Download Chapter 38

Chapter 38
Photons, Electrons
& Atoms
Need for Quantum Physics
Problems remained from classical mechanics that relativity
didn’t explain
Blackbody Radiation
Photoelectric Effect
Emission of electrons by an illuminated metal
X Rays
The electromagnetic radiation emitted by a heated object
How & why are they produced?
Why would f changes upon collisions?
Spectral Lines
Emission of sharp spectral lines by gas atoms in an electric
discharge tube
Development of Quantum Physics
1900 to 1930
Development of ideas of quantum mechanics
Also called wave mechanics
Highly successful in explaining the behavior of
atoms, molecules, and nuclei
Involved a large number of physicists
Planck introduced basic ideas
Mathematical developments and interpretations
involved such people as Einstein, Bohr,
Schrödinger, de Broglie, Heisenberg, Born and
Dirac
Blackbody Radiation
An object at any temperature
emits electromagnetic radiation
Sometimes called thermal radiation
Stefan-Boltzmann Law describes the
total power intensity radiated I = σT4
The spectrum of the radiation
depends on the temperature and
properties of the object
Blackbody Radiation Graph
Experimental data for
distribution of energy in
blackbody radiation
As the temperature
increases, the total
amount of energy
increases
Shown by the area under
the curve
As the temperature
increases, the peak of
the distribution shifts to
shorter wavelengths
Wien’s Displacement Law
The wavelength of the peak of the
blackbody distribution was found
to follow Wein’s Displacement Law
λmax T = 0.2898 x 10-2 m • K
λmax is the wavelength at which the curve
peaks
T is the absolute temperature of the
object emitting the radiation
The Ultraviolet Catastrophe
Classical theory did not
match the experimental data
At long wavelengths, the
match is good
At short wavelengths,
classical theory predicted
infinite energy
At short wavelengths,
experiment showed no
energy
This contradiction is called
the ultraviolet catastrophe
Rayleigh’s cavity model
• EM wave in cavity (light)
• Normal Modes (resonants)
• Equipartition Principle E~kT
• Power Radiation Intensity
I = 2πckT/λ4
• Didn’t explain experimental results for small λ
Planck’s Resolution
Planck hypothesized that the blackbody radiation
was produced by electromagnetic resonators
Resonators were submicroscopic charged oscillators
(electrons in a box??)
The resonators could only have discrete energies
En = n h ƒ
n is called the quantum number
ƒ is the frequency of vibration
h is Planck’s constant, 6.626 x 10-34 J s
Key point is quantized energy states
Max Planck
1858 – 1947
Introduced a
“quantum of
action,” h
Awarded Nobel
Prize in 1918 for
discovering the
quantized nature
of energy
Planck Radiation Law
I=
2πhc2/[λ5(ehc/λkT
– 1)]
I
for large λ, I ~ 1/λ4
λ
for small λ, I ~ e-hc/λkT
Electromagnetic radiation is quantized !
Photoelectric Effect
When light is incident on certain metallic
surfaces, electrons are emitted from the
surface
This is called the photoelectric effect
The emitted electrons are called photoelectrons
The effect was first discovered by Hertz
The successful explanation of the effect
was given by Einstein in 1905
Received Nobel Prize in 1921 for paper on
electromagnetic radiation, of which the
photoelectric effect was a part
Photoelectric Effect Schematic
When light strikes E
(Cathode),
photoelectrons are
emitted
Electrons collected at C
(small Anode) and
passing through the
ammeter create a
current in the circuit
C is maintained at a
positive potential by the
power supply
Photoelectric Current/Voltage Graph
The current
increases with
intensity, but
reaches a saturation
level for large ∆V’s
No current flows for
voltages less than or
equal to –∆Vs, the
stopping potential
The stopping potential
is independent of the
radiation intensity
More About Photoelectric Effect
The stopping potential is
independent of the radiation
intensity
The maximum kinetic energy of
the photoelectrons is related to the
stopping potential: KEmax = e∆Vo
Features Not Explained by
Classical Physics/Wave Theory
No electrons are emitted if the
incident light frequency is below
some cutoff frequency that is
characteristic of the material being
illuminated (Poynting S ~ |Efield|2)
The maximum kinetic energy of
the photoelectrons is independent
of the light intensity
More Features Not Explained
The maximum kinetic energy of
the photoelectrons increases
linearly with increasing light
frequency
Electrons are emitted from the
surface almost instantaneously,
even at low intensities (E ~ P∆t)
Einstein’s Explanation
A tiny packet of light energy, called a photon,
would be emitted when a quantized oscillator
jumped from one energy level to the next
lower one
Extended Planck’s idea of quantization to
electromagnetic radiation
The photon’s energy would be E = hƒ
Each photon can give all its energy to an
electron in the metal
The maximum kinetic energy of the liberated
photoelectron is KEmax = hƒ – Φ
Φ is called the work function of the metal
Light Intensity
Therefore, a more
intense beam of light
will contain more
photons, but the
energy of each
photon does not
change.
Explanation of Classical “Problems”
The effect is not observed below a
certain cutoff frequency since the
photon energy must be greater than or
equal to the work function
Without this, electrons are not emitted,
regardless of the intensity of the light
The maximum KE depends only on the
frequency and the work function, not on
the intensity
More Explanations
The maximum KE increases with
increasing frequency
The effect is instantaneous since
there is a one-to-one interaction
between the photon and the
electron
Verification of Einstein’s Theory
Experimental
observations of a
linear relationship
between KE and
frequency confirm
Einstein’s theory
The x-intercept is
the cutoff
frequency
Cutoff Wavelength
The cutoff wavelength is related to
the work function
hc
λc =
φ
Wavelengths greater than λC
incident on a material with a work
function φ don’t result in the
emission of photoelectrons
Photocells
Photocells are an application of the
photoelectric effect
When light of sufficiently high
frequency falls on the cell, a
current is produced
Examples
Solar cells, IR remote controls …etc
X-Rays
Discovered and named by Roentgen in
1895
Later identified as electromagnetic
radiation with short wavelengths
Wavelengths less than for ultraviolet
Wavelengths are typically about 0.1 nm
X-rays have the ability to penetrate most
materials with relative ease
Production of X-rays, 1
X-rays are produced when
high-speed electrons are
suddenly slowed down
Can be caused by the electron
striking a metal target
Heat generated by current in
the filament causes electrons
to be emitted
These freed electrons are
accelerated toward a dense
metal target
The target is held at a higher
potential than the filament
X-ray Spectrum
The x-ray spectrum
has two distinct
components
Continuous broad
spectrum
Depends on voltage
applied to the tube
Sometimes called
bremsstrahlung
The sharp, intense
lines depend on the
nature of the target
material
Production of X-rays, 2
An electron passes
near a target nucleus
The electron is
deflected from its
path by its attraction
to the nucleus
This produces an
acceleration
It will emit
electromagnetic
radiation when it is
accelerated
Wavelengths Produced
If the electron loses all of its
energy in the collision, the initial
energy of the electron is
completely transformed into a
photon
The wavelength can be found from
e∆V = h ƒmax =
hc
λmin
Wavelengths Produced, cont
Not all radiation produced is at this
wavelength
Many electrons undergo more than
one collision before being stopped
This results in the continuous
spectrum produced
Diffraction of X-rays by Crystals
For diffraction to occur, the spacing
between the lines must be
approximately equal to the wavelength
of the radiation to be measured
The regular array of atoms in a crystal
can act as a three-dimensional grating
for diffracting X-rays
Schematic for X-ray Diffraction
A beam of X-rays with a
continuous range of
wavelengths is incident
on the crystal
The diffracted radiation
is very intense in certain
directions
These directions correspond
to constructive interference
from waves reflected from
the layers of the crystal
The diffraction pattern is
detected by
photographic film
Photo of X-ray Diffraction Pattern
The array of spots is called a Laue
pattern
The crystal structure is determined
by analyzing the positions and
intensities of the various spots
Bragg’s Law
The beam reflected from the
lower surface travels farther
than the one reflected from
the upper surface
If the path difference equals
some integral multiple of the
wavelength, constructive
interference occurs
Bragg’s Law gives the
conditions for constructive
interference
2 d sin θ = m λ, m = 1, 2, 3…
Arthur Holly Compton
1892 – 1962
Discovered the
Compton effect
Worked with
cosmic rays
Director of the lab
at U of Chicago
Shared Nobel
Prize in 1927
The Compton Effect
Compton directed a beam of x-rays
toward a block of graphite
He found that the scattered x-rays had
a slightly longer wavelength that the
incident x-rays
This means they also had less energy
The amount of energy reduction
depended on the angle at which the xrays were scattered
The change in wavelength is called the
Compton shift
Compton Scattering
Compton assumed
the photons acted
like other particles
in collisions
Energy and
momentum were
conserved
The shift in
wavelength is
h
∆λ = λ − λo =
(1 − cos θ )
mec
Compton Scattering, final
The quantity h/mec is called the
Compton wavelength
Compton wavelength = 0.002 43 nm
Very small compared to visible light
The Compton shift depends on the
scattering angle and not on the
wavelength
Experiments confirm the results of
Compton scattering and strongly
support the photon concept
Photons and Electromagnetic Waves
Light has a dual nature. It exhibits
both wave and particle characteristics
The photoelectric effect and Compton
scattering offer evidence for the particle
nature of light
Applies to all electromagnetic radiation
Different frequencies allow one or the other
characteristic to be more easily observed
When light and matter interact, light
behaves as if it were composed of particles
Interference and diffraction offer evidence
of the wave nature of light
Rutherford Scattering
Rutherford did a very clever experiment
with thin Au foil and alpha particles.
Scattering was nearly absent or very
dramatic, leading him to conclude that
the atom was mostly empty space
around a dense (+) center.
(Muffin - raisin model)
Bohr Atom (in Rutherford’s Lab)
Classical physics predicts that a moving electron would emit
radiation and decay. (not observed!)
Bohr postulated stable orbits at quantized energy levels that
do not decay. hf=hω=Ei – Ef. (single photon-electron)
Ln= mvnrn = nh. rn = n2ao. ao=Bohr radius = 0.53Å
Reduced mass
Hydrogen Atom
Hydrogen has one set of lines in the visible spectrum (Balmer), three sets of lines in the
IR (Paschen, Braclett and Pfund series) and one in the UV (Lyman series).
Hydrogen Spectrum
1/λ = R(1/nf2 – 1/n2)
R = Rydberg Constant = 1.097 x 107 m-1
nf = 1 (Lyman Series) - UV
nf = 2 (Balmer Series) – visible
nf = 3, 4, 5 (Paschen, Brackett, & Pfund Series) – IR
Hydrogen-like atoms
•
Solutions for H apply to He+ because everything hinges on the
system being a “two-body problem.” Li++ works also, but soon
things become overly artificial.
Maxwell-Boltzmann Statistics
Population of the orbital state ~ e-E /kT
=#of electrons in the energy state Ei
=Probability to find an electron at the
energy state Ei
Probability of an electron goes thru a
transition from Ei to Ej is proportional to
e-(E - E )/kT
k is the Boltzmann’s constant = 1.38 x 10-23 J/K
i
i
j
Example
(#38.21)
A beam of alpha particles is incident on a target of lead. A particular
alpha particle comes in "head-on" to a particular lead nucleus and
stops 6.50×10−14 m away from the center of the nucleus. (This point
is well outside the nucleus). Assume that the lead nucleus, which has
82 protons, remains at rest. The mass of the alpha particle is
6.64×10−27 kg.
(a) Calculate the electrostatic potential energy at the instant that the
alpha particle stops.
U(r) = k q1q2/r =(9x109)(2)(82)(1.6x10-19)2/(6.5x10-14) = 5.82x10-13 J
1 eV = 1.6x10-19 J, so U(r) = 3.63 MeV
(b) What initial kinetic energy did the alpha particle have?
Conservation of energy: initial KE = final PE = 3.63 MeV=5.82x10-13 J
(c) What was the initial speed of the alpha particle?
KE = ½ mv2 = 5.82x10-13 J for the alpha particle
m = 6.64×10−27 kg, so v= 1.32 x 107 m/s = 0.044c
γ = [1 – (v/c)2]-1/2 = 1.00097 non-relativistic is ok
Lasers
To achieve laser action, three conditions
must be met
The system must be in a state of population
inversion
The excited state of the system must be a
metastable state
More atoms in an excited state than the ground state
Its lifetime must be long compared to the normal
lifetime of an excited state
The emitted photons must be confined in the
system long enough to allow them to stimulate
further emission from other excited atoms
This is achieved by using reflecting mirrors
Laser Beam – He Ne Example
The energy level diagram
for Ne in a He-Ne laser
The mixture of helium and
neon is confined to a glass
tube sealed at the ends by
mirrors
A high voltage applied
causes electrons to sweep
through the tube,
producing excited states
When the electron falls to
E2 from E*3 in Ne, a 632.8
nm photon is emitted
Production of a Laser Beam
(#38.30)
Example – PRK
Photorefractive keratectomy (PRK) is a laserbased surgery process that corrects near- and
farsightedness by removing part of the lens of
the eye to change its curvature and hence
focal length. This procedure can remove
layers 0.25 µm thick in pulses lasting 12 ns
with a laser beam of wavelength 193 nm.
Low-intensity beams can be used because
each individual photon has enough energy to
break the covalent bonds of the tissue.
(a) In what part of the electromagnetic spectrum
does this light lie?
Visible light has wavelengths from about 400 nm to
about 700 nm. 193 nm is shorter than 400 nm so it
hc
is in Ultra Violet (UV)
−18
E
=
=
1.03
×
10
(b) What is the energy of a single photon?
λ
(c) If a 1.50 mW beam is used, how many photons
are delivered to the lens in each pulse?
E
NE
P = tot =
t
t
so
J = 6.44 eV
Pt (1.50 × 10 −3 W)(12.0 × 10−9 s)
N=
=
= 1.75 × 107 photons
−18
E
1.03 × 10 J
Example – recoil
An atom with mass m emits a photon of wavelength λ.
(a) What is the recoil speed of the atom?
Assume a non-relativistic velocity and conserve momentum
zero mass E = pc
(#38.52)
h
⇒ mv =
λ
or
h
v=
.
mλ
(b) What is the kinetic energy K of the recoiling atom?
2
1
1  h 
h2
2
K = mv = m 
 =
2
2  mλ 
2mλ 2
(c) Find the ratio K/E, where E is the energy of the emitted photon.
K
h2
λ
h
Recoil becomes an important concern for small m and
=
⋅
=
.
E 2mλ 2 hc 2mcλ small λ since this ratio becomes large in those limits.