Wave Properties of Particles Louis debroglie:

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1 Wave Properties of Particles Louis debroglie: If light is both a wave and a particle, why not electrons? In 194 Louis de Broglie suggested in his doctoral dissertation that there is a wave connected with the moving electron and that the electron wave had a wavelength of λ = h p.

2 For a photon: p E c hf c h = = = λ λ = h p E = hf f E = h Wave properties for an electron (or any particle): h h λ = = called the DeBroglie wavelength p mv γ and. f = E h

3 If electrons are waves, orbits are only stable if they result in constructive interference. Since they can travel either direction, they must produce standing waves. π r = nλ = n mvr= nh/ e h mv π e

4 Clint Davisson and Lester Germer had been investigating the scattering of electrons from crystal surfaces at Bell Telephone Laboratories since 1917 and getting perplexing results. At a meeting in England in 196, Davisson had conversations with other scientists which led him to realize that the preidcted wave properties of electrons could be responsible. He also realized that their experiments could be used to test DeBroglie s hypothesis.

5 Davisson-Germer Experiment They showed that interference occurs when electrons scatter from a crystal surface. The nickel crystal acts like a two-dimensional diffraction grating. The surface atoms act as an array of sources since scattering occurs only where a nickel atom sits. By varying the electron energy, they change the electron wavelength. λ = h h = p mv λ = h me λ = K 150V V 0 A o

6 When the spacing of atoms on the surface is D, constructive interference will occur at angles given by n λ = Dsin φ D

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9 Low Energy Electron Diffraction (LEED) from a Hydrogenated Diamond C(100) Surface

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12 Linear Wave Equation : Wave Packets y x = 1 v y t For waves on a string or water waves, y is the displacement of the medium from the equilibrium position. For electromagnetic waves, y is E or B. Question: For an electron, what is oscillating? Answer: Probability!

13 One solution to the wave equation is a harmonic wave: π y = y0cos x t y0cos kx t π = ω λ T ( ) where k is the wave number and is the angular frequency. ω This solution is periodic and infinite in extent. The wave speed or phase velocity is λ v = p f T = λ = ω k Most often waves occur as pulses, short in time and limited in extent, they contain a mixture of harmonic waves at different wavelengths and frequencies. This group of waves makes up a wave packet. A wave packet can be resolved into harmonic waves using Fourier analysis.

14 v g = ω k

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17 v v p g = fλ = = d ω dk ω k

18 Linear Wave Equation : Wave Functions y x 1 v The function y(x,t) that satisfies this equation is the wave function. = y t For electromagnetic waves, y is, the electric field. The energy density at any position and time is U 1 = ε 0. Since the energy is actually in packets of energy hf, the energy density is proportional to particle density. So, the probability of finding a photon in a particular place is proportional to the energy density which is proportional to the electric field (wave function) squared.

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20 Wave Functions The wave function squared tells us the probability of observing a particle (photon or electron) at a place and time. In general, P( x) dx = Ψ dx where Ψ =ΨΨ and Ψ is the complex conjugate of Ψ. To get the complex conjugate of a function, you replace i with -i everywhere it appears.

21 The Uncertainty Principle

22 For all wave packets x k 1 x p 1 h ω t 1 E t 1 h x p h E t h If we define the uncertainties as equal to the standard deviation: x p h / E t h / Equal sign holds for Gaussian wave packets only.

23 Determining the position and momentum of an electron by scattering a photon off of it This is a Compton scattering and momentum must be conserved. So the x-component of the momentum of the photon must be equal to minus the change in the x-component of the electron s momentum.

24 We can observe a photon after scattering if it reaches our lens. θ θ So the uncertainty in the momentum of the electron will be p x psinθ = h sin θ λ The uncertainty in the position of the electron is determined by the diffraction limit for resolution. λ x= sinθ x px λ sinθ h sin θ = λ h

25 A particle confined in a box of length L The particle can travel back and forth but can t leave the box. x = L So the uncertainty principle tells us that p h x = h L We ll take the standard deviation of the momentum as our measure of its uncertainty. ( ) ( ) p p p ( p pp p ) ( p ) ( pp) ( p ) = = + = + + avg ( ) ( ) p = p p p + p = p p + p = p p = p avg So the square of the average momentum is p = ( p) h L And the average kinetic energy must be p E K = m h ml avg avg avg This is called the zero-point energy. avg

26 Similarly, for an electron confined to an orbit around an atom And the average kinetic energy must be p E = ( p) h r K = p m The total energy for the hydrogen atom would be h mr p ke h E = EK + EP = m r mr ke r The minimum value of the energy is found by setting de/dr equal to zero. de h ke = 3 + = 0 dr mr r h r ke m a min = = 0 4 k e m E min = = 13. 6eV h

27 Spectral Widths The decay of any excited state to a lower energy state can be described by a lifetime, J. The uncertainty principle tells us that this lifetime leads to an uncertainty in the energy of the excited state, )E, which is named the natural line width of the state, ' 0. E t E Γ t 0 τ So if this excited state with energy E decays to a lower state of energy E 0 and emits a photon in the process, there will also be an uncertainty in the wavelength of the emitted light. hc E E0 = hf = λ λ E hc λ τ E E E λ 0 λ λ λ πcτ

28 A Z 0 particle, the most massive fundamental particle known, has a rest energy of 91. GeV. Measurements of its rest energy show a natural line width of.5 GeV. What is the lifetime of the Z 0 particle? The lifetime of the lowest electronic excited state in polyene is 1.9 ps. This excitation is performed using light at a wavelength of 610 nm. What is the energy spacing between the ground and excited states? What is the uncertainty in the energy of the excited state? What spread of wavelengths would you expect for light emitted during the decay of the exited state?