Transcript Chem-130 Test Lecture

Chemistry-140
Lecture 17
Chapter 7: Atomic Structure
 Chapter Highlights
 electromagnetic radiation
 photons & Planck’s constant
 Bohr model of the atom
 Rydberg equation

quantum mechanics

orbitals

Heisenberg uncertainty principle

quantum numbers
Chemistry-140
Lecture 17
Electronic Structure & Electromagnetic Radiation
 Electronic structure of an atom: detailed description of the
arrangement of electrons in the atom
 Electromagnetic radiation: electrical and magnetic waves
traveling at 2.9979 x 108 m/s (speed of light, c). Includes
visible light, radio waves, microwaves, infrared
(heat),ultraviolet, X-ray, and g-ray radiation….
Chemistry-140
Lecture 17
Electromagnetic Radiation
 Wavelength, l: distance between two successive peaks or
troughs of a wave. Units are length (m).
 Frequency, n: number of complete waveforms that pass
through a point in one second. Units of s-1, /s, or hertz (Hz).
Relationship:
speed of light = (wavelength) x (frequency)
c = ln
Chemistry-140
Lecture 17
Chemistry-140
Lecture 17
Electromagnetic Radiation
Question:
Yellow light of a sodium vapour lamp has a wavelength
of 589 nm. What is the frequency of this light?
Chemistry-140
Lecture 17
Answer:
Since we know:
c = ln
then
 3.00 x 10 8 m / s   1 nm 
c
  -9 
n =
=
589 nm

  10 m 
l
= 5.09 x 1014 s-1
Chemistry-140
Lecture 17
Quantum Effects & Photons
 Max Planck proposed that radiation is not continuous, but
rather consists of small pieces known as quanta (a quantum).
Frequencies, n, of these quanta were whole-number
multiples of a fundamental frequency.
Energies
where
E = hn, 2hn, 3hn,...
h = Planck's constant:
h = 6.626 x 10-34 J-s.
and
E = hn
Chemistry-140
Lecture 17
Quantum Effects & Photons
Question:
A laser emits light energy in short pulses with frequency
4.69 x 1014 Hz and deposits 1.3 x 10-2 J for each pulse.
How many quanta of energy does each pulse deposit ?
Chemistry-140
Lecture 17
Answer:
Step 1: Determine the energy of one quantum (photon).
E = hn
= (6.63 x 10-34 J-s) (4.69 x 1014 s-1)
= 3.11 x 10-19 J
Chemistry-140
Lecture 17
Step 2: Determine how many quanta are in a laser pulse.
 1.3 x 10-2 J 

Number = 
-19
 3.11 x 10 J 
=
4.18 x 1016 quanta
Chemistry-140
Lecture 17
Photoelectric Effect
 Photoelectric effect: metallic surfaces produce electricity
(electrons are ejected) when exposed to light.
There is a minimum frequency below which
no electricity is produced.
 Above the minimum frequency:
 i) number of electrons ejected depends only on
light intensity,
 ii) energy of the ejected electrons depends only
on the frequency of the light.
Chemistry-140
Lecture 17
Chemistry-140
Lecture 17
Photoelectric Effect
The packet of energy sufficient to eject an electron is called a
photon. The kinetic energy EK of the electrons is given by
EK = EP - EB
EB = binding energy
EP = photon energy = hn
Chemistry-140
Lecture 17
Photoelectric Effect
Question:
Potassium metal must absorb radiation with a
minimum frequency of 5.57 x 1014 Hz before it can emit
electrons from its surface via the photoelectric effect. If
K(s) is irradiated with light of wavelength 510 nm, what is
the maximum possible velocity of an emitted electron?
Chemistry-140
Lecture 17
Answer:
Step 1: Convert threshold frequency to binding energy.
Eb = hn
= (6.63 x 10-34 J-s) (5.57 x 1014 s-1)
= 3.69 x 10-19 J
Chemistry-140
Lecture 17
Step 2: Determine the photon energy of 510 nm light.
 hc 
EP =  
 l
 (6.63 x 10-34 J s) (3.00 x 108 m / s) 

= 
7
5.10 x 10 m


= 3.90 x 10-19 J
Chemistry-140
Lecture 17
Step 3: Determine the kinetic energy of the emitted electrons
EK = EP - EB
= (3.90 x 10-19 J) - (3.69 x 10-19 J)
= 2.10 x 10-20 J
Chemistry-140
Lecture 17
Step 4: Calculate the velocity of the emitted electrons
EK
1
= mv2 = 2.10 x 10-20 J
2
v =
2 2.10 x 1020 J 
9.11 x 1031 kg
= 2.15 x 105 m/s
Chemistry-140
Lecture 17
Bohr’s Model of the Hydrogen Atom
 A spectrum is produced when radiation from a source is
separated into its component wavelengths.
 Bohr used Planck's quantum theory to interpret the line
spectrum of hydrogen.
 Bohr's model of the hydrogen atom described a nucleus
surrounded orbits of fixed (quantized) radius,
numbered n = 1, 2, 3,...
Chemistry-140
Lecture 17
Chemistry-140
Lecture 17
Chemistry-140
Lecture 17
Bohr’s Model of the Hydrogen Atom
 Bohr concluded:
 the energy of the electron in an orbit of hydrogen is quantized
 the energy difference between two orbits must also be
quantized
 The frequency of a line in the spectrum corresponds to the
energy difference between two orbits;
DE = hn
Chemistry-140
Lecture 17
Bohr’s Model of the Hydrogen Atom
 The energy of a Bohr orbit (and an electron in it) is given by
1
En = -RH 2
n
where RH is the Rydberg constant = 2.179 x 10-18 J
Chemistry-140
Lecture 17
Transitions in the Bohr Hydrogen Atom
Chemistry-140
Lecture 17
Transitions and the Rydberg Equation
 An electron in the lowest energy orbit, n = 1, is in the
ground state
 An electron in any orbit other than n = 1, is in an
excited state
 The energy of a line is the difference in the energies of the
two orbits involved in the transition
DE = Efinal - Einitial
1
DE = hn = RH  2
ni
-
1
nf 2

Chemistry-140
Lecture 17
Transitions in the Bohr Hydrogen Atom
 The radius of a Bohr orbit is given by:
r = n2(5.30 x 10-11 m)
 The ionization energy of hydrogen is the energy required to
remove the electron from the atom, that is;
the energy of the n = 1 to n =  transition
Chemistry-140
Lecture 17
Transitions in the Bohr Hydrogen Atom
Question (similar to example 7.4):
Calculate the wavelength of light that corresponds to
the transition of the electron from the n = 4 to the n = 2
state of the hydrogen atom. Is the light absorbed or
emitted by the atom?
Chemistry-140
Lecture 17
Answer:
Step 1: Use the Rydberg equation with ni = 4 and nf = 2.
1
DE = hn = RH  2
ni
-
 RH  1
1

n =

 h  ni 2 n f 2
1
nf 2

 2.18 x 10-18 J 

=
 34
 6.63 x 10 J s 
= -6.17 x 1014 s-1

 41
2
-
1
22

Chemistry-140
Lecture 17
Step 2: Convert to wavelength of light
c = ln
c
l =
n
 3.00 x 10 8 m / s 
= 
14 -1 
 6.17 x 10 s 
= 4.86 x 10-7 m
= 486 nm
Chemistry-140
Lecture 18
Chapter 7: Atomic Structure
 Chapter Highlights

electromagnetic radiation

photons & Planck’s constant

Bohr model of the atom
 Rydberg equation
 quantum mechanics
 Heisenberg uncertainty principle

orbitals

quantum numbers
Chemistry-140
Lecture 18
Bohr’s Model of the Hydrogen Atom
 The energy of a Bohr orbit (and an electron in it) is given by
1
En = -RH 2
n
where RH is the Rydberg constant = 2.179 x 10-18 J
Chemistry-140
Lecture 18
Transitions and the Rydberg Equation
 An electron in the lowest energy orbit, n = 1, is in the
ground state
 An electron in any orbit other than n = 1, is in an
excited state
 The energy of a line is the difference in the energies of the
two orbits involved in the transition
DE = Efinal - Einitial
1
DE = hn = RH  2
ni
-
1
nf 2

Chemistry-140
Lecture 18
Transitions in the Bohr Hydrogen Atom
Chemistry-140
Lecture 18
Transitions in the Bohr Hydrogen Atom
 The radius of a Bohr orbit is given by:
r = n2(5.30 x 10-11 m)
 The ionization energy of hydrogen is the energy required to
remove the electron from the atom, that is;
the energy of the n = 1 to n =  transition
Chemistry-140
Lecture 18
Transitions in the Bohr Hydrogen Atom
Question:
Calculate the wavelength of light that corresponds to
the transition of the electron from the n = 4 to the n = 2
state of the hydrogen atom. Is the light absorbed or
emitted by the atom?
Chemistry-140
Lecture 18
Answer:
Step 1: Use the Rydberg equation with ni = 4 and nf = 2.
1
DE = hn = RH  2
ni
-
 RH  1
1

n =

 h  ni 2 n f 2
1
nf 2

 2.18 x 10-18 J 

=
 34
 6.63 x 10 J s 
= -6.17 x 1014 s-1

 41
2
-
1
22

Chemistry-140
Lecture 18
Step 2: Convert to wavelength of light
c = ln
c
l =
n
 3.00 x 10 8 m / s 
= 
14 -1 
 6.17 x 10 s 
= 4.86 x 10-7 m
= 486 nm
Chemistry-140
Lecture 18
Dual Nature of the Electron
 De Broglie proposed that particles may behave as if they
were waves. Similar to the idea that light may behave as if it
was a particle. Matter wave is the term used by de Broglie
where:
h
l =
mv
where momentum = mv,
(m is mass & v is velocity)
Chemistry-140
Lecture 18
Dual Nature of the Electron
Question:
What is the characteristic wavelength of an electron with
velocity 5.97 x 106 m/s ? (mass of an electron is 9.11 x 10-28 g)
Chemistry-140
Lecture 18
Answer:
Use de Broglie's equation for matter waves.
h
l =
mv


6.63 x 1034 J s
=

 31
6
 (9.11 x 10 kg)(5.97 x 10 m / s) 
= 1.22 x 10-10 m
= 0.122 nm
Chemistry-140
Lecture 18
Quantum Mechanics
 Heisenberg Uncertainty Principle: Werner Heisenberg
proposed the uncertainty principle, which states that it is
impossible for us to know, simultaneously, both the exact
momentum of an electron and its exact location in space
Chemistry-140
Lecture 18
Quantum Mechanics
 Schrödinger Wave Equation: Erwin Schrödinger proposed a
mathematical model of the atom using measured energies
and known forces rather than a preconceived "picture" of
the atom's structure. This is called quantum mechanics or
wave mechanics.
Chemistry-140
Lecture 18
Wave Functions & Probability
 Solutions to the wave equation are called wave functions,
symbolized y. Wave functions cannot describe the exact
position of an electron only the probability of finding it in a
given location.
 The probability of finding the electron in a given location is
the electron density and is given by the square of the wave
function for that location, y2.
Chemistry-140
Lecture 18
The Wave Equation & Orbitals
Solutions to the wave equation are called orbitals…..
Chemistry-140
Lecture 19
Chapter 7: Atomic Structure
 Chapter Highlights

electromagnetic radiation

photons & Planck’s constant

Bohr model of the atom

Rydberg equation

quantum mechanics

Heisenberg uncertainty principle
 quantum numbers
 orbitals
Chemistry-140
Lecture 19
The Wave Equation & Orbitals
 Solutions to the wave equation are called orbitals and
each has a characteristic energy.
 An orbital is a region for which there is a high probability
of finding the electron; it is not a path or trajectory.
Chemistry-140
Lecture 19
The Wave Equation & Quantum Numbers
 Variables in the wave equation are called quantum
numbers. The Bohr model used only one variable or
quantum number, n. The quantum mechanical model
uses three quantum numbers, n, l & ml to describe each
orbital
y n, l, m (r, q, f) = Rnl(r)Clm(q, f)
Chemistry-140
Lecture 19
Quantum Numbers
 The principal quantum number (n) has possible values of:
n = 1, 2, 3,...
It describes the relative size of the orbital
Chemistry-140
Lecture 19
Quantum Numbers
 The angular momentum quantum number (l)
has possible values of:
l = 0, 1, 2, ...n-1
It describes the shape of the orbital.
 The value of l is often referred to by a letter equivalent;
0 = s, 1 = p, 2 = d, 3 = f, .... (the rest are alphabetical)
Chemistry-140
Lecture 19
Quantum Numbers
 The magnetic quantum number ( ml ) has values:
ml = -l,... -1, 0, 1, ...l
It describes the orientation of the orbital in space.
Chemistry-140
Lecture 19
Electronic Shells & Sub-shells
 A collection of orbitals with the same value of n
is called an electron shell
 A collection of orbitals with the same values of n and l is
called an electronic subshell. A subshell can be referred to
using n and the letter equivalent of l,
1s, 2s, 2p, 3s, 3p 3d, 3f etc
Table 7.1 A Summary of Quantum Numbers & Orbitals
n l
ml
# & Type of Orbitals
1
0
0
1 - 1s orbital
2
0
1
0
+1, 0, -1
1 - 2s orbital
3 - 2p orbitals
3
0
1
2
0
+1, 0, -1
+2, +1, 0, -1, -2
1 - 3s orbital
3 - 3p orbitals
5 - 3d orbitals
4
0
1
2
3
0
+1, 0, -1
+2, +1, 0, -1, -2
+3, +2, +1, 0, -1, -2, -3
1 - 4s orbital
3 - 4p orbitals
5 - 4d orbitals
7 - 4f orbitals
Chemistry-140
Lecture 19
s Orbitals
 The s orbitals are those for which l = 0. All s orbitals are
spherical. There is one s orbital in each s subshell.
Chemistry-140
Lecture 19
Probability Density in Orbitals
Chemistry-140
Lecture 19
Radial Distribution in Orbitals
Chemistry-140
Lecture 19
p Orbitals
 The p orbitals are those for which l = 1. All p orbitals are
"dumbbell" or "figure-eight" shaped. There are
three p orbitals in each p subshell..
Chemistry-140
Lecture 19
d Orbitals
 The d orbitals are those for which l = 2.
There are five d orbitals in each d subshell.
Four are "four-leaf clovers"; the fifth looks like a p orbital
with the addition of a ring around the centre
Chemistry-140
Lecture 19
d Orbitals
 The d orbitals are those for which l = 2.
There are five d orbitals in each d subshell.
Four are "four-leaf clovers"; the fifth looks like a p orbital
with the addition of a ring around the centre
Chemistry-140
Lecture 19
Nodal Planes in Orbitals
z
z
xy plane
x
x
y
y
xz plane
pz orbital
yz plane
dxy orbital
Chemistry-140
f Orbitals
 The f orbitals are
those for which l = 3.
There are seven f
orbitals in each f
subshell.
Each has 8 lobes
Lecture 19
Chemistry-140
Lecture 19
Textbook Questions From Chapter # 7
EM Radiation:
22, 28, 30
Photoelectric Effect:
32
Bohr Atom:
35, 38, 40
Matter Waves:
42
Quantum Mechanics:
46, 48, 50, 57, 58, 60
General & Conceptual
68, 72, 77, 79, 91