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
h

•p = momentum (p = mv)
•h = Planck’s constant (6.626 x 10-34 Js)
• = de Broglie wavelength
TOC
Davisson-Germer
(Interference)
p
h

•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/
•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/ (9.11E-31 kg)(727332.6 m/s)2/(1.602E-19 C) = 1.504 V
2
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
404 nm
W
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
W
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
W
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/ (9.11E-31 kg)(0.001069607 m/s)2/(1.602E-19 C)
2
= 3.25293E-18 V
3.25293E-18 V
W
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
W
Applications of “matter” waves
Electron Microscopes
•Image resolution  
• = h/p
•Electric or magnetic
lenses
•Gertrude Rempfer
(PSU)
TOC
Scanning electron
microscope
Scanning tunneling
electron microscope
Actually can image
atoms and molecules
(Novellus)
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)