De Broglie wavelengths

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

Louis de Broglie

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Light is acting as both particle and waveMatter perhaps does alsoE = hf = hc/E = mc2

mc2 = hc/mc = p = h/

hp

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

Davisson-Germer(Interference)

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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/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?

hp

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

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?

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 V

Ve = 1/2mv2

•V = accelerating voltage (V)•e = elementary charge

(1.602x10-19 C)•m = particle mass (kg)•v = velocity (m/s)

Whiteboards: de Broglie Wavelength

1 | 2 | 3 | 4 | 5

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404 nm W

m = 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 nm

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

1.10 x 10-27 kg m/s W

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?

3.428E-10 m W

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/sp = 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?

3.25293E-18 V W

p = 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 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)

1.00 N W

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/stotal p change = (9.36 x 1026 )(1.0687 x 10-27 kg m/s) = 1.00 N

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)

Applications of “matter” waves

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Electron Microscopes•Image resolution = h/p•Electric or magnetic lenses•Gertrude 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 behaviour

Wave behaviour

Complementarity/DualityWave XOR Particle behaviour explains the behavior.Behaviour depends on situation. (Other particle interactions)