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AP Physics Chp 29
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Wave-Particle Theory (Duality) Interference patterns by electrons
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Blackbody E=nhf n=0,1,2,…. Quantized h= 6.626 x 10 -34 Js
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Photons E=hf “packet” of energy Photoelectric Effect
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The minimum amount of energy needed to remove an electron from a metal surface by “light” is the work function W o hf = KE max + W o
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Photons only have KE and no rest mass Use of photoelectric effect Dig Cameras and Safety Devices
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Momentum of photons? Compton Effect X-ray scatters electrons and reflects itself
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hf incident photon = hf’ photon + KE electron Momentum of photon p= h/λ Wavelenth Shift λ’ – λ = (h/mc) ( 1-cosθ) λ’ is the scattered photon
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The work function for Gold is 3.9 eV ( 1ev=1.6x10 -19 J ) What is the minimum frequency of “light” that will eject an electron?
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Wo = 3.9eV x 1.6x10 -19 J/eV =6.24x10 -19 J hf = KE + Wo KE = 0 for the minumum hf = Wo f = Wo / h = 6.24x10 -19 J / 6.626 x 10 -34 Js f = 9.4x10 14 1/s
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de Broglie Wavelength λ= h/p λ= h/mv
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Heisenberg Uncertainty Principle (∆p)(∆y) ≥ h/(4π) (∆E)(∆t) ≥ h/(4π)
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What is the wavelength of a baseball (0.145 Kg) thrown at 44.7 m/s? What about an electron traveling 0.8c?
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λ= h/mv = 6.626x10 -34 Js / (0.145Kg)(44.7m/s) = 1.02x10 -34 m
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λ= h/mv = 6.626x10 -34 Js / (9.11x01 -31 Kg)(0.8c) = 3.05x10 -12 m
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