Transcript Slide 1

Chapter 7
Quantum Theory and Atomic Structure
Electro-Magnetic radiation includes visible light,
microwave, TV, radio, x-ray, etc.
Radiation is a combination of vibrating electric and
magnetic fields in repeatable waveforms
Wavelength, λ (lambda): distance from crest to crest
Frequency, n (nu): # crests to pass a point in 1
second, units are #/s or Hz
Wave velocity = λ n
All E-M radiation in a vacuum has constant velocity
called the speed of light: c = 2.998 x 108 m/s.
Therefore c = λ n (Memorize formula and c)
Long λ => short n & vice-versa
The Wave Nature
of
Light
Figure 7.1
Frequency and
Wavelength
c=ln
Figure 7.2
Amplitude (intensity) of a wave.
Figure 7.3
Regions of the electromagnetic spectrum.
Frequency
unit is
#/s or s-1.
All waves travel at the same speed through vacuum
but differ in frequency and wavelength
Wavelength above in nanometers,
Frequency below in Hertz or #/second
Sample Problem 7.1a Interconverting Wavelength and Frequency
PROBLEM: A dental hygienist uses x-rays (l= 1.000A) to take a series of dental
radiographs. What is the frequency (in s-1) of the electromagnetic
radiation? (Assume that the radiation travels at the speed of light,
2.998 x 108 m/s.)
PLAN:
Use c = l n
wavelength in units given
o
1 A = 10-10 m
1 cm = 10-2 m
1 nm = 10-9 m
wavelength in m
n = c/l
frequency (s-1 or Hz)
SOLUTION:
o
-10
-10
1.000A 10 o m = 1.000 x 10 m
1A
2.998 x 108 m/s
n=
1.000 x 10-10 m
= 2.998 x 1018 s-1
E-M radiation was considered to be a wave/energy
phenomenon and not matter
Max Planck developed a new physics when classical
physics could not be used to interpret data from
blackbody radiation ( Blackbody is an object that
absorbs all radiation incident on it)
Blackbody radiation is emitted by solid bodies that
are heated to high T and become incandescent
Classical physics had to assume continuous
radiation, and it could not resolve the data that
there was discrete radiation
Planck developed theory of Packets of Energy called
quanta
The energy associated with quanta was proportional
to the frequency of the radiation:
E = hn
h = Planck’s constant 6.626 x 10-34 J.s
If c = ln, and E = hn, then with rearranging and
substituting:
E = hc/l
What is the energy of a photon with a
wavelength of 399.0 nm?
Figure 7.6
Blackbody radiation (4th ed.)
Solid heated
To 1000 K
It emits visible
light)
E=nhn
DE = Dn h n
DE = h n
when n = 1
smoldering coal
At 1500 K
electric heating element
At 2000K
light bulb filament
Figure 7.6
Demonstration of the
photoelectric effect.
Wave model could not explain photoelectric effect.
Flow of current when monochromatic light of suff
frequency falls on metal plate.
Photoelectric Effect: electrons are ejected from a metal's
surface if it is exposed to uv radiation
Each metal required a characteristic minimum uv
frequency to start ejecting e-s
Called Threshold freq, no
- As n increases more e-s ejected with higher vel (KE)
These data also defied classical physical explanation
Einstein reviewed data, recalled Planck's quanta
The "incident" radiation consists of quanta of energy, E =
hn, called photons(small bundle of electromagnetic
energy) - thus the PHOTOELECTRIC Effect
In order to eject an e-, a min KE is required, E = hno
If E>hno then excess KE is supplied to the e-, increasing
its velocity
For Na metal, no = 5.51 x 1014 Hz
Sample Problem 7.2
Calculating the Energy of Radiation from Its
Wavelength
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?
PLAN: After converting cm to m, we can use the energy equation, E = hn
combined with n = c/l to find the energy.
SOLUTION:
E=
E = hc/l
(6.626 x 10-34 J*s) 2.998 x 108 m/s
1.20 cm x
10-2 m
cm
= 1.66 x 10-23 J
Experiments with "excited" atoms of H produced emission
spectra
- always a discrete set of lines at certain wavelengths
White light dispersed by a prism or diffraction grating:
- we see ROYGBIV – a continuous spectrum from 750
nm to 400 nm
When a gas-filled tube is charged with current, only
certain EM l's are detected - called a line spectrum or
emission spectrum
The gas particles split into individual atoms
The e-s are excited by the current into a higher energy
level.
When they drop down, they emit energy of a certain λ,
with energy gaps at distinct intervals
Figure 7.7
The line
spectra of
several
elements.
Hydrogen atomic line spectra – also called
emission spectra –
- worked out mathematically (by several
scientists) to define the energy of the light
emitted & relationships between the lines
Balmer: red, green, blue, and violet lines
(656.3, 486.1, 434.0, 410.1nm)
1/λ = Ry(1/22 - 1/n2)
If n > 2 Ry = Rydberg constant
= 1.096776 x 107/m
If n = 3 get red, n = 4 get green, n = 5 get
blue, n = 6 get violet
Rydberg equation*
*Memorize!
Figure 7.8
1
l
=
R
1
n12
1
n22
R is the Rydberg constant = 1.096776x107 m-1
Three series of spectral lines of atomic hydrogen.
Balmer is in the visible region, and the other series,
which have names also, are in uv or ir area.
for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
1. H atoms have only certain allowable energy
levels called stationary states.
 2. Atom does not radiate energy while in a
stationary state.
 3. Atoms changes to another stationary state
by absorbing or emitting a photon.
Energy=EstateA-EstateB=hn

Bohr found En = - Rhc/n2
R = 1.097 x 107/m
h = 6.626 x 10-34 J.s
c = speed of light
Rhc = 2.178 x 10-18 J (since they are all constant)
Then En= -2.178 x 10-18 J/n2
All E is therefore < 0, and has discrete values only
Nucleus (proton) & e- are so far apart there's no
attraction anymore
Negative E is more stable than zero energy
n = 1 is the ground state, all above are excited
states
Figure 7.9
Quantum staircase.
Figure 7.10
The Bohr explanation of the three series of spectral lines.




Instrumental techniques used to obtain
information about atomic or molecular
energy levels
Emission: electrons in an atom are excited to
a higher energy state and then emit photons
as they return to lower energy states
Absorption: electrons in an atom absorb
photons of certain wavelengths and jump to
higher energy states; photons NOT absorbed
are observed!
See figure 7.11 in text: why chlorophyll looks
green
Figure B7.1
(4th ed.)
Flame tests.
strontium 38Sr
Figure B7.2
copper 29Cu
Emission and absorption spectra of sodium atoms.
A hydrogen atom has an e- excited up to level
4, and it drops back to level 2. (a) determine
the wavelength of the photon emitted and (b)
the energy difference.
(a) Use 1/l = Ry(1/22 – 1/42)
l = 4.8617 x 10-7 m
(b) Use DE = hc/l
E = 4.086 x 10-19 J
Follow-up: Answer the same questions for
the e- excited up to level 6.
Stationary wave:
- fixed at both ends
- has "nodes"
- never moves on those spots
with distance = length/2
Only certain λ's are possible for a standing
wave
Figure 7.12
Wave motion in
restricted systems.
Einstein remembered for E = mc2
m = E/c2 = (hc/λ)/c2 = h/λc
This appears to say that a photon of a certain
wavelength has mass!
Proved by Arthur Compton in 1922
E-M radiation is both waves & little packets of energy
and matter called photons
De Broglie 1923: if light has wave-particle duality,
then matter, which is particle-like, must also be
wavelike under certain conditions
Rearranged m = h/lc to get λ = h/mv
This is called the deBroglie wavelength
It means that all matter exhibits both particle and
wave properties
Sample Problem 7.4 Calculating the de Broglie Wavelength of an Electron
PROBLEM: Find the deBroglie wavelength of an electron with a speed of 1.00 x
106 m/s (electron mass = 9.11 x 10-31 kg; h = 6.626 x 10-34 J.s).
PLAN:
Knowing the mass and the speed of the electron allows to use the
equation l = h/mu to find the wavelength.
SOLUTION:
6.626 x 10-34 (kg*m2/s2)s
l=
= 7.27 x 10-10 m
9.11 x 10-31 kg x 1.00 x 106 m/s
Bohr’s Theory: 1 e- in H atom occupying
certain energy states - a certain quanta
Spherical orbitals around the nucleus
With de Broglie's hypothesis: e- must have a
certain λ to make a complete revolution like a standing wave
An integral # of complete λ's to fit the
sphere's circumference
Circumference = 2  r, therefore nλ = 2  r, n
= 1, 2, 3....
Figure 7.14
CLASSICAL THEORY
Matter
particulate,
massive
Energy
continuous,
wavelike
Summary of the major observations
and theories leading from classical
theory to quantum theory.
Since matter is discontinuous and particulate
perhaps energy is discontinuous and particulate.
Observation
blackbody radiation
Theory
Planck:
photoelectric effect
Energy is quantized; only certain values
allowed
Einstein: Light has particulate behavior (photons)
atomic line spectra
Bohr:
Energy of atoms is quantized; photon
emitted when electron changes orbit.
Figure 7.14 continued
Since energy is wavelike perhaps matter is wavelike
Observation
Davisson/Germer:
electron diffraction
by metal crystal
Theory
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
Compton: photon
wavelength increases
(momentum decreases)
after colliding with
electron
Theory
Einstein/deBroglie: Mass and energy are
equivalent; particles have wavelength and
photons have momentum.
QUANTUM THEORY
Energy same as Matter
particulate, massive, wavelike
It is impossible to know the exact position
andmomentum of a particle simultaneously.
Uncertainty: (Δx)(mΔv) > h/4
Δx is the location of the electron
mΔv is its momentum
More accurately we know the position of the
particle less accurately we know the speed.
Need Δ because we can’t know both at the
same time
Sample Problem 7.4 Applying the Uncertainty Principle
(4th ed)
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)?
PLAN:
The uncertainty (Dx) is given as ±1%(0.01) of 6 x 106 m/s. Once we
calculate this, plug it into the uncertainty equation.
SOLUTION:
Du = (0.01)(6 x 106 m/s) = 6 x 104 m/s
Dx * mDu ≥
h
4
Dx ≥
6.626 x 10-34 (kg*m2/s2).s
4 (9.11 x 10-31 kg)(6 x 104 m/s)
1 J = 1 kg*m2/s2
= 9.52 x 10-9 m
Schrodinger developed Wave Functions, Ψ(psi), where
Ψ2 is the probability of finding e- in a given space
Led to 4 quantum numbers that describe the e-'s
position in a complex equation:
1. Only certain wave functions are allowed
2. Each Ψn corresponds to an allowed energy for
e- in atom
3. Thus energy of e- is quantized
4. Ψ has no physical meaning, but Ψ2 give the
probability density
5. Allowed energy states are called orbitals
6. 3 integer #'s req'd to solve Ψ2 for 3-D space: n,
l, m l
An atomic orbital is specified by three quantum numbers.
n = principal quantum number, a positive integer = 1, 2, 3,...
- determines total E of e- in its electron shell
- gives measure of prob distance from nucleus (orbital
size)
- 2 or more e-s can be in same electron shell
l = angular momentum or shape = < n - 1 , = 0,1,2,...
- subshells w/in main shell, characterized by certain
wave shapes
0 = s, 1 = p, 2= d, 3 = f, etc.
ml = magnetic q.n. = +l, +l -1, +l - 2, … 0, ... -l
- specifies which orbital w/in a subshell e- is in
(later we’ll do ms = spin q.n., +½ or -½ for each e-)
Watch: YouTube - The Quantum Number Rag
Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol
(Property)
Allowed Values
Quantum Numbers
Principal, n
Positive integer
(size, energy)
(1, 2, 3, ...)
1
Angular
momentum, l
0 to n-1
(shape)
0
0
0
0
Magnetic, ml
-l,…,0,…,+l
(orientation)
2
3
1
0
1
2
0
-1 0 +1
-1 0 +1
-2
-1
0
+1 +2
Sample Problem 7.5
Determining Quantum Numbers for an Energy Level
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?
PLAN: Follow the rules for allowable quantum numbers found in the text.
l values can be integers from 0 to n-1; ml can be integers from -l
through 0 to + l.
SOLUTION: For n = 3, l = 0, 1, 2
For l = 0 ml = 0
For l = 1 ml = -1, 0, or +1
For l = 2 ml = -2, -1, 0, +1, or +2
There are 9 ml values and therefore 9 orbitals with n = 3.
Sample Problem 7.6
Determining Sublevel Names and Orbital Quantum
Numbers
PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals
for each sublevel with the following 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, we can find ml and the number of orbitals.
SOLUTION:
n
l
(a)
3
2
3d
-2, -1, 0, 1, 2
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
sublevel name possible ml values # of orbitals
Ψ21s – the 1s orbital is spherical.
Ψ22s – the 2s orbital has some density close to nucleus and then
another sphere farther away – a sphere within a sphere
Ψ22p – the 2p orbitals have no probabilty of e- at the nucleus called nodal plane
Can be oriented in 3 directions of 3-D graph - x, y, z.
2px, 2py, 2pz have the 3 ml “names” +1, 0 and -1
Ψ23d – the 3d orbitals have 5 ml values, and each has 2 nodal
surfaces, so they are in four sections. 3dxy, 3dxz, 3dyz, 3dx2y2, 3dz2
Ψ24f – the 4f orbitals have 7 ml values, 3 nodal surfaces
Figure 7.15
Electron probability in the
ground-state H atom.
1s
Figure 7.16 The 1s, 2s, and 3s orbitals
2s
3s
Figure 7.17
The 2p orbitals.
Figure 7.18
The 3d orbitals.
Figure 7.18 continued
Figure 7.19
One of the seven
possible 4f orbitals.


Be able to draw 1s, 2s, 2p, and 3d orbitals.
Practice now!