chapter 7 the quantum-mechanical model of the atom
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Chemistry: A Molecular Approach , 1 st Ed. Nivaldo Tro. Chapter 7 The Quantum-Mechanical Model of the Atom. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2007, Prentice Hall. The Behavior of the Very Small. electrons are incredibly small - PowerPoint PPT PresentationTRANSCRIPT
Chapter 7The Quantum-
Mechanical Model of the
Atom
2007, Prentice Hall
Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro
Roy KennedyMassachusetts Bay Community College
Wellesley Hills, MA
Tro, Chemistry: A Molecular Approach 2
The Behavior of the Very Small
• electrons are incredibly smalla single speck of dust has more electrons than the
number of people who have ever lived on earth• electron behavior determines much of the
behavior of atoms• directly observing electrons in the atom is
impossible, the electron is so small that observing it changes its behavior
Tro, Chemistry: A Molecular Approach 3
A Theory that Explains Electron Behavior• the quantum-mechanical model explains the manner
electrons exist and behave in atoms• helps us understand and predict the properties of atoms
that are directly related to the behavior of the electronswhy some elements are metals while others are nonmetalswhy some elements gain 1 electron when forming an anion,
while others gain 2why some elements are very reactive while others are
practically inertand other Periodic patterns we see in the properties of the
elements
Tro, Chemistry: A Molecular Approach 4
The Nature of Lightits Wave Nature
• light is a form of electromagnetic radiationcomposed of perpendicular oscillating waves, one for the
electric field and one for the magnetic fieldan electric field is a region where an electrically charged particle
experiences a forcea magnetic field is a region where an magnetized particle experiences
a force
• all electromagnetic waves move through space at the same, constant speed3.00 x 108 m/s in a vacuum = the speed of light, c
Tro, Chemistry: A Molecular Approach 5
Speed of Energy Transmission
Tro, Chemistry: A Molecular Approach 6
Electromagnetic Radiation
Tro, Chemistry: A Molecular Approach 7
Characterizing Waves• the amplitude is the height of the wave
the distance from node to crest or node to trough
the amplitude is a measure of how intense the light is – the larger the amplitude, the brighter the light
• the wavelength, () is a measure of the distance covered by the wavethe distance from one crest to the next
or the distance from one trough to the next, or the distance between alternate nodes
Tro, Chemistry: A Molecular Approach 8
Wave Characteristics
Tro, Chemistry: A Molecular Approach 9
Characterizing Waves• the frequency, () is the number of waves that
pass a point in a given period of timethe number of waves = number of cyclesunits are hertz, (Hz) or cycles/s = s-1
1 Hz = 1 s-1
• the total energy is proportional to the amplitude and frequency of the wavesthe larger the wave amplitude, the more force it hasthe more frequently the waves strike, the more total
force there is
Tro, Chemistry: A Molecular Approach 10
The Relationship Between Wavelength and Frequency
• for waves traveling at the same speed, the shorter the wavelength, the more frequently they pass
• this means that the wavelength and frequency of electromagnetic waves are inversely proportionalsince the speed of light is constant, if we know
wavelength we can find the frequency, and visa versa
m
cs s
m1-
Tro, Chemistry: A Molecular Approach 11
Example 7.1- Calculate the wavelength of red light with a frequency of 4.62 x 1014 s-1
the unit is correct, the wavelength is appropriate for red light
Check:
Solve:∙ = c, 1 nm = 10-9 m
Concept Plan:
Relationships:
= 4.62 x 1014 s-1
, (nm)Given:
Find:
s-1) (m) (nm)
c
m 10
nm 19
m 1049.6 1062.4
1000.3c 71-s14
-1sm8
nm 1049.6m 10
nm 1m 1049.6 29
7
Tro, Chemistry: A Molecular Approach 12
Practice – Calculate the wavelength of a radio signal with a frequency of 100.7 MHz
Tro, Chemistry: A Molecular Approach 13
m 98.2 10007.1
1000.3c1-s8
-1sm8
1-8-16
s 10007.1MHz 1
s 10MHz 00.71
Practice – Calculate the wavelength of a radio signal with a frequency of 100.7 MHz
the unit is correct, the wavelength is appropriate for radiowaves
Check:
Solve:∙ = c, 1 MHz = 106 s-1
Concept Plan:
Relationships:
= 100.7 MHz , (m)
Given:Find:
Hz) (s-1) (m)
c
MHz 1s 10 -16
Tro, Chemistry: A Molecular Approach 14
Color• the color of light is determined by its wavelength
or frequency
• white light is a mixture of all the colors of visible light a spectrumRedOrangeYellowGreenBlueViolet
• when an object absorbs some of the wavelengths of white light while reflecting others, it appears colored the observed color is predominantly the colors reflected
15
Amplitude & Wavelength
Tro, Chemistry: A Molecular Approach 16
Electromagnetic Spectrum
Tro, Chemistry: A Molecular Approach 17
Continuous Spectrum
Tro, Chemistry: A Molecular Approach 18
The Electromagnetic Spectrum• visible light comprises only a small fraction of
all the wavelengths of light – called the electromagnetic spectrum
• short wavelength (high frequency) light has high energyradiowave light has the lowest energygamma ray light has the highest energy
• high energy electromagnetic radiation can potentially damage biological moleculesionizing radiation
Tro, Chemistry: A Molecular Approach 19
Thermal Imaging using Infrared Light
Tro, Chemistry: A Molecular Approach 20
Using High Energy Radiationto Kill Cancer Cells
Tro, Chemistry: A Molecular Approach 21
Interference
• the interaction between waves is called interference
• when waves interact so that they add to make a larger wave it is called constructive interferencewaves are in-phase
• when waves interact so they cancel each other it is called destructive interferencewaves are out-of-phase
Tro, Chemistry: A Molecular Approach 22
Interference
Tro, Chemistry: A Molecular Approach 23
Diffraction• when traveling waves encounter an obstacle or opening
in a barrier that is about the same size as the wavelength, they bend around it – this is called diffraction traveling particles do not diffract
• the diffraction of light through two slits separated by a distance comparable to the wavelength results in an interference pattern of the diffracted waves
• an interference pattern is a characteristic of all light waves
Tro, Chemistry: A Molecular Approach 24
Diffraction
Tro, Chemistry: A Molecular Approach 25
2-Slit Interference
Tro, Chemistry: A Molecular Approach 26
The Photoelectric Effect• it was observed that many metals emit electrons when a
light shines on their surface this is called the Photoelectric Effect
• classic wave theory attributed this effect to the light energy being transferred to the electron
• according to this theory, if the wavelength of light is made shorter, or the light waves intensity made brighter, more electrons should be ejected remember: the energy of a wave is directly proportional to its
amplitude and its frequency if a dim light was used there would be a lag time before
electrons were emitted to give the electrons time to absorb enough energy
Tro, Chemistry: A Molecular Approach 27
The Photoelectric Effect
Tro, Chemistry: A Molecular Approach 28
The Photoelectric EffectThe Problem
• in experiments with the photoelectric effect, it was observed that there was a maximum wavelength for electrons to be emittedcalled the threshold frequencyregardless of the intensity
• it was also observed that high frequency light with a dim source caused electron emission without any lag time
Tro, Chemistry: A Molecular Approach 29
Einstein’s Explanation• Einstein proposed that the light energy was
delivered to the atoms in packets, called quanta or photons
• the energy of a photon of light was directly proportional to its frequencyinversely proportional to it wavelengththe proportionality constant is called Planck’s
Constant, (h) and has the value 6.626 x 10-34 J∙s
chhE
Tro, Chemistry: A Molecular Approach 30
Example 7.2- Calculate the number of photons in a laser pulse with wavelength 337 nm and total energy 3.83 mJ
Solve:E=hc/, 1 nm = 10-9 m, 1 mJ = 10-3 J, Epulse/Ephoton = # photons
Concept Plan:
Relationships:
= 337 nm, Epulse = 3.83 mJ
number of photonsGiven:
Find:
hcEphoton
nm 1m 10 9
J 108598.5 1037.3
1000.3 10626.6hcE 19
m7
-1sm8sJ34
photon
m 1037.3nm 1
m 10nm 1037.3 79
2
nm) (m) Ephotonnumberphotons
photon
pulse
EE
J 1083.3mJ 1
J 10mJ 83.3 33
photons 1049.6
J 108598.5J 1083.3photons ofnumber
15
3
3
Tro, Chemistry: A Molecular Approach 31
Practice – What is the frequency of radiation required to supply 1.0 x 102 J of energy from
8.5 x 1027 photons?
Tro, Chemistry: A Molecular Approach 32
1-7
sJ34J26
photon s 108.1 10626.6 10761.1
hE
What is the frequency of radiation required to supply 1.0 x 102 J of energy from 8.5 x 1027 photons?
Solve:E=h, Etotal = Ephoton∙# photons
Concept Plan:
Relationships:
Etotal = 1.0 x 102 J, number of photons = 8.5 x 1027
Given:Find:
h
Ephoton
(s-1)Ephotonnumberphotons
photons ofnumber E total
J 10761.1105.8
J 100.1E 2627
2
photon
Tro, Chemistry: A Molecular Approach 33
Ejected Electrons• 1 photon at the threshold frequency has just
enough energy for an electron to escape the atombinding energy,
• for higher frequencies, the electron absorbs more energy than is necessary to escape
• this excess energy becomes kinetic energy of the ejected electron
Kinetic Energy = Ephoton – Ebinding
KE = h -
Tro, Chemistry: A Molecular Approach 34
Spectra• when atoms or molecules absorb energy, that energy is
often released as light energy fireworks, neon lights, etc.
• when that light is passed through a prism, a pattern is seen that is unique to that type of atom or molecule – the pattern is called an emission spectrumnon-continuouscan be used to identify the material
flame tests
• Rydberg analyzed the spectrum of hydrogen and found that it could be described with an equation that involved an inverse square of integers
2
22
1
1-7
n1
n1m 10097.11
Tro, Chemistry: A Molecular Approach 35
Emission Spectra
Tro, Chemistry: A Molecular Approach 36
Exciting Gas Atoms to Emit Light with Electrical Energy
Hg He H
Tro, Chemistry: A Molecular Approach 37
Examples of Spectra
Oxygen spectrum
Neon spectrum
Tro, Chemistry: A Molecular Approach 38
Identifying Elements with Flame Tests
Na K Li Ba
Tro, Chemistry: A Molecular Approach 39
Emission vs. Absorption Spectra
Spectra of Mercury
Tro, Chemistry: A Molecular Approach 40
Bohr’s Model• Neils Bohr proposed that the electrons could only have
very specific amounts of energy fixed amounts = quantized
• the electrons traveled in orbits that were a fixed distance from the nucleusstationary states therefore the energy of the electron was proportional the
distance the orbital was from the nucleus• electrons emitted radiation when they “jumped” from
an orbit with higher energy down to an orbit with lower energy the distance between the orbits determined the energy of the
photon of light produced
Tro, Chemistry: A Molecular Approach 41
Bohr Model of H Atoms
Tro, Chemistry: A Molecular Approach 42
Wave Behavior of Electrons• de Broglie proposed that particles could have wave-like
character• because it is so small, the wave character of electrons is
significant• electron beams shot at slits show an interference
pattern the electron interferes with its own wave
• de Broglie predicted that the wavelength of a particle was inversely proportional to its momentum
)s(m(kg)s
mkg
m1-
2
2
velocitymass
h
Tro, Chemistry: A Molecular Approach 43
however, electrons actually present an interference pattern, demonstrating the behave like waves
Electron Diffraction
if electrons behave like particles, there should only be two bright spots on the target
Tro, Chemistry: A Molecular Approach 44
m 1074.2
1065.2kg 10.119
10626.6
mvh
10
m631-
mk342
2
s
s
g
Solve:
Example 7.3- Calculate the wavelength of an electron traveling at 2.65 x 106 m/s
=h/mv
Concept Plan:
Relationships:
v = 2.65 x 106 m/s, m = 9.11 x 10-31 kg (back leaf) m
Given:Find:
vmh
m, v (m)
Tro, Chemistry: A Molecular Approach 45
Practice - Determine the wavelength of a neutron traveling at 1.00 x 102 m/s
(Massneutron = 1.675 x 10-24 g)
Tro, Chemistry: A Molecular Approach 46
m 1096.3
1000.1kg 10675.1
10626.6
mvh
9
m227-
mk342
2
s
s
g
Solve:
Practice - Determine the wavelength of a neutron traveling at 1.00 x 102 m/s
=h/mv, 1 kg = 103 g
Concept Plan:
Relationships:
v = 1.00 x 102 m/s, m = 1.675 x 10-24 g m
Given:Find:
vmh
m (kg), v (m)m(g)
g 10kg 13
kg 10675.1
g 10kg 1g 10675.1
27
324
Tro, Chemistry: A Molecular Approach 47
Complimentary Properties• when you try to observe the wave nature of the
electron, you cannot observe its particle nature – and visa versawave nature = interference patternparticle nature = position, which slit it is passing
through• the wave and particle nature of nature of the
electron are complimentary propertiesas you know more about one you know less about
the other
Tro, Chemistry: A Molecular Approach 48
Uncertainty Principle• Heisenberg stated that the product of the uncertainties
in both the position and speed of a particle was inversely proportional to its massx = position, x = uncertainty in positionv = velocity, v = uncertainty in velocitym = mass
• the means that the more accurately you know the position of a small particle, like an electron, the less you know about its speedand visa-versa
m1
4hv
x
Tro, Chemistry: A Molecular Approach 49
Uncertainty Principle Demonstration
any experiment designed to observe the electron results in detection of a single electron particle and no interference pattern
Tro, Chemistry: A Molecular Approach 50
Determinacy vs. Indeterminacy• according to classical physics, particles move in a path
determined by the particle’s velocity, position, and forces acting on itdeterminacy = definite, predictable future
• because we cannot know both the position and velocity of an electron, we cannot predict the path it will follow indeterminacy = indefinite future, can only predict
probability• the best we can do is to describe the probability an
electron will be found in a particular region using statistical functions
51
Trajectory vs. Probability
Tro, Chemistry: A Molecular Approach 52
Electron Energy• electron energy and position are complimentary
because KE = ½mv2
• for an electron with a given energy, the best we can do is describe a region in the atom of high probability of finding it – called an orbitala probability distribution map of a region where the
electron is likely to be founddistance vs. 2
• many of the properties of atoms are related to the energies of the electrons
Eψψ Η
Tro, Chemistry: A Molecular Approach 53
Wave Function, • calculations show that the size, shape and
orientation in space of an orbital are determined be three integer terms in the wave functionadded to quantize the energy of the electron
• these integers are called quantum numbersprincipal quantum number, nangular momentum quantum number, lmagnetic quantum number, ml
Tro, Chemistry: A Molecular Approach 54
Principal Quantum Number, n• characterizes the energy of the electron in a particular
orbitalcorresponds to Bohr’s energy level
• n can be any integer 1• the larger the value of n, the more energy the orbital has• energies are defined as being negative
an electron would have E = 0 when it just escapes the atom• the larger the value of n, the larger the orbital• as n gets larger, the amount of energy between orbitals
gets smaller
2
18- 1J 10-2.18Enn for an electron in H
Tro, Chemistry: A Molecular Approach 55
Principal Energy Levels in Hydrogen
56
Electron Transitions• in order to transition to a higher energy state, the
electron must gain the correct amount of energy corresponding to the difference in energy between the final and initial states
• electrons in high energy states are unstable and tend to lose energy and transition to lower energy statesenergy released as a photon of light
• each line in the emission spectrum corresponds to the difference in energy between two energy states
Tro, Chemistry: A Molecular Approach 57
Predicting the Spectrum of Hydrogen• the wavelengths of lines in the emission spectrum of
hydrogen can be predicted by calculating the difference in energy between any two states
• for an electron in energy state n, there are (n – 1) energy states it can transition to, therefore (n – 1) lines it can generate
• both the Bohr and Quantum Mechanical Models can predict these lines very accurately
2
182
18
electronhydrogen releasedphoton
1J 1018.21J 1018.2hch
EEEE
initialfinal
initialfinal
nn
58
Hydrogen Energy Transitions
m 1046.7
104466.2
1000.3 10626.6Ehc 6
J20-
sm8
sJ34
Solve:
Example 7.7- Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 6 to n = 5
E=hc/En = -2.18 x 10-18 J (1/n2)
Concept Plan:
Relationships:
ni = 6, nf = 5
mGiven:
Find:
2H
1REn
ni, nf Eatom Ephoton
Ech
Eatom = -Ephoton
J 104466.261
51J 1018.2E 20
2218
atom
Ephoton = -(-2.6644 x 10-20 J) = 2.6644 x 10-20 J
the unit is correct, the wavelength is in the infrared, which is appropriate because less energy than 4→2 (in the visible)
Check:
Tro, Chemistry: A Molecular Approach 60
Practice – Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 2 to n = 1
m 1021.1
1064.1
1000.3 10626.6Ehc 7
J18-
sm8
sJ34
Solve:
Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 2 to n = 1
E=hc/En = -2.18 x 10-18 J (1/n2)
Concept Plan:
Relationships:
ni = 2, nf = 1
mGiven:
Find:
2H
1REn
ni, nf Eatom Ephoton
Ech
Eatom = -Ephoton
J 1064.121
11J 1018.2E 18
2218
atom
Ephoton = -(-1.64 x 10-18 J) = 1.64 x 10-18 J
the unit is correct, the wavelength is in the UV, which is appropriate because more energy than 3→2 (in the visible)
Check:
62
Probability & Radial Distribution Functions
• 2 is the probability density the probability of finding an electron at a particular point in
space for s orbital maximum at the nucleus?decreases as you move away from the nucleus
• the Radial Distribution function represents the total probability at a certain distance from the nucleusmaximum at most probable radius
• nodes in the functions are where the probability drops to 0
Tro, Chemistry: A Molecular Approach 63
Probability Density Function
Tro, Chemistry: A Molecular Approach 64
Radial Distribution Function
Tro, Chemistry: A Molecular Approach 65
The Shapes of Atomic Orbitals• the l quantum number primarily determines the
shape of the orbital• l can have integer values from 0 to (n – 1)• each value of l is called by a particular letter
that designates the shape of the orbitals orbitals are sphericalp orbitals are like two balloons tied at the knotsd orbitals are mainly like 4 balloons tied at the knotf orbitals are mainly like 8 balloons tied at the knot
Tro, Chemistry: A Molecular Approach 66
l = 0, the s orbital• each principal energy
state has 1 s orbital• lowest energy orbital in a
principal energy state• spherical• number of nodes = (n – 1)
67
2s and 3s2s
n = 2,l = 0
3sn = 3,l = 0
Tro, Chemistry: A Molecular Approach 68
l = 1, p orbitals
• each principal energy state above n = 1 has 3 p orbitalsml = -1, 0, +1
• each of the 3 orbitals point along a different axispx, py, pz
• 2nd lowest energy orbitals in a principal energy state• two-lobed• node at the nucleus, total of n nodes
Tro, Chemistry: A Molecular Approach 69
p orbitals
Tro, Chemistry: A Molecular Approach 70
l = 2, d orbitals• each principal energy state above n = 2 has 5 d orbitals
ml = -2, -1, 0, +1, +2• 4 of the 5 orbitals are aligned in a different plane
the fifth is aligned with the z axis, dz squareddxy, dyz, dxz, dx squared – y squared
• 3rd lowest energy orbitals in a principal energy state• mainly 4-lobed
one is two-lobed with a toroid• planar nodes
higher principal levels also have spherical nodes
Tro, Chemistry: A Molecular Approach 71
d orbitals
Tro, Chemistry: A Molecular Approach 72
l = 3, f orbitals
• each principal energy state above n = 3 has 7 d orbitalsml = -3, -2, -1, 0, +1, +2, +3
• 4th lowest energy orbitals in a principal energy state• mainly 8-lobed
some 2-lobed with a toroid
• planar nodeshigher principal levels also have spherical nodes
Tro, Chemistry: A Molecular Approach 73
f orbitals
Tro, Chemistry: A Molecular Approach 74
Why are Atoms Spherical?
Tro, Chemistry: A Molecular Approach 75
Energy Shells and Subshells