1

Chapter 27

Early Quantum Theory and Models of the Atom

2

Wave-particle duality• Young’s double slit experiment with

light yielded interference patterns that were evidence of the wave nature of light

• However if the same experiment is repeated with particles (electrons) an interference pattern forms also.

• Major concept of this unit is that waves can exhibit particle-like characteristics (photons) and particles can exhibit wave-like characteristics

3

27.1 Discovery and Properties of the Electron

In the late 19th century, discharge tubes were made that emitted “cathode rays.”

4

J.J. Thomson’s experiment used crossed E, B fields to deflect the “cathode rays”

must be negatively charged particles

Thomson’s Cathode Ray Tube experiment

5

The charge to mass ratio was calculated used the crossed fields and UCM measurements

The result is

Cathode rays were later called electrons

2mvqvB qE qvB q e

r

6

Millikan oil drop experimentRobert Millikan devised an experiment to measure the charge on the electron by measuring the electric field needed to suspend an oil droplet of known mass between parallel plates.

7

The mass and charge of each droplet were measured;

the charge was always an integral multiple of a smallest charge, e.

The currently accepted value of e is:

Knowing e allows the electron mass to be calculated:

8

27.2 Planck’s Quantum Hypothesis; Blackbody Radiation

All objects emit radiation whose total intensity is proportional to the fourth power of their temperature. This is called thermal radiation; a blackbody is one that emits thermal radiation only.

The spectrum of blackbody radiation has been measured; it is found that the frequency of peak intensity increases linearly with temperature.

9

This figure shows blackbody radiation curves for three different temperatures.

Planck observed direct relationship between amount of energy and the frequency of the radiation emitted

10

27.2 Planck’s Quantum Hypothesis; Blackbody Radiation

This spectrum could not be reproduced using 19th-century physics.

A solution was proposed by Max Planck in 1900:

The energy of atomic oscillations within atoms cannot have an arbitrary value; it is related to the frequency:

The constant h is now called Planck’s constant.

11

27.2 Planck’s Quantum Hypothesis; Blackbody Radiation

Planck found the value of his constant by fitting blackbody curves:

Planck’s proposal was that the energy of an oscillation had to be an integral multiple of hf. This is called the quantization of energy.

( ) 1,2,3...E n hf n

12

Quantum:  The smallest amount of something that can exist independently.

Continuous: stream of sugar

Quantum: individual grains

Continuous: music played on a violin or guitar

Quantum: notes on a flute or piano

Continuous: box up the ramp

Quantum: box up the stairs

13

27.3 Photon Theory of Light and the Photoelectric Effect

Einstein suggested that, given the success of Planck’s theory, light must be emitted in small energy packets:

These tiny packets, or particles, are called photons.

photons

14

Classical vs Quantum Physics

• Newtonian classical physics states that particles have a total energy E comprised of KE + PE and momentum p = mv.

• Einstein’s interpretation of the photoelectric effect experiment was that electromagnetic waves are composed of particle-like “photons” that have energy and momentum that we will define later.

• Wave-particle duality is not either wave or particle but each behavior being exhibited in different circumstances.

15

• If light strikes a metal, electrons are emitted.

• The effect does not occur if the frequency of the light is too low;

• the kinetic energy of the electrons increases with frequency.

Photoelectric Effect

16

If light is a wave, theory predicts:

1. Number of electrons and their energy should increase with intensity

2. Frequency would not matter – any color light would cause current flow.

Wave nature?

17

If light is particles, theory predicts:

• Increasing intensity increases number of electrons but not energy

• Above a minimum energy required to break atomic bond, kinetic energy will increase linearly with frequency

• There is a cutoff frequency below which no electrons will be emitted, regardless of intensity

Particle nature

18

photoelectric effect

Photoelectric effect demonstrates the particle nature of light.

Photons that are energetic enough, based on frequency not intensity of the light, will do work on electrons in metal ejecting them towards the positive electrode.

19

The particle theory assumes that an electron absorbs a single photon.

Plotting the kinetic energy vs. frequency:

This shows clear agreement with the photon theory, and not with wave theory.

20

Kinetic energy – stopping voltage

Measure KE by reversing polarity of C electrode, repelling electrons.

When voltage is enough to reduce current to zero the stopping voltage has been reached

Work KE qV eV

1 ev = 1.6 x 10-19 J

21demo

22

Energy and Work function

One photon is absorbed by one electron; If the energy of the photon (hf) exceeds the work function of the metal then electron escapes metal as a photoelectron. Remaining energy is the KE of the electron.

For the most easily removed electrons KE is maximum

max

max

hf KE

KE hf

• work function = hf0 for the electrons with enough KE to escape the metal

• slope = Planck’s constant

23

Photoelectric effect – particles not waves

• Work function of metal = 2.5 eV• Shine dim red light ( = 650 nm) on surface – no electrons• Increase intensity of light (more energy in each “wave”) – no

electrons are ejected.

• Shorten the , increase the f, and a threshold (f0)is reached where electrons are ejected.

• At this frequency even very dim light causes current to flow.• Increasing the intensity of light causes more electrons to

flow but the ALL have the SAME energy – same stopping voltage can bring current to zero.

• Increasing frequency beyond f0 causes electrons with more KE to be ejected (higher stopping voltage)

24

Photoelectric effect examples

• Example

For a certain metal surface illuminated with decreasing wavelengths electrons are first ejected when the light has a wavelength of 550 nm. Find the work function of the metal and determine the stopping voltage when light of 400 nm wavelength is incident on the surface.

• Interactive example – photon torpedo

25

27.4 Energy, Mass, and Momentum of a Photon

Clearly, a photon must travel at the speed of light. Looking at the relativistic equation for momentum, it is clear that this can only happen if its rest mass is zero.

We already know that the energy is hf; we can put this in the relativistic energy-momentum relation and find the momentum:

(27-6)

26

27.5 Compton Effect

Compton did experiments in which he scattered X-rays from different materials. He found that the scattered X-rays had a slightly longer wavelength than the incident ones, and that the wavelength depended on the scattering angle:

(27-7)

27

27.5 Compton Effect

This is another effect that is correctly predicted by the photon model and not by the wave model.

Compton effect

28

27.6 Photon Interactions; Pair Production

Photons passing through matter can undergo the following interactions:

1. Photoelectric effect: photon is completely absorbed, electron is ejected

2. Photon may be totally absorbed by electron, but not have enough energy to eject it; the electron moves into an excited state

3. The photon can scatter from an atom and lose some energy due to frequency change not speed change

4. The photon can create matter, producing an electron-positron pair.

29

In pair production, energy, electric charge, and momentum must all be conserved.

Energy is conserved

• incident photon produces 2 particles each with mass=

9.11 x 10-31 kg

• E = hf E = m0c2

• Electron – positron at rest but have energy due to mass

Einstein’s famous equation

30

• Energy will be conserved through the mass and kinetic energy of the electron and positron;

• opposite charges conserve charge;

• the interaction must take place in the electromagnetic field of a nucleus, which can contribute momentum.

– total pi = total pf

27.6 Photon Interactions; Pair Production

31

2007 FRQ• Electron and positron orbiting around their

stationary center of mass until they annihilate each other, creating 2 photons of equal energy moving in opposite directions. The amount of KE before annihilation is negligible compared to the energy of the photons created.

a) calculate, in eV, the rest energy of a positronE = m0c2 = 9.11 x 10-31 kg (3 x 108)2 = 8.2 x 10-14J E = 8.2 x 10-14J/1.6 x 10-19 J/eV = 512,500 eV =

0.51 MeVb) determine, in eV, the energy each emitted

photon must have.photon has no mass – same energy as positron or

electron = 0.51 MeV

32

2007 FRQ c) calculate the wavelength of each created

photon.E = hf = hc/ =6.63 x 10-34Js x 3 x 108 m/s/8.2 x 10-14J = 2.43 x 10-12 md) calculate the magnitude of the momentum of

each photonp = h/ = 6.63 x 10-34 Js/2.43 x 10-12 mp = 2.73 x 10-22 kgm/se) determine the total momentum of the two-

photon systemtotal p initial = 0 total p final = 0

33

Relativistic vs non-relativistic

• Einstein’s relativistic energy equation applies for objects with speed v approximately > 0.10c

20

2

21

m cE

vc

We will only be doing non-relativistic problems with velocities << c so E=m0c2 m0= rest mass

34

27.8 Wave Nature of Matter

Just as light sometimes behaves as a particle, matter sometimes behaves like a wave. De Broglie turned the wave-particle model around

The wavelength of a particle of matter is:

This is the de Broglie wavelength.

de BroglieExample 27 – 11

35

27.8 Wave Nature of Matter

• Confirmation of deBroglie’s hypothesis came from the Davisson-Germer experiment

• Bombarded nickel with beam of electrons and saw a characteristic diffraction pattern similar to Young’s double slit experiment with light.

• All moving particles have a deBroglie wavelength but the effects are observable only for small mass particles

h h

p mv

36

27.10 Rutherford Scattering Experiment

• Rutherford scattered alpha particles – helium nuclei – from a metal foil and observed the scattering angle.

• Particles were scattered at angles between 0o to almost 180o (straight back).

• Results showed that the positively charged nucleus must be extremely small compared to the rest of the atom

37

The only way to account for the large angles was to assume that all the positive charge was contained within a tiny volume

demo

38

27.10 Early Models of the Atom

Therefore, Rutherford’s model of the atom is mostly empty space:

39

Exciting gases in high voltage discharge tube

40

Emission spectra from excited gases

Line spectra indicate that photons are being emitted at distinct, specific wavelengths (energies)

41

27.12 The Bohr Atom

Bohr proposed that the possible energy states for atomic electrons were quantized – only certain values were possible. Then the spectrum could be explained as transitions from one level to another.

42

27.12 The Bohr Atom

The lowest energy level is called the ground state; the others are excited states.

energy levels

43

Electrons in the ground state have the most negative binding energy

•requires positive work to move it to zero and eject it from atom

44

Atomic transitions release energy from atom in the form of photons Hydrogen atom bombarded with

photons in 10 – 12.5 eV range.

Photons of what energy can be released? In other words, what atomic transitions are possible?

• 10.2 eV absorbed photons excite electrons up to reach n=2

• transition back n = 2 to n =1 emits photons with 10.2 eV

• 12.1 eV absorbed photons cause electrons to reach n=3

• transition n=3 to n=2 emit photons with 1.9 eV

• transition n=3 to n=1 emit photons with 12.1 eV

12.5 eV photons do not have enough energy to excite electrons up to the n = 4 energy level

45

Photons of 10 and 12 eV are incident on an atom in a gas.

a) draw atomic transitions

10 eV photons do not excite ground state electrons to any higher energy level

12 eV photons excite electrons to n=3 energy level

3 transitions possible

Calculate the frequency of the emitted photons.

(for blue transition only)E = 12 eV = 1.92 x 10-18 J

f = E/h = 2.9 x 1015 Hz

Are these photons in the visible range?

=c/f = 104 nm – below lowest visible wavelength = 400 nm

energy (eV)

E = 0

n = 1 ground state

14 eV

n = 2 9 eV

n = 3 2 eV

Photons with energy >14 eV eject electrons, ionizing the gas

ionized atom

46

Chapter 30

Nuclear Physics and Radioactivity

47

30.1 Structure and Properties of the Nucleus

Nucleus is made of protons and neutrons

Proton has positive charge:

Neutron is electrically neutral:

48

30.1 Structure and Properties of the Nucleus

•Neutrons and protons are collectively called nucleons.

•The different nuclei are referred to as nuclides.

•Number of protons: atomic number, Z

•Number of nucleons: atomic mass number, A

•Neutron number: N = A – Z

•Number of electrons = Z in electrically neutral atom

• Isotope = same element (Z) different N values

49

30.1 Structure and Properties of the Nucleus

A and Z are sufficient to specify a nuclide. Nuclides are symbolized as follows:

X is the chemical symbol for the element; it contains the same information as Z but in a more easily recognizable form.

nucleus

50

Mass and Charge Conservation

• Nuclear reactions must “balance” just like chemical reactions

• Total mass number A of reactants, products must be equal

• If number of electrons changes total charge must stay balanced with a change in the number of protons

51

Radium-226 will alpha-decay to radon-22

52

Sample Problem

Ra decays into Rn plus

(A) a proton(B) a neutron(C) an electron(D) a helium nucleus He(E) a deuteron H

226

88

222

86

4

221

Mass and charge are both conserved in “balancing” these decay reactions.

53

Beta decay occurs when a nucleus emits an electron. An example is the decay of carbon-14:

The nucleus still has 14 nucleons, but it has one more proton and one fewer neutron.

Increase in the number of protons balanced by release of electron

54

30.2 Binding Energy and Nuclear Forces

The total mass of a stable nucleus is always less than the sum of the masses of its separate protons and neutrons.

Where has the mass gone?

55

30.2 Binding Energy and Nuclear Forces

It has become energy, such as radiation or kinetic energy, released during the formation of the nucleus.

This difference between the total mass of the constituents and the mass of the nucleus is called the total binding energy of the nucleus.

Binding energy is not something a nucleus has – it is energy it “lacks” relative to the total mass of its separate constituents.

binding energy

56

30.2 Binding Energy of the Nucleus

• The energy needed to separate the nucleus into its individual protons and neutrons.

• The energy liberated when the nucleus is formed from the original protons and neutrons.

• The energy equivalent of the apparent loss of mass of the nucleon components

57

30.2 Binding Energy and Nuclear Forces

• The force that binds the nucleons together is called the strong nuclear force.

• Very strong, but short-range, force.

•The Coulomb force is long-range; this is why extra neutrons are needed for stability in high-Z nuclei.

58

30.2 Binding Energy and Nuclear Forces

More massive nuclei require extra neutrons to overcome the Coulomb repulsion of the protons in order to be stable.

59

31.1 Nuclear Reactions and the Transmutation of Elements

A nuclear reaction takes place when a nucleus is struck by another nucleus or particle.

If the original nucleus is transformed into another, this is called transmutation.

An example:

60

31.1 Nuclear Reactions and the Transmutation of Elements

Neutrons are very effective in nuclear reactions, as they nave no charge and therefore are not repelled by the nucleus.

61

31.2 Nuclear Fission; Nuclear Reactors

After absorbing a neutron, a uranium-235 nucleus will split into two roughly equal parts.

One way to visualize this is to view the nucleus as a kind of liquid drop.

fission

62

Sample Problem

Fission reaction of uranium by a neutron

Predict the A, Z values of the product

1 235 140 10 92 54 0? 2A

Zn U Xe n energy

Total A = 236 in reactants

?A = 236 – 142 = 94

Total Z = 92 in reactants

?Z = 92 – 54 = 38