de broglie wavelengths contents: de broglie wavelengths example 1 whiteboards
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DESCRIPTIONDe Broglie wavelengths Contents: de Broglie wavelengths Example 1 Whiteboards Transmission electron microscopes Scanning electron microscopes Scanning tunneling electron microscopes. Louis de Broglie. p = momentum (p = mv) h = Planck’s constant (6.626 x 10 -34 Js) - PowerPoint PPT Presentation
De Broglie wavelengths
Contents:de Broglie wavelengthsExample 1WhiteboardsTransmission electron microscopesScanning electron microscopesScanning tunneling electron microscopes
Louis de BroglieTOCLight is acting as both particle and waveMatter perhaps does alsoE = hf = hc/E = mc2mc2 = hc/mc = p = h/
TOCExample 1: What is the de Broglie wavelength of a .50 kg ball going 40. m/s?
p = mv = (.50 kg)(40. m/s) = 20. kg m/sp = h/, = h/p = (6.626 x 10-34 Js)/(20. kg m/s)= 3.31 x 10-35 mGolly - nothing is that small (atoms are 10-10 m)How would you observe the wave behaviour of that?
p = h/p = momentum (p = mv)h = Plancks constant (6.626 x 10-34 Js) = de Broglie wavelengthExample 2: Through what potential must you accelerate an electron so that it has a wavelength of 1.0 nm?
p = h/ = (6.626E-34 Js)/(1.0E-9 m) = 6.626E-25 kg m/sp = mv, v = p/m = (6.626E-25 kg m/s)/(9.11E-31 kg) = 727332.6 m/sVe = 1/2mv2, V = 1/2mv2/e = 1/2(9.11E-31 kg)(727332.6 m/s)2/(1.602E-19 C) = 1.504 VVe = 1/2mv2V = accelerating voltage (V)e = elementary charge (1.602x10-19 C)m = particle mass (kg)v = velocity (m/s)
Whiteboards: de Broglie Wavelength1 | 2 | 3 | 4 | 5TOC
404 nmWm = 9.11 x 10-31 kgp = mvp = h/p = (9.11 x 10-31 kg)(1800 m/s) = 1.6398 x 10-27 kg m/s = h/p = (6.626 x 10-34 Js)/(1.6398 x 10-27 kg m/s)= 404 nmWhat is the de Broglie wavelength of an electron going 1800 m/s? (3)
1.10 x 10-27 kg m/sWp = h/p = (6.626 x 10-34 Js)/(600. x 10-9 m) = 1.10 x 10-27 kg m/sWhat is the momentum of a 600. nm photon?
3.428E-10 mWp = h/ Ve = 1/2mv2Ve = 1/2mv2, v2 = 2Ve/m, v = (2(12.8 V)(1.602E-19 C)/(9.11E-31 kg)) = 2121739.443 m/sp = h/, = h/p = h/mv = (6.626E-34 Js)/((9.11E-31 kg)(2121739.443 m/s)) = 3.428E-10 mElectrons in a microscope are accelerated through 12.8 V. What de Broglie wavelength will they have?
3.25293E-18 VWp = h/ = (6.626E-34 Js)/(.68 m) = 9.74412E-34 kg m/sp = mv, v = p/m = (6.626E-25 kg m/s)/(9.11E-31 kg) = 0.001069607 m/sVe = 1/2mv2, V = 1/2mv2/e = 1/2(9.11E-31 kg)(0.001069607 m/s)2/(1.602E-19 C) = 3.25293E-18 VYou want to use an electron to have the same wavelength as the waves on a violin string. Through what potential do you accelerate them. (violin strings are about 34 cm long, so the fundamental is about .68 m long)
1.00 NWfor 1 photon: p = h/for 5 photons: 5p = 5h/p = (6.626 x 10-34 Js)/(620. x 10-9 m) = 1.0687 x 10-27 kg m/stotal p change = (9.36 x 1026 )(1.0687 x 10-27 kg m/s) = 1.00 NA 300. MW 620. nm laser is putting out 9.36 x 1026 photons per second. since F = p/t, and t = 1 second, what is the total thrust of the laser? (2)
Applications of matter wavesTOCElectron MicroscopesImage resolution = h/pElectric or magnetic lensesGertrude Rempfer (PSU)
Scanning electron microscope
Scanning tunneling electron microscopeActually can image atoms and molecules(Novellus)
Soooo Is/are Light/electrons a wave or particle?????Particle behaviourWave behaviourComplementarity/DualityWave XOR Particle behaviour explains the behavior.Behaviour depends on situation. (Other particle interactions)
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