Reading quiz – get out a sheet of paper and a writing utensil.
In the Davisson Germer experiment, Davisson
and Germer shot a beam of electrons at a lattice
of Nickel atoms and found that the electrons
were only detected at certain angles. Explain the
reason for this result and why it was important.
Review of Bohr and deBroglie
• Background:
– Balmer found equation for Hydrogen spectrum but
didn’t know what it meant.
– Rutherford found that atoms had a nucleus, but didn’t
know why electrons didn’t spiral in.
• Bohr postulates quantized energy levels for no good
reason, and predicts Balmer’s equation.
• deBroglie postulates that electrons are waves, and
predicts Bohr’s quantized energy levels.
• Note: no experimental difference between Bohr
model and deBroglie model, but deBroglie is a lot
more satisfying.
Models of the Atom
• Thomson – Plum Pudding
–
–
–
–
–
– Why? Known that negative charges can be removed from atom.
– Problem: just a random guess
• Rutherford – Solar System
– Why? Scattering showed hard core.
– Problem: electrons should spiral into nucleus in ~10-11 sec.
+
• Bohr – fixed energy levels
– Why? Explains spectral lines.
– Problem: No reason for fixed energy levels
+
• deBroglie – electron standing waves
– Why? Explains fixed energy levels
– Problem: still only works for Hydrogen.
• Schrodinger – will save the day!!
+
–
deBroglie Waves
•
•
•
•
This is a great story.
But is it true?
If so, why no observations of electron waves?
What would you need to see to believe that this
is actually true?
Today: Electron interference!
•Why electron waves are hard to see
•Designing experiment possible to see with early 1900s tech.
•How done by Clinton Davisson and Lester Germer
•Why their technique still used today (LEED)
•How to interpret.
Reminder- what do we mean by interference?
Phet Wave Interference Sim
1
2
Waves (water, sound, …) interfering.
Peaks where add, zilch where cancel (destructive interference)
Two slit interference with light
Question in 1920s
So can we just do same experiment but replace beam of
light with beam of electrons to check deBroglie?
Let’s work through the design to see what expect to see,
what required to do proper experiment.
step 1. Go off and play with making beams of electrons.
Find can make beams of energies between ~25-1000 eV.
V
step 2. Calculate signal would expect to see from
double slit. Typical for light: slits ~0.5 mm apart
step 1. Go off and play with making beams of electrons.
Find can make beams of energies between ~25-1000 eV.
step 2. Calculate signal would expect to see from
double slit. Typical for light: slits ~0.5 mm apart
Can we just repeat light double slit experiment with electrons?
a. yes.
(if so, precisely what would experimental results would you
expect?)
b. no.
(if so, precisely why not?)
Double-slit experiment (see textbook)
Determining the space between bright regions (H)
L
bright
r1
0.5 mm =D
H
r2

bright
r = r2-r1
r = m (where m=1,2,3…)
1
D

D=m
2
r = Dsin()=D
Screen far away so
1~2 ~ & small angle approx. sin=
bright
H= Lsin()=L
H=L
Double-slit experiment
Determining the space between bright regions (H)
L
r1
5 x10-4m = D

H
r2
r = r2-r1
r = m (where m=1,2,3…)
r=m= Dsin()=D =m
Calculating pattern for light
m = 1, = 500 nm, so angle to first bright
= λ/D = 500 x 10-9/(5 x 10-4) = 0.001 rad
if L = 3m, then H= 3 m x 0.001 = 3mm.
So what will pattern look like with electrons?
How figure out?
H= L
D =m
H= L
Steps to predict pattern for debroglie electron wave.
1. find expected wavelength =h/p, h = 6.6 x 10-34 J s
2. plug into = λ/D = (h/p)(1/D), calculate angle (radians)
Best to do experiment with
a. higher energy electron beam
b. lower energy electron beam
c. does not make any difference
b. lower. Smaller energy  smaller momentum  larger λ
means bigger angle, easier to see.
Steps to predict pattern for debroglie electron wave.
1. find expected wavelength =h/p, h = 6.6 x 10-34 J s
2. plug into = λ/D = (h/p)(1/D), calculate angle (radians)
For lowest energy electron beam (E = 25eV),
how big is electron wavelength?
A. λ ~ 1pm (10-12 m)
D. λ ~ 1mm (10-3m)
B. λ ~ 1nm (10-9 m)
E. λ ~ 1m
Compare with
-6
C. λ ~ 1m (10 m)
visible light:
Energy E = ½mv2 = p2/2m
…so p = (2Em)1/2
…so λ = h/p = h/(2Em)1/2
6.6x10-34Js
λ = (2 * 25eV * 1.6x10
-19J/eV * 9.1x10-31kg)1/2
λ = 2.4 x 10-10 m = 0.24 nm ~ 1nm
λ ~ 400-700nm
OR
λ = hc/(2Emc2)1/2
1240 eV nm
λ =(2 * 25eV
* .511 x 106 eV)1/2
1 x 103 nm
1 x 103 nm
λ = (25 x 106)1/2 = 5 x 103
λ = 1/5 nm ~ 1nm
Energy and Momentum for
Massive vs. Massless Particles
Massive Particles (e.g. electrons) Lowest energy e-s
• E = ½mv2 = p2/2m = h2/2mλ2 E = 25eV
p = 2.7x10-24 kg m/s
1/2
• p = (2Em)
v = 3x106 m/s = c/100
• λ = h/p = h/(2Em)1/2
λ = 0.24 nm
Massless Particles (e.g. photons) Typical photons
• E = pc = hc/λ
E = 2.5eV
-27 kg m/s
p
=
1.3x10
• p = E/c
v = c = 3x108 m/s
• λ = h/p = hc/E
λ = 500 nm
deBroglie relationship is universal
Steps to predict pattern for debroglie electron wave.
1. find expected wavelength =h/p, h = 6.6 x 10-34 J s
2. plug into = λ/D = (h/p)(1/D), calculate angle (radians)
lowest energy 25 eV gives λ = 2.4 x 10-10 m
So for slit separation D ~ 0.5mm,
expect  to be a. << 1, b. <1, c. >1, d. >>1.
D
= λ/D = (2.4 x 10-10 m)/(5 x 10-4m) = 4.9 x 10-7 radians!!!
Much too small an angle to see!
(If L = 3m, then H=L = 1.5 x 10-7 m = 150 nm)
Big problem, if electron has wavelength  deBroglie predicted
it is REALLY SMALL ~ 2.4 x 10-10 m. Why not seen.
predict pattern for debroglie electron wave
=h/p, h = 6.6 x 10-34 J s, = λ/D = (h/p)(1/D),
 = 2.4 x 10-10 m
 = λ/D if D = 5 x 10-4 m,  = 4.9 x 10-7 radians
Much too small an angle to see!
What now? Any way to make angle bigger?
a. make D much smaller,
c. make D much bigger,
e. b. and c.
b. make electron energy lower,
d. a. and b.
ans. a) make D smaller
(already said cannot make E smaller,
electron beam no good)
designing experiment to see debroglie electron wave
=h/p, h = 6.6 x 10-34 J s, = λ/D = (h/p)(1/D)
= 2.4 x 10-10 m
= λ/D
to make  easy to see, like ~1rad,
need D = 0.25 nm
Is that a problem?
yes, it is about the same size as one atom!
Would like to have slits separated by about
an atom diameter. Impossible.
Making lemonade out of lemons
Brilliant idea: But two slits are just two sources.
Hard to get two sources
size of atom.
Easy to get two objects
that scatter electrons
that are size of atom!
But two slits are just two sources.
What stuff in nature
is made out of things
the size of atoms
with equal spacing
between them?
hard to get only two atoms next to each other.
But multiple that are same separation just work better.
Just like reflection diffraction grating.
Davisson and Germer -- VERY clean nickel crystal.
Interference is electron scattering off Ni atoms.
e
e
e e
e
e
Ni
e e det. e
e
scatter off atoms
e
e
move detector around,
see what angle electrons coming off
See peak!!
so probability of angle where detect
electron determined by interference
of deBroglie waves!
# e’s
0
e
e
e e
500
scatt. angle 
e e det. e
e
e
Ni
Observe pattern of scattering
electrons off atoms
Looks like ….
Wave!
PhET Sim: Davisson Germer
Careful…
near field view:
D = m doesn’t work here.
For qualitative use only!
http://phet.colorado.edu/simulations/schrodinger/dg.jnlp