M6-ch-3 … · The Newton’s 2 nd Law force of attraction on the electron by the nucleus is: 2 2 2...
Transcript of M6-ch-3 … · The Newton’s 2 nd Law force of attraction on the electron by the nucleus is: 2 2 2...
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3-1 Atoms and Radiation in Equilibrium3-2 Thermal Radiation Spectrum3-3 Quantization of Electromagnetic
Radiation3-4 Atomic Spectra and the Bohr Model
CHAPTER 3Planck’s ConstantPlanck’s Constant
We have no right to assume that any physical laws exist, or if they have
existed up until now, or that they will continue to exist in a similar
manner in the future.
An important scientific innovation rarely makes its way by gradually
winning over and converting its opponents. What does happen is that
the opponents gradually die out.
- Max Planck
Max Karl Ernst Ludwig Planck (1858-1947)
Blackbody Radiation
When matter is heated, it emits
radiation.
A blackbody is a cavity with a material that only emits thermal
radiation. Incoming radiation is absorbed in the cavity.
Blackbody radiation is theoretically interesting because the
radiation properties of the blackbody are independent of the particular material. Physicists can study the properties of intensity versus wavelength at fixed temperatures.
Why is Black Body Radiation important?
Rayleigh-Jeans Formula
Lord Rayleigh used the classical
theories of electromagnetism and thermodynamics to show that the blackbody spectral distribution
should be:
4( ) 8u kTλ π λ−=
It approaches the data at longer wavelengths, but it deviates badly at
short wavelengths. This problem for small wavelengths became known as the ultraviolet catastrophe and was one of the outstanding exceptions that classical physics could not explain.
4
0
( ) 8
Integrating ( ) from 0 , we find
( ) 0 when 0
u kT
u
u d
λ π λ
λ
λ λ λ
−
∞
=
→ ∞
→ →∫
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Planck made two modifications to the classical theory:
The oscillators (of electromagnetic origin) can only have certain
discrete energies, En = nhf, where n is an integer, ν is the frequency,
and h is called Planck’s constant: h = 6.6261 × 10−34 J·s.
The oscillators can absorb or emit energy in discrete multiples of the
fundamental quantum of energy given by:
Planck’s radiation law
Planck’s Radiation Law
Planck assumed that the radiation in the cavity was emitted (and
absorbed) by some sort of “oscillators.” He used Boltzman’s statistical methods to arrive at the following formula that fit the blackbody radiation data.
58( )
1hc kT
hcu
eλ
π λλ
−
=−
E hf∆ =
Application: The Big Bang theory
predicts black body radiation. This radiation was discovered in 1965 by A. Penzias & R. Wilson. Cosmic
Background Explorer (COBE) and Wilkinson Microwave Anisotropy Probe (WMAP) detected this
radiation field at 2.725± 0.001 °K
Planck’s Radiation Law
This data supports the Big
Bang Theory
Exercise 3-1: Derive Planck’s radiation law.
Stefan-Boltzmann Law
The total power radiated increases with the temperature:
This is known as the Stefan-Boltzmann law, with the constant σexperimentally measured to be 5.6705 × 10−8 W / (m2 · K4).
The emissivity є (є = 1 for an idealized blackbody) is simply the ratio of the emissive power of an object to that of an ideal blackbody and is always less than 1.
4R Tεσ=
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Wien’s Displacement Law
The spectral intensity R(λ,Τ ) is the total power radiated per unit area per unit wavelength at a given temperature.
Wien’s displacement law: The maximum of the spectrum shifts to smaller wavelengths as the temperature is increased.
-3constant=2.898 10 .mT m Kλ = ×
Planck’s Radiation LawExercise 3-2 : Show that Stefan’s Boltzmann law, Wein’s
displacement law and Rayleigh-Jean’s law can be derived from Planck’s law
Exercise 3-3: What is the average energy of an oscillator that has a
frequency given by hf=kT according to Planck’s calculations?
Exercise 3-4: How Hot is a Star? Measurement of the wavelength
at which spectral distribution R(λ) from the Sun is maximum is found to be at 500nm, how hot is the surface of the Sun?
Exercise 3-5: How Big is a Star? Measurement of the wavelength at which spectral distribution R(λ) from a certain star is maximum indicates that the star’s surface temperature is 3000K. If the star is
also found to radiate 100 times the power Psun radiated by the Sun, how big is the star? Take the Sun’s surface temperature as 5800 K.
What is a Photon?
Planck introduced the idea of a photon or quanta. A cavity emits
radiation by way of quanta. How does the radiation travel in space? We think that radiation is a wave phenomenon however the energy content is delivered to atoms in concentrated groups of waves
(quanta).
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Properties of a Photon
If a photon is to be considered as a particle we must be able to describe its mass, momentum, energy, statistics etc.
� Energy of a photon
limiting value 1/2hf otherwise integral multiple of hf
� Interaction of photons with matter
Complete absorption or partial absorption with the photon adjusting its frequency to remain as particle
� Intensity of photon
intensity has nothing to do with the energy of photons
-34 -15; h=6.626 10 . 4.136 10 .
,
E hf J s eV s
E Energy f frequency
= × = ×
= =
Intensity number of photons∝
Properties of Photons�Constant h of photon
h defines the smallest quantum angular momentum of a particle
Exercise 6 Show that h has units of angular momentum
�Mass and momentum of photonphotons move with velocity v=c
Photons have no rest mass m0
�Photon is not a material particle since rest mass is zero, it is a wave structure that behaves like a particle
2
2 2 and
hf hf hfE hf mc m p c
c c c∴ = = ⇒ = = =
( )2 2 2 20
0 22 21 0
1
m hfE hf mc c m v c
cv c= = = ⇒ = − • =
−
Properties of Photons
�Charge of a photon
photons do not carry charge, however they can eject charge particles from matter when they impinge on atoms
�Photon StatisticsConsider the radiation as a gas of photon. Photons move randomly like
molecules in a gas and have wide range of energies but same velocity
Statistics of photons is described by Bose-Einstein. We can talk about
intensity and temperature in the same way as density and temperature.
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Photoelectric Effect
Methods of electron emission:
Thermionic emission: Applying
heat allows electrons to gainenough energy to escape.
Secondary emission: The electron gains enough energy by transfer from another high-speed particle that strikes the material from outside.
Field emission: A strong external electric field pulls the electron out of the material.
Photoelectric effect: Incident light (electromagnetic radiation) shining on the material transfers energy to the electrons, allowing them to escape. We call the ejected electrons photoelectrons.
Photo-electric Effect
Experimental Setup
Photo-electric effect
observationsThe kinetic energy of
the photoelectrons is independent of the light intensity.
The kinetic energy of
the photoelectrons, for a given emitting material, depends only
on the frequency of the light.
Classically, the kinetic
energy of the photoelectrons should increase with the light
intensity and not depend on the frequency.
Electron
kinetic
energy
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Photo-electric effect observations
There was a threshold
frequency of the light, below which no photoelectrons were
ejected (related to the work function φ of the emitter material).
The existence of a threshold frequency is completely inexplicable in
classical theory.
Electron
kinetic
energy
Photo-electric effect observations
When photoelectrons
are produced, their number is proportional to the intensity of light.
Also, the photoelectrons
are emitted almost instantly following illumination of the
photocathode, independent of the intensity of the light.
Classical theory predicted that, for
extremely low light intensities, a long time would elapse before any one electron could obtain sufficient
energy to escape. We observe, however, that the photoelectrons are ejected almost immediately.
(number of
electrons)
Einstein’s Theory: Photons
Einstein suggested that the electro-magnetic radiation field is
quantized into particles called photons. Each photon has the energy quantum:
where f is the frequency of the light and h is Planck’s constant.
Alternatively,
E hf=
where:
E hω= 2h h π=
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Einstein’s TheoryConservation of energy yields:
21
2hf mvφ= +
In reality, the data were a bit more
complex. Because the electron’s energy can be reduced by the emitter material, consider fmax (not f):
where φ is the work function of the metal (potential energy to be
overcome before an electron could escape).
2
0
max
1
2eV mv hf φ
= = −
Example – Photoelectric Effect
Exercise 3-6: An experiment shows that when electromagnetic
radiation of wavelength 270 nm falls on an aluminum surface, photoelectrons are emitted. The most energetic of these are stopped by a potential difference of 0.46 volts. Use this information
to calculate the work function of aluminum in electron volts.
Exercise 3-7: The threshold wavelength of potassium is 558 nm. What is the work function for potassium? What is the stopping
potential when light of 400 nm is incident on potassium?
Exercise 3-8 Light of wavelength 400 nm and intensity 10-2 W/m2 is
incident on potassium. Estimate the time lag expected classically.
Newton discovers the
dispersion of light
Invention of Spectroscopy
Atomic Spectra
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Three Kinds of spectra
Solid, liquid or a dense gas excited to emit a continuous
spectrum
Light passing through low density
gas excites atoms to produce emission spectra
Light passing through cool low density gas results in absorption
spectra
Spectra
Chemical elements were observed to produce unique
wavelengths of light when burned or excited in an electrical discharge.
Line Spectra
Balmer Series
In 1885, Johann Balmer found an
empirical formula for the wavelength of the visible hydrogen line spectra in nm:
2
2364.6
4n
nnm
nλ =
−
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Rydberg-Ritz Formula
As more scientists
discovered emission lines at infrared and ultraviolet wavelengths, the Balmer
series equation was extended to the Rydberg equation:
2 2
7 1
H
7 1
1 1 1 for n>m, R=Rydberg constant
For Hydrogen R=R 1.096776 10
For Heavy atom R=R 1.097373 10
mn
Rm n
m
m
λ−
−
∞
= −
= ×
= ×
The Classical Atomic ModelConsider an atom as a planetary
system.
The Newton’s 2nd Law force of
attraction on the electron by the nucleus is:
2 2
2
kZe mvF
r r= =
22 2 21
22
kZev mv kze r
mr= ⇒ =
where v is the tangential velocity of the
electron:
The total energy is then:This is negative, so
the system is bound,
which is good.
2 2 2
2 2
kZe kZe kZeE K V
r r r= + = − = −
The Classical Atomic ModelExercise 7: Show that in the classical model the frequency of
radiation for an accelerating electron is
3/ 2
1f
r≈
and the Energy is
1E
r≈ −
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The Planetary Model is DoomedFrom classical E&M theory, an accelerated electric charge radiates
energy (electromagnetic radiation), which means the total energy must decrease. So the radius r must decrease!!
Physics had reached a turning point in 1900 with Planck’s
hypothesis of the quantum behavior of radiation, so a radical solution would be considered possible.
Electron
crashes
into the
nucleus!?
Electron
does not
crash in
the Bohr
model
The Bohr Model of the Hydrogen Atom
Bohr’s general assumptions:
1. Stationary states, in which orbiting electrons do not radiate energy, exist in atoms and have well-defined energies, E.
Transitions can occur between them, yielding light of energy: Bohr frequency condition
E = Ei − Ef = hf
2. Classical laws of physics do not apply to transitions between stationary states, but they do apply elsewhere.
3. The angular momentum of the nth state is: where n is called the Principal Quantum
Number.
hn
n = 1
n = 3
n = 2
Angular
momentum is
quantized!
Consequences of the Bohr Model
The angular momentum is:
1 22kZe
vmr
=
2 2 2
2 2
n kZe
m r mr=
h
hnrmL == v
mrn /v h=
So:
Solving for r:
So the velocity is:
a0 is called the Bohr radius.
a0
22
0
2n
n anr
mkZe Z= =
h
2 2
2From;
kZe mvF
r r= = ⇒
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Bohr Radius
The Bohr radius,
is the radius of the unexcited hydrogen atom.
The “ground” state Hydrogen atom diameter is:
2
0 20.0529a nm
mke≡ =
h
Energy of an electronExercise 3-9: Show that the energy of an electron in any atom at orbit n
is quantized and that it gives the ground state energy of Hydrogen atom to be -13.6eV.
2
0 2
2 4
0 2
1,2,3,...
where 2
n
ZE E n
n
mk eE
= − =
= −h
Rydberg-Ritz formula
Exercise 3-10: Derive the Ryderberg-Ritz formula
4
0
2 2 3
7 1
H
7 1
1 1 1 , =Rydberg constant
4
For Hydrogen R=R 1.096776 10
For Heavy atom R=R 1.097373 10
f i
E mkeR R
n n hc c
m
m
λ π
−
−
∞
= − = =
= ×
= ×
h
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Transitions in the Hydrogen AtomThe atom will remain in the excited state for a short time before
emitting a photon and returning to a lower stationary state. In equilibrium, all hydrogen atoms exist in n = 1.
Successes and Failures of the Bohr Model
Why should the nucleus of the atom be
kept fixed?
The electron and hydrogen nucleus
actually revolve about their mutual
center of mass.
Conservation of momentum require that
the momenta of nucleus and electron
equal in magnitude. The total kinetic
energy is then2 2 2
2
2 2 2 2
where 1
k
p p M m pE p
M m mM
mM m
M m m M
µ
µ
+= + = =
= =+ +
Success:
The electron mass is replaced
by its reduced mass:
Limitations of the Bohr Model
Works only for single-electron (“hydrogenic”) atoms.
Could not account for the intensities or the fine structure of the spectral lines (for example, in magnetic fields).
Could not explain the binding of atoms into molecules.
Failures:
The Bohr model was a great
step in the new quantum theory, but it had its limitations.
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Rydberg-Ritz Formula -Example
Exercise 3-11 Use the Rydberg-Ritz formula to calculate the first
line of Balmer, Lyman and Paschen series for the Hydrogen atom.
The Correspondence Principle
In the limits where classical and
quantum theories should agree,
the quantum theory must reduce
the classical result.
Bohr’s correspondence
principle is rather obvious:
When energy levels are
very close quantization
should have little effect
The Correspondence PrincipleExercise 3-12: Show that in the limit of large quantum number the Bohr
frequency is the same as the classical frequency.
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Fine Structure Constant
� In Bohr’s theory we know that transitions can
occur for ∆n≥1, for small n values. If we allow this
for large n and calculate the classical and Bohr
frequencies as in previous exercise, we will find
that they do not agree.
� To avoid this disagreement, A. Sommerfeld
introduced special relativity and elliptical orbits.
� From Bohr orbit in hydrogen for n=1, we have
( )
2
2 21
21.44 . 1
197.3 . 137
mvr n
kev
mr m mke
v ke eV nm
c c eV nmα
=
= = =
∴ = = ≈ =
h
h h
hh
h
α is called the fine structure constant
We will skip the
mathematical treatment of A. Sommerfeld work
Fine Structure Constant
� The fine structure constant can be understood in the following way.
For each circular orbit rn and energy En a set of n elliptical orbits exist whose major axis are the same but they have different
eccentricities and thus different velocities and Energies.
Electron transitions depend on the eccentricities of the initial and
final orbits and on the major axes, thus resulting in splitting of energy levels of n called fine-structure splitting.
Fine structure constant leads to the notion of electron spin
ExampleExercise 3-13: Show that the energy levels of oscillators in simple
harmonic motion are quantized.
Do it yourself exercise: solve the differential equation
22
20
dxx
dtω+ =
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ExampleExercise 3-14: Derive the Bohr quantum condition from Wilson-
Sommerfeld quantization rule
X-Ray Production: Theory
An energetic electron
passing through matter will radiate photons and lose kinetic energy, called bremsstrahlung.
Since momentum is conserved, the nucleus absorbs very little energy, and it can be ignored.
The final energy of the electron is determined from the conservation of energy to be:
fE
iE
hν
f iE E hf= −
X-Ray Production: Theory
If photons can transfer energy to electrons, can
part or all of the kinetic energy of electron be converted into photons?
“The Inverse photoelectric effect”
This was discovered before the work of Planck and Einstein
fE
iE
hν
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Observation of X Rays
1895: Wilhelm Röntgen studied the
effects of cathode rays passing through various materials. He noticed that a phosphorescent
screen near the tube glowed during some of these experiments. These new rays were unaffected by
magnetic fields and penetrated materials more than cathode rays.
He called them x rays and deduced
that they were produced by the cathode rays bombarding the glass walls of his vacuum tube. Wilhelm Röntgen
X-Ray Production: Experiment
Current passing through a filament produces copious numbers of
electrons by thermionic emission. If one focuses these electrons by a cathode structure into a beam and accelerates them by potential differences of thousands of volts until they impinge on a metal
anode surface, they produce x rays by bremsstrahlung as they stop in the anode material.
Electromagnetic theory Predicts X-Ray
Accelerated charges produce electromagnetic waves, when fast
moving electrons are brought to rest, they are certainly accelerated
1906: Barkla found that X-Ray show Polarization, this establishing that X-Rays are waves.
X-Rays have wavelength range 0.1 nm – 100 nm
Even though classical theory predicts x-ray’s, the experimental data
is not explainable.
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Two distinctive features
1.Some targets have enhanced peaks. For example Molybdenum shows two peaks at specific wavelengths.
a) This is due to rearrangement of electrons of the target material after bombardment.
b) X-ray ‘s have a continuous spectrum
2.No matter what the target , the threshold wavelength depends on the accelerating potential
Inverse Photoelectric Effect
Since the work function of the target
is of the order of few eV, whereas the
accelerating potential is thousand of
eV
max
min
o
hceV hf
λ= =
2
0
max
1
2eV mv hf φ
= = −
From photoelectric effect
Duane-Hunt rule
Shells have letter names:
K shell for n = 1
L shell for n = 2
The atom is most stable in its
ground state.
When it occurs in a heavy atom, the radiation emitted is an x-ray.
It has the energy E (x ray) = Eu − Eℓ.
X-Ray Spectra
An electron from higher
shells will fill the inner-shell vacancy at lower energy.
Bohr-Rutherford picture of the atom
can also be applied to heavy elements
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Atomic Number and Moseley
The x-rays have names:
L shell to K shell: Kα x-ray
M shell to K shell: Kβ x-ray
etc.
G.J. Moseley studied x-ray
emission in 1913.
Atomic number Z = number
of protons in the nucleus.
Moseley found a
relationship between the
frequencies of the
characteristic x-ray and Z.
1 2 ( )n
f A Z b= −
Moseley found this relation
holds for the Kα x-ray with b=1 and different An values (from quantum mechanics):
Moseley’s Empirical ResultsThe Kα x-ray is produced from the n = 2 to n = 1 transition.
In general, the K series of x-ray frequencies are: Form the Bohr model with Z=Z-1
Moseley’s research clarified the importance of Z and the electron
shells for all the elements, not just for hydrogen.
We use Z-1 instead of Z because one electron is already
present in the K-shell and so shields the other's from the
nucleus’ charge.
2 42 2
3 2 2 2
1 1 1( 1) ( 1) 1
4 1
mk ef Z cR Z
n nπ∞
= − − = − −
h
2
2
11nA cR
n∞
= −
where
Moseley’s Empirical ResultsFor the L series of x-ray wavelength the
frequencies are
2
2 2
1 1( 7.4)
2f cR Z
n∞
= − −
a) Neodymium Z=60 and
Samarium Z=62
b) Promethium Z=61
c) All three elements together
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Example X-Ray SpectraExercise 3-15: Calculate the wavelength of Kα line of molybdenum, Z=42,
and compare with the value λ=0.0721nm measured by Moseley.
Aguer (oh-zhay) EffectIn 1923 Pierre Auger discovered an alternative to X-ray emission.
The atom may eject a third electron from a higher-energy outer shell via radiationless process called Auger effect
3 2 1
3
E E E E
KE E E
< ∆ = −
= ∆ −
EdN(e)/dE plot:
a) Auger spectrum of Cu
b) Al and Al2O3 togethernote the energy shift in the larges peak due to
adjustment in Al electron shell energies
X-Ray ScatteringMax von Laue suggested that if x-rays were a form of electromagnetic
radiation, interference effects should be observed.
Crystals act as three-dimensional gratings, scattering the waves and producing observable interference effects.
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Bragg’s Law
William Lawrence Bragg
interpreted the x-ray scattering as the reflection of the incident x-ray beam from
a unique set of planes of atoms within the crystal.
There are two conditions for constructive interference of the scattered x rays:
1) The angle of incidence must equal the
angle of reflection of the outgoing wave.
2) The difference in path lengths must be
an integral number of wavelengths.
Bragg’s Law: nλ = 2d sin θ (n = integer)
A Bragg spectrometer scatters x
rays from crystals. The intensity of the diffracted beam is determined as a function of scattering angle
by rotating the crystal and the detector.
When a beam of x rays passes
through a powdered crystal, the dots become a series of rings.
The Bragg Spectrometer
Examples
Exercise 3-16: What is the shortest wavelength present in a radiation if
the electrons are accelerated to 50,000 volts?
Exercise 3-16:The spacing of one set of crystals planes in common salt is
d=0.282nm. A monochromatic beam of X-rays produces a Bragg
maximum when its glancing angle is 7 degrees. Assuming that this is the
first order maximum (n=1), find the wavelength of the X-rays, what is the
minimum possible accelerating voltage Vo that produced the X-rays?