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Louis de Broglie

Louis de Broglie. p = momentum (p = mv) h = Planck’s constant (6.626 x 10 -34 Js)  = de Broglie wavelength. Light is acting as both particle and wave Matter perhaps does also. Davisson-Germer (Interference). p = momentum (p = mv) h = Planck’s constant (6.626 x 10 -34 Js)

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Louis de Broglie

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  1. Louis de Broglie • p = momentum (p = mv) • h = Planck’s constant (6.626 x 10-34 Js) •  = de Broglie wavelength Light is acting as both particle and wave Matter perhaps does also

  2. Davisson-Germer (Interference)

  3. p = momentum (p = mv) • h = Planck’s constant (6.626 x 10-34 Js) •  = de Broglie wavelength Example 1: What is the de Broglie wavelength of a .145 g ball going 40. m/s? (1.1 x 10-34 m) (too small)

  4. Ve = 1/2mv2 • V = accelerating voltage (V) • e = elementary charge • (1.602x10-19 C) • m = particle mass (kg) • v = velocity (m/s) • p = h/ • p = momentum (p = mv) • h = Planck’s constant • (6.626 x 10-34 Js) •  = de Broglie wavelength Example 2: Through what potential must you accelerate an electron so that it has a wavelength of 1.0 nm? (1.504 V)

  5. Applications of “matter” waves • Electron Microscopes • Image resolution  •  = h/p • Electric or magnetic lenses • Gertrude Rempfer (PSU)

  6. Scanning electron microscope

  7. Scanning tunneling electron microscope Actually can image atoms and molecules (Novellus)

  8. Soooo – Is/are Light/electrons a wave or particle????? Wave behaviour Particle behaviour Complementarity/Duality Wave XOR Particle behaviour explains the behavior. Behaviour depends on situation. (Other particle interactions)

  9. What is the de Broglie wavelength of an electron going 1800 m/s? (3) m = 9.11 x 10-31 kg p = mv p = 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 nm 404 nm

  10. What is the momentum of a 600. nm photon? p = h/ p = (6.626 x 10-34 Js)/(600. x 10-9 m) = 1.10 x 10-27 kg m/s 1.10 x 10-27 kg m/s

  11. Electrons in a microscope are accelerated through 12.8 V. What de Broglie wavelength will they have? p = h/ Ve = 1/2mv2 Ve = 1/2mv2, v2 = 2Ve/m, v = √(2(12.8 V)(1.602E-19 C)/(9.11E-31 kg)) = 2121739.443 m/s p = h/,  = h/p = h/mv = (6.626E-34 Js)/((9.11E-31 kg)(2121739.443 m/s)) = 3.428E-10 m 3.428E-10 m

  12. You 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) p = h/ = (6.626E-34 Js)/(.68 m) = 9.74412E-34 kg m/s p = mv, v = p/m = (6.626E-25 kg m/s)/(9.11E-31 kg) = 0.001069607 m/s Ve = 1/2mv2, V = 1/2mv2/e = 1/2(9.11E-31 kg)(0.001069607 m/s)2/(1.602E-19 C) = 3.25293E-18 V 3.25293E-18 V

  13. A 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) for 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/s total p change = (9.36 x 1026 )(1.0687 x 10-27 kg m/s) = 1.00 N 1.00 N

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