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

2. Davisson-Germer (Interference)

3. 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) p = momentum (p = mv) h = Planck’s constant (6.626 x 10 -34 Js) = de Broglie wavelength

4. 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) Ve = 1 / 2 mv 2 V = accelerating voltage (V) e = elementary charge (1.602x10 -19 C) m = particle mass (kg) v = velocity (m/s)

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

7. Scanning electron microscope

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

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

19. 404 nm 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 What is the de Broglie wavelength of an electron going 1800 m/s? (3)

20. 1.10 x 10 -27 kg m/s p = h/ p = (6.626 x 10 -34 Js)/(600. x 10 -9 m) = 1.10 x 10 -27 kg m/s What is the momentum of a 600. nm photon?

21. 3.428E-10 m p = h/ Ve = 1 / 2 mv 2 Ve = 1 / 2 mv 2, v 2 = 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 Electrons in a microscope are accelerated through 12.8 V. What de Broglie wavelength will they have?

22. 3.25293E-18 V 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 / 2 mv 2, V = 1 / 2 mv 2 /e = 1 / 2 (9.11E-31 kg)(0.001069607 m/s) 2 /(1.602E-19 C) = 3.25293E-18 V 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)

23. 1.00 N 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 10 26 )(1.0687 x 10 -27 kg m/s) = 1.00 N A 300. MW 620. nm laser is putting out 9.36 x 10 26 photons per second. since F =  p/  t, and  t = 1 second, what is the total thrust of the laser? (2)