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Published byJameson Sly Modified over 2 years ago

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Dilemma Existence of quanta could no longer be questioned e/m radiation exhibits diffraction => wave-like photoelectric & Compton effect => localized packets of energy => particle-like “wave-particle duality”

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Wave properties Photon detector clicks when a photon is absorbed

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Probability Waves Photon detector might be a photoelectric device at any point, the clicks will be randomly spaced in time cannot predict when a photon will be detected at any point on the screen if we move the detector, the click rate increases near an intensity maximum the relative probability that a single photon is detected at a particular point in a specified time Intensity at that point intensity E m 2 Probability of detection E m 2

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Probability Wave Probability (per unit time) that a photon is detected in some small volume is proportional to the square of the amplitude of the wave’s electric field in that region Postulate that light travels not as a stream of photons but as a probability wave photons only manifest themselves when light interacts with matter photons originate in the source that produces the light wave (interaction) photons vanish on the screen (interaction)

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Single-photon version (1909) Light source is so weak that it emits only one photon at a time at random intervals interference fringes still build up raises the question: if the photons move through the apparatus one at a time, through which slit does the photon pass? How does a given photon know that there is another slit? Can a single photon pass through both slits and interfere with itself?

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Electrons and Matter Waves Light is a wave but can transfer energy and momentum to matter in “photon” sized lumps can a particle have the same properties? can it behave as a wave? “a matter wave” de Broglie (1924) suggested =h/p “de Broglie wavelength” a beam of electrons has a wavelength and should diffract if “slit width” comparable to =h/p 1927 Davisson and Germer observed diffraction of electrons from crystals

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Diffraction Recall for a single slit, the diffraction minima are at a sin = m if < no diffraction

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Diffraction eg. Electrons with kinetic energy of 54.0 eV if K=p 2 /2m e, then p=(2m e K) 1/2 = (2 x 9.11x10 -31 kg x 54eV x 1.6x10 -19 J/eV) 1/2 =3.97 x 10 -24 kg. m/s =h/p = (6.63x10 -34 J.s)/(3.97x10 -24 kg.m/s) = 1.67 x 10 -10 m =.167 nm smaller mass => larger typical atom has diameter of 10 -10 m

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Diffraction regular arrays of atoms (crystals) should diffract electrons! Neutrons can also diffract => study structures of solids and liquids

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Double Slit Experiment with electrons (1989)

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X-ray beam (light wave) Electron beam (matter wave)

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Problem Singly charged Na ions are accelerated through a potential difference of 300 V. What is (a) the momentum acquired? (b) what is the de Broglie wavelength? Solution: K = qV = (1.6x10 -19 C)(300V) = 4.80 x10 -17 J = 300 eV m=(22.9898 g/mole)/(6.02x10 23 atom/mole) = 3.819 x10 -23 g = 3.819 x 10 -26 kg

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Solution (a) p = (2mK) 1/2 = [2(3.819x10 -26 )(4.8x10 -17 )] 1/2 = 1.91 x10 -21 kg.m/s (b) =h/p = (6.63x10 -34 J.s)/(1.91x10 -21 kg.m/s) = 3.46 x10 -13 m

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What is Waving? If particles behave as waves, what is waving? Wave on a string => particles in string execute SHM sound wave in air => air molecules oscillate in SHM light wave => electric and magnetic fields oscillate e.g. E(x,y,z,t) electric field varies from place to place and with time intensity |E| 2 what varies from place to place for a matter wave? Wave function (x,y,z,t) “psi”

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Schrödinger Equation Schrödinger Equation 1926H =E (x,t) is a solution of this equation the wave equation for matter waves probability waves probability density is P(x,t)= (x,t) *(x,t) = | (x,t)| 2

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A harmonic wave has a definite value of k but extends to infinity A wave packet has a spread of k- values but is localized k. x 1 . t 1

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Example Radar transmitter emits pulses of electromagnetic radiation which last 0.15 s at a wavelength of = 1.2 cm (a) to what central frequency should the radar receiver be set? (b) what is the length of the wave packet? (c ) how much bandwidth should the receiver have? (a) f 0 = c/ 0 = 3 x 10 8 m/s /1.2 x 10 -2 m = 26 GHz (b) x = c t = (3 x 10 8 m/s)(.15 x 10 -6 s) = 45 m (c ) f = /2 = 1/(2 t) = 1/(6.28 x.15 x 10 -6 s)= 1.1 MHz

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