? Powerful red laser No electrons released Photoelectron Energy …and some energy is given to the electron as kinetic energy. - Some energy is needed to release the electron (the work function φ)… Photon Energy = work function + kinetic energy of electron Determining Planck’s constant • Add different filters under the light source Photoelectric experiment • Take measurements of stopping potential and wavelength to determine Planck’s constant and the threshold frequency Plot a graph of stopping potential versus frequency Photoelectric Effect: Vstop vs. Frequency eVstop hf Vstop 0 hfmin hf min Slope = h = Planck’s constant Determining “h” from the graph hf EK max Photoelectric Effect: IV Curve Dependence Intensity I dependence Vstop= Constant f1 > f2 > f3 Frequency f dependence f1 f2 Vstop f f3 Is light a wave or a particle? • http://www.schoolphysics.co.uk/age1619/Quantum%20physics/text/Photoelectric _effect_animation/index.html =Φ E max= V= Stopping voltage 1. The work function for lithium is 4.6 x 10-19 J. (a) Calculate the lowest frequency of light that will cause photoelectric emission. (6.9 x 1014 Hz ) (b) What is the maximum energy of the electrons emitted when light of 7.3 x 1014 Hz is used? (0.24 x 10-19 J ) 2. A frequency of 2.4 x 1015 Hz is used on magnesium with work function of 3.7 eV. (a) What is energy transferred by each photon? (b) Calculate the maximum KE of the ejected electrons. (c) The maximum speed of the electrons. (d) The stopping potential for the electrons. (a) 1.6 x 10-18 J (b) 1.0x 10-18 J (c) v = 1.5 x 106 m s-1 (d) Vs = 6.3 V Questions Tsokos page 396 q’s 1-7. Review of Bohr and deBroglie • Background: – Balmer found equation for Hydrogen spectrum but didn’t know what it meant. – Rutherford found that atoms had a nucleus, but didn’t know why electrons didn’t spiral in. • Bohr postulates quantized energy levels for no good reason, and predicts Balmer’s equation. • deBroglie postulates that electrons are waves, and predicts Bohr’s quantized energy levels. • Note: no experimental difference between Bohr model and deBroglie model, but deBroglie is a lot more satisfying. Davisson and Germer -- VERY clean nickel crystal. Interference is electron scattering off Ni atoms. e e e e e e Ni e e det. e e scatter off atoms e e move detector around, see what angle electrons coming off See peak!! so probability of angle where detect electron determined by interference of deBroglie waves! # e’s 0 e e e e 500 scatt. angle e e det. e e e Ni Observe pattern of scattering electrons off atoms Looks like …. Wave! Electron diffraction Diffraction rings Calculating the De Broglie λ λ = h/p (= h/(2Ekm)1/2 ) h = Planck’s constant p = Momentum In 1923, French Prince Louis de Broglie, generalised Einstein's work from the specific case of light to cover all other types of particles. This work was presented in his doctoral thesis when he was 31. His thesis was greeted with consternation by his examining committee. Luckily, Einstein had received a copy in advance and vouched for de Broglie. He passed! de Broglie questions • Calculate the wavelengths of the “deBroglie” waves associated with • a)a 1kg mass moving at 50ms-1 • b)an electron which has been accelerated by a p.d. of 500V. a)Discuss briefly deBroglie’s hypothesis and mention one experiment which gives evidence to support it. b)Calculate the wavelength of the “deBroglie wave” associated with an electron in the lowest energy Bohr orbit. (The radius of the lowest energy orbit according to the Bohr theory is 5·3×10-11m.) Questions Tsokos page 396 q’s 8-10 History of Quantum Mechanics Max Planck's work on the 'Black Body' problem started the quantum revolution in 1900. He showed that energy cannot take any value but is arranged in discrete lumps – later called photons by Einstein. In 1913, Niels Bohr proposed a model of the atom with quantised electron orbits. Although a great step forward, quantum physics was still in its infancy and was not yet a consistent theory. It was more like a collection of classical theories with quantum ideas applied. Starting in 1925 a true 'quantum mechanics' – a set of mathematically and conceptual 'tools' – was born. At first, three different incantations of the same theory were proposed independently and were then shown to be consistent. Quantum mechanics reached its final form (essentially unchanged from today) in 1928. Participants of the 5th Solvay Congress, Brussels, October 1927 E. Schrödinger W. Pauli W. Heisenberg N. Bohr M. Planck M Curie A. Einstein L.V. de Broglie Models of the Atom • Thomson – Plum Pudding – – – – – – Why? Known that negative charges can be removed from atom. – Problem: just a random guess • Rutherford – Solar System – Why? Scattering showed hard core. – Problem: electrons should spiral into nucleus in ~10-11 sec. + • Bohr – fixed energy levels – Why? Explains spectral lines. – Problem: No reason for fixed energy levels + • deBroglie – electron standing waves – Why? Explains fixed energy levels – Problem: still only works for Hydrogen. • Schrodinger – will save the day!! + – Different view of atoms The Bohr Atom Electrons are only allowed to have discrete energy values and these correspond to changes in orbit. The Schrodinger Atom Electrons behave like stationary Amplitude waves. Only certain types of wave fit the atom, and these correspond to fixed energy states. The square of the amplitude gives the probability of finding the electron at that point + 0eV Consider a ball in a hole: When the ball is here it has its lowest gravitational potential energy. Spectra 5J We can give it potential energy by lifting it up: If it falls down again it will lose this gpe: 5J 30J 20J Spectra A similar thing happens to electrons. We can “excite” them and raise their energy level: 0eV -0.85eV -1.5eV -3.4eV -13.6eV An electron at this energy level would be “free” – it’s been “ionised”. These energy levels are negative because an electron here would have less energy than if its ionised. This is called “The ground state” Spectra If we illuminate the atom we can excite the electron: Q. What wavelength of light would be needed to excite this electron to ionise it? 0eV -0.85eV -1.5eV -3.4eV Light Energy change = 3.4eV = 5.44x10-19J. Using E=hc/λ wavelength = 3.66x10-7m -13.6eV (In other words, ultra violet light) Example questions 1) State the ionisation energy of this atom in eV. 2) Calculate this ionisation energy in joules. 3) Calculate the wavelength of light needed to ionise the atom. 0eV -0.85eV -1.5eV -3.4eV 4) An electron falls from the -1.5eV to the -3.4eV level. What wavelength of light does it emit and what is the colour? 5) Light of frequency 1x1014Hz is incident upon the atom. Will it be able to ionise the atom? -13.6eV Spectra Continuous spectrum Absorption spectrum Emission spectrum Emission Spectra Hydrogen Helium Sodium Observing the Spectra Microscope (to observe the spectrum) Light source Collimator Gas Diffraction grating (to separate the colours) Questions Tsokos page 405 q’s 1-7. Models of the Atom • Thomson – Plum Pudding – – – – – – Why? Known that negative charges can be removed from atom. – Problem: just a random guess • Rutherford – Solar System – Why? Scattering showed hard core. – Problem: electrons should spiral into nucleus in ~10-11 sec. + • Bohr – fixed energy levels – Why? Explains spectral lines. – Problem: No reason for fixed energy levels + • deBroglie – electron standing waves – Why? Explains fixed energy levels – Problem: still only works for Hydrogen. • Schrodinger – will save the day!! + – Schrödinger model Schrödinger set out to develop an alternate formulation of quantum mechanics based on matter waves, à la de Broglie. At 36, he was somewhat older than his contemporaries but still succeeded in deriving the now famous 'Schrödinger Wave Equation.' The solution of the equation is known as a wave function and describes the behavior of a quantum mechanical object, like an electron. At first, it was unclear what the wave function actually represented. How was the wave function related to the electron? At first, Schrödinger said that the wave function represented a 'shadow wave' which somehow described the position of the electron. Then he changed his mind and said that it described the electric charge density of the electron. He struggled to interpret his new work until Max Born came to his rescue and suggested that the wave function represented a probability – more precisely, the square of the absolute magnitude of the wavefunction is proportional to the probability that the electron appears in a particular position. So, Schrödinger's theory gave no exact answers… just the chance for something to happen. Even identical measurements on the same system would not necessarily yield the same results! Born's key role in deciphering the meaning of the theory won him the Nobel Prize in Physics in 1954. Quantum Mechanical tunneling In the classical world the positively charged alpha particle needs enough energy to overcome the positive potential barrier which originates from protons in the nucleus. In the quantum world an alpha particle with less energy can tunnel through the potential barrier and escape the nucleus. Electron in a box model Electrons will form standing waves of wavelength 2L/n Kinetic Energy of an electron in a box • When the momentum expression for the particle in a box : • • is used to calculate the energy associated with the particle Heisenberg uncertainty principle Heisenberg made one fundamental and long-lasting contribution to the quantum world – the uncertainty principle. He showed that quantum mechanics implied that there was a fundamental limitation on the accuracy to which pairs of variables, such as (position and momentum) and (energy and time) could be determined. If a 'large' object with a mass of, say, 1g has its position measured to an accuracy of 1 , then the uncertainty on the object's velocity is a minute 10-25 m/s. The uncertainty principle simply does not concern us in everyday life. In the quantum world the story is completely different. If we try to localize an electron within an atom of diameter 10-10 m the resulting uncertainty on its velocity is 106 m/s! Heisenberg uncertainty principle Nuclear physics Determining the size of the nucleus Approach of alpha particle to nucleus Z = 79 (gold) 25 20 15 10 5 0 0 2 4 6 distance from nucleus / 8 10–14 10 m 1. Make an arithmetical check to show that at distance r = 1.0x10–14 m, the electrical potential energy, is between 20 MeV and 25 MeV, as shown by the graph. 2.How does the electrical potential energy change if the distance r is doubled? 3.From the graph, at what distance r, will an alpha particle with initial kinetic energy 5 MeV colliding head-on with the nucleus, come to rest momentarily? 1. Substituting values gives EP = 2. 3. 2 79 1.6 10 19 C 4 8.85 10 12 C 2 J 1 m 1 1.0 10 14 m 10 6 22 .7 MeV. Halves, because the potential energy is proportional to 1/r. About 4.6x10–14 m, where the graph reaches 5 MeV. Recall: + + Circular paths 2 protons, 2 neutrons, therefore charge = +2 - 1 electron, therefore charge = -1 Because of this charge, they will be deflected by magnetic fields: These paths are circular, so Bqv = mv2/r, or r =mv Bq Bainbridge mass spectrometer Ions are formed at D and pass through the cathode C and then through a slit S1 A particle with a charge q Hyperlink and velocity v will only pass through the next slit S2 if the resultant force on it is zero – that is it is traveling in a straight line. That is if: Therefore In the region of the Mag field r = Mv/(Bq) Bqv = Mv2/r Therefore Nuclear energy levels There are 2 distinct length of tracks in this Alpha decay Therefore, the energy levels in the nucleus are discrete The existence of Neutrinos How can a 2 body system create a spectrum of energies? There must be a 3rd particle The Neutrino was postulated A 2 body system only has one solution A 3 body system has many solutions Changes in Mass and Proton Number Beta - decay: 90 Sr 38 90 Y 39 + 0 β -1 “positron” Beta + decay: 11 6 C 11 B 5 + 0 +1 β Radioactive Decay Law dN/N = -λdt which when integrated, gives Taking antilogs of both sides gives: Half life and the radioactive decay constant When N = No/2 the number of radioactive nuclei will have halved Therefore when t = T1/2 N = No/2 = Noe-λT1/2 and so 1/2 = e-λT1/2 . Taking the inverse gives 2 = eλT1/2 and so: Measuring long half lives • If the half life is very long, then the activity (A) is constant • Analysis of a decay curve cannot give the half life. • If the mass of the substance is measured, then • A = -λN, so a measurement of the activity enables Measuring long half lives to be calculated (N from mass). • T1/2 can be calculated from λ. Measuring short half lives • Each decay can cause an ionisation • This can generate an electric current • If the current is displayed on an oscilloscope, then • The limit is the response time of the oscilloscope (typically µs). Questions Tsokos page 412 q’s 1-20