Download From Last Time… Wavelength of 1 eV electron Question Can this be

HW8: M Chap 15: Question B, Exercises 2, 6
M Chap 16: Question B, Exercises 1
M Chap 17: Questions C, D
Wavelength of 1 eV electron
From Last Time…
• For an electron,
• Essay topic and paragraph due Friday, Mar. 24
"=
• Light waves are particles
and matter particles are waves!
!
• Matter particles are waves
with wavelength λ=h/p.
Phy107 Lect 23
1240 eV # nm
2 $ 0.511 MeV
1
kinetic energy
• 1 eV electron,
• 10 eV electron
• 100 eV electron
Wed. Mar. 22, 2006
λ=1.23 nm
λ=0.39 nm
λ=0.12 nm
Phy107 Lect 23
2
Can this be correct?
Question
A 10 eV electron has a wavelength of ~ 0.4 nm.
What is the wavelength of a 40 eV electron?
A. 0.2 nm
• If electrons are waves, they should
demonstrate wave-like effects
– e.g. Interference, diffraction
• A 25 eV electron has wavelength 0.25 nm,
similar to atomic spacings in crystals
B. 0.4 nm
C. 0.8 nm
Wavelength =
Wed. Mar. 22, 2006
1
1.23 eV 1/ 2 # nm
=
E kinetic
E kinetic
rest energy
• Electromagnetic radiation (e.g. light)
made up of photon PARTICLES
with energy E=hf and momentum p=E/c=h/ λ
Wed. Mar. 22, 2006
constant
constant
rest energy Kinetic energy
Phy107 Lect 23
3
Wed. Mar. 22, 2006
Phy107 Lect 23
4
!
Wave reflection from crystal
Reflection
from next
plane
Constructive & Destructive
Interference
Reflection from
top plane
• Interference arises when waves change their
‘phase relationship’.
• Can vary phase relationship of two waves by
changing physical location of speaker.
‘1/2 λ phase diff’
‘in-phase’
side view
• Interference of waves reflecting from different
atomic layers in the crystal.
• Difference in path length ~ spacing between atoms
Wed. Mar. 22, 2006
Phy107 Lect 23
5
Constructive
Wed. Mar. 22, 2006
Destructive
Phy107 Lect 23
6
1
X-ray diffraction
Davisson-Germer
experiment
Molecular
structure
• Diffraction spot arrangement indicates
atomic arrangement
• Used to determine atomic
arrangements of complex molecules.
• Diffraction of
electrons from a
nickel single crystal.
Bright spot:
constructive
interference
• Established that
electrons are waves
– e.g. DNA
X-ray
diffraction
pattern
Davisson:
Nobel
Prize 1937
54 eV
electrons
(λ=0.17nm)
Wed. Mar. 22, 2006
Phy107 Lect 23
7
Wed. Mar. 22, 2006
Phy107 Lect 23
8
Analogy with sound
Suppose an electron is a wave…
• Sound wave also has the same characteristics
• But we can often locate sound waves
• Here is a wave:
"=
h
p
– E.g. echoes bounce from walls. Can make a sound pulse
• Example:
x
–
–
–
–
λ
!
…where is the electron?
Hand clap: duration ~ 0.01 seconds
Speed of sound = 340 m/s
Spatial extent of sound pulse = 3.4 meters.
3.4 meter long hand clap travels past you at 340 m/s
– Wave extends infinitely far in +x and -x direction
Wed. Mar. 22, 2006
Phy107 Lect 23
9
Beat frequency: spatial localization
Wed. Mar. 22, 2006
Phy107 Lect 23
10
Making a particle out of waves
• What does a sound ‘particle’ look like?
440 Hz +
439 Hz
– One example is a ‘beat frequency’ between two notes
– Two sound waves of almost same wavelength added.
440 Hz +
439 Hz +
438 Hz
Constructive
interference
Large
amplitude
Wed. Mar. 22, 2006
Destructive
interference
Small
amplitude
Phy107 Lect 23
Constructive
interference
Large
amplitude
11
Wed. Mar. 22, 2006
Phy107 Lect 23
440
439
438
437
436
Hz
Hz
Hz
Hz
Hz
+
+
+
+
12
2
Spatial extent
of localized sound wave
Same occurs for a matter wave
• Construct a localized particle by adding together
waves with slightly different wavelengths.
8
4
• Since de Broglie says λ = h /p, each of these
components has slightly different momentum.
0
– We say that there is some ‘uncertainty’ in the momentum
-4
-8
Δx
-15
-10
-5
0
5
10
• And still don’t know exact location of the particle!
15
J
– Wave still is spread over Δx (‘uncertainty’ in position)
– Can reduce Δx, but at the cost of increasing the spread in
wavelength (giving a spread in momentum).
• Δx = spatial spread of ‘wave packet’
• Spatial extent decreases as the spread in
included wavelengths increases.
Wed. Mar. 22, 2006
Phy107 Lect 23
Interpreting
Wed. Mar. 22, 2006
13
Phy107 Lect 23
14
8
Heisenberg Uncertainty Principle
4
0
• Using
-4
– Δx = position uncertainty
– Δp = momentum uncertainty
-8
-15
-10
-5
0
5
10
15
J
• For sound, we would just say that the sound pulse is
centered at some position, but has a spread.
• Heisenberg showed that the product
• Can’t do that for a quantum-mechanical particle.
( Δx ) • ( Δp ) is always greater than ( h / 4π )
• Many measurements indicate that the electron is
indeed a point particle.
Often write this as
• Interpretation is that the magnitude of electron ‘wavepulse’ at some point in space determines the
probability of finding the electron at that point.
Wed. Mar. 22, 2006
Phy107 Lect 23
Planck’s
constant
15
where h =
!
Wed. Mar. 22, 2006
("x )("p) ~ h /2
h
is pronounced ‘h-bar’
2"
Phy107 Lect 23
16
!
Thinking about uncertainty
Uncertainty principle question
("x )("p) ~ h /2
For a classical particle, p=mv, so an
uncertainty in momentum corresponds to an
uncertainty in velocity. !
("x )("v ) ~ h /2m
This says that the uncertainty is small for massive objects,
but becomes important for very light objects, such as
electrons.
!
Suppose an electron is inside a box 1 nm in width.
There is some uncertainty in the momentum of
the electron. We then squeeze the box to make
it 0.5 nm. What happens to the momentum?
A. Momentum becomes more uncertain
B. Momentum becomes less uncertain
C. Momentum uncertainty unchanged
Large, massive objects don’t show effects of quantum
mechanics.
Wed. Mar. 22, 2006
Phy107 Lect 23
17
Wed. Mar. 22, 2006
Phy107 Lect 23
18
3
Planetary model of atom
Using quantum mechanics
• Positive charge is concentrated in
the center of the atom ( nucleus )
• Quantum mechanics makes astonishingly
accurate predictions of the physical world
• Atom has zero net charge:
• Can apply to atoms, molecules, solids.
– Positive charge in nucleus cancels
negative electron charges.
• An early success was in understanding
– Structure of atoms
– Interaction of electromagnetic radiation with atoms
Wed. Mar. 22, 2006
Phy107 Lect 23
electrons
19
nucleus
• Electrons orbit the nucleus like
planets orbit the sun
• (Attractive) Coulomb force plays
role of gravity
Wed. Mar. 22, 2006
Phy107 Lect 23
20
Difference between atoms
• No net charge to atom
– number of orbiting negative electrons same as
number of positive protons in nucleus
– Different elements have different number of
orbiting electrons
•
•
•
•
•
Hydrogen: 1 electron
Helium:
2 electrons
Copper: 29 electrons
Uranium: 92 electrons!
Organized into periodic table of elements
Wed. Mar. 22, 2006
Phy107 Lect 23
Elements in same
column have similar
chemical properties
21
Wed. Mar. 22, 2006
Phy107 Lect 23
22
Atoms and photons
Planetary model and radiation
• Circular motion of orbiting electrons
causes them to emit electromagnetic radiation
with frequency equal to orbital frequency.
• Experimentally, atoms do emit electromagnetic
radiation, but not just any radiation!
• Same mechanism by which radio waves are emitted
by electrons in a radio transmitting antenna.
• In fact, each atom has its own ‘fingerprint’ of
different light frequencies that it emits.
• In an atom, the emitted electromagnetic wave
carries away energy from the electron.
400 nm
600 nm
500 nm
700 nm
Hydrogen
– Electron predicted to continually lose energy.
– The electron would eventually spiral into the nucleus
– However most atoms are stable!
Mercury
Wavelength (nm)
Wed. Mar. 22, 2006
Phy107 Lect 23
23
Wed. Mar. 22, 2006
Phy107 Lect 23
24
4
Hydrogen emission spectrum
• Hydrogen is simplest atom
n=4
– One electron orbiting around
one proton.
Hydrogen emission
n=3
• The Balmer Series of
emission lines empirically
given by
n = 4, λ = 486.1 nm
• This says hydrogen emits only
photons of a particular wavelength, frequency
• Photon energy = hf,
so this means a particular energy.
• Conservation of energy:
n = 3, λ = 656.3 nm
– Energy carried away by photon is lost by the
orbiting electron.
Hydrogen
Phy107 Lect 23
25
Wed. Mar. 22, 2006
The Bohr hydrogen atom
• Radiation emitted only when
electron jumps from one
stable orbit to another.
26
Energy levels
• Retained ‘planetary’ picture:
one electron orbits around
one proton
• Only certain orbits are stable
Phy107 Lect 23
• Instead of drawing orbits, we can just indicate the energy
an electron would have if it were in that orbit.
Zero energy
Einitial
n=4
n=3
Photon
E3 = "
13.6
eV
32
E2 = "
13.6
eV
22
E1 = "
13.6
eV
12
Efinal
n=2
• Here, the emitted photon has
an energy of
Einitial-Efinal
!
!
Stable orbit #2
n=1
Stable orbit #1
Wed. Mar. 22, 2006
Energy axis
Wed. Mar. 22, 2006
Phy107 Lect 23
27
Wed. Mar. 22, 2006
Phy107 Lect 23
28
!
Emitting and absorbing light
Photon emission question
Zero energy
n=4
n=3
n=2
Photon
emitted
hf=E2-E1
n=1
!
E3 = "
13.6
eV
32
E2 = "
13.6
eV
22
!
n=2
Photon
absorbed
hf=E2-E1
E1 = "
13.6
eV
12
n=1
!
E3 = "
13.6
eV
32
E2 = "
13.6
eV
22
!
E1 = "
13.6
eV
12
A. 10.2 eV
B. 3.4 eV
C. 1.7 eV
Absorbing a photon of correct
energy makes!electron jump to
higher quantum state.
Photon is emitted when
!
electron drops
from one
quantum state to another
Wed. Mar. 22, 2006
n=4
n=3
An electron can jump between the allowed quantum states
(energy levels) in a hydrogen atom. The lowest three
energy levels of an electron in a hydrogen atom are
-13.6 eV, -3.4 eV, -1.5 eV.
These are part of the sequence En = -13.6/n2 eV.
Which of the following photons could be emitted by the
hydrogen atom?
Phy107 Lect 23
29
Wed. Mar. 22, 2006
The energy carried away by the photon must be
given up by the electron. The electron can give
up energy by dropping to a lower energy state.
So possible photon energies correspond to
differences between electron orbital energies.
The 10.2 eV photon is emitted when the electron
jumps from the -3.4 eV state to the -13.6 eV
state, losing 10.2 eV of energy.
Phy107 Lect 23
30
5
Example: the Balmer series
Energy conservation for Bohr atom
• Each orbit has a specific energy
En=-13.6/n2
• Photon emitted when electron
jumps from high energy to low
energy orbit.
Ei – Ef = h f
• Photon absorption induces
electron jump from
low to high energy orbit.
Ef – Ei = h f
• Agrees with experiment!
Wed. Mar. 22, 2006
• All transitions terminate
at the n=2 level
• Each energy level has
energy En=-13.6 / n2 eV
• E.g. n=3 to n=2 transition
– Emitted photon has energy
## 13.6 & # 13.6 &&
E photon = %%" 2 ( " %" 2 (( = 1.89 eV
$$ 3 ' $ 2 ''
– Emitted wavelength
!
Phy107 Lect 23
E photon = hf =
Wed. Mar. 22, 2006
31
hc
hc
1240 eV # nm
, "=
=
= 656 nm
"
E photon
1.89 eV
Phy107 Lect 23
32
!
Spectral Question
Compare the wavelength of a photon produced from
a transition from n=3 to n=1 with that of a photon
produced from a transition n=2 to n=1.
A. λ31 < λ 21
n=3
n=2
B. λ31 = λ 21
C. λ31 > λ 21
E31 > E21
Wed. Mar. 22, 2006
so
λ31 < λ 21
n=1
Phy107 Lect 23
33
6