1 / 23

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)

lindalopez
Download Presentation

Louis de Broglie

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related