Transcript 7.25 x 10 20 Hz

AP Chapter 5
Structure of the Atom
Review Quiz Chapter 5
• Net Ionic Equations
Radioactivity
• Radioactivity is the spontaneous
breakdown of unstable atoms into more
stable atoms with the simultaneous
emission of particles and rays.
Radioactivity
• Different radioactive elements emit
different amounts of three kinds of
radiation (alpha, beta, and gamma).
• Alpha rays are helium nuclei (He2+).
• Beta rays are electrons.
• Gamma rays are photons having high
energy.
Radioactivity and Half - Life
• The half-life of carbon-14 is 5730 years.
• How old is a bone that has about 12.5% of the
carbon-14 that a living organism would have in it?
Carbon Dating
Nuclear Chemistry
• a nuclear reaction is the process in which
two nuclei, or else a nucleus of an atom
and a subatomic particle (such as a
proton, or high energy electron) from
outside the atom, collide to produce
products different from the initial particles.
Particles found in Nuclear
Reactions
• Alpha
4
2
He 2
• Beta  - or
• Positron 
• Proton
p+
+
0
1
or
e
0
1
e
or a hydrogen nucleus
1
(1
H
• Neutron n0 ( 01 n )
• Gamma rays (not truly a particle)

)
Electron Capture
• A proton is converted into a neutron when one of the
electrons in an atom is captured by the nucleus.
1
1
H  e n
0
1
1
0
p   n


1
0
Electron Capture
• You may see energy released in this reaction in
0
the form of a neutrino ( 0 vor eV ). Energy is
always released however sometimes it is not
written in the equation
Beta Emission
• In beta emission a neutron in the nucleus
changes to a proton (that remains in the
nucleus) and an electron (beta particle) is
ejected.
• Technetium (element 43), a radioactive
element that does not occur naturally on
the earth was first prepared in 1937. It
decays by beta emission.
Beta Emission
Tc Ru  
98
43
98
44

Tc Ru  e
98
43
98
44
0
1
Positron Emission
+
Beta Positive (β decay)
• A proton is converted to a neutron
and releases a positron.
1
1
H  n e
1
0
0
1
p  n

1
0

Positron Emission (β+ decay)

+
Positron Emission
Tomography (PET Scans)
Positron Emission
Tomography (PET Scans)
• Positron emission tomography (PET) is a nuclear medicine imaging
technique that produces a three-dimensional image or picture of
functional processes in the body. The system detects pairs of gamma
rays emitted indirectly by a positron-emitting radionuclide (tracer), which
is introduced into the body on a biologically active molecule. Threedimensional images of tracer concentration within the body are then
constructed by computer analysis.
• As the radioisotope undergoes positron emission decay (also known as
beta positive decay), it emits a positron, an antiparticle of the electron
with opposite charge. The emitted positron travels in tissue for a short
distance (typically less than 1 mm, but dependent on the isotope), during
which time it loses kinetic energy, until it decelerates to a point where it
can interact with an electron. The encounter annihilates both electron
and positron, producing a pair of (gamma) photons. These are detected
by the PET scanning device.
Alpha Particles in nuclear reactions
• The first radioactive element that was found,
polonium, was found in 1898 by Marie and
Pierre Curie.
• It decays by alpha emission.
• Alpha emission is the release of an alpha
particle. ( 24 He 2 )
• This causes the nucleus to lose 2 protons and 2
neutrons. Therefore the atomic number is
reduced by 2 and the mass number is reduced
by 4.
Alpha Emission
212
84
Po Pb  He
208
82
4
2
2
Gaining an alpha particle in a
nuclear reaction
Alpha Particles in nuclear reactions
• The first element prepared by artificial means
was prepared in 1919 by bombarding nitrogen
atoms with alpha particles.
• This caused the nitrogen nucleus to increase in
mass.
Alpha Particles in nuclear reactions
14
7
N  He  O  H
4
2
17
8
1
1
• The N and He have a total of 9 neutrons and 9
protons.
• In forming O a proton is lost leaving us with 8
protons and the number of neutrons stays at 9.
Lord Ernest Rutherford (1871 – 1937)
• Discovered the
nucleus of the atom.
Rutherford’s Gold Foil Experiment
Rutherford’s Gold Foil Experiment
Rutherford’s Nuclear Model of the Atom
• The nucleus is very
small, dense, and
positively charged.
• Electrons surround
the nucleus which
contains the protons
and “neutrons”.
• Most of the atom is
empty space
Subatomic Particles
PARTICLE
SYMBOL CHARGE
MASS
(amu)
LOCATION
electron
e-
-1
0
orbit nucleus
proton
p+
+1
1
inside nucleus
neutron
n0
0
1
inside nucleus
Electromagnetic Waves
• Properties of waves include speed, frequency,
wavelength and energy
• All electromagnetic waves including light travel at a
speed of 3 x 108 m/s.
• However the frequency, wavelength and energy of
the waves vary.
Wavelength ()
Measured in
units of length:
m, nm, Aº
Frequency ()
Measured in cycles/second = hertz (Hz)
Visible Light
Electromagnetic Radiation
• For all waves
• = c
c = the speed of light = 3.00 x 10
8
m/s
A photon of red light has a wavelength of
665 nm. What is the frequency of this light?
A photon of red light has a wavelength of
665 nm. What is the frequency of this light?
• 665 nm = 665 x 10-9 m
A photon of red light has a wavelength of
665 nm. What is the frequency of this light?
• 665 nm = 665 x 10-9 m
• c=•
A photon of red light has a wavelength of
665 nm. What is the frequency of this light?
• 665 nm = 665 x 10-9 m
• c=•
•  = c ÷  = 3.00 x 108 m/s ÷ 665 x 10-9 m
A photon of red light has a wavelength of
665 nm. What is the frequency of this light?
•
•
•
•
665 nm = 665 x 10-9 m
c=•
 = c ÷  = 3.00 x 108 m/s ÷ 665 x 10-9 m
 = 4.511278 x 1014/s or Hz
A photon of red light has a wavelength of
665 nm. What is the frequency of this light?
•
•
•
•
•
665 nm = 665 x 10-9 m
c=•
 = c ÷  = 3.00 x 108 m/s ÷ 665 x 10-9 m
 = 4.511278 x 1014/s or Hz
 = 4.51 x 1014/s or Hz
An x-ray has a frequency of 7.25 x 1020 Hz.
What is the wavelength?
An x-ray has a frequency of 7.25 x 1020 Hz.
What is the wavelength?
• 7.25 x 1020 Hz = 7.25 x 1020/s
An x-ray has a frequency of 7.25 x 1020 Hz.
What is the wavelength?
• 7.25 x 1020 Hz = 7.25 x 1020/s
• c=•
An x-ray has a frequency of 7.25 x 1020 Hz.
What is the wavelength?
• 7.25 x 1020 Hz = 7.25 x 1020/s
• c=•
•  = c ÷  = 3.00 x 108 m/s ÷ 7.25 x 1020/s
An x-ray has a frequency of 7.25 x 1020 Hz.
What is the wavelength?
•
•
•
•
7.25 x 1020 Hz = 7.25 x 1020/s
c=•
 = c ÷  = 3.00 x 108 m/s ÷ 7.25 x 1020/s
 = 4.137931 x 10-13 m
An x-ray has a frequency of 7.25 x 1020 Hz.
What is the wavelength?
•
•
•
•
•
7.25 x 1020 Hz = 7.25 x 1020/s
c=•
 = c ÷  = 3.00 x 108 m/s ÷ 7.25 x 1020/s
 = 4.137931 x 10-13 m
 = 4.14 x 10-13 m
Energy of Electromagnetic
Radiation
• For all waves:
h = 6.63 x 10
E = h•
-34
J • s or J/Hz
A photon of red light has a wavelength of
665 nm. What is the energy of this light?
A photon of red light has a wavelength of
665 nm. What is the energy of this light?
•
•
•
•
•
665 nm = 665 x 10-9 m
c=•
 = c ÷  = 3.00 x 108 m/s ÷ 665 x 10-9 m
 = 4.511278 x 1014/s or Hz
 = 4.51 x 1014/s or Hz
A photon of red light has a wavelength of
665 nm. What is the energy of this light?
•
•
•
•
•
•
665 nm = 665 x 10-9 m
c=•
 = c ÷  = 3.00 x 108 m/s ÷ 665 x 10-9 m
 = 4.511278 x 1014/s or Hz
 = 4.51 x 1014/s or Hz
E = h •  = (6.63 x 10-34 J • s)(4.51 x 1014/s)
A photon of red light has a wavelength of
665 nm. What is the energy of this light?
•
•
•
•
•
•
•
665 nm = 665 x 10-9 m
c=•
 = c ÷  = 3.00 x 108 m/s ÷ 665 x 10-9 m
 = 4.511278 x 1014/s or Hz
 = 4.51 x 1014/s or Hz
E = h •  = (6.63 x 10-34 J • s)(4.51 x 1014/s)
E = 2.99013 x 10-19 J
A photon of red light has a wavelength of
665 nm. What is the energy of this light?
•
•
•
•
•
•
•
665 nm = 665 x 10-9 m
c=•
 = c ÷  = 3.00 x 108 m/s ÷ 665 x 10-9 m
 = 4.511278 x 1014/s or Hz
 = 4.51 x 1014/s or Hz
E = h •  = (6.63 x 10-34 J • s)(4.51 x 1014/s)
E = 2.99013 x 10-19 J = 2.99 x 10-19 J
An x-ray has a frequency of 7.25 x 1020 Hz.
What is it’s energy?
An x-ray has a frequency of 7.25 x 1020 Hz.
What is it’s energy?
• E = h •  = (6.63 x 10-34 J/Hz)(7.25 x 1020Hz)
An x-ray has a frequency of 7.25 x 1020 Hz.
What is it’s energy?
• E = h •  = (6.63 x 10-34 J/Hz)(7.25 x 1020Hz)
• E = 4.80675 x 10-13 J
An x-ray has a frequency of 7.25 x 1020 Hz.
What is it’s energy?
• E = h •  = (6.63 x 10-34 J/Hz)(7.25 x 1020Hz)
• E = 4.80675 x 10-13 J
• E = 4.81 x 10-13 J
wavelength, frequency and energy
Red Light
X-ray
•  = 665 x 10-9 m
•  = 4.51 x 1014 Hz
• E = 2.99 x 10-19 J
•  = 4.14 x 10-13 m
•  = 7.25 x 1020 Hz
• E = 4.81 x 10-13 J
Wavelength, frequency and energy
• Wavelength and frequency have an indirect
relationship.
• Energy and frequency have a direct
relationship.
• Electromagnetic radiation of short wavelength
will have high frequency and high energy.
• Electromagnetic radiation of long wavelength
will have low frequency and low energy.
Bohr Model of the Atom
• The Bohr atom
The Bohr Atom
• Electrons orbit the
nucleus in orbits that
represent specific
quantities of energy.
• The energies of the
electrons in the atom are
quantized.
• Only certain electron
orbits (energy levels) are
allowed.
Ground State
• The lowest energy
state of an atom.
Excited State
• Any energy state of
an atom that is of
higher in energy
than the ground
state.
Energy Absorbed
Absorption (Dark – Line) Spectra
Energy Emitted
Electron jumps
to a lower orbit
Emission (Bright – Line) Spectra
Emission Spectra
The lines present in an emission
spectrum are the lines missing in
an absorption spectrum.
Star Finder Video
Electromagnetic Spectrum
Star Finder Video
Fingerprints of Light
Heisenberg’s Uncertainty Principle
• It is impossible to accurately determine the
momentum (velocity) and location of a
particle simultaneously.
• Introduced probability to atomic structure
which lead to the development of the
quantum – mechanical (electron cloud)
model of the atom.
Quantum – Mechanical Model
(Electron Cloud)
• The electron cloud is a
visual representation of
the most probable
locations for an
electron within an
atom.
• “Clarity through
fuzziness”
Energy Levels – Sublevels - Orbitals
• Electrons in an atom are within atomic
orbitals which are within sublevels which
are within energy levels.
• Chemistry uses quantum numbers to
describe these electrons.
The Principal Quantum Number (n)
• n = 1, 2, 3, 4 . . .
• Electrons with the same value of “n” within an atom
are in the same energy level or shell.
• The principal quantum number n represents the
relative overall energy of an electron and the energy of
each electron increases as the distance from the
nucleus increases.
• Example: An electron with n = 2 is further from the
nucleus and therefore has more energy than an
electron with n = 1.
The Azimuthal Quantum Number (l)
• l = 0…(n – 1).
• Orbitals with the same value of “n”
may have different shapes. The “l” value indicates the
shape of the orbitals.
• Electrons with the same value of “l” within an atom are
in the same sublevel or subshell.
• Example: In the fourth energy level (n = 4) there are
four different orbital shapes possible designated l = 0,
1, 2 or 3.
The Azimuthal Quantum Number (l)
Orbital Shapes
Page 143
Magnetic Quantum Number (m)
• m = -l…0…+ l.
• Orbitals within an energy level with the
same value of l have the same shape and energy
(degenerate) but differ in their orientation.
• Each possible orientation of the orbital has a specific
value of “m”.
• Electrons with the same value of “m” are in the same
atomic orbital (the region of space that an electron is
most likely to be found within an atom).
Magnetic Quantum Number (m)
• Example: If l = 1 then m has three possible
orientations designated:
m = -1, 0 or +1.
Possible Orientations of a
“p” atomic orbital
Page 143
Possible Orientations of a
“d” atomic orbital
Page 144
Possible Orientations of a
“f” atomic orbital
Page 144
Pauli Exclusion Principle
• No two electrons in the same atom can have
the same set of four quantum numbers.
Spin Quantum Number (s)
• S = +1/2 or -1/2
• Specifies the direction of spin of the
electron on its axis.
• Spins are designated up or down.
Electron Spin
• Opposite spins
produce opposite
magnetic fields.
+1/2
-1/2
The Maximum Number of Electrons
Possible in an Energy Level
2
2n
Electron Configuration
7
4d
7 electrons are in the d sublevel in the 4th energy level
SUBLEVEL
s
NUMBER OF
ORBITALS
1
MAX. # OF
ELECTRONS
2
p
3
6
d
5
10
f
7
14
Arrow Diagram
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
7s
7p
Write the
electron
configuration
for lead
(Z = 82).
The periodic
table and
electron
configuration.
Periodic Table and Electron Configuration
s
p
1
2
d (period-1)
3
4
5
6
7
6
7
6
7
f (period-2)
© 1998 by Harcourt Brace & Company
C. Periodic Patterns
• Example - Germanium
1
2
3
4
5
6
7
[Ar]
2
10
4s 3d
2
4p
Write the abbreviated electron configuration
for lead (Z = 82) using the periodic table.
O
6
7
© 1998 by Harcourt Brace & Company
Orbital Filling Diagrams
A. General Rules
• Pauli Exclusion Principle
– Each orbital can hold TWO electrons with
opposite spins.
A. General Rules
• Hund’s Rule
– Within a sublevel, place one e- per orbital
before pairing them.
– All electrons in singly filled orbitals have the
same direction of spin.
Incorrect
Correct
B. Notation
• Orbital Diagram
O
8e-
1s
2s
• Electron Configuration
2
2
4
1s 2s 2p
2p
Write the quantum numbers that
represent each of the electrons
within an oxygen atom
Write the quantum numbers that
represent each of the electrons
within an oxygen atom
Write the quantum numbers that
represent each of the electrons
within an oxygen atom
• The first two
electrons
designated 1s2
would have the
quantum
numbers.
1, 0, 0, +1/2
1, 0, 0, -1/2
D. Stability
• Full energy level
• Full sublevel (s, d, f). The “p” is not listed here
because it is part of the full energy level
• Half-full sublevel (only for p, d, f)
1
2
3
4
5
6
7
D. Stability
• Electron Configuration Exceptions
– Copper
EXPECT:
[Ar] 4s2 3d9
ACTUALLY:
[Ar] 4s1 3d10
– Copper gains stability with a full
d-sublevel which is of lower energy and is
therefore the ground state of copper.
D. Stability
• Electron Configuration Exceptions
• Chromium
EXPECT:
[Ar] 4s2 3d4
ACTUALLY:
[Ar] 4s1 3d5
• Chromium gains stability with a half-full
d-sublevel which is of lower energy and is
therefore the ground state of chromium.