6 to 2 violet

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6 to 2 violet 5 to 2 blue 4 to 2 green 3 to 2
red
Notice how the absorption matches emission
spectrum
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?E hf Only certain ?Es are allowed because
only certain orbits (energy levels) are allowed.
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Ionization Energy (Energy that must be absorbed
for an electron to escape the atom) ?E
EIonization hfIonization c ?Ionization x
fIonization ?Ionization c/fIonization
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Photoelectric Effect
Read section 27.2 on pages 889 - 891 Work
problems 13, 15, 17 on page 914 Study the
following slides
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The Photoelectric Effect
E hf E energy of a photon of light h
Planks Constant f frequency of the light
The photon is entirely absorbed by a single
electron in a metal.
c ?f c speed of light 3 x 108m/s ?
wavelength of the light f frequency of the
light
f c/? E hc/?
1 electron volt (eV) 1.6 x 10-19C x 1 volt 1 eV
1.6 x 10-19C x 1 Joule/C 1 eV 1.6 x 10-19
Joules
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Light impacting on the metal surface, E, causes
electrons to be emitted from the surface of the
metal. Electrons flow across the gap due to a
potential difference.

• Assumptions
• Light Source is Monochromatic (one ?)
• Low and High Intensity Light are Same ?
• Low and High Intensity Light are Same f

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• Observations
• Higher Intensity ? More photons/second
• More electrons ejected from metal/second
• More electrons ejected causes higher current
• Increasing voltage above zero does little to
• to increase the current
• Applying negative voltage does reduce
• the current

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At some negative voltage, (the stopping
potential ,?Vs), the current goes to zero.
Reason There is a maximum KE of the electrons,
independent of the radiation intensity. KEmax
e ?Vs KEmax hf F hf energy of photon F
work function minimum
energy with which electron is bound to
the metal
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KEmax increases linearly from zero at fc
(cut-off frequency) Higher frequency
more kinetic energy of the ejected electron KE
hf F F constant for a given metal
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e?Vs KEmax hf - F ?Vs (h/e)f F/e F/e
a constant Slope of the above graph
is h/e F hfthreshold
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Compton Scattering Effect Read
section 27.6 on pages 898 to 899 Work problems
27 and 29 on page 914 Study the following
slides.
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?? ? - ?o h/(mec)(1-cos?) h/(mec) Compton
Wavelength .00243 nm
Visible light 400-700 nm Difficult to detect
Compton scattering effect with visible light.
High energy x-rays (small ?) are used to verify
the theory.
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de Broglie Wavelengths and the Davisson Germer
Experiment Read Section 27.9 on pages 901
904 Work Problems 37 and 39 on page 915 One
Formula to know ? h/p h/mv
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Nuclear Physics Read Sections
29.3 - 29.4 pages 964 - 974 Work Problems 13 -
31 odd except 21 29 Study the following
slides.
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Radioactivity Nuclear Decay (nucleus
splits into two or more particles) Rate of
Decay ?N -?N?t N number of nuclei ?
decay constant (for a particular type
of nucleus ?t
elapsed time Calculus N Noe-?t 1 currie 1
ci 3.7 x 1010 decays/sec (decay rate of a gram
of radium)
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Look at figure 29.6 on page 965
Half-life The time required for ½ of the
original number of nuclei to decay
(change to other nuclei or other
particles) No/2 Noe-?t1/2 ? t1/2 ln(1/2)/?
t1/2 half -life Students to see example
29.3
t1/2.692/?
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Decay Process
Alpha (?) Decay Produces an alpha particle AX
? A-4Y 4He A atomic mass (pn) Z
Z-2 2 Z atomic number
(p) The particle 4He is a bare helium atom
(no e-). 2 It
is commonly called an alpha (?)
particle. Example 238U ? 234Th 4He 92
90 2 Note that mass is conserved
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Beta (?) Decay Produces an positron e
and a neutrino ?
OR
Produces an electron e-
and an anti - neutrino ??
AX ? AY e ? Z Z-1
OR AZ ? AY e- ?? Z Z1 Note
that protons are lost or gained but the atomic
mass does not change because of a corresponding
change in the number of neutrons
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Gamma (?) Decay Produces High Energy
Photons A nucleus of a atom can exist only for a
short time in an excited state. This
excited state arises when the nucleus has
recently undergone either Alpha or Beta
decay. Example 12B ? 12C e - ??
? Decay 5 6 12C ? 12C ?
? Decay 6 6
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Nuclear Energy Read Section
30.1 pages 991 - 993 Work Problems 1 and 3
on page 1026 Study the following slides.
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Mass is not conserved in nuclear reactions. The
missing (lost) mass in a fission or
fusion reaction appears as energy according to
Einsteins famous equation E mc2 m
missing mass or as it is written on the AP
Formula Worksheet ?E (?m)c2
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Fission nuclear reactions occur when a large
nucleus decays into two smaller nuclei.
Example Uranium decays to Thorium and
Helium Fusion nuclear reactions occur when two
small nuclei combine to form a larger
nucleus. Example Two Hydrogen atoms combine to
form one Helium atom. The most
stable nuclei have an atomic number of
approximately 60. Iron is thought by some to be
the most stable nucleus.
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Nuclear Force Short range force. Acts like an
atomic glue that holds protons and neutrons
together. Think of protons and neutrons as having
very short, but strong, hooks that can engage to
hold them together when the protons and
neutrons are within range of each others
hooks. Once this force is overcome, the electric
force causes protons to repel from each
other. Question Why do elements with a large
number of protons seem to have an even larger
number of neutrons?
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The Last Slide. Really!! Power and Photons For
a monochromatic source Power P hf x
photons/second