17 PROBING DEEP INTO MATTERCreation and Annihilation

• Describe differences between matter and antimatter

• Apply conservation laws to annihilation and materialisation events

• Explain how antimatter is used in medical imaging (PET)

Starter: Can you give reasons why imaging the brain using x rays may have limitations or drawbacks.

• Conventional x ray imaging does not show soft tissue well

• Conventional x ray imaging is a “2 dimensional” technique

• Computerised axial tomography can give 3D information but x ray dose is higher than conventional x ray imaging

• X ray techniques cannot map brain activity, only show structures

Read pages 189-191

Answer the following:1) What does PET stand for? 2) Try and summarise in you own words how it works.3) What is the main difference between matter and antimatter? What is the one thing that is the same?

Ext:4) What is meant by creation, annihilation and pair production?

Conserved quantities

Energy is conserved

= kinetic energy of particles+ rest energy of particles

total energy before

minimum value of energybefore is rest energy:

= 2 mc2 = 2 0.511 MeV

total energy after

energy after isenergy of gammaphotons= 2 0.511 MeV

=

Momentum is conserved

=total linear momentum before total linear momentum after

Electric charge is conserved

e– e+

same mass; equal andopposite velocities

energy E,momentum p = E/c

photons identical,momentums opposite

total momentum before = 0 total momentum = 0

=total charge before total charge after

charge

(–e) + (+e) = 0charge

0 + 0 = 0

e–Simplify:assume head-oncollision withequal speeds

e+

Energy, momentum and electric charge are always conserved inelectron–positron annihilation

Conserved quantities

Energy is conserved

= kinetic energy of particles+ rest energy of particles

total energy before

minimum value of energybefore is rest energy:

= 2 mc2 = 2 0.511 MeV

total energy after

energy after isenergy of gammaphotons= 2 0.511 MeV

=

Momentum is conserved

=total linear momentum before total linear momentum after

Electric charge is conserved

e– e+

same mass; equal andopposite velocities

energy E,momentum p = E/c

photons identical,momentums opposite

total momentum before = 0 total momentum = 0

=total charge before total charge after

charge(–e) + (+e) = 0

charge0 + 0 = 0

e–Simplify:assume head-oncollision withequal speeds

e+

Energy, momentum and electric charge are always conserved inelectron–positron annihilation

Conserved quantities

Energy is conserved

= kinetic energy of particles+ rest energy of particles

total energy before

minimum value of energybefore is rest energy:

= 2 mc2 = 2 0.511 MeV

total energy after

energy after isenergy of gammaphotons= 2 0.511 MeV

=

Momentum is conserved

=total linear momentum before total linear momentum after

Electric charge is conserved

e– e+

same mass; equal andopposite velocities

energy E,momentum p = E/c

photons identical,momentums opposite

total momentum before = 0 total momentum = 0

=total charge before total charge after

charge(–e) + (+e) = 0

charge0 + 0 = 0

e–Simplify:assume head-oncollision withequal speeds

e+

Energy, momentum and electric charge are always conserved inelectron–positron annihilation

Energy, momentum and Electric Charge

conserved

Making PET scans

A pair of gamma rays are emitted in oppositedirections as a result of electron/positronannihilation inside the patient

Photomultiplier –incoming photon createsa cascade of electrons,giving an electrical pulseoutput

Scintillator – captures gamma rayphoton and emits lower energyphotons into photomultiplier tubes

signal processing

Scintillators are arrangedin a grid on the insidesurface of the scanner. Inany short period of timemany detectors willrespond to gamma raysfrom many differentannihilations inside thebody. A computerproduces a slice-by-slicemap of activity in the brain.

One pair of detectors willrespond almostsimultaneously. This nearcoincidence shows thatthe two gamma rays camefrom a common source.The tiny time differencebetween the two signalsis then used to work outwhere they came fromalong the line between thedetectors.

Annihilation and CreationPair annihilation and creation

Pair creation

gamma energy= 2 0.511 MeV (minimum)nucleus carries awaymomentum, to conservemomentum and energy

Creation?

extremely rare(cannot bring twoidentical photonstogether)

Annihilation

gamma energy= 2 0.511 MeV pluskinetic energy of electrons

e– e+

e– e+

e– e+

close tonucleus

Pair Production

Pair annihilation and creation

Pair creation

gamma energy= 2 0.511 MeV (minimum)nucleus carries awaymomentum, to conservemomentum and energy

Creation?

extremely rare(cannot bring twoidentical photonstogether)

Annihilation

gamma energy= 2 0.511 MeV pluskinetic energy of electrons

e– e+

e– e+

e– e+

close tonucleus

Starter…

An electron and a positron with negligible kinetic energy annihilate and produce two identical gamma ray photons.(Rest mass of electron =9.11x10-31kg, h=6.63x10-34Js, c = 3x108 ms-1)

Calculate a) the energy released (in J and MeV) b) the frequency of the gamma-photons (in Hz).

Particle interactions• Describe how charged particles interact

via virtual photon exchange• Explain how to describe these processes

using Feynman diagrams and the “try all paths” approach

• Discuss consequences of the differences between fermions and bosons

Starter: What are the maximum and minimum amplitudes that can result from adding two phasors, each of length 1 unit, and what are the phase differences in each case?

Starter: Explain the point this diagram is making….

SourceDetector

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

The same idea is applied to interactions of particles. Richard Feynman invented a type of diagram to help

physicists keep track of all the possible ways that particles can interact. The rule “try all paths” changes to “try

everything allowed” or more technically “everything that is not forbidden is compulsory”.

Quantum fields create and destroy particles

cannot happen

photon carriesaway energyand momentum

charge e

cannot happen

Electromagnetic quantumfield creates a photon

Electromagnetic quantum fielddestroys a photon

photon deliversenergy andmomentum

charge e

Forces between charges arise from the exchange of momentum throughexchange of virtual photons

combined process can happen

charge e

charge e

photon exchanged:charges exchange energy andmomentum

virtual(unobservable)photon

energy and momentum conserved overall

energy and momentum notboth conserved

Feynman diagrams show possibilities to be combined

Propagation and interaction of a pair of electrons

Electrons at A and B arrive at C and D

no photon exchange

Electrons areidentical,so there is no wayto tell these apart:

add the phasorarrows for bothdiagrams

electrons just travel A to C, andB to D

electrons just travel A to D,and B to C

one photon exchange

C

A

D

B

C D

A B

C

D

A

D

A

B

C

B

No way to tellthese apart:

add the phasorarrows for bothdiagrams

For each diagram, add phasor arrows for all possible space-time locations A, B, C, D.Add total phasor arrows for each type of diagramQuantum rule: Try all possible ways to interact

electrons exchange a photon electrons exchange aphoton

.... plus diagrams with more photons....

Ways for an electron to scatter a photon

B

A

e–

e–

B

e–

A

e–

e–

A

B

e–

e+ plus othermorecomplexdiagrams

Simple electron–photon interactions

Electron: absorbs photon at A travels to B emits photon at B

Electron: emits photon at A travels to B absorbs photon at B

photon creates e–, e+ pair at Be+ goes to Ae– and e+ annihilate at A,emitting photon

In each diagram one electron and one photon come in, and one electron and one photon go out.All diagrams represent the same process.

Phasor arrows for all diagrams are added to find total amplitude and phase for scattering

interaction with virtual paircreation and annihilation

To find out..........What is a fermion? Give an example of a particle that is a fermion.

What happens if you try to squeeze two fermions into the same region of space?

Can you describe oneimportant consequence of this?

What is a boson? Give an example of a particle that is a boson.

What happens if two identical bosons encounter each other?

Can you give one important practical application of this?

Electron spin Photon spin

Fermions and BosonsFermions

½ integer spin (1/2, 3/2,....)Electrons, Protons etc.

Never occupy same quantum state(Avoid each other always)

Consequently.....• Two electrons in same orbital of

an atom must have opposite spin (Pauli exclusion principle)

• Matter is “hard”: atoms are difficult to squash!

Bosons

Integer spin (0,1,...)Photons

Can occupy same quantum state(Can “flock” together in step)

Consequently.....• In a laser, lots of photons join

together to produce beam of photons all of identical phase and polarisation

Identical particles in the same state

A B

photons are indistinguishable

Photons (bosons)

these possibilities cannot be distinguished

C D

A B

X Y

photon X goes to Cphoton Y goes to D

photon X goes to Dphoton Y goes to C

bring points C and D together

Phasors for the twodiagrams becomethe same.Adding phasorsgives 2 amplitude4 intensity

Photons in a given state increase the chance of others joining them inthe same state

E

add the phasors for the two diagrams

+ =

C D

A B

X Y

Identical particles in the same state

A B

electrons are indistinguishable

Electrons (fermions)

these possibilities cannot be distinguished

C D

A B

X Y

electron X goes to Celectron Y goes to D

C D

A B

X Y

electron X goes to Delectron Y goes to C

bring points C and D together

Exchanging particlesreverses phase.Adding phasors giveszero amplitude

There is zero probability for two electrons to be at the same space-timepoint (be in same state)

E

This is the Pauli exclusion principle

add the phasors for the two diagrams

+ = 0

Fermions and BosonsFermions

½ integer spin (1/2, 3/2,....)Electrons, Protons etc.

Never occupy same quantum state(Avoid each other always)

Consequently.....

• Two electrons in same orbital of an atom must have opposite spin (Pauli exclusion principle)

• Matter is “hard”: atoms are difficult to squash!

Bosons

Integer spin (0,1,...)Photons

Can occupy same quantum state(Can “flock” together in step)

Consequently.....

• In a laser, lots of photons join together to produce beam of photons all of identical phase and polarisation

Light Amplification by Stimulated Emission of Radiation

Starter

For each of the following statements, say whether it applies to FERMIONS or BOSONS:Q1. A photon is an example of this class of particle.Q2. These particles have integer spin values.Q3. Particles which cannot occupy the same quantum state.Q4. “Matter” particles, like protons, neutrons and electrons belong to

this class.Q5. Virtual particles which are exchanged between interacting matter

particles belong to this class.Q6. These particles can have spin values of 1/2, 3/2, 5/2 etc.

Q7. (For chemistry students): Two electrons cannot occupy the same quantum state. How is it then possible to get two electrons into the same orbital? Hint: in what way are the two electrons distinguishable when in the same orbital?

Conservation in nuclear processes

• Explain why a new particle was needed to account for the energy spectrum of beta particles

• Balance nuclear equations: charge, mass-energy, baryon number, lepton number

The Baryon Family

Baryons contain three quarks (we come to them later).Protons and neutrons are baryons! Baryon number must be conserved. Baryon number is +1 for all protons and -1 for anti-protons.

Note: Protons and neutrons are also described as nucleons.

Leptons – fundamental particles

Leptons are fundamental particles. As far as we are aware they are not made up of anything smaller. Examples are electrons and neutrinos.

Leptons are given a property called lepton number. Electrons and neutrinos are given lepton number 1. Where as the antiparticles are given lepton number -1.

All hadrons (non-leptons, which we learn more about later) have lepton number 0.

‘Missing’ energy in beta decay

Beta decay of strontium–90

restenergy

strontium–90

0.546 MeV –

yttrium–90

38 protons52 neutrons

39 protons51 neutrons

Energy spectrum of beta decay of strontium–90

fraction of betaparticles perunit energyrange

0.546energy/MeV

0

Beta decay of strontium–90, includingantineutrino emission

restenergy

strontium–90

0.546 MeV

–

yttrium–90

energy E

energy 0.546 MeV – E

Rutherford’s experiment• Describe how alpha

scattering changed our view of atomic structure

• Explore effects of changing alpha particle energy and nuclear charge on scattering

• Estimate upper limit on nuclear size

Starter:

Q1. I have a charge of +1 and a lepton number of -1. What could I be?

Q2. I have a charge of zero and a lepton number of +1. What could I be?

Q3. I have a baryon number of +1, a lepton number of zero and a charge of +1. What could I be?

Q4. I have a baryon number of +1 and a charge of zero. What could I be?

Q5. I have a baryon number of -1 and a charge of -1. What could I be?

Starter:

Q1. I have a charge of +1 and a lepton number of -1. What could I be? ANTIELECTRON (POSITRON)

Q2. I have a charge of zero and a lepton number of +1. What could I be? NEUTRINO

Q3. I have a baryon number of +1, a lepton number of zero and a charge of +1. What could I be? PROTON

Q4. I have a baryon number of +1 and a charge of zero. What could I be? NEUTRON

Q5. I have a baryon number of -1 and a charge of -1. What could I be? ANTIPROTON

Gravitational and electric fields comparedQ1. (a) Write down the expression for the force F between two masses, M and

m, separated by a distance R.Q1. (b) Write down the corresponding expression for the force F between two

charges Q and q, separated by a distance R. (The constant in the equation is known as the electric force constant, and is denoted by k.)

Q2. (a) Write down the expression for the gravitational potential energy for an object of mass m at a distance R from the centre of a planet of mass M.

Q2. (b) Write down the corresponding expression for the gravitational potential energy of a charge +q at a distance R from another charge +Q.

Q3. (a) How much kinetic energy would you need to give the object in Q2. (a) for it to be able to “climb out” of the potential well of the planet?

Q3. (b) How much kinetic energy would the charge +q in Q3. (a) have if it was released and allowed to coast far away from +Q?

Rutherford’s scattering experiment

scattered alphaparticles

microscope to view zincsulphide screen and countalpha particles

vary angle ofscatteringobserved

radium source ofalpha particles

thin goldfoil

alpha particlebeam

zinc sulphide screen;tiny dots of light wherestruck by alpha particle

lead block to selectnarrow beam of alphaparticles

Rutherford’s experiment

Note: an alpha particle

Is a helium nucleus with the

electrons removed. So it is

positively charged!

Rutherford’s observation

Explanation Model of the atom

A. Most alpha particles passed straight through the foil, undeflected

B. Some were deflected off course as they passed through

C. Some bounced right back from the foil

Copy and complete the table using the statements provided

there must be centres of+ charge in the atom

the centres of + chargemust be much heavierthan the alpha particles

the atom is mostly empty space

the nucleus is positively charged, whilethe electrons areoutside it, far away

nearly all the mass of the atom is in the nucleus

the nucleus is tinycompared to theoverall size of theatom

Rutherford’s observation

Explanation Model of the atom

A. Most alpha particles passed straight through the foil, undeflected

The atom is mostly empty space.

The nucleus is tinycompared to theoverall size of theatom

B. Some were deflected off course as they passed through

There must be centres of + charge in the atom.

The nucleus is positively charged, while the electrons are outside it, far away

C. Some bounced right back from the foil

The centres of +charge must be much heavier than the alpha particles

Nearly all of the mass of the atom is in the nucleus

Rutherford’s picture of alpha scattering

nucleus paths of scatteredalpha particles

105

0 30 60 90 120 150 180scattering angle/degree

Rutherford’spredic tion for asmall, massivecharged nuc leus

Number scattered decreases with angle

104

103

102

101

100

Careful investigation of alpha scattering supported the nuclear model ofthe atom

As sumptions: a lpha pa rtic le is the He nucleus, cha rge + 2e gold nucleus h as charge + Ze, and is much more massive than alpha part icles scattering force is inverse sq uare el ectrical repuls ion

scattering angle

aiming error bgold nuc leuscharge + Ze

a lpha part iclescattered

e qual force F b utnucleus is massive,so litt le r ecoil

charge +2e

For calculations

d

fo rce F = 2Ze2

40d2

TEST: Are slowed-down alpha particlesscattered more?

less energetic alphaparticle turnedaround further fromthe nucleus

reduce alphaenergy withabsorber

Z

Z

lower speed

TEST: Does using nuclei of smaller chargescatter alpha particles less?

alpha particle gets closerto nucleus of smallercharge and is deflectedless

replace foil bymetal of smalleratomic number

smaller nucleus with lesscharge, e.g. aluminium

Z

Rutherford’s picture of alpha scattering

nucleus paths of scatteredalpha particles

105

0 30 60 90 120 150 180scattering angle/degree

Rutherford’spredic tion for asmall, massivecharged nucleus

Number scattered decreases with angle

104

103

102

101

100

Careful investigation of alpha scattering supported the nuclear model ofthe atom

As sumptions: a lpha pa rtic le is the He nucleus, cha rge +2e gold nucleus h as charge + Ze, and is much more m assive than alpha part icles scattering force is inverse sq uare el ectrical repuls ion

scattering angle

aiming error bgold nucleuscharge + Ze

a lpha part iclescattered

e qual force F b utnucleus is massive,so litt le r ecoil

charge +2e

For calculations

d

fo rce F = 2Ze2

40d2

TEST: Are slowed-down alpha particlesscattered more?

less energetic alphaparticle turnedaround further fromthe nucleus

reduce alphaenergy withabsorber

Z

Z

lower speed

TEST: Does using nuclei of smaller chargescatter alpha particles less?

alpha particle gets closerto nucleus of smallercharge and is deflectedless

replace foil bymetal of smalleratomic number

smaller nucleus with lesscharge, e.g. aluminium

Z

Rutherford’s picture of alpha scattering

nucleus paths of scatteredalpha particles

105

0 30 60 90 120 150 180scattering angle/degree

Rutherford’spredic tion for asmall, m assivecharged nucleus

Number scattered decreases with angle

104

103

102

101

100

Careful investigation of alpha scattering supported the nuclear model ofthe atom

As sumptions: a lpha pa rtic le is the He nucleus, cha rge + 2e gold nucleus h as charge + Ze, and is much m ore m assive than alpha part icles scattering force is inverse sq uare el ectrical repuls ion

scattering angle

aiming error bgold nucleuscharge + Ze

a lpha part iclescattered

e qual force F b utnucleus is massive,so litt le r ecoil

charge +2e

For calculations

d

fo rce F = 2Ze2

40d2

TEST: Are slowed-down alpha particlesscattered more?

less energetic alphaparticle turnedaround further fromthe nucleus

reduce alphaenergy withabsorber

Z

Z

lower speed

TEST: Does using nuclei of smaller chargescatter alpha particles less?

alpha particle gets closerto nucleus of smallercharge and is deflectedless

replace foil bymetal of smalleratomic number

smaller nucleus with lesscharge, e.g. aluminium

Z

Distance of closest approach

5 MeV

variation of potential1r

d

alpha particle stopswhere potential hill is5 MeV high

alpha partic le with5 MeV initial kineticenergy

alpha particlescattered through 180

charge+ Ze(Z = 79 for gold)

Initial kinetic energy

= 5 MeV= 5 106 eV 1.6 10–19 J eV–1

= 8.0 10–13 J

Alpha particle stops whereinitial kinetic energy = electrical potential energy

8.0 10–13 J =+ 2 Ze2

40d

substitute values of Z, e, 0:

d = 4.5 10–14 m

Electrical potential energy

V = + 2 Ze2

40d

Z = 79, e = 1.6 10–19 C,

0 = 8.9 10–12 C2 N–1 m–2

The radius of a gold nucleus must be less than 10–14 m. Atoms are10000 times larger than their nuclei.

Where does the alpha particle stop?

initial kinetic energy= electrical potential energy

distance r

Measuring the size of nuclei• Explain how electron diffraction can be used

to measure nuclear size accurately and precisely

• Determine nuclear diameter from scattering curves

• Describe and explain the relationship between nuclear size and nucleon number

Starter: From your AS physics waves knowledge, explain the appearance of all features of the single-slit diffraction curve shown below:

JJ and GP Thomson: father and son Nobel physics prize winners

Listen sonny, I got the Nobel prize for showing that the electron is a particle...

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

Wave-particle dualityEvidence for wave-like character

Evidence for particle-like character

Light

Electrons

This eerie green glow is caused by low energy electrons in a cathode ray tube striking the phosphorescent coating on the inside of the glass bulb just behind the ruler. 

In this case the diffraction is caused by the electrons passing through a thin layer of polycrystalline graphite (pencil "lead"). The regular array of carbon atoms in the crystals is responsible for the diffraction effects.

Electron diffraction

How does the electron energy affect what is seen in an electron diffraction experiment?

Diffraction of low energy electronsDiffraction by planes of atoms of low energy electrons gives a diffraction pattern that reveals the inter atomic spacing. Here, the de Broglie wavelength of the electrons is quite large, as their momentum is small. The wavelength is comparable to the inter atomic spacing, so we get a lot of diffraction by the planes of atoms in the manner of a diffraction grating diffracting light.

Diffraction of high energy electronsWith very high energy electrons, the de Broglie wavelength is comparable to the size of a nucleus, and the diffraction effects seen are essentially the same as single slit diffraction. The atomic nucleus behaves as an obstacle for the electrons to diffract around, and the diffraction patterns seen is essentially the same as we get when light passes through a single slit.

Electron diffraction

Q1. Explain why there is a minimum in the curve.Q2. What would be the effect on the curve of using higher energy electrons?Q3. What would be the effect on the curve of using a sample of argon-40 in place of neon-20?

Density of nuclear matterVolume of nucleus increases linearly with number of nucleons

500

0

1000

1500

59Co 88Sr28Si16O12C

0 150 200100504He

1H number of nucleons

122Sb

197Au

Electron scattering measuresradius r of nucleuscalculate volume = r34

3

Estimate from graph:100 nucleons involume 700 10–45 m3

Calculate density:

density = massvolume

= 2.4 1017 kg m–3

density = 1.7 10–27 kg

7 10–45 m3

Data:mass per nucleon u= 1.7 10–27 kg

Density of ‘nuclear matter’ is roughly 2 1017 kg m–3

A matchbox-full of nuclear matter would have a mass of five billiontonnes

volume per nucleon= 7 10–45 m3

Density of nuclear matterVolum e of nucleus increases linearly with number of nucleons

500

0

1000

1500

59Co 8 8Sr2 8Si

16O12C

0 150 200100504 He

1H number of nucleons

122Sb

1 97Au

Electron scattering measuresradius r of nucleus

calculate v olume = r343

Estimate from graph:

100 nucleons involume 700 10–45 m3

Calculate density:

density = massvolum e

= 2.4 1017 kg m–3

density = 1.7 10–27 kg

7 10–45 m3

Data:

mass per nucleon u= 1.7 10–2 7 kg

Density of ‘nuclear matter ’ is roughly 2 101 7 kg m –3

A matchbox full of nuclear matter would have a mass of five billiontonnes

volum e per nucleon= 7 10–45 m3

KEY POINTVolume of nucleus is

proportional to number of particles it contains.

Nuclear density questions...Q1. If nuclear volume is proportional to the number of

nucleons (A) in the nucleus, explain why nuclear radius r is given by: r = r0 A1/3 where r0 is a constant.

Q2. The nucleon number of a gold nucleus is 197.a) The radius of the nucleon is 1.2x10-15m. Calculate the

radius r of the gold nucleus.b) Calculate volume of the gold nucleus.c) The mass of a nucleon is 1.67x10-27kg, calculate the

nuclear density.

Q3. Silver has a mass number of 108. What is its nuclear density?

The structure of nucleons

• Explain why electrons are well suited to be a probe of nucleon structure

• Deduce the quark composition of a range of hadrons (baryons and mesons)

Starter: Explain why alpha particles and protons are not well suited to probing the size of nuclei

Deep inelastic scattering

Medium energy: elastic scattering

Quarks move rapidly inside proton.The interaction time is long enoughfor the proton to behave like a blur ofcharge.

electron

proton

Deep inelastic scattering

High energy: deep inelastic scattering

electron

electronscatteredat large angle

Electron can hit one quark, and be scattered. Exchangeof high energy photons leads to the creation of a jet ofparticles and antiparticles.

‘jet’ of particles, mainly mesons

u

d u

Particle Family Tree

All particles

Fundamental particles (no internal structure)

Leptons(electron,

muon, tau, neutrinos)

Exchange Particles

Gluon, W, Z, photon

Non-fundamental Particles Hadrons

(made of quarks)

Baryons(contains 3

quarks)e.g. Proton,

neutron

Mesons(quark +

antiquark)e.g. Pion,

kaon

As we currently understand!!

Quarks The building blocks of protons and neutrons, and other fundamental particles.

Two flavours of quark…The up quark (+ 2/3 e) and the down quark (– 1/3 e).

The first direct evidence for quarks was obtained when very high-energy electrons (approx. 20Gev) in a beam were scattered from a stationary target as if there were point-like scattering centres in each proton or neutron.Quarks do not exist in isolation. They are bound together by the exchange of gluons (since they are the glue that hold the quarks together).

Quark–gluon interaction

Quarks interact by exchanging gluons, which change the quark colours. Here a red quark and a blue quarkexchange a red–blue gluon. The red quark becomes blue and the blue quark becomes red. The quarks exchangeenergy and momentum.

red quark

blue quark

blue quark

red quark

red–blue gluon

Colour charge

Quarks have a new type of charge called colour charge.

The charge comes in 3 types.Red, Green and Blue.

They aren't actually coloured.Reason for colour charge is you need one of each colour to make a proton or neutron just like you need all colours to make white.

Quarks pulled apart make more quarks

two quarks held together bythe gluon field...

...pull the quarks apart. Thegluon field increases inenergy...

...a quark–antiquark pairmaterialises from the gluonfield

quark quark

gluon field

quark antiquark

quark quark

quark quark

You cant really

separate quarks!!!

Starter: Copy the table and fill in each box with example(s) of each phenomenon

WAVES PARTICLESLight

Electrons

Chemistry questions that only physics can answer.....DISCUSS IN PAIRS

o Why are atoms mostly empty space?

o Why don’t the electrons just fall into the nucleus?

o Why are only some energy levels allowed for electrons in an atom or a molecule?

o Why does delocalisation of electrons stabilise a molecule?

o Why are carrots orange?

The classical atom dethroned

• Explain the flaws in the “solar system” model of the atom

• Investigate an electron-wave atomic model

• Use an electron-wave model to explain both the stability of aromatics and the origin of colour in chemistry

What is the problem?!?Maxwell’s theory which states that an electron being accelerated in an electric field will emit

radiation.

Hint: Think about the Diamond visit!

Wave-Particle Duality (for electrons)AS-RECAP

What is the evidence?!?

de Broglieλ = h/p = h/mv

m=9.1x10-31kgh=6.6x10-34Js

Standing Waves…AS-RECAP

harmonic n = 3

harmonic n = 2

at rest

fundamental n = 1

Strings of a guitar

Allowed discrete values of wavelengths only….

L = λ/2L= λ L= 3λ/2

Ie L = n λ /2

L-length of stringn – harmonic

Electrons in atoms… (QM)Atoms can be thought of as a box in which electrons are trapped.Results of model:Electrons can only occupy certain discrete energy levels in atoms.Each level has a different associated standing wave of a particular λ.A guitar–string atom

Simplify:

Change the 1/rpotential wellof the nucleusinto a pair offixed high walls

E4 = 2m2

h2= 42E1

energy

nucleus +

trappedelectron

1/rpotentialwell

energy

d

trappedelectron

E3 = 2m2

h2= 32E 1

E2 = 2m2

h2= 22E 1

E1 = 2m2

h2

4

3

2

1

n = 4

n = 3

n = 2

n = 1

levels increaseenergy n2

Each level has a quantum number n. The energy depends on the quantumnumber.

d

no levels at allbelow n = 1

For the nth level

En = n2 h2

2m21

En = n2 h2

2m(2d)2

In general:En = n2E1

Energy

4 = 1/4

3 = 1 /3

2 = 1/2

1 = 2d

Putting de Broglie and standing waves together….

Beta carotene…..….gives carrots and pumpkins their orange colour….in humans, is converted into retinal, which is essential for the vision process

• A conjugated system with alternating single and double bonds• Average carbon-carbon bond length = 140 pm (1 pm = 10-12 m)• 2 delocalisable electrons per double bond

ChallengeCan you use the equation for electron delocalisation in one dimension (particle in a box) to :(a) determine the pattern of energy levels in beta carotene,(b) calculate the wavelength at which it absorbs light.

Beta-carotene: a chromophore modelled as a particle in a box

• A conjugated system with alternating single and double bonds

• Average carbon-carbon bond length = 140 pm

• Calculate the length over which electrons are delocalised

• Identify the quantum numbers of the highest-occupied and lowest unoccupied energy levels

• Use the formula E=n2h2/8mL2 to calculate the energies of these levels

• Calculate the energy and wavelength of the photon that would be absorbed if an electron was excited from the highest-occupied to lowest-unoccupied level

Chemistry questions that only physics can answer.....

o Why are atoms mostly empty space?

o Why don’t the electrons just fall into the nucleus?

o Why are only some energy levels allowed for electrons in an atom or a molecule? EXPLAINED WITH BASIC MODEL

o Why does delocalisation of electrons stabilise a molecule? EXPLAINED WITH BASIC MODEL

o Why are carrots orange? EXPLAINED WITH BASIC MODEL

StarterIn the last lesson, we modelled the electron in an atom by treating the electron as a standing wave confined to a potential well.

Q1. Which of the following feature(s) of the atom did this model reproduce?- Quantised electron energies (electrons can only have certain energies

within the atom)?- Energy levels becoming closer together with increasing energy, as is

observed in the spectra of atoms?

Q2. The model could be used to describe delocalised electrons in molecules such as benzene and carotene, and thereby explain why delocalisation (resonance) stabilises these molecules, and how colour arises in them.

(a) Using the energy level expression derived from the model, explain why allowing electrons to spread out in a molecule leads to stabilisation.

(b) Explain how to calculate the energy of a photon absorbed by a molecule with delocalised electrons.

Finally, a model for the atom that fits experimental data!

• Develop an electron-wave model that incorporates potential energy

• Use the model to investigate atom stability

• Calculate energy levels and atomic radii with the model, and see if they match experiment

Considering Potential energy properly…

How small could a hydrogen atom be?

Replace 1/r potential by a box ofwidth d = 2rCalculate kinetic energy for waves = 2d = 4rCalculate potential energy at r

Find the min imum radius of an atom, for total energy < 0

potential energy Ep = – e2

40r

standing wave /2 = dmomentum p = h/

kinetic energy = p2/2m

kinetic energy Ek = h2

2m2

d = 2rimaginarybox

= 4r

nucleus1/r potential

If size is too small, the kinetic energy is too large for electrical potential energy tobind the electron

short wavelength

medium wavelength

long wavelength

unstable

juststable

stable

small radiusEk + Ep > 0

potentialenergy

kineticenergy

potentialenergy

kineticenergy

medium radiusEk + Ep = 0

potentialenergy

kineticenergy

large radiusEk + Ep < 0

120

100

80

60

40

20

0

–20

–40

total energy > 0unstable

total energy = 0just stable

total energy < 0bound

0.02 0.04 0.06 0.08radius r/nm

minimumradius ofbound atom

potential energy = – e2

40r

–60

+

+

+

+

2m2kinetic energy = h2

How small could a hydrogen atom be?

Replace 1/r potential by a box ofwidth d = 2rCalculate kinetic energy for waves = 2d = 4rCalculate potential energy at r

Find the minimum radius of an atom, for total energy < 0

potential energy Ep = – e2

40r

standing wave /2 = dmomentum p = h /

kinetic energy = p2/2m

kinetic energy Ek = h2

2m2

d = 2rimaginarybox

= 4r

nucleus1/r potential

If size is too small, the kinetic energy is too large for electrical potential energy tobind the electron

short wavelength

medium wavelength

long wavelength

unstable

juststable

stable

small radiusEk + Ep > 0

potentialenergy

kineticenergy

potentialenergy

kineticenergy

medium radiusEk + Ep = 0

potentialenergy

kineticenergy

large radiusEk + Ep < 0

120

100

80

60

40

20

0

–20

–40

total energy > 0unstable

total energy = 0just stable

total energy < 0bound

0.02 0.04 0.06 0.08radius r/nm

minimumradius ofbound atom

potential energy = – e2

40r

–60

+

+

+

+

2m2kinetic energy = h2

KEY POINT:If size is too small,

the kineticenergy is too large

for electrical potential to bind

the electron.

Considering Potential energy properly…

The Bohr model of the atom

Total Energy = Kinetic Energy + Potential Energy Q1. Which of the terms in the equation above is always positive, and which is always negative?Q2. For a stable atom, one where the electron is bound, to the nucleus, what must be true about the sum of KE+PE?

Kinetic Energy

KE = h2 / 2mλ2

where λ is the wavelength of the standing wave describing the electron

What will happen to the KE if the electron is confined in a smaller space closer to the nucleus?

Potential energy

PE = -Zke2 / r

where r is the nucleus-electron distance, e is the charge on the electron and k is a constant (=1/4πε0)

What will happen to the PE if the electron is confined in a smaller space closer to the nucleus?

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45

-3E-18

-2E-18

-1E-18

0

1E-18

2E-18

3E-18

4E-18

KE + PE for hydrogen atom, with electron modelled as a "particle on a ring", 2-dimensional standing wave

r / nm

Tota

l ene

rgy

/ J

At r = 0.059 nm, the total energy is a minimum. This is the dis-tance at which the electron settles, and determines the radius of the atom. The value of r agrees exactly with experiment in this case, showing that the model is a valid one.

At very small radii, the KE is very large as the electron is confined to a very small orbit. The result is that KE + PE > 0. The atom is unbound.

At large radii, the KE term is small. The atom is bound because of the negative PE term, but not as stable as it could be if the electron approached the nucleus more closely.

Chemistry questions that only physics can answer.....

o Why are atoms mostly empty space? EXPLAINED WITH BOHR MODEL (electron wave + PE)

o Why don’t the electrons just fall into the nucleus? EXPLAINED WITH BOHR MODEL (electron wave + PE)

o Why are only some energy levels allowed for electrons in an atom or a molecule? EXPLAINED CORRECTLY WITH BOHR MODEL (electron wave + PE)

o Why does delocalisation of electrons stabilise a molecule? EXPLAINED WITH BASIC MODEL (particle in a box)

o Why are carrots orange? EXPLAINED WITH BASIC MODEL (particle in a box)

Starter : Explaining atomic structure Explain the following statements carefully, using the key words in italics provided if need be. • An electron in orbit around a nucleus does not radiate energy and fall into

the nucleus as predicted by classical physics. • Matter is mostly empty space, with the atomic nucleus occupying a tiny

fraction of the volume of the atom. • The energy levels occupied by an electron in atom can have only certain

values described by a quantum number n, where n = 1, 2, 3 etc. • Atoms absorb and emit light only at certain specific wavelengths. wave-particle duality standing wave destructive interferencede Broglie wavelength kinetic energy potential energy stability photon energy

Energy levels and spectra

• Use atomic energy level data to calculate emission and absorption spectra

• Interpret results of the Franck-Hertz experiment in terms of atomic energy levels

Energy levels in hydrogen

En = -13.6/n2

Questions….

Q1. What is the wavelength of the Red and blue Balmer lines?Q2. If the wavelength of an emitted photon was 102 nm, which transition from the Lyman series caused it?

Sketch of Franck-Hertz Apparatus

The Franck-Hertz experiment

The Franck-Hertz experiment

• Why does the current suddenly drop sharply?

• Why does the drop occur at different beam energies for different elements?

A hydrogen atom has energy levels at -13.6 eV (ground state, containing one electron), -3.4 eV, -1.5 eV, -0.9 eV. The energy levels are measured with respect to the ionisation limit, 0.0 eV.

(a) Calculate the wavelength of the lowest-energy photon that could be absorbed when an electron in the ground state of a hydrogen atom is excited to an upperenergy level. [3](Data: h = 6.6 x 10-34 Js, c = 3 x 108 ms-1, e = 1.6 x 10-19 C)(b) Electrons of energy 9 eV are fired through hydrogen gas in its ground state. The electrons are scattered without loss of energy.When the experiment is repeated with electrons of 11 eV energy, the electrons are scattered inelastically, emerging with energies of about 1 eV. Explain these observations. [3]