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PhysicsHSC Course

Stage 6

Ideas to implementation

Part 3: Extraordinary science

Part 3: Extraordinary science 1

Contents

Introduction................................................................................ 3

Extraordinary science................................................................ 5

Faraday and Maxwell ...........................................................................5

Hertz......................................................................................................6

Antennae................................................................................. 13

Receiving with an antenna ............................................................... 13

Broadcasting with an antenna ......................................................... 14

Generating electromagnetic signals ................................................ 14

Planck: Is light a wave?........................................................... 16

Light: waves versus quanta .............................................................. 16

The quantum theory .......................................................................... 17

Explaining blackbody radiation ........................................................ 21

The photoelectric effect .......................................................... 27

Einstein and the photoelectric effect ................................................ 28

Studying the photoelectric effect ...................................................... 28

Photoelectric effect characteristics .................................................. 32

Einstein’s explanation ...................................................................... 34

Photon energy, frequency, c and l................................................... 38

Summary ................................................................................. 40

2 From ideas to implementation

Suggested answers .................................................................43

Exercises – Part 3....................................................................47


Introduction

In the 1880s, before the discovery of the electron, when the

electromagnetic spectrum was only a theoretical construct, brilliant

scientists and experimenters set forth on a journey of discovery that

ushered in the information age you learned about in The world

communicates.

The basic research these giants of science initiated and performed took

the world on the journey toward the information age and enabled

communication like at no other time in history. The opportunities made

available to ordinary people with increased access to information and

experiences changed the world forever. Enjoy this part of the module

and consider the impact of the implementation of the ideas presented in

the learning material on your daily life.

At the end of Part 3, you will have had opportunities to learn to:

• explain qualitatively Hertz’s experiments in measuring the speed of

radio waves and how they relate to light waves

• describe Hertz’s observation of the effect of a radio wave on a

receiver and the photoelectric effect he produced but failed to

investigate

• outline applications of the production of electromagnetic waves by

oscillating electric charges in radio antennae

• identify Planck’s hypothesis that radiation emitted and absorbed by

the walls of a black body cavity is quantised

• identify Einstein’s contribution to quanta and its relation to black

body radiation

• explain the particle model of light in terms of photons with particular

energy and frequency

• identify the relationships between photon energy, frequency, speed

of light and wavelength:

E = h f and c = f¥ ¥ l .


At the end of Part 3, you will have had opportunities to:

• perform an investigation to demonstrate the production and reception

of radio waves

• perform a first-hand investigation to demonstrate the photoelectric

effect

• solve problems and analyse information using:

E = h f and c = f¥ ¥ l

• identify data sources, gather, process and analyse information and

use available evidence to assess Einstein’s contribution to quanta and

their relation to black body radiation.

Extracts from Physics Stage 6 Syllabus © Board of Studies NSW, originally

issued 1999. The most up-to-date version can be found on the Board's website

at http://www.boardofstudies.nsw.edu.au/syllabus99/syllabus2000_list.html


Extraordinary science

Faraday and Maxwell

Early in the 19th century, Michael Faraday, an English physicist,

demonstrated that an electric current can produce a magnetic field and

that the energy in this field returns to the circuit when the current is

stopped or changed such as when the current is reversed in its direction.

James Maxwell, a Scottish physicist, summarised and expressed

mathematically the physical laws that relate light, electricity, magnetism

and other electromagnetic phenomena in 1873. Maxwell expressed those

laws in the form of mathematical equations.

A centre point of the equations was Maxwell’s assertion that light is an

electromagnetic oscillation. He argued that propagation of waves at a

very high frequency can only occur as a self sustaining mutual generation

of an electric field and magnetic field. The magnetic field produced an

electric field that produced a magnetic field and so on, forever. This

explained why light was able to travel through the vast distances of

space.

The significance of Maxwell’s analysis of the mutually generating

magnetic and electric fields was, not only that it explained the

propagation of electromagnetic waves, but also that it predicted

electromagnetic waves could have a full range of frequencies.

This implied the existence of the full electromagnetic spectrum.

Only light and infra-red radiation as discovered by Herschel in 1800 was

known to exist when Maxwell completed his great work.

The recognition of the relationship of the magnetic field and

complimentary electric field also enabled Maxwell to determine the

velocity of electromagnetic waves. (Remember that these waves were

not even known to exist at this time.) The velocity theoretically

calculated by Maxwell was 3 108¥ ms-1 which was the known speed of

light in air or vacuum.


The significance of this point is that for a wave to belong to the

electromagnetic spectrum it must therefore have a velocity of 3 108¥ ms-1.

Hertz

The first example of an electromagnetic wave other than light being

discovered, and shown to have the theoretically predicted properties of an

electromagnetic wave (that is a speed of 3 108¥ ms-1), was provided by

the German physicist Heinrich Hertz.

About 1887, Hertz began his famous experiments to see whether

electromagnetic waves of low frequency could be produced and detected

in the laboratory.

Hertz’s experiments

induction coil to generatehigh voltages in the loop

parabolic reflector A

metal parabolic reflectors tofocus the low frequencyelectromagnetic waves

parabolic reflector B

When spark jumped the gap in loopA low frequency electromagneticwaves were generated. They werefocussed by the parabolic reflectorA into a parallel beam aimed atparabolic reflector B. Parabolicreflector B focussed theelectromagnetic waves at the loop.

The second wire loop B with sparkgap has a voltage induced in it bythe electromagnetic radiation fromthe wire loop A. This causes a sparkto jump the gap.

loop A loop B

Hertz’s experiment equipment set up.


Hertz first demonstrated that Maxwell's predictions were true over short

distances. Hertz set up a spark gap (two conductors separated by a short

gap) at the centre of a parabolic metal mirror. He used an oscillator made

of polished brass knobs separated by a tiny gap over which sparks could

leap. This was connected to an induction coil capable of producing high

voltages to generate low frequency electromagnetic radiation. A wire

ring (able to act as an antenna to receive any low frequency

electromagnetic waves) was connected to another spark gap about 1.5 m

away at the focus of another parabolic metal mirror in line with the first.

According to Maxwell’s theory, if electromagnetic waves were spreading

from the oscillator, they would induce a current in the second loop. This

would send sparks across the gap not connected to an induction coil.

Hertz reasoned that, if Maxwell's predictions were correct, electromagnetic

waves would be transmitted during each series of sparks that jumped the

gap attached to the induction coil. The idea of using the parabolic

reflectors was that any electromagnetic waves produced by the first spark

gap would be collected and reflected by the first parabolic reflector in a

parallel beam at the second parabolic reflector. The second parabolic

reflector would then concentrate these beams at its focus aimed directly at

the wire ring antenna.

Hertz found that a spark jumping across the first gap caused a smaller

spark to jump across the gap in the disconnected ring 1.5 m away.

With this experiment Hertz had shown that:

• electromagnetic waves could be generated

• the waves travelled in straight lines

• the waves could be reflected by a metal sheet just as light waves are

reflected by a mirror.

This experiment resulted in the discovery of what were first called Hertz

waves. These are now called radio waves.

When examining Maxwell's equations, Hertz logically concluded that

electromagnetic waves of different frequencies should be able to travel

through space but also through solid material like wood. He did

experiments placing materials in between wire loops to test this

proposition. He found that non-conductors allowed most of the waves to

pass through. Conductors stopped the waves.

Hertz then continued to experiment using oscillating (high frequency

AC) circuits involving combinations of capacitors and inductors to

transmit and receive radio waves. By measuring the wavelength of the

waves reflecting off a sheet of metal forming standing waves by

superposition, Hertz was able to calculate a wavelength of 0.66 m for the

waves that were produced in that particular set up.


Hertz knew the frequency of oscillation of the sparks in his apparatus

because he could calculate the frequency of the radio waves emitted from

the plate size of the capacitors and the circuit inductance. Hence, Hertz

was able to calculate the velocity of the waves from the equation

v f= ¥ l . He found that the velocity of the radio waves he was able to

generate were in line with Maxwell's theoretical prediction that

electromagnetic waves should travel at 3 108¥ ms-1.

Hertz also was able to demonstrate that the waves he discovered were

able to be polarised. He discovered that the spark would only be

stimulated in the second loop if the gap in the second loop was parallel

with the gap in the first loop. If the second loop was at 90° to the first

loop no spark could be induced in that loop. A schematic circuit of

Hertz’s apparatus to perform this experiment is shown below.

receiver loop

spark jumps

transmitter

E

Hertz’s experimental apparatus.

As well as proving that the electromagnetic waves that Hertz discovered

could be polarised, this experiment clearly demonstrated the transverse

nature of the waves. Only transverse waves can be polarised!

(You should recall this from the preliminary module The world

communicates.)

Complete the table below to produce a summary that describes how Hertz

proved radio waves had properties like light waves.

Property Experimental evidence

polarisation

velocity

reflection

Check your answer.


To see figures of Hertz’s original equipment used in his experiments go to

pages on the physics websites page at:

http://www.lmpc.edu.au/science

Do Exercise 3.1 now.

Radio wave reception

In this activity you will test the reception of a number of different

wavelength radio waves in a tunnel or car park.

Electromagnetic waves have difficulty passing through tubes or into

buildings if the diameter of the openings to the tube or space in the

building approaches the wavelength of the wave or is smaller than the

wavelength. Radio stations on the AM band have kilohertz frequencies.

For example, Sydney ABC news radio operates on a band around

630 kHz. The wavelength of waves on this broadcast signal can be

calculated as follows.

l =

=

=

vf

300 000 000 ms630 000 Hz

476.2 m

-1

To do this activity you will need:

• a cheap AM/FM radio (the cheaper, the better)

• access to a tunnel or underground car park or building basement.

Procedure

1 Tune your radio on to an AM station about the middle of the dial and

listen for the broadcast frequency to be announced.

2 Calculate the wavelength of the frequency the radio station is

broadcasting.

3 Find a tunnel or underground car park and starting at the entrance

walk into the tunnel or entrance while listening to the radio. At the

same time pace out the distance you have walked.

4 Estimate, from your pacing, the distance into the tunnel where the

radio reception drops off.

5 Try tuning your radio to an AM station higher on the dial then one

lower on the dial to see if the reception is better.


6 Try tuning in to FM stations high and low on the dial and repeat the

experiment.

Results

• Frequency of the AM station_______________________________

• Calculated wavelength of the AM station _____________________

• Depth into the opening where the station reception fails _________

• It was easier to get reception from an AM station (higher/lower) on

the dial.

• The reception of FM radio was affected as follows _____________

______________________________________________________

• Did you notice that the FM stations were affected much less than the

AM radio stations? (You should have.)

• Calculate the wavelength of the signal of one of the FM stations you

tuned into. ____________________________________________

1 Are longer wavelength radio waves more or less penetrating into

buildings than shorter wavelength radio waves?

_____________________________________________________

2 Mobile phones use microwave radio frequencies to receive and send

so will generally continue to work until they are well into a cavity or

building. Explain why the use of these high frequency radio waves

was preferred over lower frequency waves based on the result of

your experiment above.

______________________________________________________

______________________________________________________

______________________________________________________

Check your answers.

Producing and receiving radio waves

To do this you will need access to:

• an AM/FM personal stereo radio. The type with headphones works

best.

• a car with the bonnet open to expose the coil

• a responsible friend or adult with a licence to drive a car to assist

you.


Procedure

1 Lift the bonnet and locate the ignition coil. The ignition coil of a car

produces voltages around 25 000 V and sparks regularly to enable

ignition of the petrol in cylinders by the spark plugs.

2 Turn the stereo radio to receive AM. Tune it off a station.

3 Place part of the headphone leads about 10 cm from the coil. The

headphone leads act as the radio receiver aerial in this experiment.

Make sure you cannot burn yourself or catch the headphone leads in any

moving parts of the car such as the fan.

4 Have your assistant turn the ignition of the car on while you are

wearing the headphones. You should hear the clicks of the AM radio

waves produced by the coil as it sparks. This should sound like a low

frequency continuous buzz in the earphones when the car is idling.

5 Have your assistant increase the revolutions of the engine by

pressing slightly on the accelerator while the car is in neutral. If the

car has a tachometer you may like to record the number of

revolutions the engine is doing at different idling speeds and relate

the engine revolutions to the radio wave signal heard through the

headphones.

6 Listen to the change in tone in the earphones as your assistant idles

the engine up (increase the frequency of sparking) and down

(decrease the frequency of sparking). Describe what you hear in the

results section below.

Results and observations

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

1 Identify the source producing the radio waves in this experiment.

_____________________________________________________

2 Identify the part of your radio that has a role analogous to the second

detecting ring in Hertz’s original experiment.

_____________________________________________________

_____________________________________________________

_____________________________________________________

Check your answers.


Extension activityIf your personal stereo contains a recordable cassette tape facility, tape

the sound of the radio waves at different car engine idling speeds. Play

the tape back into microphone and through a CRO program as discussed

in the preliminary topic The world communicates. Observe how the

frequency of the radio waves you recorded changes on the CRO screen.

You should see the regular pattern of the signal produced by the sparkingcoil.

Using the ideas of Hertz's work, Guglielmo Marconi worked out how to

use radio waves for communication with his invention, the radio.

Television and mobile phones and other Wireless Application Protocol

devices use radio waves as well. All of these became possible because of

the work of Maxwell through his theoretical equations and Hertz through

his brilliantly devised experiments.

To see webpages that outline the history of radio wave use see the physics

websites page at:



Antennae

Antennae, or aerials, are devices for radiating or receivingelectromagnetic waves. There is little difference between antennae used

for transmitting and receiving radio waves. Often the same antenna isused for both purposes, for example, when using radar or communicating

between mobile phones.

You may recall from the preliminary module The world communicates

that in radio transmission, for AM or FM waves, the wave transmitted

from the station uses two components; the carrier wave that actually

carries the information; and the modulating signal.

Receiving with an antenna

The quality of reception of every antenna depends on the wavelength of

the carrier wave. The antenna has the best reception when it has a size of

half the wavelength, or some value of n/2 times the wavelength of the

wave it is receiving where n is a whole number. This length requirement

enables the signal to produce a resonating alternating current effect in the

aerial. The signal then has the least interference with itself.

In order to receive higher frequencies, the antenna length required is

therefore shorter. This explains why the aerial for a mobile phone that

uses microwaves as a carrier wave is short. If you want to receive long

wave radio waves, you need to use an aerial that is somewhat longer.

Ham radio enthusiasts use aerials that require them to put up a mast.

What sort of radio wave would require the receiving and broadcast antenna

to be of the order of 15 m high?

_________________________________________________________

_________________________________________________________

Check your answer.


Broadcasting with an antenna

The most important property of an antenna is its radiation pattern or polar

pattern. In the case of a transmitting antenna, the pattern is a graphical

plot of the power or field strength radiated by the antenna in different

angular directions.

The fundamental broadcast radio antenna is a metal wire, rod or tube.

It works best if it has a physical length equal to half a wavelength l2

of

the radio wave it will broadcast. This type of antenna can be called either

a half-wave dipole, doublet or Hertzian dipole.

Example problem

For TV station SBS which transmits at a frequency of 648 MHz,

calculate the wavelength of its carrier wave. Determine the length of the

antenna required to broadcast this signal.

Solution

c = f

cf

ms s m

Since the length of the antenna is a half wavelength

the antenna required to generate the signal for SBS is 0.23 m long.

-1

-1

¥

=

=

=

l

l

300 000 000648 000 000

0 46.

Do Exercises 3.2 and 3.3 now.

Generating electromagnetic signals

The electromagnetic signal broadcast from the antenna is produced by an

alternating current operating in the antenna. That is, a body of electrons

in the antenna are moving backward and forward. The electromagnetic

wave is a representation of this back and forth movement of electrons.

If the wave has a frequency of 600 000 Hz, the electrons in the wire are

moving back and forth 600 000 times a second.


When the electrons move in the antenna, an electromagnetic field is

created around the antenna. This is exactly the same concept as the

magnetic field that surrounds a current carrying wire you learned about in

the preliminary module Electrical energy in the home.

The electromagnetic field generated by the current of electrons flowing

back and forth in the antenna travels from the antenna in all directions at

3 ¥ 108 ms

-1.

It travels until it hits a receiving antenna where, just as a current in a wire

produces an electromagnetic field, the electromagnetic field produces an

electric current in the receiving antenna. This is exactly the same as the

Hertz experiment. The weak signal causes a weak synchronous

alternating current in the aerial of the device receiving the signal.

A device such as a mobile phone, radio or television then uses amplifier

circuits to enhance the signal, and other components such as cathode ray

tubes in televisions to decode that signal and convert it into a form that

can be heard or seen.

To see websites that detail how an antenna produces electromagnetic waves

see the physics websites page at:


To see how an electromagnetic radio wave has a sound signal added to it see

the physics websites page at:


Another common use of radio waves is radar. Air and space are ideal

transmitters of radio waves. But as Hertz proved, radio waves are also

reflected by certain objects. It is this reflection that makes the use of

radio waves for radar possible. The antenna that sends out the radio

wave also receives the radio wave reflections. The timing of the receipt

of that reflected signal and its intensity defines the distance of the

reflecting object.



Planck: Is light a wave?

Light: waves versus quanta

At the close of the 19th century and the beginning of the 20th century,

light was thought to be a wave in nature because of the seemingly

conclusive experimental results of Thomas Young and Augustin Fresnel

that occurred in the first few years of the 19th century.

Young’s description of the interference of light waves showed

conclusively the superposition of light waves as addition and annulment

effects in the form of bright and dark fringes on a screen. These

interference effects occurred following the diffraction and interference of

monochromatic light passing through a pair of closely spaced pinholes in

a screen.

The ensuing concept of light as a wave came to be known as the classical

wave theory of light. This is basically the theory of light that describes

the properties of light that you learned about in the module The world

communicates.

To learn more about the experiments of Thomas Young that lead to the

development of the classical wave theory of light visit pages on the Physics

website page at:


After around half a century of accepting the classical wave theory of light

as the ultimate answer some unexplained phenomena began to crop up

that couldn’t be explained by the classical wave theory. These dealt

initially with the way light was produced. These studies eventually gave

birth to modern physics.

How is light produced other than in a discharge tube?

_________________________________________________________

_________________________________________________________

Check your answer.


Light doesn't always behave

In the last few months of 1859 on October 20, Gustav Kirchhoff

submitted an observation of the strange behaviour of light spectra under

certain conditions.

Some of the spectral absorption lines in the visible spectrum from

sunlight that you learned about in the module The cosmic engine falling

on a screen are darkened more by placing a flame from some burning

sodium metal between the initial light source causing the spectra and the

screen. Kirchhoff was unable to explain why this should occur.

Kirchhoff posed a challenge to see who could solve this mystery. The

response to that challenge eventually led to the discovery of the quantum

theory and the birth of modern physics.

The quantum theory

In the module, The cosmic engine you learned that light is commonly

emitted by heated solids and gases. For example, the tungsten filament

of an incandescent lamp. By analysing the light emitted from any source

with a spectrometer, it is possible to measure how strongly the source

emits radiation (spectral radiancy) at various wavelengths when at a

particular temperature.

tungstenat 2000K

0 1.0 2.0 3.0 4.0 5.0

50

40

30

20

10

0

spec

tral

rad

ianc

y

wavelength (x10–6 m)

Variation of spectral radiancy with wavelength of emitted light.


The diagram above shows the results of spectrometer measurements for

tungsten (the metal from which light globe filaments are made) at a

typical operating temperature of 2000K. In this case, the tungsten

filament is acting as a radiator and the curve shows the intensity of

different wavelengths of electromagnetic radiation that are emitted or

radiated by the filament at 2000K.

For every material there exists a family of spectral radiancy curves like

that of tungsten. Every substance has a unique curve for every

temperature. Now, if families of curves are compared, it is found that

there are no obvious formulas that can describe their shape.

The blackbody model

To overcome the problem of this uniqueness and the infinite number of

curves possible, scientists prefer to work with a model substance or

radiator. This model radiator is used to explain the behaviour of

materials in terms of the radiation emitted from the radiator at different

temperatures. To do this, scientists chose to base their work on an

idealised heated solid. The spectral curve for such a radiator is shown

below.

cavity radiatorat 2000K

0 1.0 2.0 3.0 4.0 5.0

50

40

30

20

10

0

spec

tral

rad

ianc

y

wavelength (x10–6 m)

Variation of spectral radiancy with wavelength for a blackbody model radiator.


Do you see any similarities between the shapes of the spectral radiancy

curves for tungsten and that of the model cavity radiator shown above? If

so, what are the similarities?

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Check your answer.

They chose the cavity radiator or blackbody that you learned about in the

preliminary module, The cosmic engine. A perfect blackbody simply

absorbs all radiation that falls upon it. This body then radiates energy as

a result of its temperature and the interaction of the energy its atoms

absorb from the radiation falling upon the body. The quantity and peak

wavelength of the released energy reflects that temperature.

The electromagnetic radiation emitting properties of this theoretical black

body are independent of any particular material but can be used to model

the behaviour of all materials. The radiation emitting properties of this

model solid are found to vary in a relatively simple way with

temperature.

To check how the performance of a theoretical blackbody performs

against some real data, experimental models of the ideal blackbody are

often constructed. An experimental cavity radiator is typically a metal

block with an internal cavity that has a very small opening through one

wall, as shown in the figure below.

more intense, shorter wavelengthradiation is emitted fromhere

An experimental cavity radiator, or blackbody. The radiation emitted from thecavity opening is more intense and of a higher frequency than that from thesides of the cavity radiator block.


If block like the one shown above is heated to a uniform temperature, for

example 4000K, and the light emitted by the block is examined in a darkroom, the following properties of the radiation are observed.

• The radiation from the cavity that escapes from the opening is more

intense than that from the outside surfaces of the block.

• For a given temperature, the radiation emitted from the hole in the

side of the blackbody is identical for all radiators no matter that the

material composition of the block may vary.

That is, the radiation emitted is independent of the material from which

the block is made. It is independent of the size of the cavity.

For example, at a temperature of 2000K, the radiant intensity from the

cavity is 90.0 W cm–2 for different cavity radiators made from the

different metals: tungsten, molybdenum and tantalum. This behaviour

may be summarised as follows.

• The radiant intensity RT , from the cavity, varies with temperature, T,

according to:

R TT4= s

(T in kelvin, K = °C + 273) where s = 5.67 10 Wm K-8 -2 -4¥ (This is

the Stefan-Boltzmann constant, a universal constant.)

• The spectral radiancy varies with the temperature of the solid, as

shown in the figure below.

12108642

6000 K

12 000 K

24000K

sp

ec

tra

l ra

dia

nc

y

wavelength

(x10–7 m)infra red

light

blue yellow redvisiblelight

ultravioletlight

note how the peak of the curve isshifted to shorter wavelengths athigher temperatures

Radiant intensity of an ideal solid for different temperatures. Note how thecurve peak is shifted at higher temperatures.


Kirchhoff challenged theorists and experimentalists alike to find an

explanation for the nature of cavity radiation. It was expected from the

classical wave theory that this cavity radiation should be easily

explained:

‘It is a highly important task to find this function. Great difficulties

stand in the way of its experimental determination. Nevertheless, there

appear grounds for the hope that it has a simple form.’

At the time no satisfactory explanation existed.

Explaining blackbody radiation

Kirchhoff's challenge was met with a flurry of research to devise a law

that governed the emission of radiation from a hot body. The first to

come up with a law that appeared to work was Wilhelm Wien.

Wien

In 1893, Wien proved that the radiant intensity emitted from the hole in

the blackbody is a function of the temperature of the solid and of the

frequency of the emitted radiation.

This relationship between temperature, frequency of the emitted radiation

and intensity was summarised in the Wien displacement law. That law

basically stated that as the temperature of the solid increases, the peak

intensity of radiation emitted and therefore the peak of the curve is

displaced to shorter wavelength (or higher frequency) electromagnetic

radiation.

In 1896 Wien extended his work and produced his radiation law that was

based largely on experimental evidence of emissions at visible

wavelengths (400–700 nm)

Rayleigh and Jeans

In 1900, Rayleigh observed that Wien’s radiation law made no sense at

low frequencies. It simply didn’t fit the experimental data that people

were finding for low frequency radiation.


Rayleigh and Jeans then produced their version of a radiation law.

Their law correctly predicted the behaviour of the experimental data at

low frequency radiation but produced data that didn’t fit the experimental

data at high frequencies. This failure to fit the experimental data at short

wavelengths became known as the UV catastrophe.

The Rayleigh-Jean’s law predicted that, as the wavelength of the

electromagnetic radiation emitted and absorbed into the cavity of a

blackbody was reabsorbed by the walls of the cavity only to be

re-emitted as shorter wavelength radiation, there was no limit to the

shortness the re-emitted radiation could achieve.

In an experimental apparatus such as a cavity radiator, the radiation

emitted from the walls of the cavity could be expected to be constantly

reabsorbed by the walls of the cavity. As such the radiation in the cavity

would become shorter and shorter in wavelength as time went on. This

presented a problem because short wavelength radiation is of higher

energy.

The big problem with this theory was that the predicted energy emission

from the hole in the cavity would in fact become infinite with an

infinitely short wavelength. The resulting catastrophic release of energy

from the cavity opening would be so intense and energetic it would be

capable of destroying the universe with a blast of high energy radiation

every time such an object was heated.

Obviously this didn’t happen. The experimental data wasn’t in line with

the theoretical model. The model had to be modified.

At the close of the 19th century, the situation was simply that scientists

had no way to explain what was a relatively simple phenomena, the

emission of light from a hot body.

Planck explains all

On 19 October, 1900, Max Planck guided by experimental data,

announced the Planck radiation law. This law actually predicted the

experimental results for radiation emitted at all wavelengths from cavity

radiators. This law, appropriately released in the first year of the new

century, ushered in a new paradigm in physical explanation of the nature

of light and matter.

Planck, with his radiation law perfectly fitting the experimental data, then

looked for a model of the atomic processes taking place in the walls of

the cavity that would explain the data. In doing this he developed one

where he assumed that each atom behaves as an electromagnetic

oscillator that is essentially a small antenna.


Rayleigh

Wien

T = 5000 K

106

104

102

11013 1014 1015

lightinfra-red ultraviolet

Frequency, f (Hz)

radi

ant i

nten

sity

Planck

The relationship between the radiation laws of Rayleigh, Wien and Planck, for acavity at a temperature of 5000K.

Each small atomic antenna has a characteristic frequency of oscillation

that emitted electromagnetic radiation into the cavity but, like a real

antenna, could also receive electromagnetic radiation by absorbing

radiation from the cavity. This led Planck to two devise two very radical

assumptions about these atomic oscillators that were acting like small

antennas in the walls of the container.

These oscillators can only have energies given by: E = nhf

where E is the energy

f is the frequency of the oscillator (Hz)

h is Planck’s constant (6.624 ¥10–34 J s)n = 0, 1, 2, ... (integers)

This had some very serious and revolutionary implications.

• The oscillators’ energies were quantised. That means they could

not have any energy value other than the energies given by this

relationship and that each of these was a whole number multiple of a

quantum of energy represented by hf.

In other words the atom could release one quantum of energy or two

quanta of energy or a million quanta of energy depending upon its

temperature but could not release any fraction of a quantum of

energy. Each energy release had to be a whole number of complete

‘energy packets’.

• These atomic oscillators do not radiate or absorb energy in

continuously variable amounts, but only in quanta, that is, small

packets of energy.


These quanta are emitted or absorbed when an oscillator changes

from one quantised energy state to another quantised energy state. If

n changes by only one unit, then the amount of energy absorbed or

emitted is given by:

E = hf

If an oscillator remains in one of its quantised energy states, then it will

neither emit nor absorb energy.

Using these assumptions, Planck was able to derive his radiation law that

was based on experimental data entirely from theory. He described it to

the Berlin Physical Society on December 14, 1900.

So Kirchhoff’s challenge was met. The era of modern quantum physics

had begun. The funny thing was, though, that Planck didn’t really

believe in quanta or packets of electromagnetic energy of a particular

size. He assumed that really all he had done was to create a

mathematical trick that made the equation describing the experimental

data work. He thought of quanta as a sort of ‘fudge factor’.

Quanta were essentially giving electromagnetic radiation emitted from

blackbodies a particle like nature. Planck was still firmly of the opinion

at this stage that electromagnetic radiation was a wave phenomena and

waves were not created from little bursts of energy!

Using Planck's energy equation

Sample problem 1

Determine the energy of a quantum of radio radiation of wavelength

1000 m.

Solution

E = hf

c = f

so f =c

therefore

E = hc

J

= 2 J

l

l

l

=¥ ¥ ¥

¥= ¥

¥

-

-

-

6 624 10 3 101 10

1 987 10

10

34 8

3

28

28

.

.


1 What is the quantum of energy of a photon of red light with a

wavelength of 700 nm?

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

2 The use of the joule as an energy unit in dealing with energy changes

that occur in atoms is often inappropriate because the joule is such a

relatively large unit. Instead the unit that is often used is the electron

volt or eV. One eV is the electric potential energy change that

occurs when one electronic charge (either an electron or a proton)

moves through one volt.

The value of an electronic charge is 1.6 ¥ 10-19

C.

Given the relationship that: 1V 1J

1C = determine the value of an

electron volt in joules.

_____________________________________________________

_____________________________________________________

_____________________________________________________

3 The energy of a photon of light is 2.41 eV. What is its wavelength?

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

4 What was revolutionary about Planck's idea about the radiation from

cavity radiators?

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

Check your answers.



Max Planck, in his explanation for blackbody radiation had quantised the

atomic oscillators acting as tiny aerials emitting electromagnetic radiation in

the walls of the cavity of a blackbody radiator. He hadn’t considered that

light itself could really be quantised!

Planck had ushered in the era of the quantum with his ingenious

explanation of blackbody radiation. Planck’s concept of the quantum

was radical, and he had doubts about the accuracy of his theory before

and after he made it public. It was apparently a matter of great anguish

for the great man. He was filled with self doubt. He had invented

modern physics but was sceptical of his own invention.


The photoelectric effect

In 1886 and 1887, Hertz discovered that UV can cause electrons to be

ejected from a metal surface. Hertz discovered this when he set up a loop

of wire with a small spark gap in a circuit with a high voltage across the

spark gap. A separate loop was placed some distance away. A spark

jumping in the first loop caused a spark to jump in the second loop.

This was the same as the experiment Hertz performed to generate radio

waves. However, Hertz discovered that the spark could be made to jump

more easily in the second loop if the second loop was illuminated with

UV. The source of the UV was the spark emitted from the first loop.

Hertz reasoned that both light and electricity were in some way

connected in this phenomena, so he called it the photoelectric effect.

Hertz didn’t follow up the photoelectric phenomenon before he died of

severe blood poisoning at age 36. That was left for later experimentalists

and perhaps the greatest scientist of the 20th century, Albert Einstein.

Despite this failure to follow up the photoelectric effect Hertz had still

proven himself to be one of the 19th century’s greatest experimental

scientists.

Describe Hertz’s observation of the effect of a radio wave on a receiver and

the photoelectric effect.

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Check your answer.


Einstein and the photoelectric effect

It was not too long before other evidence surfaced in support of Planck’s

quantum theory. First came the explanation of the photoelectric effect by

Albert Einstein.

The classical wave theory of light was in vogue prior to the work of

Einstein on the photoelectric effect. The classical theory predicted the

intensity of the light required to cause the ejection of electrons from a

metal surface should be dictated by the amplitude (or intensity) of the

incident light waves. The classical theory stated that if light could cause

the emission of electrons, a greater light intensity should cause the

electrons on the metal to be ejected from the metal surface with a greater

kinetic energy.

Einstein tackled the problem of these features of the photoelectric effect

in 1905. His explanation of these phenomena was complete.

Testing the classical theory

The experimental result of tests of whether the ejected electrons from a

metal do have more kinetic energy if monochromatic light of various

frequencies and intensities was shone on metals found a number of

things.

• The kinetic energy of the electrons ejected depends on the frequency

of the light falling on the metal.

• The kinetic energy of the electrons emitted was not dependent on the

intensity of the light.

• The light intensity was found to affect the number of ejected

electrons.

• Higher intensities of light caused the ejection of more electrons from

the metal surface but did not change the kinetic energy with which

the electrons were ejected.

• If the frequency of the light hitting the metal was not of a threshold

frequency no electrons were released at all.

Studying the photoelectric effect

The photoelectric effect can be studied using the simple arrangement that

uses a photocell as shown in the figure following. Results of such

studies are discussed and interpreted.


- +

receiverswitch

G

B

V

A

incident light

evacuatedglasstube

switch

DC power supply

galvanometer tomeasure smallelectric currents

variable resistorenables the stoppingvoltage to be set

photocell

A photoelectric circuit. The evacuated tube in this figure is a photocell.Photocells are also used in devices such as light meters in cameras.

Changing the light intensity

When monochromatic (single frequency) light falls on a curved metal

plate at A in the diagram above, electrons are liberated from the metal

because of the photoelectric effect. This means that the curved metal

plate is acting as a cathode. It is hence often referred to as a

photocathode.

These emitted electrons can be detected as a current if they are attracted

to a positive electrode B by applying a potential difference, V, between

the plates A and B. The electrode B is acting as an anode in this

photocell. A sensitive meter called a galvanometer that is capable of

measuring small electric currents measures this photoelectric current that

flows between the photocathode and the anode.

As the light intensity falling on the photocathode in the photocell is

increased the current is also increased for a particular voltage. This

suggests more electrons are emitted from the photocathode as the light

intensity is increased.


Increasing the voltage across plates A and B

As the voltage is increased between the photocathode and the anode in

the photocell, the photoelectric current reaches a limiting value for any

particular monochromatic light intensity. This maximum is interpreted to

occur when all of the photoelectrons emitted from A are collected by B.

Reversing the polarity of plates A and B

When the polarity or direction of the voltage between the photocathode

and the anode is reversed, the current does not immediately drop to zero

but rather takes a finite time. This indicates that the electrons being

emitted from A have a maximum velocity and kinetic energy. It also

implies that some of these electrons are actually reaching B in spite of the

fact that the electric field present between A and B is opposing their

motion.

These photoelectrons are doing work in moving against the electric field.

They can do that work because they have energy. If this reverse potential

is large enough, a value is reached at which the current is zero. This

voltage is called the stopping potential and is given the symbol V0.

Increasing the voltage for reversed polarity

If the electric field between A and B is strong enough, it will stop the

most energetic of the electrons crossing from A to B. The maximum

kinetic energy of the photoelectrons can then be calculated from theproduct of the electric charge on the electron, qe and the stopping

potential, V0.

That is: 0e

2

emaximumK Vqvm2

1E ==

The maximum kinetic energy in such a situation is therefore independent

of the intensity of the monochromatic light incident on the photocathode

surface.

When these measurements are repeated for monochromatic light of a

particular intensity but of different frequencies, then a graph similar to

the one shown on the next page is obtained.

In this situation the current flowing in the circuit through the

galvanometer reaches a maximum value as shown on the figure.

This value is called the saturation current.


O

f3

–V3

I

f2f1

–V2–V1

f1 > f2 > f3

V

Photoelectric current vs stopping potential difference for light of constantintensity but at three different frequencies f1, f2 and f3.

The saturation current is independent of the frequency of the incident

light. This saturation current occurs when every single electron emitted

from A is collected by the electrode B. Under that condition the current

cannot increase in size because there is no other source of electrons to

increase the current size.

In conclusion, it is apparent that for negative potential differences

between A and B, the stopping potential, V0, is dependent on the

frequency of the incident light. Higher frequency light incident on the

photocathode means that the stopping the photoelectrons requires a larger

stopping potential V0.

It is therefore logical to conclude that the maximum kinetic energy of the

ejected photoelectrons depends on the frequency of the incident light

falling on the photoelectric material. It is not dependent on the light

intensity.

Changing the cathode

What happens if a different metal is used for the photocathode?

Notice from the figure below that for each photoelectric material there is

a definite cut off frequency, below which no photoelectric effect occurs

(f0 value). That is, the KEmax of the electrons is zero, therefore they do

not leave the surface. In fact, each different type of metal has its own

characteristic cutoff frequency below which no photoelectrons will be

emitted. This is called the threshold frequency of the metal.

If the incident light hitting the metal surface has a frequency below the

cut off frequency, there is no photoelectric current and therefore no

photoelectric effect is observed, no matter how intense the incident light.


This presented a severe problem to classical wave mechanics. That

theory predicted electron emissions with kinetic energies increasing as

the intensity of the light increased. There was at this time no theory that

explained how this situation could occur.

A B

foA foB0

frequency, f(Hz)

EKmax (J)

below the threshold frequencyno photoelectrons are emitted

as the frequency increasesso does the maximum KE ofthe photoelectrons.

Maximum kinetic energy versus frequency for two different photoelectricmetals A and B.

Photoelectric effect characteristics

The characteristics of the photoelectric effect can therefore be

summarised as the following list of points.

• A photon (light quantum) carries energy, hf, into the photoelectric

surface.

• Part of this energy, (E0), is used to get an electron to escape from the

metal surface.


• Any photon energy remaining is given to the electron in the form of

kinetic energy, EK. If the electron does not lose all its energy by

internal collisions within the material as it escapes from the material,

it will have all that excess energy as kinetic energy. That is the

maximum kinetic energy the free electrons can possess.

• If the intensity of the incident light is increased, the number of

photons ejected increases to a maximum value (saturation current).

It does not increase the energy of the photons of a set frequency, nor

the energy of the photoelectrons ejected from the surface.

If the photon has just enough energy to eject the photoelectrons from the

photoelectric substance and none extra to appear as kinetic energy then

that energy value is the quantity (E0). That value is called the work

function of the material.

If the frequency of the light is less than some threshold frequency

required to overcome the work function and liberate electrons the

photons do not have enough energy to eject any electrons. This is

irrespective of the intensity of the incident light. This is because

intensity is a measure of the number of photons, not the energy of the

individual photons.

When light with a frequency at or above the threshold frequency is

incident on some photoelectric material, the photoelectric effect is

instantaneous, because the required energy to eject the electrons comes in

a small packet of light energy, the photon.

e

e-

E'

photonE = hf

photoelectricmaterial

kinetic energyof the electronEK

The photoelectric effect. The electron absorbs all of the energy of the photon.Some of this energy is used in escaping from the material, E', and the balance,hf – E', shows up as EK of the emitted electron, the photoelectron.


Einstein’s explanation

Planck assumed that the electronic oscillators were acting like small

aerials to produce electromagnetic waves of a particular frequency.

These waves had quantum energies given by Planck's formula.

The photoelectric effect phenomena were impossible to explain using the

classical wave theory of light. Einstein did explain them, though by

making the assumption that the energy in light travels in little bundles or

photons. These were in effect quanta of light energy! Einstein had taken

the energy quanta that Planck had used to explain blackbody radiation

and related them to light!

The energy, E, of a single photon is given by: E = hf

Einstein applied Planck’s quantum concept to the photoelectric effect andhe described the process with the equation: hf = E E0 Kmax+

This says that the energy of the light photon is partly used to:

• liberate the electron from the metal surface (E0); and then

• supply the kinetic energy of the free electron enabling it to travel

away from the metal surface with some velocity.

Using the concept of quanta, Einstein successfully explained the

photoelectric effect using this assumption. By doing so he calculated a

value of close to the value obtained by Planck for these quanta of energy.

Einstein had effectively quantised light.


Looking for evidence of the photoelectric effect

To do this activity you will require the following equipment:

• two electroscopes

• a sheet of zinc metal or a large galvanised (zinc coated) metal washer.

Whichever you use, the metal surface should be cleaned with a weak

acid such as in Coca Cola® or vinegar

Or, if you don’t have access to an electroscope you can make your own

with:

• two identical soft drink bottle made from glass but having plastic lids

• two 10 cm nails

• the light thin silver foil wrapper from some sweets such as a roll of

candy


• cellotape

• a nail

• an inflated balloon

• a woolen jumper

• a piece of polystyrene packing or a polystyrene drinking cup

• a small sheet of glass or a jar to place over the zinc plate of the

electroscope

spine

twogoldleaves

earthedcase

cap

How to make your own electroscope if you don’t have access to a real

electroscope.

1 Hammer a nail through the washer or zinc sheet and then through the

top of the lid of the bottle. Take care not to destroy the lid. You will

need to screw it on the bottle.

2 Attach two strips of foil wrapper to the bottom of the nail to act as

the leaves of the electroscope.

3 Screw the lid of the bottle back on with the nail and two thin foil

leaves inside the bottle and you have a home made electroscope.

You will need to make two identical electroscopes for this activity.

Procedure

1 Set up your two electroscopes with the zinc sheet on the cap of each

electroscope.

2 Charge both electroscopes negatively by rubbing a balloon against a

woollen jumper then touching the balloon to each electroscope at the

sheet of zinc.

3 Place one in the shade and the other in direct strong sunlight.

You will know the electroscope is charged because your foil leaves

will separate.


If you do not recall how an electroscope works you may need to

refer back to the preliminary module Electrical energy in the home.

__

___

___

___

___

___

_

Two negatively charged electroscopes. Both have their leaves repelled byaround the same amount.

3 Observe which of the two electroscopes, the one in the shade or the

one in the sunlight, discharges the fastest. Record your observation

in the space below.

4 Discharge both your electroscopes completely by touching them to a

water pipe or earthing them to ground with a length of copper wire.

5 Recharge both your electroscopes with a positively charged balloon.

You can make the balloon positively charged if you rub the balloon

against a piece of polystyrene foam.

charged bodytouching cap

charged bodytouching cap

Two positively charged electroscopes.

6 Place one electroscope in the shade and the other in direct strong

sunlight.

7 Observe which of the two electroscopes, the one in the shade or in

the sunlight, discharges the fastest. Record your observation in the

space below.

8 Charge up one electroscope with a positive charge and the other

with a negative charge. Place both electroscopes in full sunlight.

Observe which electroscope discharges fastest.


9 Repeat the above experiment procedure with two negatively charged

electroscopes but this time shade one negatively charged

electroscope with a sheet of glass or cover the zinc plate with a large

glass jar. The glass will absorb most of the UV from the Sun but

allows the visible light to penetrate unhindered. Record your

observations in the space below.

Observations

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Results

Cross out the incorrect word to complete your results in the sentences

below.

• For the negatively charged electroscope: The electroscope that

discharged first was the electroscope in the sunlight/shade.

• For the positively charged electroscope: The electroscope that

discharged first was the electroscope in the sunlight/shade.

• For the comparison between the positively and negatively charged

electroscope: The positively/negatively charged electroscope

discharged the fastest.

• The electroscope with the full amount of UV exposure discharged

faster/slower.

• The negatively charged electroscope in full sunlight without the

glass UV filter effect discharged the fastest/ slowest.


Explaining the results of the experiment

The following is an account of what you should have observed in the

activity above and an explanation for those observations based on the

photoelectric effect acting on the zinc plate.

When the sunlight and associated UV hits the zinc charging plate on the

negatively charged electroscope photoelectrons are emitted from the

insulated zinc plate. This effectively discharges the electroscope. The

result that is the negatively charged electroscope in the sunlight should

have been the fastest to discharge.

When the sunlight and associated UV hits the zinc charging plate on the

positively charged electroscope, photoelectrons are emitted from the

plate but are almost immediately attracted back to the zinc charging plate.

Both positively charged electroscopes will discharge at approximately

the same rate.

When the positively and negatively charged electroscopes are both in full

sunlight and associated UV, the photoelectrons emitted from the

negatively charged electroscope cause it to discharge steadily. The

positively charged electroscope remains charged for much longer.

The UV is necessary for the photoelectric effect, as the UV filtered light

passing through the jar didn’t stimulate the steady discharge of the

negatively charged electroscope. Therefore UV, not visible light, is

necessary to stimulate the photoelectric effect in zinc.

Photon energy, frequency, c and llll

Recall that Planck believed that light travelled through space as an

electromagnetic wave, although it was being emitted and absorbed by the

walls of the cavity resonator in discrete amounts or quanta of

electromagnetic energy.

It was Einstein’s idea that light, and all electromagnetic radiation,

travelled through space as photons, and not as waves. This appears to

conflict with the wave properties of light, for example diffraction and

interference you learned about in The world communicates.

Physicists have now come to a modern view of the nature of light that

accepts it has a dual character, behaving like a wave under certain

conditions and behaving as a particle or photon under other conditions.


Example

The work function of tungsten is 8.6 ¥10–19 J . What is the maximum

energy of the photoelectrons ejected from the tungsten surface by ultra

violet radiation of wavelength 1.8 ¥10–7 m?

Solution

Energy of incoming photon:

E = h f

Since

l

lc

f

fc

=

=

then

J101.1

108.1

10 3 1063.6

hcE

18

7

834

-

-

-

¥=¥

¥¥¥=

=l

Energy required to release an electron from tungsten (work function):

E0= 8.6 ¥ 10-19

J

So, the energy remaining for the ejected electron to have as kinetic

energy is:

KE = E - E

J - 8.6 10 J

= 1.099 J

0

-19 = ¥ ¥

¥

-

-

1 1 10

10

12

12

.



Summary

Complete this outline to produce a summary of the main points covered

in this part.

Hertz determined the speed of radio waves by_____________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Hertz observed the photoelectric effect when _____________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

A radio signal is produced in an antenna when ____________________

_________________________________________________________

Planck’s hypothesis to explain the emission of radiation from a

blackbody was _____________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Einstein’s contribution to quanta and its relation to blackbody was

_________________________________________________________

_________________________________________________________

_________________________________________________________


The particle model of light in terms of photons says _______________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

The relationship between photon energy, frequency, speed of light and

wavelength is _____________________________________________

_________________________________________________________

_________________________________________________________


Suggested answers

Hertz’s experiments

Property Experimental evidence

polarisation the spark could only be stimulated in the second loop if itwas parallel to the first loop

velocity Hertz determined wavelength by interference and thefrequency of the sparking from the experimental set up (platesize in the capacitor) and used the wave equation todetermine the velocity

reflection Used parabolic metal mirrors to focus radio wave energy to apoint

Radio wave reception1 Shorter wavelength radiation (higher FM stations on the dial) have

the best reception inside buildings.

2 Very short wavelength radio waves penetrate the best. This is

because they can more easily penetrate into openings as complete

waves.

Producing and receiving radio waves1 The sparking engine coil is the source of the radio waves.

2 The aerial has the same function as the second detecting ring in

Hertz’s experiment.

Receiving with an antenna

Long wavelength radio with the wavelength on the order of 30 m would

need a 15 m antenna.


Light: waves versus quanta

Heating a solid or gas causes the electrons to get excited then jump down

to a lower energy level. This releases light energy in the process.

The blackbody model

The spectral radiancy for tungsten and the model cavity radiator both

have higher spectral radiancy at similar wavelength radiation. The

curves almost parallel.

Using Planck’s energy equation1

J1084.2

s1029.4 Js10626.6

hfE

Now

Hz1029.4

m10 700

ms10 3

cf

determined bemust light theoffrequency First the

19

1-1434

14

9-

18

-

-

-

¥=

¥¥¥=

=

¥=¥

¥=

=l

2 1 eV 1V 1.6 10 C

1.6 10 J

Remember the eV is a unit of energy!

-19

-19

= ¥ ¥

= ¥

3 First convert the eV to J.

2.41 eV 2.41 1.6021 10 J

3.9066 10 J

-19

-19

= ¥ ¥

= ¥

Now E hf

fEh

3.861 10 J6.626 10 Js

s

-19

-34

=

=

=¥¥

= ¥ -5 827 1014 1.


Since l =

=¥¥

=

-

cf

ss

nm

-13 105 896 10514

8

14 1.

The light has a wavelength of 514 nm.

4 It was emitted in small packets of energy called quanta rather than as

a continuous energy stream of wavelengths of continuously varying

frequency.

The photoelectric effect

Hertz saw the spark in the second loop. That meant that the radio wave

had travelled through the air to the second loop. The fact that the spark

was caused to jump more easily when the second loop was illuminated

by the UV suggested a link between light and the ability to generate a

spark.


Exercises – Part 3

Exercises 3.1 to 3.6 Name: _________________________________

Exercise 3.1

Explain how the experiments conducted by Hertz clearly related radio

waves to light waves.

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Exercise 3.2

Explain why the radio antenna used to produce microwaves is much

shorter than the antenna used to produce long wavelength radio waves.

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________


Exercise 3.3

Calculate the optimum length antenna that would receive the best radio

reception of a 104.9 MHz radio station.

_________________________________________________________

_________________________________________________________

_________________________________________________________

Exercise 3.4

Outline four of the applications of the production of electromagnetic

waves by oscillating electric charges in radio antennae.

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

_________________________________________________________

Exercise 3.5a) What was Planck's hypothesis to explain the emission of radiation

from the walls of a blackbody cavity?

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________

______________________________________________________


b) What was Einstein's contribution to quanta and its relationship to

blackbody radiation?

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

Exercise 3.6

A metal has a work function of 9 ¥10-19

J. It is irradiated with UV

radiation with a wavelength of 180 nm.

a) Will photoelectrons be emitted from the surface? Explain your

answer.

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

b) If the same metal is irradiated with green light of wavelength 550 nm

will photoelectrons be emitted from the metals surface? Explain

your answer.

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

_____________________________________________________

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