### CHAPTER 7 LEARNING OBJECTIVES - crypt

```7 QUANTUM BEHAVIOUR
The quantum nature of light
• Observe phenomena where light displays a particle-like
nature
• Observe and explain the photoelectric effect in terms of
the quantum theory of light
Light is a wave
motion!
Nein, light is
particles!
the quantised nature of
Gamma source
The Electron Volt….
Energy in Joules = energy in eV x 1.6x10-19 JeV-1
This is the work done when an electron is moved through
a potential difference of 1 volt. Since the charge of the
electron is 1.6x10-19 C, then 1 eV = 1.6x10-19 J.
The Photoelectric Effect….
Evidence for the graininess of lig
Photoelectricity
The photoelectric effect
photons
metal
electron absorbs
photon and
leaves the metal
–
High-frequency photons eject electrons
from clean metal surfaces. Some of the
energy transferred by the photon extracts
the electron, some ends up as kinetic
energy of the electron.
E/J
E = e Vblue
Note:
Energy to extract an electron = Φ (the work function)
Evidence for the graininess of light
The photoelectric effect
High-frequency photons eject electrons
from clean metal surfaces. Some of the
energy transferred by the photon extracts
the electron, some ends up as kinetic
energy of the electron.
Find the kinetic
energy of the
electron from the p.d.
needed to stop them.
A
E/J
E = e Vblue
+
E = e  V green
energy () to extract
one electron from
the metal (the work
function).
The energy of the photon
is measured from this
the energy to extract the
electron from the metal.
Constant slope, E/f. The
number of joules per hertz
is constant for all
h, the gradient, is 6.634 
10–34 J Hz–1
More often written as h =
6.634  10–34 J s
+
Note:
+
constant!
f0
fgreen
f/Hz
f
blue
Four key observations of the Photoelectric Effect
1. The number of photoelectrons emitted per second is
proportional to the intensity of the incident radiation.
2. For a given metal, there is a certain minimum frequency of light
below which no emission occurs no matter how intense the
3. Photoelectrons are emitted with a range of kinetic energies up
to a certain maximum value. This maximum value increases if light
of higher frequency (shorter wavelength) is used. The maximum
kinetic energy is completely independent of the radiation
intensity.
4. Even with a source of low intensity, there is no time delay
between starting to illuminate the metal and the production of
photoelectrons.
Evidence for the graininess of light
The photoelectric effect
High-frequency photons eject electrons
from clean metal surfaces. Some of the
energy transferred by the photon extracts
the electron, some ends up as kinetic
energy of the electron.
Find the kinetic
energy of the
electron from the p.d.
needed to stop them.
A
E/J
E = e Vblue
+
E = e  V green
energy () to extract
one electron from
the metal (the work
function).
The energy of the photon
is measured from this
the energy to extract the
electron from the metal.
Constant slope, E/f. The
number of joules per hertz
is constant for all
h, the gradient, is 6.634 
10–34 J Hz–1
More often written as h =
6.634  10–34 J s
+
Note:
+
Planck's
constant!
f0
fgreen
f/Hz
f
blue
Planck’s Equation
E = hf = hc/λ
Einstein’s Photoelectric effect equation
Photon energy = Work function + Kinetic Energy
hf
=
Φ
+
Ek
Key:
h – Planck’s constant (6.63x10-34Js)
f – frequency of photon (Hz)
Φ – work function (min energy required to eject a photon (J))
c – speed of light (3x108ms-1)
λ – wavelength (m)
Starter:
Annotate the graph with the answers to the following questions.
Q1. Add a line to the graph for a metal with a lower work function than the one depicted.
Q2. Show how you would determine the work function for a metal from the graph.
Q3. Highlight the point on the graph corresponding to irradiation of the metal with the
shortest wavelength light. What colour is the light in this case?
Q4. Highlight the point on the graph corresponding to irradiation of the metal with the
longest wavelength light. What colour is the light in this case?
Q5. Light of wavelength 488 nm irradiates the metal surface. Use the graph to determine
the maximum kinetic energy electrons could have when the surface is irradiated in this
way.
Electron KE against light frequency
12
Electron KE (10^-20 J)
10
8
6
4
2
0
0
1
2
3
4
frequency (10^14 Hz)
5
6
7
8
Example questions….
h is Planck’s constant (6.63x10-34Js)
1) Blue light has a frequency of 7.7x1014Hz. What energy
does a blue photon have? (in both J and eV)
2) What is the energy of a red photon (in J and eV) in
red light with a wavelength of 7x10-7m.
3) Light of frequency 6.7x1014Hz shines on to clean
caesium metal. What is max kinetic energy of
electron emitted? (Φ for caesium = 3.43x10-19J)
4) Light of wavelength 4.2x10-7m is incident on
potassium metal. If the kinetic energy of the emitted
electron is 0.78eV what is the work function?
Two lasers have identical output powers.
Q1. What can you say about the amount of energy
emitted by each laser every second?
Q2. Laser A emits red light, laser B emits blue light.
Which laser emits more
photons every second?
Estimating the photons emitted from a
The power of reading lamp is 40W
Average wavelength of light emitted from the
lamp is 5x10-7m.
1. Calculate the energy transferred by each
photon.
2. Calculate the number of photons emitted by
the lamp in each second.
Evidence for the graininess of light
Light-emitting diodes
Particular LEDs are engineered to
drop each electron by a particular
striking potential difference (p.d.),
when just glowing, and so to emit a
photon of a particular colour.
Striking p.d. fixes
energy
E/J
E = qV
E = e Vblue
+
E = e  V green
energy
energy transfered
transferred
to a photon
=
by one
of particular
electron
frequency
E = e  Vred
+
+
e V = hf
f/Hz
fred light
fgreen light
f
blue light
Constant slope, E/f. The number of joules per hertz is uniform for all
h, the gradient, is 6.634  10–34 J Hz–1. More often written as
h = 6.634  10–34 J s
Quantum calculations
• Learn about the “Try all
paths” algorithm for
quantum calculations
• Explain light phenomena in
all possible paths
Starter: Find the
length of the
the 3 phasor arrows
shown. Assume that
each one has length 1
unit.
“How do the photons know which
path to take?”
“Try all paths” is a quantum rule obeyed by allSource
photons and
electrons.
Detector
Richard Feynman argued that instead of just taking one
route between the source and the detector, a photon will
take all of the possible paths to the detector in one go. You
can keep track of this photon going on along every possible
route using phasors.
Quanta and phasors:
E = hf and c = f can be used to work out the number of
phasor rotations for a photon taking a particular path.
1) What is the rate of rotation of a phasor associated a
photon with energy 5.1x10-19J?
2) What is the rate of rotation of a phasor with light of
wavelength of 7x10-7m?
3) A photon of wavelength 5.2x10-7m explores a path
4m long. What is its travel time and how many times
does the phasor rotate?
Now try questions 3 and 4 on page 157
Calculating the probability of arrival
P1
P2
Resultant
Probability α (resultant phasor)2
Question…
Question: The resultant phasors for a photon
reaching points A and B have magnitudes of
6.3 and 4.5 respectively. How many times more
likely is it that a photon will arrive at point A
than at point B?
(2 marks)
Does the rule ‘explore all paths’ really work in all
situations.
On computers explore the following software activities
to check out this rule in three crucial situations….
reflection with Activity 110S, 120S, 140S, 170S;
refraction with 190S and 200S
diffraction with 210S.
Trip times, phase differences
Paths for the
mirror
D
S
t
x
large
Trip time t for
path
t
x
t
x
large
t
x
small
t
x
t
x
x, position along mirror
Arrows from
each path
Sum of
arrows gives
amplitude
large differences in trip time
small differences in trip time large differences in trip time
arrows curl up
arrows line up
arrows curl up
small contribution to
amplitude
large contribution to
amplitude, from small part
of mirror
small contribution to
amplitude
Arrows line up where the trip time hardly changes as the path varies
Wave properties of electrons
• Describe the evidence for
wave behaviour of entities
we normally regard as
particles*
(* electrons, neutrons, large
molecules, even bigger
Starter: Describe a physical
stuff...........)
situation/observation where electrons
are best described as particles.
How does the particle description of
electrons help to explain it?
Copy and complete...
Evidence for light as
particles
ANIMATION
Evidence for light as
waves
Evidence for
Evidence for electrons as
electrons as particles
waves
Kinetic energy (J) = electronic charge × accelerating voltage
Electronic charge = -1.6 x 10 -19 C
KE = e × V
1. If the accelerating voltage in a computer monitor is 400V what is the
total kinetic energy acquired by a single electron?
2. An electron acquires a Kinetic Energy of 600 x 10-17J in a cathode ray
tube. What voltage was required to provide this energy?
3. A kinetic energy of 0.02 mJ per second was transferred to a screen of a
television during a TV programme. If the voltage was 500V how many
electrons per second would be hitting the screen.
4. (a) How many electrons would hit the screen in 5 seconds in an
oscilloscope if a current of charge of 0.01 A is flowing? (Hint I = Q/t)
(b) If the voltage across the same oscilloscope was 100V then what would
be the total Kinetic Energy transferred to the screen by the electrons?
Kinetic energy = electronic charge × accelerating voltage
KE = e × V
To diffract
electrons you
need very small
slits – the gaps
between atoms
Electron diffraction
This eerie green glow is
caused by electrons in a
cathode ray tube striking
the phosphorescent
coating on the inside of
the glass bulb just behind
the ruler.
In this case the diffraction is
caused by the electrons
passing through a thin layer
of polycrystalline graphite
array of carbon atoms in the
crystals is responsible for
the diffraction effects.
Prince LouisVictor-PierreRaymond, 7th
duc de Broglie
λ=
____h____
momentum
Everything acts with quantum
behaviour – if we give particles
enough energy they will have a
wavelength
λ=
____h____
momentum
The distance it
takes to make one
complete turn of
the phasor
momentum =
mass x velocity
The de Broglie wavelength can be thought of as the distance
travelled by the photon for one turn of the phasor associated
with it.
The de Broglie wavelength is given by λ = h/p, where p is the
momentum of the particle (=mass x velocity, mv)
If you know the particle’s kinetic energy, the speed can be
worked out using KE = ½ mv2, and hence momentum (=mv) can
be found. The de Broglie wavelength can then be calculated.
Note: The energy expression used for photons (E=hc/ λ) can also
be used to find the de Broglie wavelength of particles, but this
method is normally reserved for very high energy electrons (with
energies of 100s of MeV) where the relationship E = pc holds.
NOW DO Q1-4 FROM P167
Solutions:
Q1. KE = e x V = 1.6 x 10-19 C x 100 V = 1.6 x 10-17 J
Q2. KE = ½ mv2 = 1.6 x 10-17 = ½ x 9 x 10-31 x v2
v = 5.9 x 106 ms-1
Q3. momentum = mv = 9 x 10-31 x 5.9 x 106
= 5.3 x 10-24 kg ms-1
Q4. λ = h/p = 6.6 x 10-34 / 5.3 x 10-24 = 1.2 x 10-10 m
Two slit interference with electrons
The greatest
physics
experiment…
…in the
world.
..\doubleslite-n.wmv
JJ and GP Thomson: father and son Nobel
physics prize winners
Listen sonny, I got the
Nobel prize for
showing that the
electron is a particle!
Yeah whatever