Post on 03-Jul-2018
transcript
What is LIGHT??
4:38 minutes
3:58 minutes
2
31:08
35:34
4
5
At the end of the 19th century (late 1800’s)
scientists had it all figured out.
What is the physics called that describes the motion
of matter?
Newtonian
It explained the motion of objects on Earth as well as
heavenly bodies.
What theory did Newton use explained the nature
of light?
Isaac Newton thought light was made of little
particles (he called them corpuscles) emitted by
hot objects (such as the sun or fire).
(Or did they??)
However,
his contemporary, the Dutch physicist
Christian Huygens, thought light was a kind
of wave vibrating up and down as it moved
forward.
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7
What evidence do we have that light acts
as a wave?
It matches the kinematics' wave formula:
Where c is the speed of light
f is the frequency
l is the wavelength
The interference patterns of waves was used to
explain different light phenomena:
The colours on the surface of oil films was explained as
interference patterns by Christian Huygens and Robert
Hooke.
The double slit experiment by Thomas Young in 1803.
In 1815, Augustin Fresnel explained the diffraction
of light by using this wave theory of light.
c f l83.00 10 m
sc
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In 1864, James Clerk Maxwell completed “classical
physics” by amalgamating electricity and magnetism and
in the process combined light into the electromagnetic
wave family.
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So life was good. They could explain it all
with only minor problems left to solve!!
UNTIL:
Wilhelm Roentgen discovered X-rays.
The UV catastrophe in Black body radiation curves
Light behaving like a particle in the photoelectric
effect
X-ray photons with momentum
Particles that exhibit wave properties
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In this unit
We study these phenomena and the theories
that resulted.
We discuss the transition from the long-
standing Newtonian Physics to the Modern
Age of Physics
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The study of how MATTER and Energy are
interrelated is called Quantum Theory
This energy is Light Energy.
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Paradigm Shifting Time
Matter does not always behave in the way Newton
thought it should ( sometimes it behaves like a
wave)
Light always behave in the way Maxwell thought it
should (sometimes it behaves as a particle)
THE INFAMOUS DOUBLE SLIT
EXPERIMENT The double slit experiment as previously
studied is one way to illustrate the difference
between the classical world and the quantum
world.
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Water Waves Light
If we look at the results of the double slit
experiment with marbles we see that they
behave like we expect particles to behave.
If, however, we perform the double slit
experiment with subatomic particles such as
electrons, we see quite different behaviour with
the electrons showing wave patterns of
interference.
This discrepancy is characteristic of what
happens in the quantum world.16
The clear cut distinction between particles
and waves breaks down in the quantum
world.
This dual nature of exhibiting both wave and
particle properties is known as wave-particle
duality.
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In this unit we learn about the evidence that
supports the dual (wave-particle) nature of light.
(aka photons)
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Question: Does the sun cause
sunburn because it is so bright?
A sunburn is a burn to living tissue such as skin
or leaves produced by overexposure to ultraviolet
(UV) radiation, commonly from the sun's rays.
A similar burn can be produced by overexposure
to other sources of UV such as from tanning
lamps, or occupationally, such as from welding
arcs.
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Why is it necessary to wear protective
goggles, and clothing when welding?
The welder light contains ultraviolet (UV) light that has a
higher energy than visible light which makes the light
penetrate skin cells instead of reflecting off them.
The brightness of the light is not related to
its penetrating power.
Brightness is related to the amount of light present.
The penetrating power is related to the
lights energy level.
This is related to its frequency (or wavelength).
The Quantum Mechanical model of light can
be used to explain this quantized energy level
and other phenomena such as black-body
radiation and the photoelectric effect that
could not be explained by the classical wave
theory of light.
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A Brief History Of Quantum
Mechanics https://www.youtube.com/watch?v=B7pACq_
xWyw
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Black - Body Radiation
An ideal black-body is any opaque object that
absorbs all light (from all of the EM spectrum)
that is incident upon it.
Therefore it reflects no radiation and appears
perfectly black.
In practice no material has been found to
absorb all incoming radiation, but carbon in its
graphite form absorbs all but about 3%.
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Black - Body Radiation
This object will also become a perfect
radiator, emitting the full spectrum of light.
In quantum mechanics, the term black body
radiation is used to describe the release of
energy.
This energy can be from different parts of
the EM spectrum including infrared or
thermal energy (hence, this is why we feel
things as being hot).
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Black - Body Radiation
Any object that is above 0o K, will emit
radiation in this form.
Black body radiation helps to explain why
objects will glow when they are at a certain
temperature.
For example, the element on a stove will glow
red when it is at a certain temperature. Or, an
incandescent light will glow yellow at a certain
temperature.
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Classical theory suggests that as the
frequency of emitted energy increase, the
intensity of radiation will increase as shown.
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Draw the corresponding curve in terms
of wavelength
The problem with classical theory is it implied
there is no limit to the intensity of radiation as
frequency increases, which contradicted
experimental results (as shown below)
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Max Planck suggested that objects release energy
in packets called quanta, meaning energy can only
be emitted in discrete packets of specific
frequency.
But it was also was considered to behave as a
wave because it had a frequency.
These quanta later became known as photons.
When objects are at a certain temperature, it will
release quanta at a specific frequency.
If an object is at a higher temperature, it will release
energy at a different (higher) frequency.
Those different frequencies are the reason why objects
will emit different colours at different temperatures.29
Therefore, the hotter the object is, the bluer
the light will be emitted.
The cooler the object, the redder it is, thus
giving the experimental pattern above.
Planck was able to use the idea of quanta to
produce the graph below, thus verifying the
new model of understanding.
30
Planck used data from blackbody radiation
experiments and then “fit” the formula:
E = hfto match the data.
This led the scientific community to quantum
theory.
The photoelectric effect was Einstein’s test of the
theory, which turned out to validate Planck’s
work.
There was a lot of back and forth between
Einstein and Planck during the development of
this theory.31
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Planck’s equation:
Planck’s constant:
Also, the total
energy is found
using the formula
where
n = 1, 2, 3, ...
(number of photons)
E hf
346.626 10h J s
E nhf
Problems
1. Calculate the energy of a single photon of
blue light if it has a frequency of
7.0588X1014 Hz.
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What is E and what is the equation
to calculate this based on the
wavelength of light?
E is the smallest amount or quantum of energy that can be transferred for a given wavelengthof electromagnetic radiation
The equation is:
This is obtained by replacing f in Planck’s equation:
hcE
l
cf
l
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2. What is the energy of a photon that
has a wave length of 460 nm?
hcE
l
34 8
9
6.626 10 3.00 10
460 10
ms
J s
m
194.3 10 J
NOTE:
At the atomic level, energy is commonly
measured in electronvolts (eV).
Which is the energy gained or lost as a single
electron moves across an electric potential
difference of 1 volt.
RECALL:
1eV = 1.602x10-19 C
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3. Calculate the energy of a single photon if it
has a frequency of 1.50 × 1015 Hz. Express
your answer in Joules and electron volts, eV.
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5. A) What is the energy of a photon with a
wavelength of 525 nm?
B) If the source releases 100.0 J, how many
photons were released?
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Black lights and other UV light
UVA: sun tanning, testing, inspection, insect traps, stage
effects, blacklight, phototherapy (315-400nm)
UVB: sun burning, inspection, analysis, testing,
phototherapy (280-360nm)
UVC: germicidal (253.7nm), ozone producing (185nm)40
Consider:
GTL3 Bulb - 3W Germicidal
3 W at 10.5V
UV power: 0.16 W at 254 nm
How many photons of UV light is emitted in 120
seconds?
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“PHOTO ELECTRIC EFFECT”?
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What is the “photo electric effect”? It is a phenomenon that sometimes occurs when light is
shone on a metal surface causing electrons to be
emitted from the surface.
The surface may become positively charged.
This is because electrons gain energy from the light,
and are able to leave the metal's surface.
PHOTOELECTRIC EFFECT APPARATUS
If we illuminate one of the
metal plates with the proper
light, electrons will absorb
the light and escape from
the surface of the metal.
These electrons will move
off the surface with
whatever kinetic energy
they possess.
Some will move across the
tube and hit the metal plate
at the opposite end of the
tube and the ammeter will
register a current.45
The current gives no indication
of the kinetic energy of the
photoelectrons only the number
of them released per second.
If the photoelectrons’ kinetic
energy is transferred into
potential energy, by applying an
electric field to oppose their
motion, we can work out what
that initial kinetic energy was.
We therefore apply an electric
voltage across the tube to
oppose the travel of the
electrons between electrodes.
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We make the illuminated
electrode positive and the one
they are travelling to negative
and therefore repulsive to the
electrons.
The electrons are repelled by
the negative plate and some
electrons go back.
But some of them with high
kinetic energy are still able to
reach the plate.
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As the negative potential
increases, less and less electrons
are able to reach the other side of
the tube and finally when none can
reach the photocurrent becomes
zero.
The stopping potential (voltage)
Vstop is related to the maximum
kinetic energy of the ejected
photoelectrons.
Recall:
so for the electrons:
or
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EV
q
maxk
stop
EV
e
maxstop keV E
If the brightness/intensity of the photon beam
is increased the photocurrent will increase
and vice versa.... but this does not affect the
voltage needed to stop the photoelectrons.
This cannot be explained by the wave theory
of light!!
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There are some strange effects with the photoelectric effect that
can not be explained by considering light as a wave.
The frequency of the light must be above a certain threshold value for that metal for electrons to be emitted.
For example: blue light causes sodium to emit electrons but red light does not.
If light was a wave the electrons would eventually absorb enough energy to be emitted, regardless of frequency.
Also, incredibly weak beams of light can cause electrons to be emitted.
If light was a wave and spread out you would not expect any electrons to obtain the energy to escape.
Finally, it was discovered that the kinetic energy of the electrons depends not on the intensity (brightness) of the light but on its frequency.
A very weak ultraviolet beam will give electrons a higher kineticenergy than a very bright blue beam of light.
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SO, if you were asked:
1. How does the photoelectric effect
contradict the wave theory of light?
Your answer should be something like this:
According to wave theory the energy of light,
i.e. its amplitude, or brightness, should
increase the energy of the ejected electrons.
However, this is not the case.
It is found that by increasing the frequency of
the light to a different colour, the energy of
the electrons was immediately increased.51
2. Why doesn’t red light eject
electrons when shined on a metal
plate? What happens to the red light?
The photons of the red light do not have enough
energy to liberate any electrons.
The light is absorbed by the metal.
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There is a minimum, or threshold
frequency, f0, of incident light below which
no photoemission would occur.
If the light shining on the metal is of a
frequency lower than this threshold
frequency, no electrons are emitted,
regardless of the brightness of the light
shining on the metal.
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Define the work function, Wo.
It is the minimum energy required to evict an
electron from a metal.
This energy varies depending on the type of
metal.
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The energy of the incoming photons of light is
analogous to the energy of the incoming player’s
foot.
To kick the ball the incoming foot has two jobs to
do:
One - It must do work to loosen the ball from the
sand. (Work function)
Two - Any remaining energy moves the ball on its way
or gives the ball kinetic energy.
Similarly, the energy of the incident photons
must be sufficient to:
One - Free the electron from the metal plate (Wo)
Two - Get the electron moving or give it kinetic energy.
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How does the higher energy light
eject electrons from a metal plate?
If the light is not reflected, it transfers energy to
the electrons in the metal.
If the energy transferred is greater than Wo, the
electron will be emitted.
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The liberated electron’s kinetic
energy is determined by the equation:
This shows that the kinetic energy of the electron
is the energy of the photon that liberated the
electron minus the energy required to release the
electron.(Wo)
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max k photon oE E W
When we solve the equation above
for Ephoton and apply Planck’s
equation we get:
hcE hf
l
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maxk o
hcE hf E W
l
max photon k oE E W
Note:
E=hf is the energy of the incident photon
Ekmax is the maximum amount of energy of the
liberated electron
Wo is the work function.
This is Einstein’s photoelectric effect
equation:
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max k ohf E W
SUMMARY (thus far)
Planck used data from blackbody radiationexperiments and then “fit” the formula E =hf to match the data!
This match lead the scientific community to quantum theory.
The photoelectric effect was Einstein’s test of the theory, which turned out to validate Planck’s work.
Quantum theory is the theory of how atoms, matter, and energy are interrelated.
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#1. Light, with a frequency of 5.0 x 10 14 Hz
illuminates a photo electric surface that has a work
function of 2.3 x 10-19 J.
What is the maximum energy of the ejected electrons?
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max
34 14 196.626 10 5.0 10 / 2.3 10 kE J s s J
191.0 10 J
Example:
max k ohf E W
max k oE hf W
#2. When electromagnetic radiation with a
wavelength of 350 nm falls on a metal, the
maximum kinetic energy of the ejected
electrons is 1.20 eV. What is the work
function of the metal, in Joules and eV?
Since we are given wavelength we use:
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maxk o
hcE W
l max
o k
hcW E
l34 8
9
6.626 10 3.00 101.20
350 10
ms
J seV
m
195.68 10 1.20J eV
In Joules
In eV
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19 195.68 10 1.20 1.602 10 JeV
J eV
193.758 10 J
193.8 10 J
Wo 3.55eV 1.20eV 2.35eV
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195.68 10 1.20
1.602 10
eVJ eV
J
#3. What is the stopping voltage of
an electron that has 7.4 x 10 -19 J of
kinetic energy?
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197.4 10 stopeV J197.4 10
stop
JV
e
19
19
7.4 10
1.602 10
J
C
maxstop keV E
4.6V
4. Experiments show that the work function for cesium metal is 2.10 eV.
Determine the threshold frequency and wavelength for photons capable of producing photoemission from cesium.
Recall: The threshold frequency, f0, is found by
the equation:
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o oW hf
Then do: Page 720: #1-7
And QUESTIONS 8-10 are to be passed in next
day
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PRACTICE PROBLEMS:
p. 716 #1-3; Page 718 #1,2
Wavelength (nm) Kinetic Energy (eV) Frequency(Hz) f=c/l
500 0.36 6E+14490 0.41 6.12E+14440 0.70 6.82E+14390 1.05 7.69E+14340 1.52 8.82E+14290 2.14 1.03E+15240 3.025 1.25E+15
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This graph shows how the kinetic energy of
the ejected electrons varies with frequency
for multiple metal surfaces.
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•The threshold
frequency is the x-
intercept of this graph.
•The work function is
the y-intercept.
•The slope of the graph
represents Planck’s
constant.
2. What is the work function for platinum?
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1. Which metal would
required the greatest
amount of light energy to
evict an electron?
The Compton Effect and Photon
Momentum
What is momentum is defined as?
p = mv.
So, does this mean that a photon (which
is electromagnetic energy) has mass?!?
What famous equation relates energy to
mass?
E = mc2
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In 1923, Arthur Holly Compton shot a
beam of x-rays at a thin foil and a target
made from carbon.
After the target was struck, he noticed
that it emitted other x-rays and some
electrons.
He discovered that the second lot of x-
rays had lower energy than the incident x-
rays. This phenomenon is called the
Compton Effect.
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Notice the high-energy x rays hitting the target.
Notice lower-energy x-rays and electronsbeing scattered from the target.
The word scattering is used because the x-rays are deflected in many different directions.
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Can you see that the picture resembles a
collision of pool balls on a pool table?
Compton must have had the same
impression because he decided to use the
laws of conservation of energy and
momentum that you have already learned
about in Physics 2204.
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The conservation of energy law:
Energy can not be created or destroyed, only
transformed from one form into another.
Ex-ray = Escattered + Eelectron
Note:
Ex-ray = hfx-ray is the energy of the incoming x-ray
photon
Escattered = hfscattered is the energy of the scattered x-ray
Eelectron =½ mev2 is the kinetic energy of the electron
Thus the equation above can be written as:
hfx-ray = hfscattered + ½ mev2
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Conservation of momentum:
The momentum of a system before a collision is
equal to the momentum after.
In order for this equation to be true, photons
would have to have mass, or some mass-like
property.
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p xray
p scattered
p electron
Compton boldly hypothesize that a photon
has mass and used Einstein's Special Theory
of Relativity and mass-energy equivalency
formula: E = mc2
Solve E = mc2 for mass
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2
Em
c
Substitute this for mass into the equation for
momentum - p=mv (ignoring direction)
But for all photons, v = c.
Also,
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2
Ep v
c
2
Ep c
c
Ec
hcE hf
l
which reduces to
Compton has succeeded in deriving an
elegant expression for the momentum of a
photon!! His work gave enormous support
to the notion that light possessed both
wave and particle properties.
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Ep
c
hc
cl
hp
l
Practice:
1.In a Compton experiment a 82 eV x-ray photon collides with an electron causing the scattering of a lower energy x-ray of wavelength 295 nm. The mass of an electron is 9.11 x 10-31 kg. Determine
a) the momentum of the original x-ray.
b) the momentum of the scattered x-ray
c) the energy imparted to the electron
d) the increase in speed of the electron
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Given:
Ex-ray = 82 eV = 82 eV x 1.6 x 10-19 J/eV
= 1.3 x 10-17 J
lscattered = 295 nm = 295 x 10-9 m = 2.95 x 10-7 m
me = 9.11 x 10-31 kg
h = 6.626 x 10-34 J·s c = 3.00 x 108 m/s
Note that the last two givens did not appear
in the exercise, but they should always be in
the back of your mind.
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Solution:
a) the momentum of the original x-ray.
(Also, be mindful of the useful formulae
below): E = hf, f = c /l, E = (hc)/l, p = E/c,
p = h/l Ek = ½ mv2 (particle)
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px-ray = E/c = 1.3 x 10-17 J ÷ (3.00 x 108 m/s)
= 1.3 x 10-17 Nm ÷ (3.00 x 108 m/s)
= 4.3 x 10-26 N·s
Solution:
b) the momentum of the scattered x-ray
pscattered = h/l = 6.626 x 10-34 J·s ÷ (2.95 x 10-7 m)
= 2.2 x 10-27 N·s
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c) the energy imparted to the electron
Eelectron = Ex-ray - Escattered
= 1.3 x 10-17 J - hfscattered
= 1.3 x 10-17 J - (hc)/lscattered
= 1.3 x 10-17 J - (6.626x10-34 J·s x 3.00 x108 m/s
÷(2.95x10-7m))
= 1.3 x 10-17 J - 6.7 x 10-19 J
= 1.2 x 10-17 J
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d)the increase in speed of the electron
Assume the electron was
initially at rest. Find the
new speed from
Eelectron = Ee = ½ mev2.
=5.1x106m/s
2 e
e
Ev
m
17
31
2 1.2 10
9.11 10
Jv
kg
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Practice2. A scattered x-ray photon has 75% of the
momentum of the incident photon. If the wavelength of the incident photon is 150 nm, what is the wavelength of the scattered photon?
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Given: Let p1 = momentum of incident photon
and p2 = momentum of scattered photon
Then p2 = 0.75 p1
l 1 = 150 nm = 150 x 10-9 m = 1.5 x 10-7 m
h = 6.626 x 10-34 J·s l 2 = ?
Solution:
p2 = 0.75 p1 and p = h/l.
Therefore,
Cancelling h and solving
for l 2
2 1
0.75h hl l
12 0.75
ll
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-7
2
1.5 x 10 m
0.75l
72 2.0 10 ______m nml
Matter Waves
Louis de Broglie noticed that sometimes physics
ideas are reversible. For example, we have already
seen how changing electric fields produce
magnetism (Oersted) and changing magnetic fields
produce electricity (Faraday).
After Compton showed that electromagnetic
radiation (photons) exhibited properties of mass
(i.e., momentum), de Broglie wondered if mass
might have wave properties. He rearranged
Compton's equation.
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Compton's equation:
De Broglie's
rearrangement:
But p = momentum
= mass x velocity
= mv.
Therefore,
hp
l
hp
l
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hmv
l
Practice exercise 1
If you have a mass of 70 kg, find your
associated wavelength when you are running
at 8.0 km/hr (2.2 m/s).
Solution:
l= h/(mv)
=6.626 x 10-34 J·s ÷ (70 kg x 2.2 m/s)
= 4.3 x 10-36 m
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Practice exercise 2
Find the wavelength associated with an electron that
is moving at 1% the speed of light.
Solution:
Right away, you can see two differences in this exercise
and the first one: here we have an object that is
(i) extremely small
(ii) moving extremely fast.
At the end of this exercise you will realize that it is the first
difference that is the important one.
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1% the speed of light means that
v = 0.01 x 3.00 x 108 m/s = 3.00 x 106 m/s
and the mass of an electron,
m = 9.11 x 10-31 kg.
Computing the wavelength:
l= h/(mv)
= 6.63 x 10-34 J·s ÷ (9.11 x 10-31 kg x 3.00 x106 m/s)
= 2.43 x 10-10 m
= 0.243 nm.
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This wavelength is still small, but is enormous
compared to your own wavelength in
exercise 1.
2.43 x 10-10 m / (4.3 x 10-36 m )
= 6 x 10 25 (times as large as your wavelength)
The reason why this wavelength is so large
by comparison is because the mass of the
electron is so tiny.
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Can you remember the test for wave
properties from Physics 2204?
Answer: If wave properties are present, an interference pattern can be produced.
To produce such patterns the openings through which the moving matter passes must be about the
same size as the wave itself.
Right away you can see why your own associated wave cannot be detected, even if you had a few clones to interfere with!
There is no way you and your clones are going to squeeze through a hole that is 4.3 x 10-36 m in diameter!
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But for an electron, the story is different. In fact, as
amazing as it sounds, electrons do interfere constructively
and destructively, and hence they do create interference
patterns.
De Broglie's prediction that matter has an associated wave
characteristic is true.
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Practice:1. A) Calculate the wavelength for an electron that is moving at a
velocity of 42 000 m/s.
B) Calculate the wavelength for a proton that is moving at a velocity of
42 000 m/s.
C) How do the electron and protons wavelength compare?
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3. In your television set, an electron is accelerated through
a potential difference of 21 000 V.
(a) How much energy does the electron acquire?
(b) What is the wavelength of an electron of this energy?
Emission/Absorption Spectra
What happens when
light shines through a
prism?
Different frequencies of
light waves will bend by
varying amounts upon
passage through a
prism.
The separation of visible
light into its different
colors is known
as dispersion
Refraction is the bending
of the path of a light wave
as it passes from one
material into another
material.
The refraction occurs at
the boundary and
is caused by a change in
the speed of the light
wave
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Emission/Absorption Spectra
How is this related to rainbows?
To view a rainbow, your back must be
to the sun as you look at an
approximately 40 degree angle above
the ground into a region of the
atmosphere with suspended droplets
of water or even a light mist.
Each individual droplet of water acts
as a tiny prism that both disperses the
light and reflects it back to your eye.
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A continuous spectrum of light is produced when
white light passes through a prism. All colors of the
rainbow are produced.
An emission spectrum is observed when light is
emitted from excited gases. It was discovered that
only certain set frequencies or wavelengths of light
were given off (bright line emission spectra).
Now for a BOHRing Topic Niels Bohr (1885-1962) said a shocking thing: he said
that the classical ideas of physics cannot be applied to
orbiting electrons.
Classical physics theory demanded that an
accelerating charge emit electromagnetic radiation.
Electrons are certainly accelerating because they were
going around in circles.
But, if accelerating charges emit energy, the electrons
must, after some time, lose all of theirs and therefore
collapse into the nucleus. Sounds like our world
should come to a disastrous end.
112
Also, gases can be made to emit a spectrum that
does not appear as fuzzy coloured bands, but
rather as bright, sharp lines.
Some examples are below
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These sharp coloured lines at specific places in
the spectrum suggest that there is something
going on in the atom that is allowing only certain
frequencies to be emitted.
It appears that instead of the electron gradually
spiraling into the nucleus and emitting a spectrum
of light (like the bottom portion above), it makes
certain jumps only.
It moves from one orbit to another, giving out a
particular frequency as it "falls".
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Here is a summary of what Bohr said:
Of all the possible orbits that electrons might
take around a nucleus, only certain orbits are
"allowed". For each of those allowed orbits
the electron has a specific amount of energy.
While it stays in an orbit, the electron will not
radiate energy even though it is accelerating!
Each orbit is a stationary state.
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Electrons lose energy only when they "jump"
from a higher energy orbit to a lower energy
orbit.
The difference in the energies of the two
orbits is radiated as one photon of
electromagnetic radiation.
Similarly, an electron can be "bumped" up to
a higher energy level only if the atom absorbs
a photon whose energy is exactly equal to
the energy difference between the two orbits.
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What makes the allowed orbits so special?
Recall waves on a string that was fastened at both ends. The waves had to fit nicely so that each end of the string became a
node.
The picture below is a slight variation of this. That is, here we start with a fixed wave and choose the strings that work.
Notice that the strings are exact multiples of the chosen wave. No other strings would work if we wish to put one wavelength, two wavelengths and three wavelengths on them.
117
Now think back to de Broglie. He said that
matter has an associated wave.
In fact, we have calculated some waves
associated with moving electrons.
Using de Broglie's idea, Bohr proposed that
the only electron orbits allowed would be
those for which the electron wave would fit
exactly on the circumference of the orbit.
118
You can see from the picture below that the
smallest allowed orbit has a circumference of
2pr1 =l; the next largest allowed orbits have
circumferences of 2pr2 =2l ; 2pr3 =3l and so on,
up to the 4th orbit with a circumference of
2pr4 =4l.
119
In general for the nth orbit:
the radius will be rn
the circumference of the orbit will be 2prn
the de Broglie wavelength of the electron in
this orbit will be such that n wavelengths fit
on the circumference:
2prn = nl.............eq. 1.
"n" is the quantum number for the orbit.
Bohr has applied Planck's quantum idea to
determine the allowable orbits for the
electron!
120
Don't forget de Broglie's equation which
expresses the matter wave of the electron
in terms of the momentum of the electron:
l= h/p where p = meve.
Therefore, ..............eq. 2 .
121
e e
hmv
l
Use eq. 2 to substitute for l in eq. 1:
2prn = n ( h/meve), and rearrange to get
mevern = n(h/2p) ...........eq. 3.
Eq. 3 is called "Bohr's quantum condition" for allowed orbits.
In fact the left-hand side of eq. 3 has a special name: angular
momentum.
Using this and centripetal force ( caused by attractive
electric forces - Coulombs law) we get:
2 2
2 24 en
hk
nm e
rp
122
Look at the coloured bits. They are all
constants.
h = 6.626 x 10-34 J·s; p = 3.14;
k = 9.0 x 109 N·m2/C2
me = 9.11 x 10-31 kg and e = 1.6 x 10-19 C.
When you combine these constants, the equation
becomes:
rn = (5.3 x 10-11 m) n2
This is known as the Bohr radius equation.
123
2 2
2 24 en
hk
nm e
rp
As you can see a Bohr radius is directly
proportional to the square of the quantum
number associated with the radius.
Isn't it neat that the units work out to be
metres?
124
What is the orbital radius (in nm) of an
electron in a hydrogen atom in the third
energy level?
Solution:
Given n = 3 , find rn
rn = (5.3 x 10-11 m) n2
= (5.3 x 10-11 m) x (3)2
= 4.77 x 10-10 m
= 0.477 x 10-9 m = 0.477 nm
The orbital radius at the 3rd energy level is 0.477 nm.
125
Now let's get back to the real purpose of this part of the lesson:
Finding the energy of an electron energy level as a function of the quantum (n) of the level.
It was a high school teacher named J.J. Balmer (1885) who very cleverly wrote a formula to describe what he was seeing when he looked at visible line spectra like the ones described earlier.
126
The down arrows represent different
transitions of electrons as they "fall"
from higher levels to a lower level.
In each case a different colour of
visible light was emitted.
In the series of transitions shown, no
electrons fall lower than the n=2 orbit
or energy level. Why?
The jump from the second level to the
first is so great that uv light is emitted.
(which you can’t see)
127
When Balmer determined the wavelength of the
various lines in the spectrum, he discovered (by
fiddling with the numbers, not by theory), that the
inverse of the wavelength was directly proportional
to
That is,
2 2
1 1
2 n
2 2
1 1 1
2 n
l
128
Along comes Bohr, and using his theoretical model he,
too, developed a formula to describe the spectral lines.
Recall the expression for the radius of the electron orbit as
a function of the quantum number (n).
rn = (5.3 x 10-11 m) n2
Also there is an expression for the energy of the electron
as a function of the Bohr orbit radius (rn).
2
2e
keE
r
129
Substitute rn for r from the first equation into the second
equation:2
2en
keE
r
2
11 22(5.3 10 )e
keE
m n
130
29 19 2
2
11 2
9 10 (1.6 10 )
2(5.3 10 )e
N mC
CEm n
18
2
2.18 10e
N mE
n
18
2
2.18 10e
JE
n
This result must have been very pleasing to Bohr. His
theory exactly matches the experimental evidence
described by Balmer!
2
13.6e
eVE
n
2
13.6c eVh
nl
2
1 1
n
l
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which means that (ignoring constants) the
reciprocal of the wavelength is directly proportional
to the reciprocal of the energy level of the electron (n)
Why negative energy??
question: In the picture drawn earlier how many possible orbits were shown?
answer:
Four.
133
question: In which orbit or energy level does the
electron experience the greater force? Why?
answer:
In the first orbit which is the smallest orbit (n=1).
It experiences the greater force here because of the
Coulomb force caused by the positive proton in the
nucleus.
The Coulomb force is inversely proportional to the
square of the separation between the proton and the
revolving electron--that's the distance r, the radius of
the orbit.
The smaller the r, the greater the force.
134
question: In which orbit or energy level does the electron
have the greatest energy?
answer:
This is not a trick question.
It will have its greatest energy at the first energy level
(the smallest orbit).
That makes sense because it is in this orbit that thecentripetal force is greatest (due to the Coulomb force).
question: What happens to the Coulomb force or the
centripetal force as the electron moves in orbits further
and further from the positive proton in the nucleus?
answer:
The force becomes smaller.
135
question (This is the biggie): What will be the
energy of the electron if it moves very far away,
say to infinity?
answer:
It will have zero energy!
question: Can you remember how an electron can
be bumped up from one level to another?
answer:
Energy must be put into (added) the atom to "excite"
the electron to higher levels.
136
Now, let's sum up.
Energy has to be put into the atom to excite the
electrons to higher orbits. And if the electron is
bumped far enough away from the nucleus, its
energy goes to zero.
question: If adding energy makes the energy of
the electron go to zero, what kind of energy did
the electron have to begin with?
answer:
THE ENERGY TO BEGIN WITH MUST HAVE
BEEN NEGATIVE!!
137
Think about a simple case: if adding 3.0 J to an
electron makes the energy go to zero, then the
energy that the electron already had must have
been -3.0 J.
138
We will sum up by putting our new
knowledge on a picture that was drawn
earlier in the lesson.
The simplest energy calculation for an electron is
for n=1 (the lowest level allowed, the ground
state).
This value for the ground state is represented
by:
139
Ee =-13.6 eV / (1)2 = -13.6 eV.
2
13.6e
eVE
n
13.6oE eV
In the Pashen series no
electrons fall lower than
the n=3 level, while in
the Lyman series, the
electrons make it all the
way to the ground state,
n=1.
142
question: In the Balmer series a photon is
emitted as an electron falls from the n=5
state to the n=2 state.
A) What is the energy of the photon?
answer: In dropping from the upper n=5 to the lower
n=2 levels the electron loses energy lEe :
lEe = Eupper - Elower = E5 – E2
=-0.54 eV - (-3.40 eV)
= 2.86 eV.
143
B) What is the wavelength of the photon?
cE h
l
hcE
l
144
34 8
19
6.626 10 3.00 10
1.602 102.86
mJ s
sJ
eVeV
l
74.34 10 m
C) What type of light is
this?
Answer:
Visible violet light
The colors of the visible light spectrum
colorwavelength
interval
frequency
interval
red ~ 625–740 nm ~ 480–405 THz
orange ~ 590–625 nm ~ 510–480 THz
yellow ~ 565–590 nm ~ 530–510 THz
green ~ 500–565 nm ~ 600–530 THz
cyan ~ 485–500 nm ~ 620–600 THz
blue ~ 450–485 nm ~ 670–620 THz
violet ~ 380–450 nm ~ 790–670 THz
145
http://en.wikipedia.org/wiki/Color
question: In the Lyman series a photon is
emitted as an electron falls from the n=2
state to the n=1 state.
A) What is the energy of the photon?
B) What is the frequency of the photon?
C) What type of light is this?
D) Why was this jump not listed in the Balmer series?
146
Here's what is really good about Bohr's
model:
it matches the observed spectrum of hydrogen gas
it accounts for the stability of atoms.
That is, once the electron is in its lower orbit or ground state, it will go no further.
It will not lose energy while it is in a fixed orbit and therefore will not spiral into the nucleus.
the model can be applied to other one-electron atoms, such as a helium ion that has lost one of its two electrons.
147
From two of the above statements, you will not be surprised to learn that Bohr's model of the atom did not work for atoms with more than one electron.
Also, remember those bright discrete lines that we got so excited about earlier in the lesson? It turns out that they are not so discrete after all. Sophisticated equipment has shown that each of the lines is, in fact, composed of separate finer lines.
To explain this "splitting of the levels" the model has to be modified to include orbits which are not circular and electrons that spinas well as revolve.
148
Moreover, in the latest theory, electrons are
not considered to be particles revolving in a
single orbit.
Brace yourself for the next sentence:
electrons are now considered to be a
wave found in a probability cloud of negative
charge distributed around the nucleus.
The cloud is densest in areas of high
probability, and less dense in areas where
the electron is less likely to be found. ....but
that's a topic for another time (college?
university?)
149
Practice exercises 1. What is the ionization energy of hydrogen?
Solution:
This is not a fair question if you don't know what ionization means. An atom becomes ionized when it gains or loses
electrons.
An ion is a charged atom.
In this exercise you are asked how much energy is required to strip the electron from a hydrogen atom so that only the positive nucleus remains.
Assume that the atom is in an unexcited state--that is, the electron is at n=1. What is its energy there?
-13.6 eV.
150
In order for the electron to be removed it must be
bumped to such a high orbit number that its
energy goes to 0 eV. How can -13.6 eV be
changed to 0 ?
Answer: just add 13.6 eV
The ionization energy of hydrogen is 13.6 eV.
In other words, the ionization energy is the same
as the energy at the ground state.
151
2. An electron drops from the n=4 to the n= 1 energy
level. In what region of the electromagnetic spectrum is
this photon found?
Solution
When the electron falls from n=4 to n=1, a photon is emitted.
The energy of the photon (Ep) will be equal to the difference between the energies at n=4 and n=1.
152
E4 = -13.6/n2 eV = -13.6/42 eV = -0.85 eV.
E1 = -13.6/12 eV = -13.6 eV.
Therefore Ep = E4 - E1
= -0.85 eV - (-13.6 eV)
= 12.8 eV
Before applying Planck’s equation, change 12.8 eV
to joules by multiplying by 1.6 x 10-19 J/eV.
This gives
Ep = 2.0 x 10-18 J .
153
Phosphorescence and Fluorescence
There are examples of natural luminosity such
as phosphorescence and fluorescence that can
be explained by the Bohr model of the atom.
Phosphorescent and fluorescent materials both
contain atoms that are easily excited.
Phosphorescence and fluorescence differ in the
amount of time it takes their excited atoms to
return to their normal energy levels.
What does an electron give off as it returns to its
normal energy level?
Photons
155
Phosphorescence
Phosphorescence occurs when atoms are excited
by absorption of a photon to an energy level said to
be metastable.
Normally when an atom is raised to a normal excited
state, it drops back down quickly and releases a
photon of energy at a lower energy level than the
incident photon.
Metastable states last much longer.
In a collection of atoms, many of these atoms will
remain in the excited state for over an hour.
Hence, light will be emitted even after long periods.
156
Examples of phosphorescence are:
luminous watch dials
glow in the dark stickers
glow in the dark tent lines
157
Fluorescence
Fluorescence occurs when an atom is excited
from one energy state to a higher one by the
absorption of a photon.
The atom may return to the lower level in a
series of two or more jumps.
This happens almost immediately (within about
10−8 seconds ( 10 ns)).
The emitted photons will have lower energy
and frequency than the absorbed photon.
158
Fluorescence Because reemission occurs so quickly, the
fluorescence ceases as soon as the exciting
source is removed, unlike phosphorescence,
which persists as an afterglow.
Examples of fluorescence include:
trail marker tape used in hiking and road signs.
A fluorescent lightbulb is coated on the inside with a
powder and contains a gas; electricity causes the gas
to emit ultraviolet radiation, which then stimulates the
tube coating to emit visible light.
159
Tooth whiteners and toothpaste may contain
molecules that make your teeth glow brightly
under black light.
Optical brighteners in laundry soap
This is how laundry detergents can get things "whiter than white": by absorbing non-visible UVlight and fluorescing in the visible spectrum.
160
IT’S ALIVE… AND IT GLOWS!
FLUORESCENCE IN MEDICAL RESEARCH
161
Remember,
the key difference between both is the length
of time they will last.
Phosphorescence lasts longer due to the
metastable state of atoms and will remain
luminous after the stimulating radiation is
absent.
Fluorescence will only be luminous while the
stimulating radiation is present.
162
Final Objective for Unit 4:Students should be able to summarize the evidence for
the wave and particle models of light
That means
define wave-particle duality
The property of matter and electromagnetic
radiation that is characterized by the fact that
some properties can be explained best by wave
theory and others by particle theory
163
give evidence of light being both a wave:
behaviour of long wavelengths - AM radio waves
interference - In Young’s double slit experiment
where coherent light (light which is in the same
phase, and has the same frequency and
wavlength) interferes to produce an interference
pattern similar to those found in water waves.
diffraction - Rainbows
164
a particle:
blackbody radiation
photoelectric effect
Compton effect
line spectra
behaviour of short wavelengths like x-rays
165