Quantum and Nuclear Physics

1901: Max Plank found that atoms can only adsorb and emit energy in distinct quantities; this showed that energy displayed particle-like properties. 1905: Albert Einstein suggested that energy itself is quantized and can be viewed as a string of particles called photons. He established that energy has mass.

The quantum nature of radiation (particle nature of waves)

The Photoelectric effect

The Photoelectric effect

Coconut shy

In this analogy, how do we model the;The photon energy? Work function?The max kinetic energy of the photoelectrons?

Bowling ball

Cricket ball

Table tennis ball

How are the electrons released?

Powerful red laser

No electrons released

Photoelectron EnergyPhoton

-Some energy is needed to release the electron (the work function φ)…

…and some energy is given to the electron as kinetic energy.

Photon Energy = work function + kinetic energy of electron

Photoelectron emission

Stopping potential

http://www.yteach.com/page.php/resources/view_all?id=p5_emission_electron_photoelectric_effect_wavelength_kinetic_energy_Einstein_Planck&from=search

Determining Planck’s constant

• Add different filters under the light source

                                                               

                                                            

Photoelectric experiment

• Take measurements of stopping potential and wavelength to determine Planck’s constant and the threshold frequency

Plot a graph of stopping potential versus frequency

Photoelectric Effect: Vstop vs. Frequency

stopeV hf

min0stopV hf

Slope = h/e Planck’s constantfmin

e

Determining “h” from the graph

maxKhf E

Photoelectric Effect: IV Curve Dependence

Intensity I dependence

Frequency f dependence

Vstop= Constant

Vstop f

f1 > f2 > f3

f1

f3

f2

Is light a wave or a particle?

E max=

=Φ

V= Stopping voltage

1. The work function for lithium is 4.6 x 10-19 J.(a) Calculate the lowest frequency of light that will cause photoelectric emission. (6.9 x

1014 Hz )(b) What is the maximum energy of the electrons emitted when light of 7.3 x 1014 Hz is

used? (0.24 x 10-19 J )

2. A frequency of 2.4 x 1015 Hz is used on magnesium with work function of 3.7 eV.(a) What is energy transferred by each photon?(b) Calculate the maximum KE of the ejected electrons.(c) The maximum speed of the electrons.(d) The stopping potential for the electrons.(a) 1.6 x 10-18 J(b) 1.0x 10-18 J(c) v = 1.5 x 106 m s-1 (d) Vs = 6.3 V

Questions

Hamper HL page 231 Q’s 1-4.IB revision pack Q’s 1,7,9,13,15,17,20.Tsokos page 396 q’s 1-7.

The wave nature of matter

Can particles be waves?

Review of Bohr and deBroglie• Background:

– Balmer found equation for Hydrogen spectrum but didn’t know what it meant.

– Rutherford found that atoms had a nucleus, but didn’t know why electrons didn’t spiral in.

• Bohr postulates that the orbit can fit an integral number of wavelengths associated with the electron, therefore has quantized energy levels, and predicts Balmer’s equation.

• deBroglie postulates that electrons are waves, and predicts Bohr’s quantized energy levels.

• Note: no experimental difference between Bohr model and deBroglie model, but deBroglie is a lot more satisfying.

Davisson and Germer -- VERY clean nickel crystal. Interference is electron scattering off Ni atoms.

ee

ee

e

e

e ee

e e

scatter off atoms

e det.

move detector around,see what angle electrons coming offNi

ee

ee ee

e

e

e det.

Ni

Observe pattern of scattering electrons off atomsLooks like …. Wave!

# e’s

scatt. angle 5000

See peak!!

so probability of angle where detectelectron determined by interferenceof deBroglie waves!

An Accident at the Phone Company Makes Everything Crystal Clear There was an accident at the Bell Telephone Laboratories in April 1925. Clinton Davisson and L. H. Germer, looking for ways to improve vacuum tubes, were watching how electrons from an electron gun in a vacuum tube scattered off a flat nickel surface. Suddenly, while the experiment was running and the nickel target was very hot, a bottle of liquid air near the apparatus exploded, smashing one of the vacuum pipes, and air rushed into the apparatus. The hot nickel target oxidized immediately. The layer of oxide made their target useless for further investigations. They decided to clean off the oxide by heating the nickel in a hydrogen atmosphere then in vacuum. After doing this for a prolonged period, the nickel looked good, and they resumed the investigation. To their amazement, the pattern of electron scattering from the newly cleaned nickel target was completely different from that before the accident. What had changed? On examining their newly cleaned crystal carefully, they found a clue. The original target was polycrystalline -- made up of a multitude of tiny crystals, oriented randomly. During the prolonged heating of the cleaning process, the nickel had re-crystallized into a few large crystals.

Electron diffraction

Diffraction rings

Calculating the De Broglie λ

λ = h/p (= h/(2Ekm)1/2 )

h = Planck’s constant p = Momentum In 1923, French Prince Louis de

Broglie, generalised Einstein's work from the specific case of light to cover all other types of particles. This work was presented in his doctoral thesis when he was 31. His thesis was greeted with consternation by his examining committee. Luckily, Einstein had received a copy in advance and vouched for de Broglie. He passed!

de Broglie questions

• Calculate the wavelengths of the “deBroglie” waves associated with

• a)a 1kg mass moving at 50ms-1

• b)an electron which has been accelerated by a p.d. of 500V.

a)Discuss briefly deBroglie’s hypothesis and mention one experiment which gives evidence to support it.

b)Calculate the wavelength of the “deBroglie wave” associated with an electron in the lowest energy Bohr orbit. (The radius of the lowest energy orbit according to the Bohr theory is 5·3×10-11m.)

Questions

Hamper HL page 235 Q’s 9,10.IB revision pack Q’sTsokos page 396 q’s 8-10

History of Quantum MechanicsMax Planck's work on the 'Black Body' problem started the quantum revolution in 1900. He showed that energy cannot take any value but is arranged in discrete lumps – later called photons by Einstein.

In 1913, Niels Bohr proposed a model of the atom with quantised electron orbits. Although a great step forward, quantum physics was still in its infancy and was not yet a consistent theory. It was more like a collection of classical theories with quantum ideas applied.

Starting in 1925 a true 'quantum mechanics' – a set of mathematically and conceptual 'tools' – was born. At first, three different incantations of the same theory were proposed independently and were then shown to be consistent. Quantum mechanics reached its final form (essentially unchanged from today) in 1928.

Participants of the 5th Solvay Congress, Brussels, October 1927

A. Einstein

M Curie

M. Planck

N. Bohr

L.V. de Broglie

W. HeisenbergW. PauliE. Schrödinger

• Thomson – Plum Pudding– Why? Known that negative charges can be removed from atom.– Problem: just a random guess

• Rutherford – Solar System– Why? Scattering showed hard core.– Problem: electrons should spiral into nucleus in ~10-11 sec.

• Bohr – fixed energy levels– Why? Explains spectral lines.– Problem: No reason for fixed energy levels

• deBroglie – electron standing waves– Why? Explains fixed energy levels– Problem: still only works for Hydrogen.

• Schrodinger – will save the day!!

Models of the Atom–

–

––

–

+

+

+ –

Visualising the models

Hyperlink

Different view of atomsThe Bohr Atom

The Schrodinger Atom

Electrons are only allowed to have discrete energy values and these correspond to changes in orbit. The Bohr postulate

Electrons behave like stationary waves. Only certain types of wave fit the atom, and these correspond to fixed energy states. The square of the amplitude gives the probability of finding the electron at that point

+

0eV

Amplitude

• Thomson – Plum Pudding– Why? Known that negative charges can be removed from atom.– Problem: just a random guess

• Rutherford – Solar System– Why? Scattering showed hard core.– Problem: electrons should spiral into nucleus in ~10-11 sec.

• Bohr – fixed energy levels– Why? Explains spectral lines.– Problem: No reason for fixed energy levels

• deBroglie – electron standing waves– Why? Explains fixed energy levels– Problem: still only works for Hydrogen.

• Schrodinger – will save the day!!

Models of the Atom–

–

––

–

+

+

+ –

Schrödinger set out to develop an alternate formulation of quantum mechanics based on matter waves, à la de Broglie. At 36, he was somewhat older than his contemporaries but still succeeded in deriving the now famous 'Schrödinger Wave Equation.' The solution of the equation is known as a wave function and describes the behavior of a quantum mechanical object, like an electron.At first, it was unclear what the wave function actually represented. How was the wave function related to the electron? At first, Schrödinger said that the wave function represented a 'shadow wave' which somehow described the position of the electron. Then he changed his mind and said that it described the electric charge density of the electron. He struggled to interpret his new work until Max Born came to his rescue and suggested that the wave function represented a probability – more precisely, the square of the absolute magnitude of the wavefunction is proportional to the probability that the electron appears in a particular position. So, Schrödinger's theory gave no exact answers… just the chance for something to happen. Even identical measurements on the same system would not necessarily yield the same results! Born's key role in deciphering the meaning of the theory won him the Nobel Prize in Physics in 1954.

Schrödinger model

Quantum Mechanical tunneling

In the classical world the positively charged alpha particle needs enough energy to overcome the positive potential barrier which originates from protons in the nucleus. In the quantum world an alpha particle with less energy can tunnel through the potential barrier and escape the nucleus.

Electron in a box model

Electrons will form standing waves of wavelength 2L/n

Standing waves and energy levels

Kinetic Energy of an electron in a box

• When the momentum expression for the particle in a box :

•

• is used to calculate the energy associated with the particle

SpectraConsider a ball in a hole:

When the ball is here it has its lowest gravitational potential energy.

We can give it potential energy by lifting it up:

If it falls down again it will lose this gpe:

20J

5J

5J

30J

SpectraA similar thing happens to electrons. We can “excite” them and raise their energy level:

0eV

-0.85eV

-1.5eV

-3.4eV

-13.6eV

An electron at this energy level would be “free” – it’s been “ionised”.

These energy levels are negative because an electron here would have less energy than if its ionised.

This is called “The ground state”

SpectraIf we illuminate the atom we can excite the electron:

0eV

-0.85eV

-1.5eV

-3.4eV

-13.6eV

Q. What wavelength of light would be needed to excite this electron to ionise it?

Light

Energy change = 3.4eV = 5.44x10-

19J.Using E=hc/λ wavelength = 3.66x10-7m(In other words, ultra violet light)

Example questions1) State the ionisation energy of this atom in

eV.

2) Calculate this ionisation energy in joules.

3) Calculate the wavelength of light needed to ionise the atom.

4) An electron falls from the -1.5eV to the -3.4eV level. What wavelength of light does it emit and what is the colour?

5) Light of frequency 1x1014Hz is incident upon the atom. Will it be able to ionise the atom?

0eV

-0.85eV

-1.5eV

-3.4eV

-13.6eV

2.17x10-18J

94ηm

654ηm

SpectraContinuous spectrum

Absorption spectrum

Emission spectrum

Emission SpectraHydrogen

Helium

Sodium

Observing the Spectra

Light source

Gas

Collimator (to produce parallel light)

Diffraction grating(to separate the colours)

Microscope

(to observe the spectrum)

Emission spectrum

Absorption spectrum

Questions

Hamper HL page 232 Q’s 5-8.IB revision pack Q’sTsokos page 405 q’s 1-7.

Outline the Heisenberg uncertainty principle with regard to position–momentum and time–energy.

Students should be aware that the conjugate quantities, position–momentum and time–energy, cannot be known precisely at the same time. They should know of the link between the uncertainty principle and the de Broglie hypothesis. For example, students should know that, if a particle has a uniquely defined de Broglie wavelength, then its momentum is known precisely but all knowledge of its position is lost.

Heisenberg uncertainty principleHeisenberg made one fundamental and long-lasting contribution to the quantum world – the uncertainty principle. He showed that quantum mechanics implied that there was a fundamental limitation on the accuracy to which pairs of variables, such as (position and momentum) and (energy and time) could be determined.

If a 'large' object with a mass of, say, 1g has its position measured to an accuracy of 1 , then the uncertainty on the object's velocity is a minute 10-25 m/s. The uncertainty principle simply does not concern us in everyday life. In the quantum world the story is completely different. If we try to localize an electron within an atom of diameter 10-10 m the resulting uncertainty on its velocity is 106 m/s!

Heisenberg uncertainty principle

Questions

Hamper HL page 232 Q’s 5-8.IB revision pack Q’s 3,11,16,22.

Nuclear physics

Determining the size of the nucleus

25

20

15

10

5

0

0 2 4 6 8 10

distance from nucleus / 10–14 m

Approach of alpha particle to nucleus

Z = 79 (gold)

1. Make an arithmetical check to show that at distance r = 1.0x10–14 m, the electrical potential energy, is between 20 MeV and 25 MeV, as shown by the graph.

2.How does the electrical potential energy change if the distance r is doubled? 3.From the graph, at what distance r, will an alpha particle with initial kinetic

energy 5 MeV colliding head-on with the nucleus, come to rest momentarily?

1. Substituting values gives

.MeV7.2210m 100.1m J C 1085.84

C 106.1792 = 6

1411212

19

P

E

2. Halves, because the potential energy is proportional to 1/r.3. About 4.6x10–14 m, where the graph reaches 5 MeV.

Charge in a magnetic field

Circular pathsRecall:

++ -2 protons, 2 neutrons,

therefore charge = +2

1 electron, therefore charge = -1

Because of this charge, they will be deflected by magnetic fields:

+

These paths are circular, so Bqv = mv2/r, orr =mv

Bq

Bainbridge mass spectrometerIons are formed at D and pass through the cathode C and then through a slit S1

A particle with a charge q and velocity v will only pass through the next slit S2 if the resultant force on it is zero – that is it is traveling in a straight line. That is if:

Therefore

In the region of the Mag field

Bqv = Mv2/r

Therefore

r = Mv/(Bq)

Hyperlink

Velocity selector

Mass selector

Nuclear energy levelsThere are 2 distinct length of tracksin this Alpha decay

Therefore, the energy levels in the nucleus are discrete

The existence of Neutrinos

How can a 2 body system create a spectrum of energies?

There must be a 3rd particle

The Neutrino was postulated

A 2 body system only has one solution

A 3 body system has many solutions

Changes in Mass and Proton Number

11

5

0

+1C

11

6B β+ ν+

90

39Sr

90

38Y β + ν

0

-1+

Beta - decay:

Beta + decay:

“positron”

Questions

Hamper HL page 239 Q’s 11-17.IB revision pack Q’s 14,21.

Radioactive Decay Law

dN/N = -λdt which when integrated, gives

Taking antilogs of both sides gives:

Half life and the radioactive decay constant

When N = No/2 the number of radioactive nuclei will have halved

Therefore when t = T1/2

N = No/2 = Noe-λT1/2 and so 1/2 = e-λT1/2 . Taking the inverse gives 2 = eλT1/2 and so:

Measuring long half lives

• If the half life is very long, then the activity (A) is constant

• Analysis of a decay curve cannot give the half life.• If the mass of the substance is measured, then• A = -λN, so a measurement of the activity enables

Measuring long half lives to be calculated (N from mass).

• T1/2 can be calculated from λ.

Measuring short half lives

• Each decay can cause an ionisation• This can generate an electric current• If the current is displayed on an oscilloscope,

then• The limit is the response time of the

oscilloscope (typically µs).

Questions

Hamper HL page 255. Q’s 27-31.IB revision pack Q’s 2,4,10,23.