Chapter 7. Quantum Theory and Atomic Structure. Quantum Theory and Atomic Structure. 7.1 The Nature of Light. 7.2 Atomic Spectra. 7.3 The WaveParticle Duality of Matter and Energy. 7.4 The QuantumMechanical Model of the Atom. The Wave Nature of Light.
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Chapter 7
Quantum Theory and Atomic Structure
Quantum Theory and Atomic Structure
7.1 The Nature of Light
7.2Atomic Spectra
7.3 The WaveParticle Duality of Matter and Energy
7.4The QuantumMechanical Model of the Atom
The Wave Nature of Light
Wave properties are described by two interdependent variables:
Frequency: n(nu) = the number of cycles the wave undergoes
per second (units of s1 or hertz (Hz)) (cycles/s)
Wavelength: l (lambda) = the distance between any point on a
wave and a corresponding point on the next wave (the distance
the wave travels during one cycle) (units of meters (m), nanometers
(109 m), picometers (1012 m) or angstroms (Å, 1010 m) per cycle)
(m/cycle)
Speed of a wave = cycles/s x m/cycle = m/s
c= speed of light in a vacuum = nl =3.00 x 108 m/s
Figure 7.1
Frequency and Wavelength
Amplitude (intensity)
of a Wave
(a measure of the
strength of the wave’s
electric and magnetic
fields)
Figure 7.2
Regions of the Electromagnetic Spectrum
Figure 7.3
Light waves all travel at the same speed through a vacuum but
differ in frequency and, therefore, in wavelength.
PROBLEM:
A dental hygienist uses xrays (l = 1.00 Å) to take a series of dental radiographs while the patient listens to a radio station (l = 325 cm) and looks out the window at the blue sky (l = 473 nm). What is the frequency (in s1) of the electromagnetic radiation from each source? Assume that the radiation travels at the speed of light, 3.00 x 108m/s.
1.00 Å
1010m
1 Å
325 cm
102m
1 cm
109m
473 nm
1 nm
3.00 x 108m/s
473 x 109m
Sample Problem 7.1
Interconverting Wavelength and Frequency
PLAN:
Use c = ln
x
= 1.00 x 1010m
3.00 x 108m/s
SOLUTION:
n =
= 3 x 1018s1
1.00 x 1010m
x
= 325 x 102m
3.00 x 108m/s
n =
= 9.23 x 107s1
325 x 102m
= 473 x 109m
x
n =
= 6.34 x 1014s1
Distinguishing Between a Wave and a Particle
Refraction: the change in the speed of a wave when it
passes from one transparent medium to another
Diffraction: the bending of a wave when it strikes the
edge of an object, as when it passes through a slit;
an interference pattern develops if the wave passes
through two adjacent slits
Different behaviors of waves and particles
Figure 7.4
The diffraction pattern caused by light passing through two adjacent slits
Figure 7.5
Blackbody Radiation
Changes in the intensity and
wavelength of emitted light when
an object is heated at different
temperatures
Figure 7.6
Planck’s equation
E = nhn
E = energy of radiation
h = Planck’s constant
n = frequency
n = positive integer (a
quantum number)
h = 6.626 x 1034 J.s
The Notion of Quantized Energy
If an atom can emit only certain quantities of energy, then the atom can have only certain quantities of energy. Thus, the energy of an atom is quantized.
Each energy packet is called a quantum and has
energy equal to hn.
An atom changes its energy state by absorbing or
emitting one or more quanta of energy.
DEatom = Eemitted(or absorbed) radiation= Dnhn
DE = hn (Dn = 1)energy change between adjacent energy states
Demonstration of the photoelectric effect
Wave model could not explain the
(a) presence of a threshold frequency,
and (b) the absence of a time lag.
Led to Einstein’s photon theory of
light:
Ephoton = hn = DEatom
Figure 7.7
PROBLEM:
A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20 cm. What is the energy of one photon of this microwave radiation?
102m
cm
Calculating the Energy of Radiation from its Wavelength
Sample Problem 7.2
PLAN:
After converting cm to m, we use the energy equation, E = hn combined with n = c/l to find the energy.
SOLUTION:
E = hc/l
(6.626 x 1034J.s)
x
(3.00 x 108m/s)
E =
= 1.66 x 1023J
1.20 cm
x
Atomic Spectra
Line spectra of several elements
Figure 7.8
1
1
1
l
n12
n22
=
R

Rydberg equation:
R is the Rydberg constant = 1.096776 x 107 m1
Three series of spectral lines of atomic hydrogen
for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
Figure 7.9
The Bohr Model of the Hydrogen Atom
Postulates:
1. The H atom has only certain allowable energy levels
2. The atom does not radiate energy while in one of its
stationary states
3. The atom changes to another stationary state
(i.e., the electron moves to another orbit) only by
absorbing or emitting a photon whose energy equals
the difference in energy between the two states
The quantum number is associated with the radius of an electron
orbit; the lower the n value, the smaller the radius of the orbit and
the lower the energy level.
ground state and excited state
Quantum staircase
Figure 7.10
Limitations of the Bohr Model
Can only predict spectral lines for the hydrogen atom
(a one electron model)
For systems having >1 electron, there are additional
nucleuselectron attractions and electronelectron
repulsions
Electrons do not travel in fixed orbits
The Bohr explanation of the three series of spectral lines for atomic hydrogen
Figure 7.11
Figure B7.1
Flame tests
strontium 38Sr
copper 29Cu
Emission and absorption spectra of sodium atoms
Figure B7.2
Fireworks emissions
from compounds
containing specific
elements
Samplein compartment absorbs characteristic amount of each incoming wavelength.
Computer converts signal into displayed data.
Lenses/slits/collimaters narrow and align beam.
Monochromator (wavelength selector) disperses incoming radiation into continuum of component wavelengths that are scanned or individually selected.
Sourceproduces radiation in region of interest. Must be stable and reproducible. In most cases, the source emits many wavelengths.
Detector converts transmitted radiation into amplified electrical signal.
Figure B7.3
The main components of a typical spectrophotometer
The Absorption
Spectrum of
Chlorophyll a
Absorbs red and
blue wavelengths
but no green and
yellow wavelengths;
leaf appears green.
The WaveParticle Duality of Matter and Energy
De Broglie: If energy is particlelike, perhaps matter is
wavelike!
Electrons have wavelike motion and are restricted to orbits
of fixed radius; explains why they will have only certain possible
frequencies and energies.
Wave motion in restricted systems
Figure 7.13
h
/mu
l
=
The de Broglie Wavelength
Table 7.1 The de Broglie Wavelengths of Several Objects
Substance
Mass (g)
Speed (m/s)
l (m)
slow electron
9 x 1028
1.0
7 x 104
fast electron
9 x 1028
5.9 x 106
1 x 1010
alpha particle
6.6 x 1024
1.5 x 107
7 x 1015
onegram mass
1.0
0.01
7 x 1029
baseball
142
25.0
2 x 1034
Earth
6.0 x 1027
3.0 x 104
4 x 1063
PROBLEM:
Find the deBroglie wavelength of an electron with a speed of 1.00 x 106m/s (electron mass = 9.11 x 1031kg; h = 6.626 x 1034kg.m2/s).
Calculating the de Broglie Wavelength of an Electron
Sample Problem 7.3
PLAN:
Knowing the mass and the speed of the electron allows use of the equation, l = h/mu, to find the wavelength.
SOLUTION:
6.626 x 1034kg.m2/s
= 7.27 x 1010m
l =
9.11 x 1031kg
x
1.00 x 106m/s
Comparing the diffraction patterns of xrays and electrons
Figure 7.14
Electrons  particles with mass and charge  create diffraction
patterns in a manner similar to electromagnetic waves!
Since matter is discontinuous and particulate, perhaps energy is discontinuous and particulate.
blackbody radiation
Planck: Energy is quantized; only certain values
are allowed
photoelectric effect
Einstein: Light has particulate behavior (photons)
atomic line spectra
Bohr: Energy of atoms is quantized; photon
emitted when electron changes orbit.
Figure 7.15
CLASSICAL THEORY
Matter particulate, massive
Energy continuous, wavelike
Summary of the major observations and theories leading from classical theory to quantum theory
Observation
Theory
Figure 7.15 (continued)
Since energy is wavelike, perhaps matter is wavelike.
Davisson/Germer:
electron diffraction
by metal crystal
deBroglie: All matter travels in waves; energy of
atom is quantized due to wave motion of
electrons
Since matter has mass, perhaps energy has mass
Observation
Theory
Compton: photon
wavelength increases (momentum decreases) after colliding with electron
Einstein/deBroglie: Mass and energy are
equivalent; particles have wavelength and
photons have momentum.
QUANTUM THEORY
Energy same as Matter:
particulate, massive, wavelike
Observation
Theory
h
4p
The Heisenberg Uncertainty Principle
It is impossible to know simultaneously the exact
position and momentum of a particle
DxxmDu≥
Dx = the uncertainty in position
Du = the uncertainty in speed
A smaller Dx dictates a larger Du, and vice versa.
Implication: cannot assign fixed paths for electrons; can know
the probability of finding an electron in a given region of space
PROBLEM:
An electron moving near an atomic nucleus has a speed 6 x 106 ± 1% m/s. What is the uncertainty in its position (Dx)?
h
D xx mD u ≥
4p
Sample Problem 7.4
Applying the Uncertainty Principle
PLAN:
The uncertainty in the speed (Du) is given as ±1% (0.01) of 6 x 106m/s. Once we calculate this value, the uncertainty equation is used to calculate Dx.
SOLUTION:
Du = (0.01)(6 x 106m/s) = 6 x 104m/s (the uncertainty in speed)
6.626 x 1034kg.m2/s
≥ 1 x 109m
Dx ≥
4p (9.11 x 1031kg)(6 x 104m/s)
d2Y
d2Y
d2Y
wave function
mass of electron
potential energy at x,y,z
dx2
dy2
dz2
8p2mQ
+
+
+
(EV(x,y,z)Y(x,y,z) = 0
h2
how y changes in space
total quantized energy of the atomic system
The Schrödinger Equation
HY = EY
Each solution to this equation is associated with a given
wave function, also called an atomic orbital
Electron probability in the groundstate hydrogen atom
Figure 7.16
Quantum Numbers and Atomic Orbitals
An atomic orbital is specified by three quantum numbers.
nthe principal quantum number  a positive integer (energy level)
lthe angular momentum quantum number  an integer from 0 to (n1)
mlthe magnetic moment quantum number  an integer from l to +l
0
1
2
0
1
0
+1
1
0
+1
2
1
0
+1
+2
Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol
(Property)
Allowed Values
Quantum Numbers
Principal, n (size, energy)
Positive integer (1, 2, 3, ...)
1
2
3
Angular momentum, l (shape)
0 to n1
0
0
1
0
0
Magnetic, ml (orientation)
l,…,0,…,+l
PROBLEM:
What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3?
Anthony S. Serianni:
Sample Problem 7.5
Determining Quantum Numbers for an Energy Level
PLAN:
Follow the rules for allowable quantum numbers.
l values can be integers from 0 to (n1); ml can be integers from l through 0 to + l.
SOLUTION:
For n = 3, l = 0, 1, 2
For l = 0 ml = 0 (s sublevel)
For l = 1 ml = 1, 0, or +1 (p sublevel)
For l = 2 ml = 2, 1, 0, +1, or +2 (d sublevel)
There are 9 mlvalues and therefore 9 orbitals with n = 3
PROBLEM:
Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers:
Sample Problem 7.6
Determining Sublevel Names and Orbital Quantum Numbers
(a) n = 3, l = 2
(b) n = 2, l = 0
(c) n = 5, l = 1
(d) n = 4, l = 3
PLAN:
Combine the n value and l designation to name the sublevel. Knowing l, find ml and the number of orbitals.
SOLUTION:
n
l
sublevel name
possible ml values
no. orbitals
(a)
2
3d
2, 1, 0, 1, 2
3
5
(b)
2
0
2s
0
1
(c)
5
1
5p
1, 0, 1
3
(d)
4
3
4f
3, 2, 1, 0, 1, 2, 3
7
S orbitals
1s
3s
2s
spherical
nodes
Figure 7.17
2p orbitals
Figure 7.18
nodal planes
3d orbitals
Figure 7.19
perpendicular nodal planes
Figure 7.19 (continued)
One of the seven possible 4f orbitals
Figure 7.20
The energy levels
in the hydrogen atom
The energy level
depends only on the
n value of the orbital
Figure 7.21