PHY293 Lecture #15 November 27, 2017 1. Quantum Mechanics and the Atom The Thompson/Plum Pudding Model Thompson discovered the electron in 1894 (Nobel Prize in 1906) Heating materials (metals) causes corpuscles of electric charge to emitted This is how X-ray tubes and electron microscopes source their electrons in the first place It was also known that electric charges are emitted by some atoms in radioactive decay Thompson showed corpuscles of charge were the same in each case: Had the same charge/mass ratio: e/m Were very much lighter than the atoms that emitted them Presuming atoms were partially made of electrons but electrically neutral Thompson developed the plum pudding model: electrons embedded in a spherical cloud of positive charge The Rutherford/Nuclear Model Standard physics trick is to scatter particles off a target, to learn about the structure of the target 1909 Geiger/Marsden (working in Rutherford s lab) scattered α particles off a gold foil Their expectation, based on Thompson model: Very little scattering The α particles were 8000 times heavier than the electron so electrons should not deflect them much The positive charge was equal in magnitude to the total charge from the electrons, but spread out over the whole atom, so hard to scatter much either Expected little scattering or only at small angles This was not at all what was observed In fact did see that most of the α particles passed straight through the foil: undeflected by the gold atoms But some were scattered at large angles A few were deflected straight back towards source (took a while to find these because apparatus not ready to detect αs in backwards direction) This was interpreted (but only 5 years later in 1911 so they were quite puzzled) as evidence that the positive charge in the atom was concentrated in a small region of the atom at the centre Improves the picture of the atom, but doesn t add any information about where the electrons are Problems with the Rutherford/Planetary Model of the atom Classically, an electron in a circular orbit radiates energy (accelerating around in a circle) This lowers the energy, reducing the radius of the orbit Eventually the electron loses all of its energy and hits the nucleus Classical calculation predicts this happens very quickly ( 10 12 s for an 0.1 nm diameter atom) Planetary model of the atom (with Rutherford nucleus at core) can t explain stable atoms In addition there were experimental observations of atomic spectra that could not be explained classically Hydrogen atom absorbs and emits light at a set of specific frequencies Moreover, these frequencies have a very specific pattern 2. Understanding the Hydrogen Atom with the Uncertainty Principle The electron in the Hydrogen atom is held near the proton/nucleus by electrostatic attraction If the total energy is the sum of its kinetic energy and electrostatic potential energy And the two charges are a distance r apart E classical = 1/2mv 2 + 1 (+e)( e) 4πɛ 0 r An orbiting electron obeys F = ma so attractive electric force balances the circular centripetal acceleration: 1 e 2 4πɛ 0 r 2 = mv2 r v 2 = e 2 4πɛ 0 mr ( )
Can substitute this back into the total energy expression E classical = 1/2m ( e 2 ) 1 e 2 4πɛ 0 mr 4πɛ 0 r = e2 8πɛ 0 r So the classical potential energy exceeds the kinetic energy there are bound states But there is no minimum energy as r 0 the energy becomes more and more negative Problem with the planetary model the electrons will always want to spiral in to very small r losing energy as they go. As this happens v and mv become smaller, so energy is radiated at the expense of the electron, but still not stable. 3. Solution (suggested by Bohr) was to treat the electron as a orbiting wave As the electron kinetic energy shrinks it will approach the uncertainty bound discussed last lecture If, at some point, we take the actual radius of the electron to be r and the actual momentum of the electron at that point to be p r the uncertainty principle requires that pr h or mvr h (it can t be less than this) Plugging this in to the electron energy equation gives: E wave = 1/2mv 2 1 e 2 4πɛ 0 r = 1/2m( h ) 2 1 e 2 mr 4πɛ 0 r = h2 2mr 2 e2 4πɛ 0 r Now as r decreases wave becomes more compact, the speed associated with the electron increases inversely (mvr h), so the kinetic energy increases (first term) For the classical solution the energy decreases monotonically as r decreases For the matter wave solution the energy reaches a minimum and then increases as r 0 The system will seek the minimum energy state which is a finite r in the matter wave solution The classical atom is unstable (bad news for us) The uncertainty principle inspired quantum mechanical treatment is stable. Also provides and order of magnitude prediction of electron energy levels Detailed measurements of the emission and absorption lines of hydrogen revealed an interesting pattern: n2 λ(nm) = 364.6 n 2 4 n = 3, 4, 5,... A pattern like this, based on integers, with a single constant measured to one part in 1000 demands theoretical understanding Turns out the wave-atom can explain this too Bohr Atom 4. The Bohr (Hydrogen) Atom Start from Rutherford model, assume electrons orbit a heavy, positively charged nucleus Leads to all the conclusions we had before including E = 1/8 e2 πɛ 0r Classically r can take on any value and eventually radiate all electron energy away down to r = 0 Bohr knew the energies appeared to take on discrete values, implying radii might be quantised Bohr suggested electron angular momentum was quantised in units of h, ie. L = n h n = 1, 2, 3,... L = mvr = n h v 2 = n2 h 2 His physical interpretation of this was that if the orbiting electron really did correspond to a standing wave, then the orbits should have an integral number of wavelengths Otherwise the electron wave would interfere with itself (get all the way around the orbit with a different phase/amplitude and start to cancel out probability). So the other way to look at this is: m 2 r 2 2πr = nλ λ = h/p = h/mv 2πr = n h mv
Which we can re-arrange to mvr = n h... Bohr s original hypothesis If we equate Bohr s expression for v 2 to the classical centripetal force balancing version (equation above) we find e 2 4πɛ 0 mr = n2 h 2 m 2 r 2 r = 4πɛ 0 h 2 e 2 m n2 n = 1, 2, 3... So Bohr s hypothesis predicts discrete radii at r = a 0 n 2 with a 0 = 0.0529 nm this is referred to as the Bohr radius The energies of the electrons in these orbits must also be discretised (or quantised): E = e2 8πɛ 0 r = e2 ( e 2 m ) 1 8πɛ 0 r 4πɛ 0 h 2 n 2 = 13.6 ev 1 n 2 n = 1, 2, 3,... The lowest energy state (ground state) is given by n = 1 and has binding energy of -13.6 ev The first excited state (n = 2) has binding energy -3.4eV and so on These also explain the Balmer absorption and emission lines Consider the n = 4 state with energy (-13.6/16) = -0.85 ev Transition from n = 4 to n = 2 releases -0.85 - (-3.4) = 2.55 ev (E = hc/λ = 1240/λ = 486 nm) Transition from n = 4 to n = 3 gives 1.89 ev or a wavelength of 656 nm Transitions to the ground-state (ie. 2 1 give energies like 10.2 ev, λ = 122 nm) not in the visible, so had to be measured after Bohr prediction Started a whole cottage industry of spectroscopy to find the lines predicted by Bohr See last figure, but note that the colours of the lines are not the colours of the light, all of the transitions have slightly different energies and hence slightly different wavelengths. But they are (kind of confusingly) coloured more red towards the right (the Pfund lines which are more infra-red transitions (but all different wavelengths of infra-read)), more blue/green (visible near the middle in the Balmer and Paschen series) and then really UV (coloured black in the picture!) for the Lyman lines. Eventually they were all found
Thompson Atomic Model
Rutherford Scattering Experiment
Rutherford Expectation
Rutherford Observation
Hydrogen Absorption/Emission Lines
Electron Energy
Hydrogen Balmer Series
Bohr Orbits
Bohr Predictions for Balmer Lines
Bohr Predictions