Chapter 27

Quantum Physics

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Conceptual Physics Text

Need for Quantum Physics

• Problems remained from classical mechanics that relativity didn’t explain.

• Blackbody Radiation– The electromagnetic radiation emitted by a heated

object• Photoelectric Effect– Emission of electrons by an illuminated metal

• Spectral Lines– Emission of sharp spectral lines by gas atoms in an

electric discharge tube

Introduction

Development of Quantum Physics

• 1900 to 1930– Development of ideas of quantum mechanics

• Also called wave mechanics• Highly successful in explaining the behavior of atoms,

molecules, and nuclei

• Involved a large number of physicists– Planck introduced basic ideas.– Mathematical developments and interpretations

involved such people as Einstein, Bohr, Schrödinger, de Broglie, Heisenberg, Born and Dirac.

Introduction

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Black Body Radiation(Start to minute 4:00 and 6:50 to end)

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Pgs. 870-872 from College Physics text

Blackbody Radiation

• An object at any temperature emits electromagnetic radiation.– Also called thermal radiation.– Stefan’s Law describes the total power radiated.– The spectrum of the radiation depends on the

temperature and properties of the object.• The spectrum shows a continuous distribution of

wavelengths from infrared to ultaviolet.

Section 27.1

Blackbody Radiation – Classical View• Thermal radiation originates

from accelerated charged particles.

• Problem in explaining the observed energy distribution

• Opening in a cavity is a good approximation

• The nature of the radiation emitted through the opening depends only on the temperature of the cavity walls.

Section 27.1

Blackbody Radiation Graph• Experimental data for

distribution of energy in blackbody radiation

• As the temperature increases, the total amount of energy increases.– Shown by the area under the

curve• As the temperature

increases, the peak of the distribution shifts to shorter wavelengths.

Section 27.1

Wien’s Displacement Law

• The wavelength of the peak of the blackbody distribution was found to follow Wein’s Displacement Law.– λmax T = 0.2898 x 10-2 m • K• λmax is the wavelength at which the curve peaks.• T is the absolute temperature of the object emitting the

radiation.

Section 27.1

The Ultraviolet Catastrophe• Classical theory did not match

the experimental data.• At long wavelengths, the

match is good.• At short wavelengths, classical

theory predicted infinite energy.

• At short wavelengths, experiment showed no energy

Section 27.1

Planck’s Resolution

• Planck hypothesized that the blackbody radiation was produced by resonators.– Resonators were submicroscopic charged oscillators.

• The resonators could only have discrete energies.– En = n h ƒ

• n is called the quantum number• ƒ is the frequency of vibration• h is Planck’s constant, 6.626 x 10-34 J s

• Key point is quantized energy states

Section 27.1

Max Planck

• 1858 – 1947• Introduced a “quantum

of action,” h• Awarded Nobel Prize in

1918 for discovering the quantized nature of energy

Section 27.1

Quantized Energy

• Planck’s assumption of quantized energy states was a radical departure from classical mechanics.

• The fact that energy can assume only certain, discrete values is the single most important difference between quantum and classical theories.– Classically, the energy can be in any one of a

continuum of values.

Section 27.1

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Photoelectric Effect

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Photoelectric Effect

• When light is incident on certain metallic surfaces, electrons are emitted from the surface.– This is called the photoelectric effect.– The emitted electrons are called photoelectrons.

• The effect was first discovered by Hertz.• The successful explanation of the effect was given

by Einstein in 1905.– Received Nobel Prize in 1921 for paper on

electromagnetic radiation, of which the photoelectric effect was a part

Photoelectric Effect Schematic• When light strikes E,

photoelectrons are emitted.• Electrons collected at C and

passing through the ammeter create a current in the circuit.

• C is maintained at a positive potential by the power supply.

Section 27.2

Photoelectric Current/Voltage Graph

• The current increases with intensity, but reaches a saturation level for large ΔV’s.

• No current flows for voltages less than or equal to –ΔVs, the stopping potential.

Section 27.2

More About Photoelectric Effect

• The stopping potential is independent of the radiation intensity.

• The maximum kinetic energy of the photoelectrons is related to the stopping potential: KEmax = eDVs

Section 27.2

EXAMPLE 27.1 Photoelectrons from Sodium

Goal Understand the quantization of light and its role in the photoelectric effect. Problem A sodium surface is illuminated with light of wavelength 0.300 µm. The work function for sodium is 2.46 eV. Calculate (a) the energy of each photon in electron volts, (b) the maximum kinetic energy of the ejected photoelectrons, and (c) the cutoff wavelength for sodium. Strategy Parts (a), (b), and (c) require substitution of values into the energy of a photon equation, Photoelectric effect equation, and the cutoff wavelength equation, respectively.

(a) Calculate the energy of each photon. Obtain the frequency from the wavelength:

c = fλ → f = c

= 3.00 108 m/s

λ 3.00 10-7 m

f = 1.00 1015 Hz

Calculate the photon's energy: E = hf = (6.63 10-34 J·s)(1.00 1015 Hz) = 6.63 10-19 J

= (6.63 10-19 J)( 1.00 eV ) = 4.14 eV 1.60 10-19 J

(b) Find the maximum kinetic energy of the photoelectrons.

Substitute into the photoelectric effect equation:

KEmax = hf - = 4.14 eV - 2.46 eV = 1.68 eV

(c) Compute the cutoff wavelength.

Convert from electron volts to joules:

= 2.46 eV = (2.46 eV)(1.60 10-19 J/eV) = 3.94 10-19 J

Find the cutoff wavelength:

λc = hc

= (6.63 10-34 J·s)(3.00 108 m/s)

3.94 10-19 J

= 5.05 10-7 m = 505 nm

LEARN MORE

Remarks The cutoff wavelength is in the yellow-green region of the visible spectrum. Question Suppose in a given photoelectric experiment the frequency of light is larger than the cutoff frequency. If ΔVs is the stopping potential and e the electron charge, the magnitude of eΔVs is then equal to: (Select all that apply.)

the energy of the incident photons. the difference between the

photon energy and the work function. the work function. the maximum kinetic energy of emitted electrons.

Features Not Explained by Classical Physics/Wave Theory

• No electrons are emitted if the incident light frequency is below some cutoff frequency that is characteristic of the material being illuminated.

• The maximum kinetic energy of the photoelectrons is independent of the light intensity.

Section 27.2

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Light Quantum or Photons

More Features Not Explained

• The maximum kinetic energy of the photoelectrons increases with increasing light frequency.

• Electrons are emitted from the surface almost instantaneously, even at low intensities.

Section 27.2

Einstein’s Explanation• A tiny packet of light energy, called a photon, would be emitted

when a quantized oscillator jumped from one energy level to the next lower one.– Extended Planck’s idea of quantization to electromagnetic radiation

• The photon’s energy would be E = hƒ• Each photon can give all its energy to an electron in the metal.• The maximum kinetic energy of the liberated photoelectron is

KEmax = hƒ – φ• φ is called the work function of the metal

Section 27.2

Explanation of Classical “Problems”

• The effect is not observed below a certain cutoff frequency since the photon energy must be greater than or equal to the work function.– Without this, electrons are not emitted, regardless of the

intensity of the light

• The maximum KE depends only on the frequency and the work function, not on the intensity.– The absorption of a single photon is responsible for the

electron’s kinetic energy.

Section 27.2

More Explanations

• The maximum KE increases with increasing frequency.

• The effect is instantaneous since there is a one-to-one interaction between the photon and the electron.

Section 27.2

Verification of Einstein’s Theory

• Experimental observations of a linear relationship between KE and frequency confirm Einstein’s theory.

• The x-intercept is the cutoff frequency.

Section 27.2

Cutoff Wavelength

• The cutoff wavelength is related to the work function.

• Wavelengths greater than lC incident on a material with a work function f don’t result in the emission of photoelectrons.

Section 27.2

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Pgs. 875-876 from College Physics text

X-Rays

• Discovered and named by Rӧntgen in 1895• Later identified as electromagnetic radiation

with short wavelengths– Wavelengths lower (frequencies higher) than for

ultraviolet– Wavelengths are typically about 0.1 nm.– X-rays have the ability to penetrate most materials

with relative ease.

Section 27.3

Production of X-rays, 1• X-rays are produced when

high-speed electrons are suddenly slowed down.– Can be caused by the electron

striking a metal target• Heat generated by current in

the filament causes electrons to be emitted.

• These freed electrons are accelerated toward a dense metal target.

• The target is held at a higher potential than the filament.

Section 27.3

X-ray Spectrum• The x-ray spectrum has two

distinct components.• Continuous broad spectrum

– Depends on voltage applied to the tube

– Sometimes called bremsstrahlung

• The sharp, intense lines depend on the nature of the target material.

Section 27.3

Production of X-rays, 2• An electron passes near a

target nucleus.• The electron is deflected

from its path by its attraction to the nucleus.– This produces an acceleration

• It will emit electromagnetic radiation when it is accelerated.

Section 27.3

Wavelengths Produced

• If the electron loses all of its energy in the collision, the initial energy of the electron is completely transformed into a photon.

• The wavelength can be found from

Section 27.3

Wavelengths Produced, Cont.

• Not all radiation produced is at this minimum wavelength.– Many electrons undergo more than one collision

before being stopped.– This results in the continuous spectrum produced.

Section 27.3

Arthur Holly Compton

• 1892 – 1962• Discovered the

Compton effect• Worked with cosmic

rays• Director of the lab at U

of Chicago• Shared Nobel Prize in

1927

Section 27.5

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Compton Scattering

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The Compton Effect

• Compton directed a beam of x-rays toward a block of graphite.

• He found that the scattered x-rays had a slightly longer wavelength that the incident x-rays.– This means they also had less energy.

• The amount of energy reduction depended on the angle at which the x-rays were scattered.

• The change in wavelength is called the Compton shift.

Section 27.5

Compton Scattering

• Compton assumed the photons acted like other particles in collisions.

• Energy and momentum were conserved.

• The shift in wavelength is

Section 27.5

Compton Scattering, Final

• The quantity h/mec is called the Compton wavelength.– Compton wavelength = 0.002 43 nm– Very small compared to visible light

• The Compton shift depends on the scattering angle and not on the wavelength.

• Experiments confirm the results of Compton scattering and strongly support the photon concept.

Section 27.5

Photons and Electromagnetic Waves

• Light has a dual nature. It exhibits both wave and particle characteristics.– Applies to all electromagnetic radiation– Different frequencies allow one or the other characteristic

to be more easily observed.• The photoelectric effect and Compton scattering offer

evidence for the particle nature of light.– When light and matter interact, light behaves as if it were

composed of particles.• Interference and diffraction offer evidence of the wave

nature of light.

Section 27.6

Louis de Broglie

• 1892 – 1987• Discovered the wave

nature of electrons• Awarded Nobel Prize in

1929

Section 27.6

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Pgs. 880- 881 from College Physics text

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De Broglie Wavelength

Wave Properties of Particles

• In 1924, Louis de Broglie postulated that because photons have wave and particle characteristics, perhaps all forms of matter have both properties.

• Furthermore, the frequency and wavelength of matter waves can be determined.

Section 27.6

de Broglie Wavelength and Frequency

• The de Broglie wavelength of a particle is

• The frequency of matter waves is

Section 27.6

Dual Nature of Matter

• The de Broglie equations show the dual nature of matter.

• Each contains matter concepts.– Energy and momentum

• Each contains wave concepts.– Wavelength and frequency

Section 27.6

EXAMPLE 27.4 The Electron Versus the Baseball

Goal Apply the de Broglie hypothesis to a quantum and a classical object. Problem (a) Compare the de Broglie wavelength for an electron (me = 9.11 10-31 kg) moving at a speed equal to 1.0 107 m/s with that of a baseball of mass 0.145 kg pitched at 45.0 m/s. (b) Compare these wavelengths with that of an electron traveling at 0.999c. Strategy This problem is a matter of substitution into the equation for the de Broglie wavelength. In part (b) the relativistic momentum must be used.

SOLUTION

(a) Compare the de Broglie wavelengths of the electron and the baseball. Substitute data for the electron into the De Broglie wavelength equation:

λe = h

= 6.63 10-34 J·s

= 7.28 10-11 m mev (9.11 10-31 kg)(1.00 107 m/s)

Repeat the calculation with the baseball data:

λb = h

= 6.63 10-34 J·s

= 1.02 10-34 m mbv (0.145 kg)(45.0 m/s)

(b) Find the wavelength for an electron traveling at 0.999c.

Replace the momentum in de Broglie's equation with the relativistic momentum:

λe = h

= h√1- v2/c2

mev/√1- v2/c2 mev Substitute:

λe = (6.63 10-34 J·s)√1 - (0.999c)2/c2

= 1.09 10-13 m (9.11 10-31 kg)(0.999·3.00 108 m/s)

LEARN MORE

Remarks The electron wavelength corresponds to that of x-rays in the electromagnetic spectrum. The baseball, by contrast, has a wavelength much smaller than any aperture through which the baseball could possibly pass, so we couldn't observe any of its diffraction effects. It is generally true that the wave properties of large-scale objects can't be observed. Notice that even at extreme relativistic speeds, the electron wavelength is still far larger than the baseball's. Question How does doubling the speed of a particle affect its wavelength? (Select all that apply.)

At ordinary speeds where v/c << 1, it halves the wavelength of the

particle. At ordinary speeds where v/c << 1, it doubles the wavelength of

the particle. It halves the wavelength at all speeds. It less than doubles

the wavelength at relativistic speeds where v/c gets close to 1. It doubles

the wavelength of the particle at all speeds. It reduces the wavelength to less than half at relativistic speeds where v/c gets close to 1.

The Davisson-Germer Experiment

• They scattered low-energy electrons from a nickel target.

• They followed this with extensive diffraction measurements from various materials.

• The wavelength of the electrons calculated from the diffraction data agreed with the expected de Broglie wavelength.

• This confirmed the wave nature of electrons.

Section 27.6

The Electron Microscope• The electron microscope

depends on the wave characteristics of electrons.

• Microscopes can only resolve details that are slightly smaller than the wavelength of the radiation used to illuminate the object.

• The electrons can be accelerated to high energies and have small wavelengths.

Section 27.6

Werner Heisenberg• 1901 – 1976• Developed an abstract

mathematical model to explain wavelengths of spectral lines– Called matrix mechanics

• Other contributions– Uncertainty Principle

• Nobel Prize in 1932– Atomic and nuclear models– Forms of molecular hydrogen

Section 27.8

Chapter 28

Atomic Physics

Quantum Numbers and Atomic Structure

• The characteristic wavelengths emitted by a hot gas can be understood using quantum numbers.

• No two electrons can have the same set of quantum numbers – helps us understand the arrangement of the periodic table.

• Atomic structure can be used to describe the production of x-rays and the operation of a laser.

Introduction

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Atomic Emissions Spectra

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Emission Spectra

• A gas at low pressure has a voltage applied to it.• The gas emits light which is characteristic of the gas.• When the emitted light is analyzed with a

spectrometer, a series of discrete bright lines is observed.– Each line has a different wavelength and color.– This series of lines is called an emission spectrum.

Section 28.2

Examples of Emission Spectra

Section 28.2

Emission Spectrum of Hydrogen – Equation

• The wavelengths of hydrogen’s spectral lines can be found from

– RH is the Rydberg constant• RH = 1.097 373 2 x 107 m-1

– n is an integer, n = 1, 2, 3, …– The spectral lines correspond to different values of n.

Section 28.2

Spectral Lines of Hydrogen

• The Balmer Series has lines whose wavelengths are given by the preceding equation.

• Examples of spectral lines– n = 3, λ = 656.3 nm– n = 4, λ = 486.1 nm

Section 28.2

Absorption Spectra

• An element can also absorb light at specific wavelengths.

• An absorption spectrum can be obtained by passing a continuous radiation spectrum through a vapor of the element being analyzed.

• The absorption spectrum consists of a series of dark lines superimposed on the otherwise continuous spectrum.– The dark lines of the absorption spectrum coincide with

the bright lines of the emission spectrum.

Section 28.2

Absorption Spectrum of Hydrogen

Section 28.2

Application of Absorption Spectrum

• The continuous spectrum emitted by the Sun passes through the cooler gases of the Sun’s atmosphere.– The various absorption lines can be used to

identify elements in the solar atmosphere.– Led to the discovery of helium

Section 28.2

Specific Energy Levels

• The lowest energy state is called the ground state.– This corresponds to n = 1– Energy is –13.6 eV

• The next energy level has an energy of –3.40 eV.– The energies can be compiled in an energy level

diagram.

Section 28.3

Specific Energy Levels, Cont.

• The ionization energy is the energy needed to completely remove the electron from the atom.– The ionization energy for hydrogen is 13.6 eV

• The uppermost level corresponds to E = 0 and n

Section 28.3

Energy Level Diagram

Section 28.3

EXAMPLE 28.1 The Balmer Series for Hydrogen

Transitions responsible for the Balmer series for the hydrogen atom. All transitions terminate at the n = 2 level. Goal Calculate the wavelength, frequency, and energy of a photon emitted during an electron transition in an atom. Problem The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n = 2, as shown

in the figure (a) Find the longest-wavelength photon emitted in the Balmer series and determine its frequency and energy. (b) Find the shortest-wavelength photon emitted in the same series. Strategy This problem is a matter of substituting values. The frequency can then be obtained from c = fλ and the energy from E = hf. The longest-wavelength photon corresponds to the one that is emitted when the electron jumps from the ni = 3 state to the nf = 2 state. The shortest-wavelength photon corresponds to the one that is emitted when the electron jumps from ni = to the nf = 2 state.

SOLUTION

(a) Find the longest-wavelength photon emitted in the Balmer series and determine its frequency and energy. Substitute, with ni = 3 and nf = 2:

1 = RH (

1 -

1 ) = RH (

1 -

1 ) =

5RH

λ nf2 ni2 22 32 36

Take the reciprocal and substitute, finding the wavelength:

λ = 36

= 36

= 6.563 10-7 m = 656.3 nm 5RH 5(1.097 107 m-1)

Now use c = fλ to obtain the frequency:

f = c

= 2.998 108 m/s

= 4.568 1014 Hz λ 6.563 10-7 m

Calculate the photon's energy by substituting: E = hf = (6.626 10-34 J·s)(4.568 1014 Hz)

= 3.027 10-19 J = 1.892 eV

(b) Find the shortest-wavelength photon emitted in the Balmer series.

Substitute with 1/ni → 0 as ni → and nf = 2:

1 = RH (

1 -

1 ) = RH (

1 - 0 ) =

RH

λ nf2 ni2 22 4 Take the reciprocal and substitute, finding the wavelength:

λ = 4

= 4

= 3.646 10-7 m = 364.6 nm RH (1.097 107 m-1)

LEARN MORE

Remarks The first wavelength is in the red region of the visible spectrum. We could also obtain the energy of the photon in the form hf = E3 - E2, where E2 and E3 are the energy levels of the hydrogen atom. Note that this photon is the lowest-energy photon in the Balmer series because it involves the smallest energy change. The second photon, the most energetic, is in the ultraviolet region. Question What is the upper-limit energy of a photon that can be emitted from hydrogen due to the transition of an electron between energy levels? (Select all that apply.)

1.70 eV (3/4) RH · h · c 13.6 eV RH · h · c

Atomic Transitions – Energy Levels• An atom may have many

possible energy levels.• At ordinary temperatures,

most of the atoms in a sample are in the ground state.

• Only photons with energies corresponding to differences between energy levels can be absorbed.

Section 28.7

Atomic Transitions – Stimulated Absorption

• The blue dots represent electrons.

• When a photon with energy ΔE is absorbed, one electron jumps to a higher energy level.– These higher levels are called

excited states.– ΔE = hƒ = E2 – E1

– In general, ΔE can be the difference between any two energy levels.

Section 28.7

Atomic Transitions – Spontaneous Emission

• Once an atom is in an excited state, there is a constant probability that it will jump back to a lower state by emitting a photon.

• This process is called spontaneous emission.

• Typically, an atom will remain in an excited state for about 10-8 s

Section 28.7

Atomic Transitions – Stimulated Emission

• An atom is in an excited state and a photon is incident on it.

• The incoming photon increases the probability that the excited atom will return to the ground state.

• There are two emitted photons, the incident one and the emitted one.– The emitted photon is exactly

in phase with the incident photon.

Section 28.7

Chapter 29Nuclear Physics

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Mass Energy Equivalence

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College Physics Text

Nuclear Physics

• Topics in nuclear physics include– Properties and structure of atomic nuclei– Radioactivity– Nuclear reactions• Decay processes• Fission• Fusion

Introduction

Ernest Rutherford

• 1871 – 1937• Discovery that atoms

could be broken apart• Studied radioactivity• Nobel prize in 1908

Section 29.1

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Pgs. 913-914 from College Physics text

Some Properties of Nuclei• All nuclei are composed of protons and neutrons.

– Exception is ordinary hydrogen with just a proton• The atomic number, Z, equals the number of protons in

the nucleus.• The neutron number, N, is the number of neutrons in the

nucleus.• The mass number, A, is the number of nucleons in the

nucleus.– A = Z + N– Nucleon is a generic term used to refer to either a proton or a

neutron.– The mass number is not the same as the mass.

Section 29.1

Symbolism

• Symbol:

– X is the chemical symbol of the element.• Example:

» Mass number is 27» Atomic number is 13» Contains 13 protons» Contains 14 (27 – 13) neutrons

– The Z may be omitted since the element can be used to determine Z.

Section 29.1

More Properties

• The nuclei of all atoms of a particular element must contain the same number of protons.

• They may contain varying numbers of neutrons.– Isotopes of an element have the same Z but

differing N and A values.– Example, isotopes of carbon:

Charge

• The proton has a single positive charge, +e• The electron has a single negative charge, -e– e = 1.602 177 33 x 10-19 C

• The neutron has no charge.– Makes it difficult to detect

Section 29.1

Mass

• It is convenient to use unified mass units, u, to express masses.– Based on definition that the mass of one atom of C-12 is

exactly 12 u– 1 u = 1.660 559 x 10-27 kg

• Mass can also be expressed in MeV/c2

– From ER = m c2

– 1 u = 931.494 MeV/c2

Section 29.1

Summary of Masses

Section 29.1

The Size of the Nucleus

• First investigated by Rutherford in scattering experiments

• He found an expression for how close an alpha particle moving toward the nucleus can come before being turned around by the Coulomb force.

• The KE of the particle must be completely converted to PE.

Section 29.1

Marie Curie

• 1867 – 1934• Discovered new

radioactive elements• Shared Nobel Prize in

physics in 1903– For study of radioactive

substances• Nobel Prize in

Chemistry in 1911– For the discovery of

radium and polonium

Section 29.3

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Pgs. 918 from College Physics text

Radioactivity

• Radioactivity is the spontaneous emission of radiation.

• Experiments suggested that radioactivity was the result of the decay, or disintegration, of unstable nuclei.

Section 29.3

Radioactivity – Types

• Three types of radiation can be emitted– Alpha particles

• The particles are 4He nuclei.

– Beta particles• The particles are either electrons or positrons.

– A positron is the antiparticle of the electron.– It is similar to the electron except its charge is +e

– Gamma rays• The “rays” are high energy photons.

Section 29.3

Distinguishing Types of Radiation

• A radioactive beam is directed into a region with a magnetic field.• The gamma particles carry no charge and thus are not deflected.• The alpha particles are deflected upward.• The negative beta particles (electrons) are deflected downward.

– Positrons would be deflected upward.Section 29.3

Penetrating Ability of Particles

• Alpha particles– Barely penetrate a piece of paper

• Beta particles– Can penetrate a few mm of aluminum

• Gamma rays– Can penetrate several cm of lead

Section 29.3

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Pgs. 921-924, (exclude example

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Alpha Decay

• When a nucleus emits an alpha particle it loses two protons and two neutrons.– N decreases by 2– Z decreases by 2– A decreases by 4

• Symbolically– X is called the parent nucleus.– Y is called the daughter nucleus.

Section 29.4

Alpha Decay – Example• Decay of 226 Ra

• Half life for this decay is 1600 years

• Excess mass is converted into kinetic energy.

• Momentum of the two particles is equal and opposite.

Section 29.4

Decay – General Rules

• When one element changes into another element, the process is called spontaneous decay or transmutation.

• The sum of the mass numbers, A, must be the same on both sides of the equation.

• The sum of the atomic numbers, Z, must be the same on both sides of the equation.

• Conservation of mass-energy and conservation of momentum must hold.

Section 29.4

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Nuclear Reactions

• Structure of nuclei can be changed by bombarding them with energetic particles.– The changes are called nuclear reactions.

• As with nuclear decays, the atomic numbers and mass numbers must balance on both sides of the equation.

Section 29.6

Nuclear Reactions – Example

• Alpha particle colliding with nitrogen:

• Balancing the equation allows for the identification of X

• So the reaction is

Section 29.6

Radiation Damage in Matter

• Radiation absorbed by matter can cause damage.• The degree and type of damage depend on many

factors.– Type and energy of the radiation– Properties of the absorbing matter

• Radiation damage in biological organisms is primarily due to ionization effects in cells.– Ionization disrupts the normal functioning of the cell.

Section 29.7

Types of Damage

• Somatic damage is radiation damage to any cells except reproductive ones.– Can lead to cancer at high radiation levels– Can seriously alter the characteristics of specific organisms

• Genetic damage affects only reproductive cells.– Can lead to defective offspring

Section 29.7

Chapter 30Nuclear Energy

andElementary Particles

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Nuclear Fission

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Pgs. 937-938 from College Physics text

Processes of Nuclear Energy

• Fission– A nucleus of large mass number splits into two

smaller nuclei• Fusion– Two light nuclei fuse to form a heavier nucleus

• Large amounts of energy are released in either case.

Introduction

Forces and Particles

• Fundamental interactions govern the behavior of subatomic particles.

• The current theory of elementary particles states that all particles come from only two families– Quarks– Leptons

Introduction

Nuclear Fission

• A heavy nucleus splits into two smaller nuclei.

• The total mass of the products is less than the original mass of the heavy nucleus.

Section 30.1

Fission Equation

• Fission of 235U by a slow (low energy) neutron

– 236U* is an intermediate, short-lived state• Lasts about 10-12 s

– X and Y are called fission fragments.• Many combinations of X and Y satisfy the requirements

of conservation of energy and charge.

Section 30.1

More About Fission of 235U

• About 90 different daughter nuclei can be formed.

• Several neutrons are also produced in each fission event.

• Example:

• The fission fragments and the neutrons have a great deal of KE following the event.

Section 30.1

Sequence of Events in Fission

• The 235U nucleus captures a thermal (slow-moving) neutron.

• This capture results in the formation of 236U*, and the excess energy of this nucleus causes it to undergo violent oscillations.

• The 236U* nucleus becomes highly elongated, and the force of repulsion between the protons tends to increase the distortion.

• The nucleus splits into two fragments, emitting several neutrons in the process.

Section 30.1

Sequence of Events in Fission – Diagram

Section 30.1

Chain Reaction

• Neutrons are emitted when 235U undergoes fission.

• These neutrons are then available to trigger fission in other nuclei.

• This process is called a chain reaction.– If uncontrolled, a violent explosion can occur.– The principle behind the nuclear bomb, where 1

kg of 235U can release energy equal to about 20000 tons of TNT

Section 30.1

Chain Reaction – Diagram

Section 30.1

Nuclear Reactor

• A nuclear reactor is a system designed to maintain a self-sustained chain reaction.

• The reproduction constant, K, is defined as the average number of neutrons from each fission event that will cause another fission event.– The maximum value of K from uranium fission is 2.5.

• In practice, K is less than this

– A self-sustained reaction has K = 1

Section 30.1

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Nuclear Fusion

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Nuclear Fusion

• Nuclear fusion occurs when two light nuclei combine to form a heavier nucleus.

• The mass of the final nucleus is less than the masses of the original nuclei.– This loss of mass is accompanied by a release of

energy.

Section 30.2

Fusion in the Sun

• All stars generate energy through fusion.• The Sun, along with about 90% of other stars,

fuses hydrogen.– Some stars fuse heavier elements.

• Two conditions must be met before fusion can occur in a star.– The temperature must be high enough.– The density of the nuclei must be high enough to

ensure a high rate of collisions.

Section 30.2

Forces and Mediating Particles

Interaction (force) Mediating Field Particle

Strong Gluon

Electromagnetic Photon

Weak W± and Z0

Gravitational Gravitons

Section 30.3

Link to Webassign Unit 5 Atomic and Nuclear Physics Discussion Questions

Grading Rubric for Unit 5 Atomic and Nuclear

Name: ______________________ Conceptual Physics Text Pgs 596-601 ---------------------------------------------------_____

Pgs 629-634 --------------------------------------------------_____ Pgs 636-642 --------------------------------------------------_____

Advanced notes from text book:

Pgs 870-872 --------------------------------------------------_____ Pgs 872-874 --------------------------------------------------_____ Pgs 875-876 --------------------------------------------------_____ Pgs 879 --------------------------------------------------------_____ Pgs 880-881 --------------------------------------------------_____ Pgs 892-894 --------------------------------------------------_____ Pgs 859-861 --------------------------------------------------_____ Pgs 913-914 --------------------------------------------------_____ Pgs 918 --------------------------------------------------------_____ Pgs 921-924 --------------------------------------------------_____ Pgs 927 --------------------------------------------------------_____ Pgs 937-938 --------------------------------------------------_____ Pgs 941-942 --------------------------------------------------_____

Example Problems:

27.1 (a-c) -------------------------------------------------------------_____ 27.4 (a-b) -------------------------------------------------------------_____ 28.1 (a-b) -------------------------------------------------------------_____