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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.
• 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!!
Models of the Atom–
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+ –
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?
Waves (water, sound, …) interfering.Peaks where add, zilchzilch where cancel (destructive interference)
Phet Wave Interference Sim
1
2
Two slit interference with light
Question in 1920sSo 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
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?)
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
Double-slit experiment (see textbook)Determining the space between bright regions (H)
0.5 mm =D
r1
r2
H
L
Screen far away so1~2 ~ & small angle approx. sin=
D
r = Dsin()=DH= Lsin()=L
H=L
bright
bright
bright
r = r2-r1
r = m (where m=1,2,3…)
D=m
Double-slit experiment Determining the space between bright regions (H)
5 x10-4m = D
r1
r2
r = r2-r1
r = m (where m=1,2,3…)
r=m= Dsin()=D =m
H
L
H= L m = 1, = 500 nm, so angle to first bright= λ/D = 500 x 10-9/(5 x 10-4) = 0.001 radif L = 3m, then H= 3 m x 0.001 = 3mm.
Calculating pattern for light
So what will pattern look like with electrons? How figure out?
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 beamb. lower energy electron beamc. 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.
D =m H= L
OR
λ = hc/(2Emc2)1/2
λ =
λ = =
λ = 1/5 nm ~ 1nm
Energy E = ½mv2 = p2/2m
…so p = (2Em)1/2
…so λ = h/p = h/(2Em)1/2
λ =
λ = 2.4 x 10-10 m = 0.24 nm ~ 1nm
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)
Steps to predict pattern for debroglie electron wave.
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 ~ 1mC. λ ~ 1m (10-6 m)
Compare with visible light: λ ~ 400-700nm
6.6x10-34Js (2 * 25eV * 1.6x10-19J/eV * 9.1x10-31kg)1/2
1240 eV nm (2 * 25eV * .511 x 106 eV)1/2
1 x 103 nm (25 x 106)1/2
1 x 103 nm 5 x 103
Energy and Momentum for Massive vs. Massless Particles
Massive Particles (e.g. electrons)• E = ½mv2 = p2/2m = h2/2mλ2
• p = (2Em)1/2
• λ = h/p = h/(2Em)1/2
Massless Particles (e.g. photons)• E = pc = hc/λ• p = E/c• λ = h/p = hc/E
Lowest energy e-s
E = 25eVp = 2.7x10-24 kg m/sv = 3x106 m/s = c/100
λ = 0.24 nm
Typical photons
E = 2.5eVp = 1.3x10-27 kg m/sv = c = 3x108 m/s λ = 500 nm
deBroglie relationship is universal
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 mSo for slit separation D ~ 0.5mm, expect to be a. << 1, b. <1, c. >1, d. >>1.
= λ/D = (2.4 x 10-10 m)/(5 x 10-4m) = 4.9 x 10-7 radians!!!D
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.
Steps to predict pattern for debroglie electron wave.
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, b. make electron energy lower, c. make D much bigger, d. a. and b. e. b. and c.
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 aboutan atom diameter. Impossible.
Making lemonade out of lemons
Brilliant idea: But two slits are just two sources.
Hard to get two sourcessize of atom. Easy to get two objectsthat scatter electronsthat are size of atom!
But two slits are just two sources.
hard to get only two atoms next to each other.But multiple that are same separation just work better.Just like reflection diffraction grating.
What stuff in nature is made out of things the size of atoms with equal spacing between them?
Davisson and Germer -- VERY clean nickel crystal. Interference is electron scattering off Ni atoms.
ee
ee
e
e
e ee
e e
scatter off atoms
e det.
move detector around,see what angle electrons coming offNi
ee
ee ee
e
e
e det.
Ni
Observe pattern of scattering electrons off atomsLooks like …. Wave!
# e’s
scatt. angle 5000
See peak!!
so probability of angle where detectelectron determined by interferenceof deBroglie waves!
PhET Sim: Davisson Germer
http://phet.colorado.edu/simulations/schrodinger/dg.jnlp
Careful… near field view:
D = m doesn’t work here.For qualitative use only!
