1

Physics 3204 Unit 4 –

Modern Physics (or

Introduction to Quantum

Theory)Not sure of Title yet!

What is LIGHT??

4:38 minutes

3:58 minutes

2

5:39 minutes

3

31:08

35:34

4

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At the end of the 19th century (late 1800’s)

scientists had it all figured out.

What is the physics called that describes the motion

of matter?

Newtonian

It explained the motion of objects on Earth as well as

heavenly bodies.

What theory did Newton use explained the nature

of light?

Isaac Newton thought light was made of little

particles (he called them corpuscles) emitted by

hot objects (such as the sun or fire).

(Or did they??)

However,

his contemporary, the Dutch physicist

Christian Huygens, thought light was a kind

of wave vibrating up and down as it moved

forward.

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7

What evidence do we have that light acts

as a wave?

It matches the kinematics' wave formula:

Where c is the speed of light

f is the frequency

l is the wavelength

The interference patterns of waves was used to

explain different light phenomena:

The colours on the surface of oil films was explained as

interference patterns by Christian Huygens and Robert

Hooke.

The double slit experiment by Thomas Young in 1803.

In 1815, Augustin Fresnel explained the diffraction

of light by using this wave theory of light.

c f l83.00 10 m

sc

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In 1864, James Clerk Maxwell completed “classical

physics” by amalgamating electricity and magnetism and

in the process combined light into the electromagnetic

wave family.

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So life was good. They could explain it all

with only minor problems left to solve!!

UNTIL:

Wilhelm Roentgen discovered X-rays.

The UV catastrophe in Black body radiation curves

Light behaving like a particle in the photoelectric

effect

X-ray photons with momentum

Particles that exhibit wave properties

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In this unit

We study these phenomena and the theories

that resulted.

We discuss the transition from the long-

standing Newtonian Physics to the Modern

Age of Physics

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The study of how MATTER and Energy are

interrelated is called Quantum Theory

This energy is Light Energy.

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Paradigm Shifting Time

Matter does not always behave in the way Newton

thought it should ( sometimes it behaves like a

wave)

Light always behave in the way Maxwell thought it

should (sometimes it behaves as a particle)

THE INFAMOUS DOUBLE SLIT

EXPERIMENT The double slit experiment as previously

studied is one way to illustrate the difference

between the classical world and the quantum

world.

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Water Waves Light

THE INFAMOUS DOUBLE SLIT

EXPERIMENT

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If we look at the results of the double slit

experiment with marbles we see that they

behave like we expect particles to behave.

If, however, we perform the double slit

experiment with subatomic particles such as

electrons, we see quite different behaviour with

the electrons showing wave patterns of

interference.

This discrepancy is characteristic of what

happens in the quantum world.16

The clear cut distinction between particles

and waves breaks down in the quantum

world.

This dual nature of exhibiting both wave and

particle properties is known as wave-particle

duality.

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In this unit we learn about the evidence that

supports the dual (wave-particle) nature of light.

(aka photons)

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Question: Does the sun cause

sunburn because it is so bright?

A sunburn is a burn to living tissue such as skin

or leaves produced by overexposure to ultraviolet

(UV) radiation, commonly from the sun's rays.

A similar burn can be produced by overexposure

to other sources of UV such as from tanning

lamps, or occupationally, such as from welding

arcs.

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Do bright stage lights cause sunburn?

No. They emit visible light, not UV light.

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Why is it necessary to wear protective

goggles, and clothing when welding?

The welder light contains ultraviolet (UV) light that has a

higher energy than visible light which makes the light

penetrate skin cells instead of reflecting off them.

The brightness of the light is not related to

its penetrating power.

Brightness is related to the amount of light present.

The penetrating power is related to the

lights energy level.

This is related to its frequency (or wavelength).

The Quantum Mechanical model of light can

be used to explain this quantized energy level

and other phenomena such as black-body

radiation and the photoelectric effect that

could not be explained by the classical wave

theory of light.

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A Brief History Of Quantum

Mechanics https://www.youtube.com/watch?v=B7pACq_

xWyw

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Black - Body Radiation

An ideal black-body is any opaque object that

absorbs all light (from all of the EM spectrum)

that is incident upon it.

Therefore it reflects no radiation and appears

perfectly black.

In practice no material has been found to

absorb all incoming radiation, but carbon in its

graphite form absorbs all but about 3%.

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Black - Body Radiation

This object will also become a perfect

radiator, emitting the full spectrum of light.

In quantum mechanics, the term black body

radiation is used to describe the release of

energy.

This energy can be from different parts of

the EM spectrum including infrared or

thermal energy (hence, this is why we feel

things as being hot).

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Black - Body Radiation

Any object that is above 0o K, will emit

radiation in this form.

Black body radiation helps to explain why

objects will glow when they are at a certain

temperature.

For example, the element on a stove will glow

red when it is at a certain temperature. Or, an

incandescent light will glow yellow at a certain

temperature.

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Classical theory suggests that as the

frequency of emitted energy increase, the

intensity of radiation will increase as shown.

27

Draw the corresponding curve in terms

of wavelength

The problem with classical theory is it implied

there is no limit to the intensity of radiation as

frequency increases, which contradicted

experimental results (as shown below)

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Max Planck suggested that objects release energy

in packets called quanta, meaning energy can only

be emitted in discrete packets of specific

frequency.

But it was also was considered to behave as a

wave because it had a frequency.

These quanta later became known as photons.

When objects are at a certain temperature, it will

release quanta at a specific frequency.

If an object is at a higher temperature, it will release

energy at a different (higher) frequency.

Those different frequencies are the reason why objects

will emit different colours at different temperatures.29

Therefore, the hotter the object is, the bluer

the light will be emitted.

The cooler the object, the redder it is, thus

giving the experimental pattern above.

Planck was able to use the idea of quanta to

produce the graph below, thus verifying the

new model of understanding.

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Planck used data from blackbody radiation

experiments and then “fit” the formula:

E = hfto match the data.

This led the scientific community to quantum

theory.

The photoelectric effect was Einstein’s test of the

theory, which turned out to validate Planck’s

work.

There was a lot of back and forth between

Einstein and Planck during the development of

this theory.31

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Planck’s equation:

Planck’s constant:

Also, the total

energy is found

using the formula

where

n = 1, 2, 3, ...

(number of photons)

E hf

346.626 10h J s

E nhf

Problems

1. Calculate the energy of a single photon of

blue light if it has a frequency of

7.0588X1014 Hz.

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What is E and what is the equation

to calculate this based on the

wavelength of light?

E is the smallest amount or quantum of energy that can be transferred for a given wavelengthof electromagnetic radiation

The equation is:

This is obtained by replacing f in Planck’s equation:

hcE

l

cf

l

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2. What is the energy of a photon that

has a wave length of 460 nm?

hcE

l

34 8

9

6.626 10 3.00 10

460 10

ms

J s

m

194.3 10 J

NOTE:

At the atomic level, energy is commonly

measured in electronvolts (eV).

Which is the energy gained or lost as a single

electron moves across an electric potential

difference of 1 volt.

RECALL:

1eV = 1.602x10-19 C

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3. Calculate the energy of a single photon if it

has a frequency of 1.50 × 1015 Hz. Express

your answer in Joules and electron volts, eV.

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4.What is the wavelength of a photon having an

energy of 2.12 eV?

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5. A) What is the energy of a photon with a

wavelength of 525 nm?

B) If the source releases 100.0 J, how many

photons were released?

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Black lights and other UV light

UVA: sun tanning, testing, inspection, insect traps, stage

effects, blacklight, phototherapy (315-400nm)

UVB: sun burning, inspection, analysis, testing,

phototherapy (280-360nm)

UVC: germicidal (253.7nm), ozone producing (185nm)40

Consider:

GTL3 Bulb - 3W Germicidal

3 W at 10.5V

UV power: 0.16 W at 254 nm

How many photons of UV light is emitted in 120

seconds?

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“PHOTO ELECTRIC EFFECT”?

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What is the “photo electric effect”?

44

What is the “photo electric effect”? It is a phenomenon that sometimes occurs when light is

shone on a metal surface causing electrons to be

emitted from the surface.

The surface may become positively charged.

This is because electrons gain energy from the light,

and are able to leave the metal's surface.

PHOTOELECTRIC EFFECT APPARATUS

If we illuminate one of the

metal plates with the proper

light, electrons will absorb

the light and escape from

the surface of the metal.

These electrons will move

off the surface with

whatever kinetic energy

they possess.

Some will move across the

tube and hit the metal plate

at the opposite end of the

tube and the ammeter will

register a current.45

The current gives no indication

of the kinetic energy of the

photoelectrons only the number

of them released per second.

If the photoelectrons’ kinetic

energy is transferred into

potential energy, by applying an

electric field to oppose their

motion, we can work out what

that initial kinetic energy was.

We therefore apply an electric

voltage across the tube to

oppose the travel of the

electrons between electrodes.

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We make the illuminated

electrode positive and the one

they are travelling to negative

and therefore repulsive to the

electrons.

The electrons are repelled by

the negative plate and some

electrons go back.

But some of them with high

kinetic energy are still able to

reach the plate.

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As the negative potential

increases, less and less electrons

are able to reach the other side of

the tube and finally when none can

reach the photocurrent becomes

zero.

The stopping potential (voltage)

Vstop is related to the maximum

kinetic energy of the ejected

photoelectrons.

Recall:

so for the electrons:

or

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EV

q

maxk

stop

EV

e

maxstop keV E

If the brightness/intensity of the photon beam

is increased the photocurrent will increase

and vice versa.... but this does not affect the

voltage needed to stop the photoelectrons.

This cannot be explained by the wave theory

of light!!

49

There are some strange effects with the photoelectric effect that

can not be explained by considering light as a wave.

The frequency of the light must be above a certain threshold value for that metal for electrons to be emitted.

For example: blue light causes sodium to emit electrons but red light does not.

If light was a wave the electrons would eventually absorb enough energy to be emitted, regardless of frequency.

Also, incredibly weak beams of light can cause electrons to be emitted.

If light was a wave and spread out you would not expect any electrons to obtain the energy to escape.

Finally, it was discovered that the kinetic energy of the electrons depends not on the intensity (brightness) of the light but on its frequency.

A very weak ultraviolet beam will give electrons a higher kineticenergy than a very bright blue beam of light.

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SO, if you were asked:

1. How does the photoelectric effect

contradict the wave theory of light?

Your answer should be something like this:

According to wave theory the energy of light,

i.e. its amplitude, or brightness, should

increase the energy of the ejected electrons.

However, this is not the case.

It is found that by increasing the frequency of

the light to a different colour, the energy of

the electrons was immediately increased.51

2. Why doesn’t red light eject

electrons when shined on a metal

plate? What happens to the red light?

The photons of the red light do not have enough

energy to liberate any electrons.

The light is absorbed by the metal.

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There is a minimum, or threshold

frequency, f0, of incident light below which

no photoemission would occur.

If the light shining on the metal is of a

frequency lower than this threshold

frequency, no electrons are emitted,

regardless of the brightness of the light

shining on the metal.

53

Define the work function, Wo.

It is the minimum energy required to evict an

electron from a metal.

This energy varies depending on the type of

metal.

54

NOTE: The threshold frequency, f0, is found

by the equation:

55

o oW hf

A real-world analogy to the

concept of work function

56

The energy of the incoming photons of light is

analogous to the energy of the incoming player’s

foot.

To kick the ball the incoming foot has two jobs to

do:

One - It must do work to loosen the ball from the

sand. (Work function)

Two - Any remaining energy moves the ball on its way

or gives the ball kinetic energy.

Similarly, the energy of the incident photons

must be sufficient to:

One - Free the electron from the metal plate (Wo)

Two - Get the electron moving or give it kinetic energy.

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How does the higher energy light

eject electrons from a metal plate?

If the light is not reflected, it transfers energy to

the electrons in the metal.

If the energy transferred is greater than Wo, the

electron will be emitted.

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59

The liberated electron’s kinetic

energy is determined by the equation:

This shows that the kinetic energy of the electron

is the energy of the photon that liberated the

electron minus the energy required to release the

electron.(Wo)

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max k photon oE E W

When we solve the equation above

for Ephoton and apply Planck’s

equation we get:

hcE hf

l

61

maxk o

hcE hf E W

l

max photon k oE E W

Note:

E=hf is the energy of the incident photon

Ekmax is the maximum amount of energy of the

liberated electron

Wo is the work function.

This is Einstein’s photoelectric effect

equation:

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max k ohf E W

SUMMARY (thus far)

Planck used data from blackbody radiationexperiments and then “fit” the formula E =hf to match the data!

This match lead the scientific community to quantum theory.

The photoelectric effect was Einstein’s test of the theory, which turned out to validate Planck’s work.

Quantum theory is the theory of how atoms, matter, and energy are interrelated.

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#1. Light, with a frequency of 5.0 x 10 14 Hz

illuminates a photo electric surface that has a work

function of 2.3 x 10-19 J.

What is the maximum energy of the ejected electrons?

64

max

34 14 196.626 10 5.0 10 / 2.3 10 kE J s s J

191.0 10 J

Example:

max k ohf E W

max k oE hf W

#2. When electromagnetic radiation with a

wavelength of 350 nm falls on a metal, the

maximum kinetic energy of the ejected

electrons is 1.20 eV. What is the work

function of the metal, in Joules and eV?

Since we are given wavelength we use:

65

maxk o

hcE W

l max

o k

hcW E

l34 8

9

6.626 10 3.00 101.20

350 10

ms

J seV

m

195.68 10 1.20J eV

In Joules

In eV

66

19 195.68 10 1.20 1.602 10 JeV

J eV

193.758 10 J

193.8 10 J

Wo 3.55eV 1.20eV 2.35eV

19

195.68 10 1.20

1.602 10

eVJ eV

J

#3. What is the stopping voltage of

an electron that has 7.4 x 10 -19 J of

kinetic energy?

67

197.4 10 stopeV J197.4 10

stop

JV

e

19

19

7.4 10

1.602 10

J

C

maxstop keV E

4.6V

4. Experiments show that the work function for cesium metal is 2.10 eV.

Determine the threshold frequency and wavelength for photons capable of producing photoemission from cesium.

Recall: The threshold frequency, f0, is found by

the equation:

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o oW hf

Then do: Page 720: #1-7

And QUESTIONS 8-10 are to be passed in next

day

69

PRACTICE PROBLEMS:

p. 716 #1-3; Page 718 #1,2

Wavelength (nm) Kinetic Energy (eV) Frequency(Hz) f=c/l

500 0.36 6E+14490 0.41 6.12E+14440 0.70 6.82E+14390 1.05 7.69E+14340 1.52 8.82E+14290 2.14 1.03E+15240 3.025 1.25E+15

70

This graph shows how the kinetic energy of

the ejected electrons varies with frequency

for multiple metal surfaces.

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•The threshold

frequency is the x-

intercept of this graph.

•The work function is

the y-intercept.

•The slope of the graph

represents Planck’s

constant.

2. What is the work function for platinum?

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1. Which metal would

required the greatest

amount of light energy to

evict an electron?

The Compton Effect and Photon

Momentum

What is momentum is defined as?

p = mv.

So, does this mean that a photon (which

is electromagnetic energy) has mass?!?

What famous equation relates energy to

mass?

E = mc2

73

In 1923, Arthur Holly Compton shot a

beam of x-rays at a thin foil and a target

made from carbon.

After the target was struck, he noticed

that it emitted other x-rays and some

electrons.

He discovered that the second lot of x-

rays had lower energy than the incident x-

rays. This phenomenon is called the

Compton Effect.

74

Below is a picture of Compton's set-up for

the experiment.

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Notice the high-energy x rays hitting the target.

Notice lower-energy x-rays and electronsbeing scattered from the target.

The word scattering is used because the x-rays are deflected in many different directions.

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Can you see that the picture resembles a

collision of pool balls on a pool table?

Compton must have had the same

impression because he decided to use the

laws of conservation of energy and

momentum that you have already learned

about in Physics 2204.

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The conservation of energy law:

Energy can not be created or destroyed, only

transformed from one form into another.

Ex-ray = Escattered + Eelectron

Note:

Ex-ray = hfx-ray is the energy of the incoming x-ray

photon

Escattered = hfscattered is the energy of the scattered x-ray

Eelectron =½ mev2 is the kinetic energy of the electron

Thus the equation above can be written as:

hfx-ray = hfscattered + ½ mev2

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Note: Why must fscattered must be less than

fx-ray?

The scattered x-rays has lower energy

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Conservation of momentum:

The momentum of a system before a collision is

equal to the momentum after.

In order for this equation to be true, photons

would have to have mass, or some mass-like

property.

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p xray

p scattered

p electron

Compton boldly hypothesize that a photon

has mass and used Einstein's Special Theory

of Relativity and mass-energy equivalency

formula: E = mc2

Solve E = mc2 for mass

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2

Em

c

Substitute this for mass into the equation for

momentum - p=mv (ignoring direction)

But for all photons, v = c.

Also,

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2

Ep v

c

2

Ep c

c

Ec

hcE hf

l

which reduces to

Compton has succeeded in deriving an

elegant expression for the momentum of a

photon!! His work gave enormous support

to the notion that light possessed both

wave and particle properties.

83

Ep

c

hc

cl

hp

l

Practice:

1.In a Compton experiment a 82 eV x-ray photon collides with an electron causing the scattering of a lower energy x-ray of wavelength 295 nm. The mass of an electron is 9.11 x 10-31 kg. Determine

a) the momentum of the original x-ray.

b) the momentum of the scattered x-ray

c) the energy imparted to the electron

d) the increase in speed of the electron

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Given:

Ex-ray = 82 eV = 82 eV x 1.6 x 10-19 J/eV

= 1.3 x 10-17 J

lscattered = 295 nm = 295 x 10-9 m = 2.95 x 10-7 m

me = 9.11 x 10-31 kg

h = 6.626 x 10-34 J·s c = 3.00 x 108 m/s

Note that the last two givens did not appear

in the exercise, but they should always be in

the back of your mind.

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Solution:

a) the momentum of the original x-ray.

(Also, be mindful of the useful formulae

below): E = hf, f = c /l, E = (hc)/l, p = E/c,

p = h/l Ek = ½ mv2 (particle)

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px-ray = E/c = 1.3 x 10-17 J ÷ (3.00 x 108 m/s)

= 1.3 x 10-17 Nm ÷ (3.00 x 108 m/s)

= 4.3 x 10-26 N·s

Solution:

b) the momentum of the scattered x-ray

pscattered = h/l = 6.626 x 10-34 J·s ÷ (2.95 x 10-7 m)

= 2.2 x 10-27 N·s

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c) the energy imparted to the electron

Eelectron = Ex-ray - Escattered

= 1.3 x 10-17 J - hfscattered

= 1.3 x 10-17 J - (hc)/lscattered

= 1.3 x 10-17 J - (6.626x10-34 J·s x 3.00 x108 m/s

÷(2.95x10-7m))

= 1.3 x 10-17 J - 6.7 x 10-19 J

= 1.2 x 10-17 J

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d)the increase in speed of the electron

Assume the electron was

initially at rest. Find the

new speed from

Eelectron = Ee = ½ mev2.

=5.1x106m/s

2 e

e

Ev

m

17

31

2 1.2 10

9.11 10

Jv

kg

89

Practice2. A scattered x-ray photon has 75% of the

momentum of the incident photon. If the wavelength of the incident photon is 150 nm, what is the wavelength of the scattered photon?

90

Given: Let p1 = momentum of incident photon

and p2 = momentum of scattered photon

Then p2 = 0.75 p1

l 1 = 150 nm = 150 x 10-9 m = 1.5 x 10-7 m

h = 6.626 x 10-34 J·s l 2 = ?

Solution:

p2 = 0.75 p1 and p = h/l.

Therefore,

Cancelling h and solving

for l 2

2 1

0.75h hl l

12 0.75

ll

91

-7

2

1.5 x 10 m

0.75l

72 2.0 10 ______m nml

Practice: Text

Page 725

#1-4,6

Page 742

#10,11,20

92

Matter Waves

Louis de Broglie noticed that sometimes physics

ideas are reversible. For example, we have already

seen how changing electric fields produce

magnetism (Oersted) and changing magnetic fields

produce electricity (Faraday).

After Compton showed that electromagnetic

radiation (photons) exhibited properties of mass

(i.e., momentum), de Broglie wondered if mass

might have wave properties. He rearranged

Compton's equation.

93

Compton's equation:

De Broglie's

rearrangement:

But p = momentum

= mass x velocity

= mv.

Therefore,

hp

l

hp

l

94

hmv

l

Practice exercise 1

If you have a mass of 70 kg, find your

associated wavelength when you are running

at 8.0 km/hr (2.2 m/s).

Solution:

l= h/(mv)

=6.626 x 10-34 J·s ÷ (70 kg x 2.2 m/s)

= 4.3 x 10-36 m

95

Practice exercise 2

Find the wavelength associated with an electron that

is moving at 1% the speed of light.

Solution:

Right away, you can see two differences in this exercise

and the first one: here we have an object that is

(i) extremely small

(ii) moving extremely fast.

At the end of this exercise you will realize that it is the first

difference that is the important one.

96

1% the speed of light means that

v = 0.01 x 3.00 x 108 m/s = 3.00 x 106 m/s

and the mass of an electron,

m = 9.11 x 10-31 kg.

Computing the wavelength:

l= h/(mv)

= 6.63 x 10-34 J·s ÷ (9.11 x 10-31 kg x 3.00 x106 m/s)

= 2.43 x 10-10 m

= 0.243 nm.

97

This wavelength is still small, but is enormous

compared to your own wavelength in

exercise 1.

2.43 x 10-10 m / (4.3 x 10-36 m )

= 6 x 10 25 (times as large as your wavelength)

The reason why this wavelength is so large

by comparison is because the mass of the

electron is so tiny.

98

Can you remember the test for wave

properties from Physics 2204?

Answer: If wave properties are present, an interference pattern can be produced.

To produce such patterns the openings through which the moving matter passes must be about the

same size as the wave itself.

Right away you can see why your own associated wave cannot be detected, even if you had a few clones to interfere with!

There is no way you and your clones are going to squeeze through a hole that is 4.3 x 10-36 m in diameter!

99

But for an electron, the story is different. In fact, as

amazing as it sounds, electrons do interfere constructively

and destructively, and hence they do create interference

patterns.

De Broglie's prediction that matter has an associated wave

characteristic is true.

100

Practice:1. A) Calculate the wavelength for an electron that is moving at a

velocity of 42 000 m/s.

B) Calculate the wavelength for a proton that is moving at a velocity of

42 000 m/s.

C) How do the electron and protons wavelength compare?

101

Practice:2. Calculate the momentum of a 500-nm photon.

102

103

3. In your television set, an electron is accelerated through

a potential difference of 21 000 V.

(a) How much energy does the electron acquire?

(b) What is the wavelength of an electron of this energy?

The Final, most colourful topic:

104

105

106

Emission/Absorption Spectra

What happens when

light shines through a

prism?

Different frequencies of

light waves will bend by

varying amounts upon

passage through a

prism.

The separation of visible

light into its different

colors is known

as dispersion

Refraction is the bending

of the path of a light wave

as it passes from one

material into another

material.

The refraction occurs at

the boundary and

is caused by a change in

the speed of the light

wave

107

Emission/Absorption Spectra

How is this related to rainbows?

To view a rainbow, your back must be

to the sun as you look at an

approximately 40 degree angle above

the ground into a region of the

atmosphere with suspended droplets

of water or even a light mist.

Each individual droplet of water acts

as a tiny prism that both disperses the

light and reflects it back to your eye.

108

109

A continuous spectrum of light is produced when

white light passes through a prism. All colors of the

rainbow are produced.

An emission spectrum is observed when light is

emitted from excited gases. It was discovered that

only certain set frequencies or wavelengths of light

were given off (bright line emission spectra).

Different

elements have

different emission

spectrums

110

111

Now for a BOHRing Topic Niels Bohr (1885-1962) said a shocking thing: he said

that the classical ideas of physics cannot be applied to

orbiting electrons.

Classical physics theory demanded that an

accelerating charge emit electromagnetic radiation.

Electrons are certainly accelerating because they were

going around in circles.

But, if accelerating charges emit energy, the electrons

must, after some time, lose all of theirs and therefore

collapse into the nucleus. Sounds like our world

should come to a disastrous end.

112

Also, gases can be made to emit a spectrum that

does not appear as fuzzy coloured bands, but

rather as bright, sharp lines.

Some examples are below

113

These sharp coloured lines at specific places in

the spectrum suggest that there is something

going on in the atom that is allowing only certain

frequencies to be emitted.

It appears that instead of the electron gradually

spiraling into the nucleus and emitting a spectrum

of light (like the bottom portion above), it makes

certain jumps only.

It moves from one orbit to another, giving out a

particular frequency as it "falls".

114

Here is a summary of what Bohr said:

Of all the possible orbits that electrons might

take around a nucleus, only certain orbits are

"allowed". For each of those allowed orbits

the electron has a specific amount of energy.

While it stays in an orbit, the electron will not

radiate energy even though it is accelerating!

Each orbit is a stationary state.

115

Electrons lose energy only when they "jump"

from a higher energy orbit to a lower energy

orbit.

The difference in the energies of the two

orbits is radiated as one photon of

electromagnetic radiation.

Similarly, an electron can be "bumped" up to

a higher energy level only if the atom absorbs

a photon whose energy is exactly equal to

the energy difference between the two orbits.

116

What makes the allowed orbits so special?

Recall waves on a string that was fastened at both ends. The waves had to fit nicely so that each end of the string became a

node.

The picture below is a slight variation of this. That is, here we start with a fixed wave and choose the strings that work.

Notice that the strings are exact multiples of the chosen wave. No other strings would work if we wish to put one wavelength, two wavelengths and three wavelengths on them.

117

Now think back to de Broglie. He said that

matter has an associated wave.

In fact, we have calculated some waves

associated with moving electrons.

Using de Broglie's idea, Bohr proposed that

the only electron orbits allowed would be

those for which the electron wave would fit

exactly on the circumference of the orbit.

118

You can see from the picture below that the

smallest allowed orbit has a circumference of

2pr1 =l; the next largest allowed orbits have

circumferences of 2pr2 =2l ; 2pr3 =3l and so on,

up to the 4th orbit with a circumference of

2pr4 =4l.

119

In general for the nth orbit:

the radius will be rn

the circumference of the orbit will be 2prn

the de Broglie wavelength of the electron in

this orbit will be such that n wavelengths fit

on the circumference:

2prn = nl.............eq. 1.

"n" is the quantum number for the orbit.

Bohr has applied Planck's quantum idea to

determine the allowable orbits for the

electron!

120

Don't forget de Broglie's equation which

expresses the matter wave of the electron

in terms of the momentum of the electron:

l= h/p where p = meve.

Therefore, ..............eq. 2 .

121

e e

hmv

l

Use eq. 2 to substitute for l in eq. 1:

2prn = n ( h/meve), and rearrange to get

mevern = n(h/2p) ...........eq. 3.

Eq. 3 is called "Bohr's quantum condition" for allowed orbits.

In fact the left-hand side of eq. 3 has a special name: angular

momentum.

Using this and centripetal force ( caused by attractive

electric forces - Coulombs law) we get:

2 2

2 24 en

hk

nm e

rp

122

Look at the coloured bits. They are all

constants.

h = 6.626 x 10-34 J·s; p = 3.14;

k = 9.0 x 109 N·m2/C2

me = 9.11 x 10-31 kg and e = 1.6 x 10-19 C.

When you combine these constants, the equation

becomes:

rn = (5.3 x 10-11 m) n2

This is known as the Bohr radius equation.

123

2 2

2 24 en

hk

nm e

rp

As you can see a Bohr radius is directly

proportional to the square of the quantum

number associated with the radius.

Isn't it neat that the units work out to be

metres?

124

What is the orbital radius (in nm) of an

electron in a hydrogen atom in the third

energy level?

Solution:

Given n = 3 , find rn

rn = (5.3 x 10-11 m) n2

= (5.3 x 10-11 m) x (3)2

= 4.77 x 10-10 m

= 0.477 x 10-9 m = 0.477 nm

The orbital radius at the 3rd energy level is 0.477 nm.

125

Now let's get back to the real purpose of this part of the lesson:

Finding the energy of an electron energy level as a function of the quantum (n) of the level.

It was a high school teacher named J.J. Balmer (1885) who very cleverly wrote a formula to describe what he was seeing when he looked at visible line spectra like the ones described earlier.

126

The down arrows represent different

transitions of electrons as they "fall"

from higher levels to a lower level.

In each case a different colour of

visible light was emitted.

In the series of transitions shown, no

electrons fall lower than the n=2 orbit

or energy level. Why?

The jump from the second level to the

first is so great that uv light is emitted.

(which you can’t see)

127

When Balmer determined the wavelength of the

various lines in the spectrum, he discovered (by

fiddling with the numbers, not by theory), that the

inverse of the wavelength was directly proportional

to

That is,

2 2

1 1

2 n

2 2

1 1 1

2 n

l

128

Along comes Bohr, and using his theoretical model he,

too, developed a formula to describe the spectral lines.

Recall the expression for the radius of the electron orbit as

a function of the quantum number (n).

rn = (5.3 x 10-11 m) n2

Also there is an expression for the energy of the electron

as a function of the Bohr orbit radius (rn).

2

2e

keE

r

129

Substitute rn for r from the first equation into the second

equation:2

2en

keE

r

2

11 22(5.3 10 )e

keE

m n

130

29 19 2

2

11 2

9 10 (1.6 10 )

2(5.3 10 )e

N mC

CEm n

18

2

2.18 10e

N mE

n

18

2

2.18 10e

JE

n

Changing to eV

1819

2

12.18 10

1.602 10e

eVJ

JEn

2

13.6e

eVE

n

131

This result must have been very pleasing to Bohr. His

theory exactly matches the experimental evidence

described by Balmer!

2

13.6e

eVE

n

2

13.6c eVh

nl

2

1 1

n

l

132

which means that (ignoring constants) the

reciprocal of the wavelength is directly proportional

to the reciprocal of the energy level of the electron (n)

Why negative energy??

question: In the picture drawn earlier how many possible orbits were shown?

answer:

Four.

133

question: In which orbit or energy level does the

electron experience the greater force? Why?

answer:

In the first orbit which is the smallest orbit (n=1).

It experiences the greater force here because of the

Coulomb force caused by the positive proton in the

nucleus.

The Coulomb force is inversely proportional to the

square of the separation between the proton and the

revolving electron--that's the distance r, the radius of

the orbit.

The smaller the r, the greater the force.

134

question: In which orbit or energy level does the electron

have the greatest energy?

answer:

This is not a trick question.

It will have its greatest energy at the first energy level

(the smallest orbit).

That makes sense because it is in this orbit that thecentripetal force is greatest (due to the Coulomb force).

question: What happens to the Coulomb force or the

centripetal force as the electron moves in orbits further

and further from the positive proton in the nucleus?

answer:

The force becomes smaller.

135

question (This is the biggie): What will be the

energy of the electron if it moves very far away,

say to infinity?

answer:

It will have zero energy!

question: Can you remember how an electron can

be bumped up from one level to another?

answer:

Energy must be put into (added) the atom to "excite"

the electron to higher levels.

136

Now, let's sum up.

Energy has to be put into the atom to excite the

electrons to higher orbits. And if the electron is

bumped far enough away from the nucleus, its

energy goes to zero.

question: If adding energy makes the energy of

the electron go to zero, what kind of energy did

the electron have to begin with?

answer:

THE ENERGY TO BEGIN WITH MUST HAVE

BEEN NEGATIVE!!

137

Think about a simple case: if adding 3.0 J to an

electron makes the energy go to zero, then the

energy that the electron already had must have

been -3.0 J.

138

We will sum up by putting our new

knowledge on a picture that was drawn

earlier in the lesson.

The simplest energy calculation for an electron is

for n=1 (the lowest level allowed, the ground

state).

This value for the ground state is represented

by:

139

Ee =-13.6 eV / (1)2 = -13.6 eV.

2

13.6e

eVE

n

13.6oE eV

And when n = 4,

= -0.85 eV.

You can do the others yourself.

140

2

13.6

4e

eVE

You will note that the picture to the

right shows the Balmer transition

series only.

141

In the Pashen series no

electrons fall lower than

the n=3 level, while in

the Lyman series, the

electrons make it all the

way to the ground state,

n=1.

142

question: In the Balmer series a photon is

emitted as an electron falls from the n=5

state to the n=2 state.

A) What is the energy of the photon?

answer: In dropping from the upper n=5 to the lower

n=2 levels the electron loses energy lEe :

lEe = Eupper - Elower = E5 – E2

=-0.54 eV - (-3.40 eV)

= 2.86 eV.

143

B) What is the wavelength of the photon?

cE h

l

hcE

l

144

34 8

19

6.626 10 3.00 10

1.602 102.86

mJ s

sJ

eVeV

l

74.34 10 m

C) What type of light is

this?

Answer:

Visible violet light

The colors of the visible light spectrum

colorwavelength

interval

frequency

interval

red ~ 625–740 nm ~ 480–405 THz

orange ~ 590–625 nm ~ 510–480 THz

yellow ~ 565–590 nm ~ 530–510 THz

green ~ 500–565 nm ~ 600–530 THz

cyan ~ 485–500 nm ~ 620–600 THz

blue ~ 450–485 nm ~ 670–620 THz

violet ~ 380–450 nm ~ 790–670 THz

145

http://en.wikipedia.org/wiki/Color

question: In the Lyman series a photon is

emitted as an electron falls from the n=2

state to the n=1 state.

A) What is the energy of the photon?

B) What is the frequency of the photon?

C) What type of light is this?

D) Why was this jump not listed in the Balmer series?

146

Here's what is really good about Bohr's

model:

it matches the observed spectrum of hydrogen gas

it accounts for the stability of atoms.

That is, once the electron is in its lower orbit or ground state, it will go no further.

It will not lose energy while it is in a fixed orbit and therefore will not spiral into the nucleus.

the model can be applied to other one-electron atoms, such as a helium ion that has lost one of its two electrons.

147

From two of the above statements, you will not be surprised to learn that Bohr's model of the atom did not work for atoms with more than one electron.

Also, remember those bright discrete lines that we got so excited about earlier in the lesson? It turns out that they are not so discrete after all. Sophisticated equipment has shown that each of the lines is, in fact, composed of separate finer lines.

To explain this "splitting of the levels" the model has to be modified to include orbits which are not circular and electrons that spinas well as revolve.

148

Moreover, in the latest theory, electrons are

not considered to be particles revolving in a

single orbit.

Brace yourself for the next sentence:

electrons are now considered to be a

wave found in a probability cloud of negative

charge distributed around the nucleus.

The cloud is densest in areas of high

probability, and less dense in areas where

the electron is less likely to be found. ....but

that's a topic for another time (college?

university?)

149

Practice exercises 1. What is the ionization energy of hydrogen?

Solution:

This is not a fair question if you don't know what ionization means. An atom becomes ionized when it gains or loses

electrons.

An ion is a charged atom.

In this exercise you are asked how much energy is required to strip the electron from a hydrogen atom so that only the positive nucleus remains.

Assume that the atom is in an unexcited state--that is, the electron is at n=1. What is its energy there?

-13.6 eV.

150

In order for the electron to be removed it must be

bumped to such a high orbit number that its

energy goes to 0 eV. How can -13.6 eV be

changed to 0 ?

Answer: just add 13.6 eV

The ionization energy of hydrogen is 13.6 eV.

In other words, the ionization energy is the same

as the energy at the ground state.

151

2. An electron drops from the n=4 to the n= 1 energy

level. In what region of the electromagnetic spectrum is

this photon found?

Solution

When the electron falls from n=4 to n=1, a photon is emitted.

The energy of the photon (Ep) will be equal to the difference between the energies at n=4 and n=1.

152

E4 = -13.6/n2 eV = -13.6/42 eV = -0.85 eV.

E1 = -13.6/12 eV = -13.6 eV.

Therefore Ep = E4 - E1

= -0.85 eV - (-13.6 eV)

= 12.8 eV

Before applying Planck’s equation, change 12.8 eV

to joules by multiplying by 1.6 x 10-19 J/eV.

This gives

Ep = 2.0 x 10-18 J .

153

Recall:

and

Thus, l= 9.7 x 10-8 m.

This wavelength is in the: x-ray range.

154

l

cE h l

ch

E

Phosphorescence and Fluorescence

There are examples of natural luminosity such

as phosphorescence and fluorescence that can

be explained by the Bohr model of the atom.

Phosphorescent and fluorescent materials both

contain atoms that are easily excited.

Phosphorescence and fluorescence differ in the

amount of time it takes their excited atoms to

return to their normal energy levels.

What does an electron give off as it returns to its

normal energy level?

Photons

155

Phosphorescence

Phosphorescence occurs when atoms are excited

by absorption of a photon to an energy level said to

be metastable.

Normally when an atom is raised to a normal excited

state, it drops back down quickly and releases a

photon of energy at a lower energy level than the

incident photon.

Metastable states last much longer.

In a collection of atoms, many of these atoms will

remain in the excited state for over an hour.

Hence, light will be emitted even after long periods.

156

Examples of phosphorescence are:

luminous watch dials

glow in the dark stickers

glow in the dark tent lines

157

Fluorescence

Fluorescence occurs when an atom is excited

from one energy state to a higher one by the

absorption of a photon.

The atom may return to the lower level in a

series of two or more jumps.

This happens almost immediately (within about

10−8 seconds ( 10 ns)).

The emitted photons will have lower energy

and frequency than the absorbed photon.

158

Fluorescence Because reemission occurs so quickly, the

fluorescence ceases as soon as the exciting

source is removed, unlike phosphorescence,

which persists as an afterglow.

Examples of fluorescence include:

trail marker tape used in hiking and road signs.

A fluorescent lightbulb is coated on the inside with a

powder and contains a gas; electricity causes the gas

to emit ultraviolet radiation, which then stimulates the

tube coating to emit visible light.

159

Tooth whiteners and toothpaste may contain

molecules that make your teeth glow brightly

under black light.

Optical brighteners in laundry soap

This is how laundry detergents can get things "whiter than white": by absorbing non-visible UVlight and fluorescing in the visible spectrum.

160

IT’S ALIVE… AND IT GLOWS!

FLUORESCENCE IN MEDICAL RESEARCH

161

Remember,

the key difference between both is the length

of time they will last.

Phosphorescence lasts longer due to the

metastable state of atoms and will remain

luminous after the stimulating radiation is

absent.

Fluorescence will only be luminous while the

stimulating radiation is present.

162

Final Objective for Unit 4:Students should be able to summarize the evidence for

the wave and particle models of light

That means

define wave-particle duality

The property of matter and electromagnetic

radiation that is characterized by the fact that

some properties can be explained best by wave

theory and others by particle theory

163

give evidence of light being both a wave:

behaviour of long wavelengths - AM radio waves

interference - In Young’s double slit experiment

where coherent light (light which is in the same

phase, and has the same frequency and

wavlength) interferes to produce an interference

pattern similar to those found in water waves.

diffraction - Rainbows

164

a particle:

blackbody radiation

photoelectric effect

Compton effect

line spectra

behaviour of short wavelengths like x-rays

165

The Duality of Light

166

BACK

167

168