Transcript Unit 5 Atomic and Nuclear

Chapter 27
Quantum Physics
Read and Take notes on
Pages 596-601 in your
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)
Read and take notes on
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
Read and take notes on
Pgs. 872-874 from College Physics text
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 3.00 108 m/s
c = fλ → f = =
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
1.00 eV
= (6.63 10 J) 1.60 10-19 J
-19
(
) = 4.14 eV
(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.
maximum kinetic energy of emitted electrons.
the
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
Read and take notes on
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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Pgs. 879 from College Physics text
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
Read and take notes on
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:
6.63 10-34 J·s
h
=
= 7.28 10-11 m
λe =
-31
7
mev (9.11 10 kg)(1.00 10 m/s)
Repeat the calculation with the baseball data:
6.63 10-34 J·s
h
=
= 1.02 10-34 m
λb =
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
mev/√1- v /c
2
2
=
h√1- v2/c2
mev
Substitute:
λe =
(6.63 10-34 J·s)√1 - (0.999c)2/c2
-31
8
(9.11 10 kg)(0.999·3.00 10 m/s)
= 1.09 10-13 m
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.
the particle.
At ordinary speeds where v/c << 1, it doubles the wavelength of
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
Read and take notes on
Pgs. 892-894 from College Physics text
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 shortestwavelength 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 longestwavelength 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
1 1
1 1
5R H
- 2 = RH
- 2 =
= RH
2
2
2 3
36
λ
nf ni
Take the reciprocal and substitute, finding the wavelength:
36
36
-7
=
=
6.563
10
m = 656.3 nm
λ=
7
-1
5RH 5(1.097 10 m )
Now use c = fλ to obtain the frequency:
c 2.998 108 m/s
14
=
4.568
10
Hz
f= =
-7
λ 6.563 10 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
λ
= RH
1
(n
2
f
-
1
ni
2
1
RH
)=R (2 -0)= 4
H
2
Take the reciprocal and substitute, finding the wavelength:
λ=
4
RH
=
4
(1.097 107 m-1)
= 3.646 10-7 m = 364.6 nm
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 29
Nuclear Physics
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Mass Energy Equivalence
Read and take notes on
pages 859-861 in your
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
Read and take notes on
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
Read and take notes on
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
Read and take notes on
Pgs. 921-924,
from College
Physics text
(exclude example problems)
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
Read and take notes on
Pgs. 927 from College Physics text
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 30
Nuclear Energy
and
Elementary Particles
Read and Take notes on
Pages 629-634 and 636-642
in your Conceptual Physics Text
Link to Brighstorm on
Nuclear Fission
Read and take notes on
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
Link to Brighstorm on
Nuclear Fusion
Read and take notes on
Pgs. 941-942 from College Physics text
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:
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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) -------------------------------------------------------------_____