7 QUANTUM BEHAVIOURThe 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

Nein, light is made of particles!

Light is a wave motion!

What about evidence for the quantised nature of

radiation?

Gamma source

What about evidence for the quantised nature of radiation….

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….Photoelectricity

metal

photons

electron absorbsphoton andleaves the metal

–

Evidence for the graininess of light

Constant slope, E/f. Thenumber of joules per hertzis constant for allelectromagnetic radiation.

h, the gradient, is 6.634 10–34 J Hz–1

More often written as h =6.634 10–34 J s

The photoelectric effect

High-frequency photons eject electronsfrom clean metal surfaces. Some of theenergy transferred by the photon extractsthe electron, some ends up as kineticenergy of the electron.

The energy of the photonis measured from thiskinetic energy added tothe energy to extract theelectron from the metal.

Find the kineticenergy of theelectron from the p.d.needed to stop them.

energy () to extractone electron fromthe metal (the workfunction).

fblue

+

f/Hz

E/J

f0 fgreen

E = e Vblue

E = e Vgreen+

+

A

Note:Energy to extract an electron = Φ (the work function)

Evidence for the graininess of light

Constant slope, E/f. Thenumber of joules per hertzis constant for allelectromagnetic radiation.

h, the gradient, is 6.634 10–34 J Hz–1

More often written as h =6.634 10–34 J s

The photoelectric effect

High-frequency photons eject electronsfrom clean metal surfaces. Some of theenergy transferred by the photon extractsthe electron, some ends up as kineticenergy of the electron.

The energy of the photonis measured from thiskinetic energy added tothe energy to extract theelectron from the metal.

Find the kineticenergy of theelectron from the p.d.needed to stop them.

energy () to extractone electron fromthe metal (the workfunction).

fblue

+

f/Hz

E/J

f0 fgreen

E = e Vblue

E = e Vgreen+

+

A

Note:GRAD is Planck's constant!

Four key observations of the Photoelectric Effect1. 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 radiation. 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

Constant slope, E/f. Thenumber of joules per hertzis constant for allelectromagnetic radiation.

h, the gradient, is 6.634 10–34 J Hz–1

More often written as h =6.634 10–34 J s

The photoelectric effect

High-frequency photons eject electronsfrom clean metal surfaces. Some of theenergy transferred by the photon extractsthe electron, some ends up as kineticenergy of the electron.

The energy of the photonis measured from thiskinetic energy added tothe energy to extract theelectron from the metal.

Find the kineticenergy of theelectron from the p.d.needed to stop them.

energy () to extractone electron fromthe metal (the workfunction).

fblue

+

f/Hz

E/J

f0 fgreen

E = e Vblue

E = e Vgreen+

+

A

Note:gradient is Planck's constant!

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)

4.5 5 5.5 6 6.5 7 7.50

2

4

6

8

10

12

Chart Title

frequency (10^14 Hz)

Elec

tron

KE

(10

-̂20

J)

Electron KE against light frequency

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.

h is Planck’s constant (6.63x10-34Js)Example questions….

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?

Justify your answer.

The power of reading lamp is 40WAverage 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.

Estimating the photons emitted from a reading lamp…

Evidence for the graininess of light

Light-emitting diodesStriking p.d. fixesenergy

Constant slope, E/f. The number of joules per hertz is uniform for allelectromagnetic radiation.h, the gradient, is 6.634 10–34 J Hz–1. More often written ash = 6.634 10–34 J s

fblue light

+

f/Hz

E/J

fred light fgreen light

E = e Vblue

E = e Vgreen

E = e Vred

E = qV

+

+

Particular LEDs are engineered todrop each electron by a particularstriking potential difference (p.d.),when just glowing, and so to emit aphoton of a particular colour.

energy transferedto a photonof particularfrequency

energytransferredby oneelectron

=

e V = hf

Quantum calculations• Learn about the “Try all

paths” algorithm for quantum calculations

• Explain light phenomena in terms of phasor addition for all possible paths

Starter: Find the length of the resultant on adding the 3 phasor arrows shown. Assume that each one has length 1 unit.

“How do the photons know which path to take?”

SourceDetector

“Try all paths” is a quantum rule obeyed by all photons and electrons.

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:

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?

E = hf and c = f can be used to work out the number of phasor rotations for a photon taking a particular path.

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)

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.

Does the rule ‘explore all paths’ really work in all situations.

Paths for themirror

Trip time t forpath

Arrows fromeach path

Sum ofarrows givesamplitude

large differences in trip time small differences in trip time large differences in trip time

arrows curl up arrows curl uparrows line up

x, position along mirror

tx

small

tx

large

tx

larget

x

tx

xt

small contribution toamplitude

large contribution toamplitude, from small partof mirror

small contribution toamplitude

S D

Trip times, phase differences

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 stuff...........) Starter: Describe a physical

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

particlesEvidence for light as

waves

Evidence for electrons as waves

Evidence for electrons as particles

ANIMATION

Kinetic energy (J) = electronic charge × accelerating voltage

KE = e × V1. 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?

Elec

tron

ic c

harg

e =

-1.6

x 1

0 -1

9 C

Kinetic energy = electronic charge × accelerating voltage

KE = e × V

To diffract electrons you

need very small slits – the gaps between atoms

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 (pencil "lead"). The regular array of carbon atoms in the crystals is responsible for the diffraction effects.

Electron diffraction

Prince Louis-Victor-Pierre-Raymond, 7th duc de Broglie

Everything acts with quantum behaviour – if we give particles enough energy they will have a

wavelength

λ = ____h____ momentum

λ = ____h____ momentum

The distance it takes to make one complete turn of

the phasor

momentum = mass x velocity

For electr

ons only

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.

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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 Dad. I got it for showing that the electron is a wave!