Exam 2 Info
Tuesday, Nov 4, 7:30p – Duane G1B20 (HERE!)
Bring 1 cheat sheet in your own writing. - pencil, calculator
Topics just what you would expect from course calendar
and homework
E/M Waves to Wavefunctions
Roughly 10 M/C, some short answer
1/3rd long answer essayٛ
Don’t stress:
Remember: Exam gets dropped if lower than final
What to review?
- Chap. 4, 5, and 6 (Everything but 6.10)
- HW, lecture notes, ConcepTests, reading quizzes, reading
This is not particularly about memorizing, but sense making. 1
Need to know why we believe things and how to make inferences from observations
Review of main topics since Exam 1
•
•
•
E/M Waves - Wavelength and Frequency
Electric Fields put forces on charges
Complex exponential plane wave solutions.
•
•
•
Photons
Relationship between intensity, energy, photons, and light you see.
Photoelectric effect:
– Workfunction, KE of electron , current v. V, KE vs. f etc.
Electrostatic potential energy due to voltage.
Arguments for why a photon view is needed to explain observations.
Wave nature of light / Interference - probability ~ |E|2
•
•
•
•
•
•
Discharge lamps:
– Spectra, KE, model Photons emitted only when electrons jump
down in energy
– Conservation of energy:
Spectra to energy levels and energy levels to spectra
Idea of ionization energy for electron in atom … relate to photoelectric
effect
2
•
Balmer series and Bohr model-- what it does and does not explain
•
deBroglie waves- concept of what they are, energy and frequency,
relation between wavelength and momentum.
electron wave (generally matter wave) interference in both double
slit and Davisson and Germer experiment
wave functions as representing probability amplitude (Ψ* Ψ =
prob.density)
wave packets and how idea of wave packet relates to uncertainty
principle
•
•
•
•
Reasoning when you don’t already know the theory. Using large
ideas like Huygen’s Principle and DeBroglie Relations.
ٛ
3
What do we mean by wave-view:
Wave-view is the classical view (before we
knew anything about how E/M waves were
made up of photons):
Electromagnetic
Wave
(Laser light)
E
Travels
Straight
Represented light by:
Oscillating electric and magnetic field traveling as a wave.
The bigger the amplitude of the electric field,
4
the higher the intensity (the brighter the light)
Wave view
Classical view
• no knowledge of photons
• magnitude of E field
changes as total
power changes.
• only delocalized/wave
perspective
Particle view
Photon view
• photons make up E/M wave
• each photon carries discreet
amount of energy (cannot
divide),
• total power tells you how many
photons per second
• delocalized wave when traveling
• localized when interacts/detected
5
Case 1
E
Case 2
E
Classical view
of EM wave
from green laser
pointer. Each
will create green
spot.
If max E field in case 1 is 2 times max E field in case 2,
then power in case 1 is _____ of case 2
number of photons in case 1 is _____ of case 2.
A. 2x, 2x
B. same, unknown
C. 4x, 4x
D. 2x, unknown
E. some other combination.
6
Case 1
E
Case 2
E
Emax
Classically:
Intensity =
power/area ~ Emax2
Emax = max strength
of electric field!
Case 1 has 4 times power.
Ephoton is same in both cases
So need 4 times the no.
of photons/sec in case 1.
If max E field in case 1 is 2 times max E field in case 2,
then power in case 1 is _____ of case 2
number of photons/sec in case 1 is _____ of case 2.
A. 2x, 2x
B. same, unknown
C. 4x, 4x
D. 2x, unknown
E. some other combination.
7
Revisiting the idea of a photon
A photon:
I. is a chunk of light energy
II. never behaves as a wave
III. is always localized as a point particle
Which of these do you agree with:
a. I only
b. I and II
c. I and III
d. II and III
Answer is A: I only
e. all
Still struggling … reconsider your
definition of a “particle”:
A “particle” is non-divisible bit of light
with a specific amount of energy.
When a photon travels it is
delocalized and acts as a wave,
when detected it localizes to one
spot.
8
Photons
Photons travel according to a
wave theory.
Theory predicts 2-slit
interference patterns correctly
BUT, does not predict the
discrete package of energy
each photon carries.
QUANTUM PHOTON
ENERGY: Demonstrated by
the Photoelectric Effect.
9
Metal A
Metal B
Ephot
Energy
Ephot
Inside
metal
Outside
metal
Inside
metal
Outside
metal
In each case, the blue photon ejects the red electron. Consider
the following statements:
I. The work functions are the same in both cases.
II. The KE of the ejected electrons are the same in both cases.
A. I=True, II=True
B. I=True, II=False
C. I=False, II=True
D. I=False, II=False
10
Metal A
Metal B
Ephot
Ephot
Energy
work function Φ
work function Φ
Most loosely
bound e
Inside
metal
Outside
metal
Inside
metal
Outside
metal
In each case, the blue photon ejects the red electron. Consider
the following statements:
I. The work functions are the same in both cases.
II. The KE of the ejected electrons are the same in both cases.
A. I=True, II=True
B. I=True, II=False
C. I=False, II=True
D. I=False, II=False
Work function = amount of energy
needed to kick out most loosely bound
electron out of the metal.
11
Metal A
Metal B
Ephot
Ephot
Energy
work function Φ
work function Φ
Most loosely
bound e
Inside
metal
Outside
metal
Inside
metal
Outside
metal
In each case, the blue photon ejects the red electron. Consider
the following statements:
I. The work functions are the same in both cases.
II. The KE of the ejected electrons are the same in both cases.
A. I=True, II=True
B. I=True, II=False Conservation of energy
C. I=False, II=True KE of ejected = Photon – Energy needed
energy
to extract
12
D. I=False, II=False electron
electron from metal
individual atoms vs. atoms bound in solids
Energy
Highest energy electrons
are in “band” of energy
levels: Lots and lots of
energy levels spaced really
close together
(a continuum of levels)
Inside
metal
Outside
metal
Individual atom:
Specific, discrete energy
levels
Energy
Solid metal: ~1023 atoms per cm3
Quantized electron energies of
atoms appear in discharge spectra
13
2ev
3ev
5ev
0
Photon energy
100 200 300 400 500 600 700 800 nm
What energy levels for electrons are consistent
with this spectrum for “McKaganolium”?
Electron Energy levels:
A
B
10eV
5eV
3eV
2eV
0eV
0eV
C
0eV
D
0eV
-2eV
-3eV
-2eV
-5eV
-5eV
-5eV
-7eV
-8eV
-10eV
E
0eV
-5eV
-7eV
14
-10eV
2ev
3ev
5ev
0
Photon energy
100 200 300 400 500 600 700 800 900 1000 nm
What energy levels for electrons are
consistent with this spectrum for
“McKaganolium”?
Electron Energy levels:
A
B
10eV
5eV
3eV
2eV
0eV
0eV
At top (dashed), electron has
escaped. (So no transitions
from top back down)
C
0eV
D
0eV
-2eV
-3eV
-2eV
-5eV
-5eV
-5eV
-7eV
-8eV
-10eV
E
0eV
-5eV
-7eV
-10eV
15
In discharge lamps, one electron bashes into an atom.
10 V
1
-2 eV
-3 eV
2
3
-6 eV
If atoms fixed at these points in tube,
-10eV
what will you see from each atom?
A. All atoms will emit the same colors.
B. Atom 1 will emit more colors than 2 which will emit more
colors than 3
C. Atom 3 will emit more colors than 2 which will emit more
colors than 1
D. Atom 3 will emit more colors than 2. Atom 1 will emit no
colors.
16
E. Impossible to tell.
IF THIS WERE AN ESSAY QUESTION…..
10 V
ΔV=2V
1
2
-2 eV
-3 eV
3
Free electrons start with 10eV of PE
-6 eV
-10eV
Eelec= qΔV (noting ΔV ~ distance)
Atom 1: free electrons only have ~2eV of KE…
therefore cannot excite electron in atom because
- atoms start with electrons in lowest energy level (-10eV).
- Atoms can only make discrete transitions…
- If there is no -8eV level then can’t store energy into atom..
- No light
Atom 2: As above…
• free electrons only have ~5eV of KE…
• can excite electron in atom up to first level only (-6eV)
•Atoms can release energy as light, by discrete transitions
•In this case, only have -6eV to -10eV
• one color emitted (UV)
17
IF THIS WERE AN ESSAY QUESTION…..
10 V
ΔV=2V
1
2
-2 eV
-3 eV
3
Free electrons start with 10eV of PE
-6 eV
-10eV
Main ideas:
1. KE of free electron at time of
Atom 3: same as above,
collision determines maximum
• free electrons have ~10eV of
possible jump in energy of
KE… but just less than10eV to
electron in atom.
eject electron of atom.
2. KE of free electron determined
• excite electron in atom up to any by the change in voltage over
the path it travels.
level…
3. Free electron loses energy
• can emit 1ev, 3eV, 4eV, 7eV or
when it hits atom in fixed
8eV light
(discrete amounts) and
afterwards has less energy
18
4. Atom releases energy in fixed
amount in light E = hf = hc/λ
Energy levels for H-atom
Consider the following statements:
-1.5 eV
-3.4 eV
-13.6eV
I. The potential energy of the electron
in the ground state is -13.6eV
II. When an electron is excited from
n=1 to n=2, it’s PE increases by
10.2eV.
A. I=true, II=true
B. I=true, II=false
C. I=false, II=true
D. I=false, II=false
19
Consider the following
statements:
r=0.05nm
I. The potential energy of the
r=0.2nm
electron in the ground state
is
-13.6eV.
False
II. When an electron is excited
from n=1 to n=2, it’s PE
increases by 10.2eV.
False
PE=-ke2/r
Total=KE+PE=-ke2/2r
+ -
-1.5 eV
-3.4 eV
-13.6eV
PE=-27.2eV
20
Models for matter
•
•
•
Bohr Model
– Quantized energy levels as ad-hoc postulate
– Only works for Hydrogen or Hydrogen-like atoms (1 electron)
– Correctly predicted exact energy levels and spectral lines, sizes of
atoms
deBroglie Model
– Explains quantized levels as due to standing waves
– Still ad-hoc, only works for Hydrogen or H-like atoms
Schrodinger Model
– Gives exact equation for wave function that describes any matter
– In the case of Hydrogen, predicts the SAME quantized energy levels
as other two models
– Works for EVERYTHING!
21
Understanding and applying the Bohr Model
Understand basis of Bohr’s model, his assumptions,
the limitations and successes of his model
Be able to apply his model to various situations to analyze H
and H-like atoms: their electronic structure, emission
lines, physical size
22
Understanding and applying the Bohr Model
Applying Bohr model to analyze H and H-like atoms: their
electronic structure, emission lines, physical size
CQ: The ground state (n=1) energy of an electron in the H
atom is -13.6eV. If an electron is in the n=2 state, what is its
total energy?
a. -3.4 eV
b. -6.8 eV
c. -13.6 eV d. -27.2 eV e. -54.4 eV
mk 2 e 4 1
Answer is A. Hard way: En = −
2= 2 n 2
Easy way:
1
1
En = − E1 2 = −13.6eV 2
n
n
For hydrogen
= ground state energy of H
= E1 = -13.6eV
For n=2 state,
En = -13.6/4 =-3.4 eV
23
CQ: So electron in n=2 for hydrogen has total energy of 3.4eV. If you wanted to calculate n=2 energy for a hydrogenlike atom with Z protons, what would you need to multiply by?
a. 1/Z2
b. 1/Z
c. 1
d. Z
e. Z2
Answer is E. Energy goes as Z2.
Two factors:
1. Coulomb attraction is Z times larger at all distances
(Z times more protons Æ Z times larger force)
Means, at a given distance PE and KE is Z times
2. Orbital distance for each state, n=1, 2, etc
is closer by (factor of 1/Z)
Means, also closer so PE and KE is another Z times
mk 2 e 4 Z 2
En = −
2= 2 n 2
Z2
Z2
En = − E1 2 = −13.6eV 2
n
n
24
CQ: The radius of H atom in the ground state is a0 = 0.0529 nm,
what is its radius in the n=3 state?
a. a0n2
b. a0n
c. a0
d. a0/n
e. a0/n2
Answer is A.
Finding radius (size) of atoms…
Hard way:
n =
rn = −
Z mke 2
2
2
=a0 = the radius of electron orbit
in n=1 state (gives size of H-atom
in the ground state)
Easy way:
n2
n2
rn = a0 = (0.053nm)
Z
Z
25
DeBroglie Hypothesis
Matter can be described by waves! Wavelength and
frequency are related to momentum and energy:
• λ = h/p
• E=hf
DeBroglie relationship is universal
Energy and momentum are related DIFFERENTLY
for different particles
26
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 = 0.24 nm
• λ = h/p = h/(2Em)1/2
Massless Particles (e.g. photons) Typical photons
• E = pc = hc/λ
E = 2.5eV
p = 1.3x10-27 kg m/s
• p = E/c
v = c = 3x108 m/s
• λ = h/p = hc/E
λ = 500 nm
deBroglie relationship is universal
27
CQ: For an electron in the n=10 state in the H-atom, the
wavelength of the wave function is ….. than the wavelength of
the wave function for the n=1 state?
A. smaller
B. same as
C. larger
Answer is C.
h
h
λ= =
Two methods …
p
2mEK
1. DeBroglie says:
and KE of the electron at n=10
KE = kZe2/2r Æ so at larger r Æ smaller KE
Æ so larger wavelength
2. Condition that must have integral # of wavelengths
in orbit
2πr = nλ
2π (a0 n ) = nλ ⎯
⎯→ λ = 2πa0 n
2
28
Wave Equations:
∂ y 1 ∂ y
= 2 2
2
∂x
v ∂t
Wave equation for violin string:
y represents displacement of string
∂2E 1 ∂2E
= 2 2
2
∂x
c ∂t
Maxwell’s equations for EM waves:
E represents magnitude of E field.
(But intensity of light ~ E2)
2
2
∂Ψ ( x, t )
=∂ Ψ ( x , t )
−
+ V ( x , t ) Ψ ( x , t ) = i=
2
2m∂x
∂t
2
Schrodinger’s Equation:
What does Ψ represent physically?
29
Interpretation of Ψ
• |Ψ|2 tells you probability density – can be directly
measured.
• Ψ tells you probability amplitude – cannot be
directly measured.
• But Schrodinger equation describes Ψ, not |Ψ|2.
• Just like Maxwell’s equations describe E&B
(amplitude), but when you measure intensity of
light, measure |E|2.
• Dynamics, interference, come from amplitude,
not intensity.
30
Great Student Question:
Why can’t we avoid the wavefunction and just
make a theory based on probability? Why
calculate Ψ(r,t) if P(r,t) can do the job?
|Ψ1 |2 + |Ψ2|2+
Ψ1*Ψ2 + Ψ2*Ψ1
→ Interference
|Ψ1 |2 + |Ψ2|2
→ No Interference
31
Interference
Ψ1
Ψ2
Due to superposition of two waves:
• Ψtot = Ψ1 + Ψ2
• |Ψtot |2 = |Ψ1 + Ψ2|2 = |Ψ1 |2 + |Ψ2|2 + Ψ1*Ψ2 + Ψ2*Ψ1
Interference Terms:
negative → destructive interference
32
positive → constructive interference
33