2.
Problems with Planetary Atomic Model
The electron, in orbiting the nucleus,
undergoes an acceleration and should
therefore continuously emit
electromagnetic radiation
The predictions are completely wrong!!!
The successful model which came out of these
attempts is now called the Bohr model and will be
discussed in more detail later on
3.
Photoelectric Effect
When light shines on a metal
surface, the surface emits
electrons.
For example, you can measure a current
in a circuit just by shining a light on a
metal plate
4.
The current flow varies with the
wavelength of light such that there was a
sharp cutoff and no current flow for long
wavelengths.
The photocurrent increases when the
light intensity increases but the
wavelength is held constant.
5.
Einstein successful explained the
photoelectric effect within the context of
quantum physics.
Einstein proposed that light delivers its energy in chunks;
light consists of little particles, or quanta, called photons,
each with an energy of Planck's constant times its
frequency.
E=hf
h: Planck’s constant
F: Frequency of Radiation
E: Energy
h = 6.6 x 10-34 J.s.
6.
Each photon carries a specific energy related to its
wavelength, such that photons of short wavelength (blue
light) carry more energy than long wavelength (red light)
photons
Photons when they impact a metal, if their frequency is large
enough, can liberate an electron by overcoming the work
function of the metal
7.
Light made of two colors (two frequencies) shines on a metal
surface whose photoelectric threshold frequency is 6.2 x 1014
Hz. The two frequencies are 5 x 1014 Hz (orange) 7 x 1014 Hz
(violet).
(1) Find the energies of the orange and violet photons.
(2) Find the amount of energy a photon needs to knock
electrons out of this surface.
(3) Do either the orange photons or the violet photons have
this much energy?
8.
1. The energy for the orange frequency is:
E = hf = (6.6 x 10-34) x (5 x 1014)
E = 3.3 x 10-19 J.
The energy for the violet light is:
E = hf = (6.6 x 10-34) x (7 x 1014)
E = 4.6 x 10-19 J.
2. The threshold energy is
E = hf = (6.6 x 10-34) x (6.2 x 1014)
= 4.1 x 10-19 J.
3. The blue photons have sufficient energy to knock
electrons out, but the orange photons do not.
9.
The human eye can detect as few as 10,000 photons per second
entering the pupil. About how much energy is this, per second?
(Make an estimate)
Solution
10,000 = 104
The energy of 10,000 visible photons is:
(104) (hf) = 104 x (6.6 x 10-34 J-s) x (1015 Hz)
= 6.6 x 10-15 joules.
11.
Regular old classical
physics
Newtonian
Physics
When things get small
Quantum
physics
When things get fast
Special
relativity
Small &
fast
Quantum
field theory
12.
Classical Physics: Newtonian Physics
Modern Physics: Relativity, quantum physics,
and any other field that employs these theories.
13.
Quantum Physics: the “quantum” comes
from quantization: we need to understand
the origin of this.
The Photoelectric Effect
electrons
metal
light
Electron
detector
14.
Actual Experimental Observations:
[1] There is no delay between the light hitting the surface and
the electrons being ejected
[2] Electrons are ejected only if the incident light has a frequency
above some threshold value (i.e., it depend on the color of the light!!)
electrons
metal
light
Electron
detector
15.
# of electrons
ejected
Many emitted
electrons
Threshold frequency
No emitted
electrons
Red orange yellow green blue indigo violet ultraviolet
frequency
16.
Energy
Before Planck: the
particle could have
any energy
Vibrating
particle
17.
Energy
After Planck: the
particle could have
only specific,
quantized energies
Vibrating
particle
18.
Energy
E1
This means that if a
particle is in some energy
state, if it is to give up
energy in the form of light,
it has to give up a definite
amount
Light with energy
equal to E1-E2!!
E2
Vibrating
particle
Same diagram as last
slide except “blown up”
to see the spacing
better.
19.
Einstein argued that the emission of the light must occur
in instantaneous bursts of radiation.
He then took this one step further: he said that all light had the
following properties:
[1] All radiation occurs as tiny bundles (particles).
[2] The bundles always move a lightspeed (c).
[3] They have zero rest-mass.
[4] The energy of a photon is given by
E = hf
h = Planck’s constant = 6.6 x 10-34 J x s
20.
How do we explain why there is a threshold frequency?
# of electrons
ejected
Many emitted
electrons
The electron must get this
much energy (E = hf) from the
photon.
This energy is different for
all metals and is called the
work function of the metal.
Threshold frequency
No emitted
electrons
0
Red orange yellow green blue indigo violet ultraviolet
frequency
21.
Radiation: waves or particles?
Radiation is made up of particles called photons
Evidence of the wave nature from interference
Essentially, wave/particle duality employs the notion that
an entity simultaneously possesses localized (particle) and
distributed (wave) properties.
Close inspection shows that an interference pattern is formed
by individual photons hitting all over a screen!!!
Interference light were made up of waves.
Pattern "speckled“ individual particle impacts
22.
What is Light?
Is it a WAVE?
Remember the interference pattern with the laser going through
two apertures?
It definitely BEHAVES like a wave!
Is it a PARTICLE?
Photoelectric effect: It BEHAVES like a
particle.
So which one is it? Does it behave like a wave or a
particle?!!
23.
So which one is it? Does it behave like a wave or a
particle?!!
Answer: It behaves like BOTH! We call this the wave-particle duality
of radiation (light).
24.
What is the universe made of?
Matter and radiation (as far as we know
anyway).
How do matter and radiation interact with each other??
Examples:
Reflection
Refraction
Absorption
Photoelectric effect
Burning/melting (lasers) …
26.
De Broglie, a Ph.D. student at the University of Paris in
1923, felt that if radiation exhibits wave-particle duality,
matter should have a dual nature too!!!!!
This was confirmed using a double-slit experiment with
particles of matter, such as electrons.
28.
Every material particle has wave properties with a wavelength
equal to:
λ = h/ms (De Broglie)
m: mass of particle
s: its speed
h: Planck’s constant
29.
In-class Problem
Calculate the de Broglie Wavelength of a basketball (m = 1 kg) moving at a speed of 1
m/s. How does this wavelength compare with the size of an atom (~ 1 x 10-9m)?
Solution:
λ = h/mv = (6.6 x 10-34 J·s)/(1 kg · 1 m/s) = 6.6 x 10-34 m (MUCH smaller than the atom!)
What is the de Broglie wavelength of an electron (m = 9 x 10-31 kg) moving at a speed of
1 x 107 m/s?
Solution:
λ= h/mv = (6.6 x 10-34 J·s)/{(9 x 10-31 kg · 1 x 107 m/s)
= (6.6/9)(10-34)/ (10-31 · 107 ) ( J · s · kg · m/s)
= 0.7 x 10-10 m (about the same size as an atom!)
32.
sand
Two overlapping
piles of sand
electrons
Not just two
overlapping “piles”
of electrons
33.
sand
Two overlapping
piles of sand
electrons
Interference pattern
34.
Since the wavelength of the electron is so small, the apertures have to be
very narrow for us to see the interference. The experiment originally
devised to observe this interference was to force electrons through a
metallic foil. The separation of atoms in the foil was the aperture size.
Electrons
Metal
foil
35.
Note: J. J. Thompson discovered the electron in 1900 and
called it a “charged particle.”
In 1927, G. P. Thomson did the “electron interference”
experiment to show that electrons behaved like waves.
G. P. was J. J.’s son!
The wave theory of matter: Every particle has wave properties with a wavelength
equal to h/mv, where m is the particle’s mass, v its speed, and h is Planck’s
constant.
40.
Screen
Given ONE photon, we cannot predict exactly where it will hit.
We can only predict the PROBABILITY that it will hit a certain place
on the screen: i.e., we can predict the pattern that many photons will
make!!
41.
We can only predict the PROBABILITY that it will hit a certain place
on the screen: i.e., we can predict the pattern that many photons will
make!!
This is where Einstein had a problem with quantum mechanics,
and also the origin of his famous quote: “ … God does not play
dice with the universe.”
42.
The pattern we observed can be predicted through the
application of quantum theory. Erwin Schroedinger
came up with an equation that predicts the “probability
wave” Ψ (psi). We call it a psi-wave.
This mathematical term psi, is used predict all kinds of
physical phenomena at the atomic level—from interference
patterns to light emitted from atoms.
A physicist will have had 2 to 3 full years of quantum mechanics
classes just to be functional with the theory!!
43.
One of quantum theory’s main applications: describing the atom
One type of spectroscope:
aperture
screen
prism
Gas
tube
Why LINES?
44.
What energy transitions are possible?
Energy
E4—E3, E4—E2, E4—E1
If these energy differences are:
E5
E4
E3
E2
E1
2 x 10-19J, 5 x 10-19J, 9 x 10-19J,
Respectively, what are the
frequencies of the emission lines
that we should see?
E = hf so f = E/h where
(f = frequency, h = 6.6 x 10-34 J·s)
For E4—E3,
f = (2 x 10-19J) / (6.6 x 10-34 J·s)
= 0.3 x 1015 Hz
f = 3.0 x 1014 Hz
Similarly,
f = 7.5 x 1014 Hz (for E4—E2)
f = 13.5 x 1014 Hz (for E4—E1)
45.
Not only can these atom only
EMIT photons of a certain
frequency, but they can only
ABSORB light of a certain
frequency.
Energy
E5
E4
E3
E2
Photon
E1
46.
To get atoms to EMIT light, we use the discharge tube. We had to
give it energy to “pump up” the electrons in the atoms so that they
could fall back into lower rungs on the ladder and emit photons.
aperture
prism
Gas
tube
screen
47.
One of quantum theory’s main applications: describing the atom
aperture
screen
Volume of
atoms (gas)
prism
White
light
What should we
see??!!
48.
Parts of the spectrum are absorbed, i.e., those
at frequencies of allowed transitions.
49.
4
3
2
1
A certain type of atom has only four energy
levels, as shown in the diagram. The "spectral
lines" produces by this element are all visible,
except for one ultra-violet line. The quantum
jump that produces the UV line is
(a) state 2 to 1.
(b) state 4 to 1.
(c) state 4 to 3.
(d) state 1 to 4
(e) impossible to determine without further
information.
50.
4
3
2
1
Continuing the preceding question, the total
number of spectral lines produced by this
element is
(a) 3.
(b) 4.
(c) 6.
(d) 10.
(e) impossible to determine without further
information.
51.
Comparing a radio photon with an infrared photon,
(a) they both have the same wavelength, and the infrared
photon moves faster.
(b) the radio photon has the longer wavelength, and the
infrared photon moves faster.
(c) the radio photon has the shorter wavelength, but they
both move the same speed.
(d) the radio photon has the longer wavelength, but they
both move the same speed.
(e) the radio photon has the longer wavelength and it also
moves faster.
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