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Rutherford Model 1911 Positive charge is concentrated in a very small nucleus. So aparticles can sometimes approach very close to the charge Ze in the nucleus and the Coulomb force 1 ( Ze)( Ze) F= 4πε o r2 Can be large enough to cause large angle deflections. Nuclear model of the atom Detailed calculations now show that there is a non-negligible probability of large angle scattering Stability problem (I) If electrons are stationary, there is nothing to prevent them from responding to the Coulomb force of attraction and being sucked into the nucleus. But the atomic diameter would then be ~ nuclear diameter. (II) If the electrons are in orbit, they are continuously being accelerated and therefore, classically, should emit radiation. They should lose energy and spiral into the nucleus! Bohr’s postulates Bohr 1913: circumvented stability problem by making two postulates: 1. An electron in an atom moves in a circular orbit for which its angular momentum is an intergral multiple of h 2. An electron in one of these orbits is stable. But if it discontinuously changes its orbit from one where energy is E i to one whose energy is Ef , energy is emitted or absorbed in quanta satisfying: Ei − E f = hν Analysis based on Bohr postulates: for hydrogen-like atom (1 electron) Force of attraction 1 Ze 2 mv 2 F= = 2 4πε o r r m = mass of electron 1 2 1 1 Ze 2 KE = mv = 2 2 4πε o r 1 Ze 2 PE = − 4πε o r 1 Ze 2 Total E = − 4πε o 2r mvr = nh 1 Ze mv = 4πε o r 2 and 2 2 n h 1 Ze m 2 2= 4πε o r m r E =− 2 r = 4πε o 2 4 1 mZ e ( 4πε o ) 2n h 2 2 2 2 2 2 n h mZe 2 For convienience we define the “Bohr Radius” as: a= 4πε o h me 2 = 0.0529nm 2 And the energy unit 1 Rydberg (1Ry) as: me 4 ( 4πε o ) 2h 2 2 2 n r= a Z and = 13.6eV E= − Z 2 n 2 Ry n=1 n=2 n=3 “excited states” “ground state” a=radius of ground state (n=1) orbit for hydrogen (Z=1) 1 Ry=13.6eV is the ground state (n=1) binding energy for hydrogen (agrees with experiment) For larger Z-values, radius of orbit shrinks for given n. Binding energies get much bigger mvr = nh Electron velocity: obtained from 2 2 nh nhZe Ze v= = = mr 4πε o n 2h 2 4πε o nh 2 Z=1, n=1 e 6 −1 = 2.2 ×10 ms 4πε o nh This is less than 1% of c ( 3 × 108 ms −1 ) suggesting that a calculation involving non-relativistic mechanics is valid Emission Spectra Electron makes transition from initial qu. State ni to final state nf . The frequency ν of the photon emitted satisfies: mZ e 1 1 hν = Ei − E f = − 2 2 2 2 (4πε o ) 2h n f ni 2 4 1 Express this in terms of wavelength 1 1 mZ e 1 1 = − λ (4πε o ) 2 4πh 3c n 2f ni2 2 4 1 1 1 = R∞ Z 2 2 − 2 n λ n i f R∞ = 1 me 4 ( 4πε o ) 4πh c 2 3 = 1.0974 ×10 m Is the Rydberg constant This generates various families of spectral lines. 7 −1 Note that in each case there is a series limit corresponding to 1 For the Lyman series, this is at = R∞ = 91nm λ Well within the UV range The Balmer series lies in the near UV and visible region, and the others are all in the infrared. At time of Bohr’s proposal, only Balmer-Paschen series were known, and the series were therefore predicted in advance of the discovery (triumph for Bohr theory). Books Physics of Atoms and Molecules, Bransden and Joachain, ~pp 1-50 (Longman). Quantum Mechanics, Rae, 3rd edition (first chapter) NOTE: THESE ARE ADVANCED TEXTS AND I AM ONLY RECOMMENDING YOU READ FIRST SECTIONS OF EACH BOOK. Absorption spectrum General formula above also applies to the case where an electron gains the energy of a suitable photon having energy hν exactly equal to the difference between initial and final states. Normally, electron will start off in ground state so only the Lyman series is observed in absorption spectrum. Finite Nuclear Mass In the theory above, we assumed that the nucleus is so massive that it is effectively at rest. But the mass of the nucleus is finite and the classical model of the atom therefore envisages that the electron and the nucleus both revolve around their common centre of mass. Therefore, any theory should be modified by replacing the mass of the electron by the “reduced mass” of the system. mM m µ= = m + M 1+ m M 1 1 1 = R.Z 2 2 − 2 n λ n i f R = µe 1 4 ( 4πε o ) 4πh c 2 3 R µ M 1 = = = m R∞ m m + M 1 + M R∞ is the limit of R as M tends to infinity Difference between R and R∞ is small (~ 1 part in 1800) but shows up in precise measurements of spectral lines. For hydrogen R = 1.0968 ×10 m 7 Cf. −1 R∞ = 1.0974 ×10 m 7 −1 Example Positron “atom” =system containing 1 positron (like an e- with a positive charge) and 1 electron. How does emission spectrum differ from hydrogen? Set M=m, so R∞ R∞ R= = m 2 1+ M 1 R∞ 1 1 1 = − = × λ 2 n 2f ni2 2 previous formula All wavelengths are doubled Effect on energy and radii of quantum states? We replace m in the formulae above by m µ= 2 Therefore, all energies are halved in magnitude, radii doubled. Fine Structure This is the splitting of the spectral lines into several components, when measured with equipment of high resolution. Explanation: each energy level actually consists of several distinct states with almost the same energy. When viewed at high resolution, transitions split. Transitions between any two Bohr energy states involve several spectral lines. Theoretical explanation: modification of simple Bohr theory by Wilson and Sommerfeld: electron orbits can be elliptical, of which a circular orbit is a special case. Each orbit is specified by 2 parameters instead of 1. Geometrically by semi-major and semi-minor axes a,b, no just radius r. Thus, energy levels turn out to be dependent on two quantum numbers, but only when one takes relativsitic considerations into account. Without relativity, we get the same formula for E as before. Relativistic correction: electrons in very eccentric orbits have large velocities when they are near the nucleus, so v/c is NOT negligible. Can show 2 2 Z α Z 1 3 − Ry E = − 2 1 + n nθ 4n n 2 Where n is the principal quantum number = 1, 2, 3... nθ is the “azimuthal” quantum number = 1, 2, ….n 2 1 e 1 α= ≈ 4πε o hc 137 Fine structure constant But note that not all possible transitions occur in practice. The “allowed” transitions are shown in the diagram: these are the transitions for which: nθf − nθi = ±1 This is due to the operation of a “selection rule”, which states that transitions can only occur between states for which nθ changes by ± 1 The only Dane ever to win the Nobel Prize in Physics Why does the Bohr theory work? More precisely, why does the Bohr theory generate the correct formula for the energy levels in a hydrogen atom? Answer: because the Bohr condition: mvr = nh inadvertently makes use of the wave associated with an electron. Supposing the classical orbit is circular: we assume that the associated wave is a standing wave following the motion. But in order for the wave to return to its initial value (i.e. we are requiring that the wave be single valued), we require that 2πr = circumfere nce = nλ h p= λ h h λ= = p mv 2πrmv = nh mvr = nh The Bohr condition Inadequacies of the Bohr Theory 1. Does well to describe hydrogen, but can be extended only to 1-electron atoms, i.e. hydrogen-like, with higher Z values. Can treat alkali atoms with some success, but only because they have 1 electron only outside closed shells. Fails to account for spectra of other atoms. 2. Theory does not explain rate at which transitions occur between states, i.e. the relative intensities of spectral lines. 3. The theory is ad hoc and lacks a satisfying basis. Superseded by Quantum Mechanics, initiated by de Broglie (1924) and Schrodinger (1926). In Schrodinger’s theory, the electron is described by a wavefunction Ψ (r ) which is the solution of SchrÖdinger’s equation: d 2Ψ dx 2 + [E − V ( x)] Ψ ( x) = 0 2 2m h In 1D The electron does not move on well-defined orbital paths as in the Bohr theory, one can only speak of the probability density 2 Ψ (x ) of finding the electron in any region of space in the neighbourhood of the nucleus. This is the modern view of the behaviour of electrons in atoms and in similar situations on an atomic scale. The classical view of “particles” subject to Newtonian mechanics and electromagnetic forces has been replaced by a wave picture in which the behaviour of the electron is governed by the behaviour of the associated wave. Note: discrete wavelengths and energies are associated with particles which are confined to a finite region. Analogy with waves on strings. Confinement leads to standing NOT travelling waves. Particle confined to 1D Suppose particle is confined to a region between x=0 and x=l. x=0 x=l If there is no chance of the particle lying outside this region, the associated wave must vanish: nλ l= 2 n = 1, 2,3... p2 h E= and p = 2m λ 2 2 2 2 2 h n n h E= = = 2mλ2 2m 2l 8ml 2 h So, like the hydrogen atom, energy is quantised and the quantum number arises from the confinement of the particle. Energy level diagram n=2 Excited states n=1 Ground state E=0 Note minimum energy E= h2 8ml 2 = zero point energy Even at T=0, particle must have at least this energy Explained by Uncertainty Principle. E =0 ⇒ ∆x ~ l p2 = 0 ⇒ p = 0, ∆p = 0. 2m Contradiction of Heisenberg Uncertainty Principle Examples of Zero Point energy 1. Nucleus can’t contain electrons. [between 1911-1933 structure of the nucleus wasn’t known, since neutron had not been discovered. Nucleus= mixture of protons and electrons? But then each electron would have zero point energy ~20MeV whereas β-decay shows electron energies are all ~1MeV. 2. Liquid He doesn’t solidify at T=0, except at very high pressure. Wavefunctions: for particle in a 1D potential well nπx Ψn ( x) = A sin l Note: this leads to standing waves. As time proceeds, get oscillations in the usual way. Since Ψ 2 gives probability of particle being found, this also varies... Notes: n=0 not relevant (wave doesn’t exist!) Strong analogy with standing wave on a stretched string But analogy not complete since in quantum case: hν = E ⇒ ν n ∝ E n ∝ n 2 string ν n ∝ En ∝ n Example: calculate wavelength of photon emitted when electron in potential well of width 1nm undergoes a transition from n=2 to n=1. n=2 n=1 E=0 En = n 2 h 2 8ml 2 =n (6.6 × 10 2 8 × 9.1× 10 En = n 2 .0.374eV −34 2 −31 ) −9 2 × (10 ) E 2 = 1.496eV E 2 − E1 = 1.122 eV J Wavelength λ given by: hc hν = = 1.122eV ⇒ λ = 1.1µm λ Note variation with l: if we change l from 1nm to 0.5nm, due to factor of l 2 in denominator all energies are x4, so wavelengths reduced by a factor of 4. Quantum dots: practical example of an electron (or several electrons) being confined to a region 10nm thick. Such devices are increasingly important in quantum engineering, I.e. in devices which exploit quantum physics. Quantum Tunneling “Wave associated with a particle” : what equation does it satisfy? Answer: Schrodinger’s equation. No derivation here (too complex). Subject of equation in 1D is Ψ ( x , t ) in 1D 2 Ψ (x ) gives probability that particle will be found at position x at a time t. For a fixed time t, the profile of Ψ is given by: d 2Ψ dx 2 + 2m h 2 [E − V ( x)] Ψ = 0 Where m is the mass, V(x) is the potential energy function, E= total energy Note that: E − V ( x ) = KE E = KE + V (x ) Classically allowed region E − V ( x) ≥ 0 Classically forbidden region E − V ( x) ≤ 0 Nature of the solution Ψ (x ) in classically allowed region Suppose Ψ (x) is positive, then d 2 Ψ ( x ) d dΨ ( x ) = 2 dx dx dx d 2 Ψ( x) dx 2 <0 dΨ ( x ) dx slope Also, dΨ ( x) dx measures rate of change. So if this means that the slope is decreasing d dΨ ( x) <0 dx dx Ψ (x) x The REVERSE holds when Ψ(x) < 0 d 2 Ψ ( x) dx 2 >0 Therefore throughout classically allowed region Ψ (x) must be oscillatory. Ψ (x) in a classically forbidden region Ψ(x) > 0 Exponential increase d 2 Ψ ( x) dx 2 >0 Exponential decrease: this one makes sense Similarly if Ψ(x) < 0 Ψ (x) d 2 Ψ ( x) dx 2 <0 Near a potential barrier Main point is that the particle has a finite probability of being in a region which is inaccessible classically, and of “tunneling” from one classically allowed region to another. Known as “barrier penetration”. W V =0 V =W − D / 2 < x < +D / 2 x ≤ − D / 2, x ≥ + D / 2 Wavefunction oscillatory inside, decays exponentially outside dψ ψ, dx continuous Applications/Illustrations (a) 238 92 U 234 → 90Th + 4 2 He α-particle V V0 Distance from nucleus Cold electron emission Direct a strong electric field towards a metal surface V Potential energy of an electron near the surface of metal Cold electron emission: scanning tunnelling microscope http://eaps4.iap.tuwien.ac.at/www/surface/STM_Gallery/stm_ schematic.html http://eaps4.iap.tuwien.ac.at/www/surface/STM_Gallery/stm_ schematic.html Quantum puzzles and Mysteries 1. Meaning of the Uncertainty Principle. h ∆p.∆x ≥ 2 Extent to which position x is unpredictable Sometimes expressed: impossible to determine simultaneously the position and momentum of a particle with arbritary accuracy (Copenhagen Interpretation). Supported by argument involving Heisenberg microscope. Resolving power of microscope is: λ ∆x ≈ sin θ Recoil momentum of particle is uncertain to an extent: h ∆px ≈ sin θ λ h λ ∆x.∆p x ≈ sin θ . ≈h λ sin θ But is this a “simultaneous measurement”? Alternative viewpoint: (statistical interpretation) ∆x is a standard deviation for an ensemble of particles, i.e. it is a statistical property. 2 Ψ ∝ probabilit y Implies a statistical population To verify experimentally, we would have to measure x for a large number of particles and build up a histogram. Do the same for p x But: no simultaneous measurements! Check ∆x.∆p x ≥ h 2 Einstein-Podolsky-Rosen (EPR) Paradox (thought experiment) Firstly, concept of spin: non-classical property of electrons, protons etc. Nearest classical analogue: spin of the Earth. But this is not a good analogy since the electron is a point particle. Can have: correlated system of 2 electrons,such that total spin component in a certain direction is zero. (in the up-down direction say). So if one electron has a “spin up” the other has “spin-down” Now suppose we prepare such a pair of electrons and separate them. Measure spin of one - say “up” other has spin “down” Therefore, 1st measurement determines (immediately) the result of the second, no matter how far apart the electrons are! But: perhaps this is the same as for a completely classical system - e.g. 2 boxes, one with something in it, the other empty. So make hypothesis: “each electron in the EPR thought experiment has a definite spin state, even before one looks” Determined by Hidden variables Then we find Bell’s theorem hypothesis is correct. predictions if this 1. Measure one particle effects other particle 2. Measure one particle only gives us knowledge of other (true before measurement) 1. Relates to QM, 2. Is classical Hence forced to accept that QM is “non-local”, which is disturbing. 3. Schrodinger’s Cat (Thought Experiment) Logical problem. “The atom at any time has decayed or it has not” “The cat at any time, is either alive or dead” Defies common sense! Can partly evade this problem by emphasising that the wavefunction describes an ensemble of radioactive atoms, accompanied by an ensemble of cats. A certain fraction will be dead and the other fraction alive.