De Broglie wavelengths Contents: de Broglie wavelengths Example 1 Whiteboards Transmission electron microscopes Scanning electron microscopes Scanning tunneling electron microscopes

Louis de Broglie Light is acting as both particle and wave Matter perhaps does also E = hf = hc/ E = mc2 mc2 = hc/ mc = p = h/ p = momentum (p = mv) h = Planck’s constant (6.626 x 10-34 Js)  = de Broglie wavelength TOC

Davisson-Germer (Interference)

h = Planck’s constant (6.626 x 10-34 Js)  = de Broglie wavelength 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 .50 kg ball going 40. m/s? p = mv = (.50 kg)(40. m/s) = 20. kg m/s p = h/,  = h/p = (6.626 x 10-34 Js)/(20. kg m/s) = 3.31 x 10-35 m Golly - nothing is that small (atoms are 10-10 m) How would you observe the wave behaviour of that? TOC

p = h/ = (6.626E-34 Js)/(1.0E-9 m) = 6.626E-25 kg m/s p = momentum (p = mv) h = Planck’s constant (6.626 x 10-34 Js)  = de Broglie wavelength Ve = 1/2mv2 V = accelerating voltage (V) e = elementary charge (1.602x10-19 C) m = particle mass (kg) v = velocity (m/s) Example 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/s p = mv, v = p/m = (6.626E-25 kg m/s)/(9.11E-31 kg) = 727332.6 m/s Ve = 1/2mv2, V = 1/2mv2/e = 1/2(9.11E-31 kg)(727332.6 m/s)2/(1.602E-19 C) = 1.504 V

Whiteboards: de Broglie Wavelength 1 | 2 | 3 | 4 | 5 TOC

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 W 404 nm

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

Electrons in a microscope are accelerated through 12. 8 V 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 W 3.428E-10 m

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 W 3.25293E-18 V

A 300. MW 620. nm laser is putting out 9. 36 x 1026 photons per second 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 W 1.00 N

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

Scanning electron microscope

The fossilized shell of a microscopic ocean animal is magnified 392 times its actual size. The ancient creature, called Radiolarian, lived in the waters off Antarctica and is now used to study such things as climate and ocean circulation.

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

Wave behaviour Particle behaviour 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)

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