•
In the two-slit experiment, we cannot say which slit the photon
goes through - it goes through both as a wave
•
In Bragg diffraction, we could not tell which atom a photon
scattered from; we will soon see this is the case for electrons as
well.
•
If we measure a particle initially at position a and next at
position b, we generally cannot tell the path it took from a to b.
Thursday, March 14, 2013

Wave Particle Duality
1924: Louis-Victor de Broglie proposes that, since light acts as
both a particle and a wave, "particles" also act as both
particles and waves. We assume non-relativistic particles...
Photons
Electrons, etc.
E = hf
E = hf
p = h/
λ
=
ħ
k
p = mv = h/
λ
=
ħ
k
p = E/c
p = (2mK)
1/2
Thursday, March 14, 2013

Photons vs. "Particles"
We have the same relations between momentum p
and wavelength
λ
: p = h/
λ
.
We will focus on this
relation, since we will study many phenomena, such as
interference, that are directly connected to the
wavelength.
The uncertainty relations we discussed for photons -
Δ
E
Δ
t
≥
ħ
/2,
Δ
x
Δ
p
x
≥
ħ
/2, etc. also apply to particles.
But we have different relations between energy E and
momentum p since the photon is massless: E = pc vs.
E
2
= (pc)
2
+ (mc
2
)
2
(relativistically), or K = p
2
/2m (non-
relativistically).
Thursday, March 14, 2013

Experimental Confirmation
The wave nature of electrons was demonstrated within a few
years of its prediction by the Davisson-Germer experiment,
which measured diffraction of electrons by a Ni crystal, the
same phenomena as we discussed in Bragg diffraction of X-rays
by a crystal.
Thursday, March 14, 2013

The Data and Picture
The meaning of the intensity distribution
shown to the left is unclear, since we don't
know what the distribution would look like for
a disordered sample of Nickel - it gives a
smooth falloff with angle.
When the sample was heated to nearly
melting, and cooled, forming a better crystal,
a maximum appears for an angle near 50
o
.
From the figure: m
λ
= d sin
θ
. For m = 1,
λ
/d =
0.77.
Thursday, March 14, 2013

More Data
Top: diffraction pattern of 71-pm
X-rays from aluminum foil.
Bottom: diffraction pattern of 600
eV electrons from aluminum.
The scale of the two is different
- the electrons do not have 71 pm
wavelength.
What is the electron wavelength?
K = 600 eV
➮
p = (2mK)
1/2
= (2 x 9.11x10
-31
kg x 600 eV x 1.6x10
-19
J/
eV)
1/2
= 1.32x10
-23
kg m/s.
λ
= h/p = 6.626x10
-34
Js / (1.32x10
-23
kg m/s) = 5.02x10
-11
m = 50.2 pm.
Note: atomic separations typically a few hundred pm.
Thursday, March 14, 2013

iClicker
If m
λ
= d sin
θ
for the maxima,
shouldn't we get maxima evenly
spaced in sin
θ
? (Spread further
apart in angle with increasing
angle towards 90
o
.) Why doesn't
this happen? Why are there a
couple lines close together?
a) They
are
spread out even in sin
θ
.
b) There are two electron wavelengths, one
from the kinetic energy K and the other from
the total energy E.
c) There are two electron wavelengths, one
from
λ
= h/p and one from
λ
= v/f = v/(K/h).

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- Atom, Electron, Radiation, Fundamental physics concepts, Thursday, blackbody radiation