Download Quantum theory and atomic structure

Quantum theory and atomic
structure
The Wave Nature
of
Light
Frequency and
Wavelength
c=ln
Amplitude (intensity) of a wave.
Regions of the electromagnetic spectrum.
Interconverting Wavelength and Frequency
PROBLEM:
PLAN:
o
A dental hygienist uses x-rays (l= 1.00A) 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 s-1) of the electromagnetic radiation from each source?
(Assume that the radiation travels at the speed of light, 3.00x108 m/s.)
Use c = ln
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
SOLUTION:
o
-10
-10
1.00A 10 o m = 1.00x10 m
1A
= 3x1018 s-1
n=
-2
325 cm 10 m = 325x10-2 m
1 cm
n=
= 9.23x107 s-1
325x10-2 m
473nm
frequency (s-1 or Hz)
3x108 m/s = 6.34x1014 s-1
n=
473x10-9 m
Different
behaviors
of waves
and
particles.
The diffraction pattern caused by light
passing through two adjacent slits.
Blackbody radiation
E=nhn
DE = Dn h n
DE = h n
when n = 1
smoldering coal
electric heating element
light bulb filament
Hot metal gives both longer and shorter wavelenghts
compared to visible light.
Moreover the maximum intensity shifts toward lower
wavelenghts when temperature is increased. If that was
true as the temperature raises the light will decrease the
wavelenght and increases the energy (ultraviolet
catastrophe)
Planck explained this behaviour by assuming that light was
formed by photons bearing energy units
E = n hn n = positive integer
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 1.20cm. 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
6.626X10-34J*s
1.20cm
3x108m/s
10-2m
cm
= 1.66x10-23J
The line
spectra of
several
elements.
Rydberg equation
1
l
=
R
1
n12
1
n22
R is the Rydberg constant = 1.096776x107 m-1
Three series of spectral lines of atomic hydrogen.
for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
Bohr theory
• The quantum number n defines the
energies of the allowed orbits in the H
atom
• The energy of an electron in an orbit has a
negative value
• An atom in the lowest possible energy
level is said to be in the GROUND state.
State for the hydrogen atom with higher
energies are called EXCITED states
• Introduce the concept of quantization.
Quantum staircase.
En = Rhc/n2
The Bohr explanation of the three series of spectral lines.
DE = E final – E initial = - NARhc(1/n2final- 1/n2initial)
DE = hn
Flame tests.
strontium 38Sr
copper 29Cu
Emission and absorption spectra of sodium atoms.
The main components of a typical spectrophotometer.
Lenses/slits/collimaters
narrow and align beam.
Source produces radiation
in region of interest. Must
be stable and reproducible.
In most cases, the source
emits many wavelengths.
Sample in compartment
absorbs characteristic
amount of each incoming
wavelength.
Monochromator
(wavelength selector)
disperses incoming
radiation into continuum
of component
wavelengths that are
scanned or individually
selected.
Computer converts
signal into displayed
data.
Detector converts
transmitted radiation
into amplified electrical
signal.
Particle wave duality
• Photoelectric effect demonstrated that light,
usually considered to be a wave, can also have
the properties of particles , albeit without mass
• De Broglie proposed that a free electron having
a mass m moving with a velocity v should have
an associated wavelenght l, calculated by the
equation:
l = h/mv
l = h /mu
Table 7.1
The de Broglie Wavelengths of Several Objects
Substance
Mass (g)
Speed (m/s)
l (m)
slow electron
9x10-28
fast electron
9x10-28
5.9x106
1x10-10
alpha particle
6.6x10-24
1.5x107
7x10-15
one-gram mass
1.0
0.01
7x10-29
baseball
142
25.0
2x10-34
6.0x1027
3.0x104
4x10-63
Earth
1.0
7x10-4
Comparing the diffraction patterns of x-rays and electrons.
x-ray diffraction of aluminum foil
electron diffraction of aluminum foil
Wave motion in
restricted systems.
Standing waves
Standing waves features
• Standing waves can have points of zero
amplitude (called nodes)
• They can have only certain vibration
possible that is when L = n(l/2) where n =
an integer ( n= 1,2, 3…..
• Then, standing waves are quantized,
vibration are quantized and the integer n is
a quantum number.
• By defining electron as a standing wave,
quantization is introduced into the
description of electronic structure
• Schrodinger equation for an electron in a
three dimensional space requires three
quantum numbers : n, l, ml all whole
numbers.
• Each wave function is associated to an allowed
energy value. The energy is quantized because
only certain values of energy are possible for the
electron.
• The value of the wave function  at a given point
in space is the amplitude of the wave
• The square of the  defines the probability to
find the electron in a rewgion of the space.
• Schrodinger equation defines exactly the
energy of an electron . It does not define its
position
orbitals
• Define the region of the space within an
electron of a given energy is most likely to
be located
• With Schrodinger equation, we describe
the probability of a given electron to be
within a certain region in space.
Calculating the de Broglie Wavelength of an Electron
PROBLEM: Find the deBroglie wavelength of an electron with a speed of
1.00x106m/s (electron mass = 9.11x10-31kg; h = 6.626x10-34
kg*m2/s).
PLAN:
Knowing the mass and the speed of the electron allows to use the
equation l = h/mu to find the wavelength.
SOLUTION:
l=
6.626x10-34kg*m2/s
9.11x10-31kg
1.00x106m/s
= 7.27x10-10m
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 7.15 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
The Heisenberg Uncertainty Principle
Dx * mDu ≥
h
4p
Applying the Uncertainty Principle
PROBLEM: An electron moving near an atomic nucleus has a speed 6x106 ± 1%
m/s. What is the uncertainty in its position (Dx)?
PLAN:
The uncertainty (Dx) is given as ±1%(0.01) of 6x106m/s. Once we
calculate this, plug it into the uncertainty equation.
SOLUTION:
Du = (0.01)(6x106m/s) = 6x104m/s
Dx * mDu ≥
h
4p
Dx ≥
6.626x10-34kg*m2/s
4p (9.11x10-31kg)(6x104m/s)
= 9.52x10-9m
The Schrödinger Equation
HY = EY
wave function
mass of electron
potential energy at x,y,z
8p2mQ
d2Y
d2Y
d2Y
+
+
+
(E-V(x,y,z)Y(x,y,z) = 0
2
2
2
2
h
dx
dy
dz
how  changes in space
total quantized energy of
the atomic system
Electron probability in the
ground-state H atom.
Quantum Numbers and Atomic Orbitals
An atomic orbital is specified by three quantum numbers.
n the principal quantum number - a positive integer
l the angular momentum quantum number - an integer from 0 to n-1
ml the magnetic moment quantum number - an integer from -l to +l
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
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.
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
sublevel name possible ml values # of orbitals
(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
The 2p orbitals.
The 3d orbitals.
One of the seven
possible 4f orbitals.