Transcript ATOMIC STRUCTURE - Imperial Valley College

ATOMIC STRUCTURE
All Bold Numbered Problems
1
Chapter 7 Outline
• Events leading to Quantum Mechanics
–Newton
–Planck
–Einstein
–Bohr
–de Broglie
–Schrödinger
–Heisenberg
• Using Quantum Numbers
2
ATOMIC STRUCTURE
From the ERA of
Newtonian Physics to
Quantum Physics
ELECTROMAGNETIC
RADIATION
4
Electromagnetic Radiation
• Most subatomic particles behave as
PARTICLES and obey the physics of
waves.
• Define properties of waves
– Figure 7.1 and7.2.
• Wavelength, 
• Node
• Amplitude
5
Figures 7.1
Electromagnetic Frequency
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Electromagnetic Radiation
wavelength
Visible light
There are
no
Amplitude
LIMITS
to ...
Wavelength ()
Ultraviolet radiation
Node
there are
an  .
7
Electromagnetic Radiation
wavelengt
h
Visible light
Amplitude
wavelengt
h
Ultaviolet radiation
Nod
e
Node in a
standing wave
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Electromagnetic Radiation
• Waves have a frequency

• Use the Greek letter “nu”, , for frequency,
and units are “cycles per sec”
• All radiation:
• = c
where c = velocity of light = 3.00x108 m/sec
• Long wavelength ----> small frequency
• Short wavelength ----> high frequency
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Electromagnetic Radiation
Long wavelength -----> small frequency
Short wavelength -----> high frequency
increasing
frequency
increasing
wavelength
See Figure 7.3
10
Figure 7.3
Long wavelength -----> small frequency
Short wavelength -----> high frequency
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Electromagnetic Radiation
Red light has  = 700. nm.
Calculate the frequency.
1 x 10 -9 m
700 nm •
= 7.00 x 10-7 m
1 nm
Freq =
3.00 x 10 8 m/s
7.00 x 10
-7
m
= 4.29 x 1014 sec -1
Examples
12
Standing
Waves
1st vibration
1st vibration = ½
2nd vibration = 2(½)
3rd vibration = 3(½)
2nd vibration
See Figure 7.4
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Newtonian Physics Breakdown
-Quantization of EnergyIt was believed that like wave theory,
energy was also continuous.
Max Planck
(1858-1947)
Solved the
“ultraviolet
catastrophe”
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Figure 7.5
Intensity should
Increase with
Decreasing  . As you
add more energy,
atoms should vibrate
with a higher energy,
in a continuous
Objects
can gain or
fashion.
lose energy by
absorbing or
emitting radiant
energy in QUANTA.
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Quantization of Energy
Energy of a vibrating system (electromagnetic radiation) is proportional to
frequency.
Ep = h • 
h = Planck’s constant = 6.6262 x 10-34 J•s
We now MUST abandon the idea that
Energy acts as a continuous wave!
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From Planck on to Einstein
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Photoelectric Effect
A. Einstein (1879-1955)
• Experiment demonstrates the particle
nature of light. (Figure 7.6)
• Classical theory said that E of ejected
electron should increase with increase
in light intensity—not observed!
• No e- observed until light of a certain
minimum E is used &
• Number of e- ejected depends on light
intensity.
18
Photoelectric Effect
Experimental observations says that light
consists of particles called PHOTONS
having discrete energy.
• It takes a high energy particle to bump
into an atom to knock it’s electron out,
hence the use of a ½ mv2 term.
• It would take some minimum energy i.e.
critical energy to knock that electron
away from it’s atom.
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Energy of Radiation
PROBLEM: Calculate the energy of
1.00 mole of photons of red light.
 = 700. nm ( c =   )
 = 4.29 x 1014 sec-1
Ep = h•
= (6.63 x 10-34 J•s)(4.29 x 1014 sec-1)
= 2.85 x 10-19 J/photon
Notice Einstein's use of Planck's formula.
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Energy of Radiation
Energy of 1.00 mol of photons of red light.
Ep = h•
= (6.63 x 10-34 J•s)(4.29 x 1014 sec-1)
= 2.85 x 10-19 J per photon
E per mol =
(2.85 x 10-19 J/ph)(6.02 x 1023 ph/mol)
= 171.6 kJ/mol
This is in the range of energies that can break
bonds.
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Photoelectric Effect
A minimum frequency is required to cause any current
flow. Above that frequency, the current is related to
the intensity of the light used. The ejected electrons
(since we are talking about collisions between photons
and electrons) also have more kinetic energy when
higher frequencies are used.
EK =
1/2
meve2 = Einput - Eminimum
Einstein finds:
Ep = h• = 1/2 meve2, evidence that photons have both
wave/particle properties
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Photoelectric Effect
Light is used to eject an electron
from a metal. Calculate the velocity
of the ejected electron if the photon
used to eject the electron has a
wavelength of 2.35 x 10 -7 m and the
minimum frequency required to
eject an electron is 8.45 x 10 14 s-1.
Step by step!!
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The Final Crack in Classical,
Newtonian Physics
MONUMENTAL Edifice
• Planck---Energy is NOT Continuous like
waves
• Einstein---Energy comes in packets or is
Quantized and energy also has some
wave and particle behavior
• Bohr---Applies Quantized idea to atomic
particles….the H1 Atom to explain…..
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Atomic Line Spectra and
Niels Bohr
Bohr’s greatest contribution to
science was BUILDING a
SIMPLE MODEL of the ATOM.
Niels Bohr
(1885-1962)
It was based on an
understanding of the LINE
SPECTRA of excited atoms
and it’s relationship to
quantized energy.
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Line Spectra
of Excited Atoms
• Excited atoms emit light of only certain
wavelengths (Planck).
• The wavelengths of emitted light depend on the
element.
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Figure 7.7
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Figure 7.8
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Figure 7.9
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Line Spectra
of Excited Atoms
High E
Short 
High 
Low E
Long 
Low 
Visible lines in H atom spectrum are
called the BALMER series.
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Shells or Levels!!
Why??
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Figure
7.12
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Excited Atoms Emit Light
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Atomic Spectra and Bohr
One view of atomic structure in early 20th
century was that an electron (e-) traveled about
the nucleus in an orbit.
Electron
+
Orbit
1. Any orbit (like a wave-see slide 3) should be
possible and so should any energy.
2. But a charged particle would always be accelerating
from the nucleus (vector velocity is always
changing) and since it is moving in an electric field
would continuously emit energy.
End result should be destruction since the energy
mentioned in the previous step is finite!
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Atomic Spectra and Bohr
Bohr said classical (Newtonian) view is wrong!!!.
Need a new theory — now called QUANTUM or
WAVE MECHANICS.
e- can only exist in certain discrete orbits —
called stationary states.
e- is restricted to QUANTIZED energy states.
Energy of state, En = - C/n2
where n = quantum no. = 1, 2, 3, 4, ....
this describes the potential energy of an electron
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Atomic Spectra and Bohr
Energy of quantized state, En = - C/n2
• Only orbits where n = integral
numbers are permitted.
• Radius of allowed orbitals, Rn,
Rn= n2 R0 with Ro = 0.0529 nm
• Note the same equations come
from modern wave mechanics
approach.
• Results can be used to explain
atomic spectra.
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Atomic Spectra and Bohr
If e-’s are in quantized energy
states, then DE of states can
have only certain values. This
explain sharp line spectra.
2
n=2
2
n=1
E = -C ( 1/2 )
E = -C ( 1/1)
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Atomic
Spectra
and Bohr
E
N
E
R
G
Y
E = -C ( 1 / 2 2 )
E = -C ( 1 / 1 2 )
n=2
n=1
Calculate DE for e- “falling” from high energy
level (n = 2) to low energy level (n = 1).
DE = Efinal - Einitial = - C [ (1/1)2 - (1/2)2 ]
DE = - (3/4) C
Note that the process is exothermic!
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Atomic
Spectra
and Bohr
E
N
E
R
G
Y
E = -C ( 1 / 2 2 )
E = -C ( 1 / 1 2 )
n=2
n=1
DE = - (3/4)C
C has been found from experiment and is
proportional to RH, the Rydberg constant.
RHhc = C = 1312 kJ/mole.
 of emitted light = (3/4)C = 2.47 x 1015 sec-1
and  = c/ = 121.6 nm
This is exactly in agreement with experiment!
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Line Spectra
of Excited Atoms
DE = Efinal - Einitial = - RHhc [ (1/nfinal2) - (1/ninitial2) ]
A photon of light with frequency 8.02 x 1013 s-1
is emitted from a hydrogen atom when it deexcites from the n = 8 level to the n = ? level.
Calculate the final quantum number state of
the electron.
40
Atomic Line Spectra and
Niels Bohr
Niels Bohr
(1885-1962)
Bohr’s theory was a great
accomplishment.
Rec’d Nobel Prize, 1922
Problems with theory —
• theory only successful for H and
only 1e- systems He+, Li2+.
• introduced quantum idea
artificially.
• However, Bohr’s model does not
explain many e- systems….So, we
go on to QUANTUM or WAVE
MECHANICS
41
Quantum or Wave Mechanics
de Broglie (1924) proposed
that all moving objects
have wave properties.
For light: E = mc2
E = h = hc / 
L. de Broglie
(1892-1987)
h
 =
mv
Therefore, mc = h / 
and for particles
(mass)(velocity) = h /  ,
the wave-nature of matter.
Quantum or Wave
Mechanics
Baseball (115 g) at 100 mph
 = 1.3 x 10-32 cm
e- with velocity =
1.9 x 108 cm/sec
 = 0.388 nm
Experimental proof
of wave properties
of electrons
43
Quantum or Wave Mechanics
Schrödinger applied
idea of e- behaving as
a wave to the problem
of electrons in atoms.
He developed the
E. Schrödinger
1887-1961
WAVE EQUATION.
44
Quantum or Wave Mechanics
Solution of the wave
equation give a set of
mathematical
expressions called
WAVE FUNCTIONS, .
E. Schrodinger
1887-1961
Each describes an
allowed energy state of
an e-.
Quantization is
introduced naturally.
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WAVE FUNCTIONS, 
 is a function of distance and two angles.
Each  corresponds to an ORBITAL — the region of
space within which an electron is found.
 does NOT describe the exact location of the
electron.
2
 is proportional to the probability of finding an e- at
a given point.
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Uncertainty Principle
W. Heisenberg
1901-1976
Problem of defining nature
of electrons in atoms
solved by W. Heisenberg.
Cannot simultaneously
define the position and
momentum (= m•v) of an
electron.
We define e- energy exactly
but accept limitation that
we do not know exact
position.
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QUANTUM NUMBERS
Each orbital is a function of 3 quantum
numbers:
n, l, and ml
Electrons are arranged in shells(levels)
and subshells(sublevels).
n
l
ml
--> shell
--> subshell
--> designates an orbital within a subshell
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QUANTUM NUMBERS
Symbol
Values
Description
n (major)
1, 2, 3, ..
l (angular)
0, 1, 2, .. n-1
Orbital shape
or type
(subshell)
ml (magnetic)
- l..0..+ l
Orbital
orientation
Orbital size
and energy
where E = - RHhc(1/n2)
# of orbitals in subshell = 2 l + 1
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All 4 Quantum Numbers
•
•
•
•
Principle quantum number (n)
Azimuthal quantum number (l)
Magnetic quantum number (m)
Spin quantum number (s)
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Atomic Orbitals the result of
Quantum Mechanics Calculations
51
Shells and Subshells
When n = 1, then l = 0 and ml = 0 .
Therefore, if n = 1, there is 1 type of
subshell and that subshell has a
single orbital.
(ml has a single value ---> 1 orbital)
This subshell is labeled s
Each shell has 1 orbital labeled s, and it
is SPHERICAL in shape.
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1s Orbital
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2s Orbital
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3s Orbital
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Atomic Orbitals the result of
Quantum Mechanics Calculations
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p Orbitals
When n = 2, then l = 0 and 1
Therefore, in the n = 2 shell
there are 2 types of
orbitals — 2 subshells
For l = 0 ml = 0
this is an
For l = 1
s subshell
ml = -1, 0, +1
this is a p subshell
with 3 orbitals
Typical p orbital
planar node
When l = 1, there is
a
PLANAR NODE
thru
the nucleus.
See Figure 7.16
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59
p Orbitals
A p orbital
The three p
orbitals lie 90o
apart in space
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2px Orbital
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2py Orbital
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2pz Orbital
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3px Orbital
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3py Orbital
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3pz Orbital
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d Orbitals
When n = 3, what are the values of l?
l = 0, 1, 2
and so there are 3 subshells in the shell.
For l = 0, ml = 0
---> s subshell with a single orbital
For l = 1, ml = -1, 0, +1
--->
p subshell with 3 orbitals
For l = 2, ml = -2, -1, 0, +1, +2
--->
d subshell with 5 orbitals
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d Orbitals
s orbitals have no planar
node (l = 0) and so are
spherical.
p orbitals have l = 1, and
have 1 planar node,
and so are “dumbbell”
shaped.
typical d orbital
planar node
planar node
See Figure 7.16
This means d orbitals,
( l = 2) have
2 planar nodes
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3dxy Orbital
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3dxz Orbital
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3dyz Orbital
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2
3dz Orbital
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2
2
3dx - y
Orbital
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f Orbitals
When n = 4, l = 0, 1, 2, 3 so there are 4
subshells in the shell.
For l = 0, ml = 0
---> s subshell with single orbital
For l = 1, ml = -1, 0, +1
---> p subshell with 3 orbitals
For l = 2, ml = -2, -1, 0, +1, +2
---> d subshell with 5 orbitals
For l = 3, ml = -3, -2, -1, 0, +1, +2, +3
---> f subshell with 7 orbitals
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Orbitals and Quantum Numbers
n
l
ml
1
0
0
1s
2
2
2
2
0
1
1
1
0
1
0
-1
2s
2p
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Orbitals and Quantum Numbers
n
l
ml
3
3
3
3
0
1
1
1
0
1
0
-1
3
3
3
3
3
2
2
2
2
2
2
1
0
-1
-2
3s
3p
3d
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Sample Problems
No 1. Is it possible to have a d orbital in level 1?
Yes 2. Is it possible to have a 6s subshell?
One 3. How many orbitals are in a 7s sublevel?
9 4. How many orbitals are possible if n = 3?
5. What type of orbital has the quantum
numbers
5d
a) n = 5, l = 2, ml = 1
3d
b) n = 3, l = 2, ml =-1
6f
c) n = 6, l = 3, ml = -3
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Practice Problems
1. Calculate the wavelength of a photon
having an energy of 2.58 x 10-18 J.
2. In the hydrogen atom, which transition,
3 --> 2 or 2 --> 1, has the longer
wavelength?
3. Calculate the wavelength of an
object (mass = 545 lbs) with a speed of
45 miles/hour.
4. Give all possible sets of quantum
numbers for 4p, 3d, and 5s.
5. How many orbitals are in the
a. the third level?
b. l = 3 sublevel?
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Practice Problems Answers
1. 7.71 x 10-8
2. 3 --> 2
3. 1.3 x 10-37 m
5. a) 9
4. 4p
n
l
b) 7
ml
4
1
1
4
1
0
4
1
-1
Problem 4 continued on next slide.
79
Practice Problems Answers
3d
5s
n
l
ml
3
3
3
3
3
2
2
2
2
2
2
1
0
-1
-2
5
0
0
80
Sample Problem
1. Calculate the frequency of light having a
wavelength of 1 x 10-7m.
• = c
1 x 10-7m .  = 3.00 x 108 m/s
 = 3 x 1015 /s
81
Sample Problem
2. Calculate the wavelength of light having a
frequency of 1.5 x 108 hz.
• = c
. 1.5 x 108 /s = 3.00 x 108 m/s
 = 2.0 m
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Sample Problem
3. Calculate the frequency of light having a
wavelength of 1 x 103nm.
• = c
1 x 10-6m .  = 3.00 x 108 m/s
 = 3 x 1014 /s
83
Practice Problem
1. Calculate the energy of a photon having a
frequency of 3 x 1015/s.
Ep = h•
Ep = 6.63 x 10-34 Js • 3 x 1015/s
= 2 x 10-18 J
84
Practice Problem
2. Calculate the frequency of light having an
energy of 2.0 x 105 J/mole.
Ep = h•
2.0 X 105 J mole
mole
6.02 x 1023 photon
3.3 x 10-19 J = 6.63 x 10-34 Js • 
 = 5.0 x 1014 /s
85
Practice Problem
3. Calculate the energy of a photon with a
wavelength of 575 nm.
• = c
5.75 x 10-7 m •  = 3.00 x 108 m/s
 = 5.22 x 1014/s
Ep = h•
Ep = 6.63 x 10-34 Js • 5.22 x 1014/s
= 3.46 x 10-19 J
86
Calculate the energy of the photon:
Photon wavelength = 2.35 x 10 -7 m
• = c
2.35 x 10-7 m •  = 3.00 x 108 m/s
 = 1.28 x 1015/s
Ep = h•
Ep = 6.63 x 10-34 Js • 1.28 x 1015/s
= 8.49 x 10-19 J
87
Calculate the min. energy to eject an electron:
Min.  = 8.45 x 10 14 s-1.
Ep = h•
Ep = 6.63 x 10-34 Js • 8.45 x 1014/s
= 5.60 x 10-19 J
88
Calculate the extra energy of the electron:
8.49 x 10-19 J - 5.60 x 10-19 J = 2.89 x 10-19 J
Calculate the velocity of the electron:
E = 1/2 m v2
2.89 x 10-19 J = (1/2) 9.11 x 10-31 kg v2
= 7.96 x 105 m/s
89
Calculate the energy of the photon:
Ep = h•
Ep = 6.63 x 10-34 Js • 8.02 x 1013/s
5.32 X 10-20 J 6.02 x 1023 photon
photon
mole
= 3.20 x 104 J = 32.0 kJ
90
Calculate the level number :
Ep = h•
DE = -C [(1/n)2 - (1/n)2]
-32. kJ = -1312 kJ [(1/n)2 - (1/8)2]
n=5
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