Transcript Quantum Notes (Chapter 16)(Powerpoint document)

The frequency of a wave is the number of wavelengths
that pass through a given point per unit time.
For EMR, we use the symbol n and units of sec-1 (Hz).
For EMR, radiation of given wavelength has a
specific frequency.
c  n
c = 2.99792458 x 108 m/s
For red light with  = 7.00 x 10-7 m…
n=
Energy…
The electromagnetic spectrum…
Paradoxes in classical physics … the ultraviolet catastrophe.
The photoelectric effect.
Red light - no electrons even
if intensity is much greater
than blue light.
The question of the Rutherford atom.
The atomic spectrum of hydrogen…
For H, the visible lines are known as the Balmer series.
For the Balmer series:
1 1 
n    2 3.29x1015 s 1
4 n 
Where n is an integer > 2. Where the hell did that come from?
There are four other series:
Lyman (uv), Paschen, Brackett
and Pfund (IR).
Moseley noticed that metals hit with a lot of energy emitted Xrays whose frequencies correlated with their atomic number Z!
Rydberg developed a formula that calculates the position of
all lines in the H atom spectrum…
 1 1 
15 1

n  

3.29x10
s
n 2 n 2 
 2
1 
where n1 and n2 are integers, n1 >n2. Note that the Balmer
equation is just a special case of the above with n2 = 2.
Welcome to the quantum universe…
We can measure out 24 g of water (1.3 mol)…
but not 4.0 x 10-23 g of water
Because that would be 1.3 molecules.
Charge also comes in discrete packets of e:
1.602x10-19C.
Max Planck argued that energy is also available only in
discrete portions that he called quanta (singular: quantum).
For light of given frequency:
E  hn
E
Where Planck’s constant, h = 6.626x10-34J•s
hc

For red light, n = 4.3 x 1014 s-1, energy is available in quanta of:
Einstein and the photoelectric effect…
He argued that EMR exists in particle-like packets which he
called…
photons. Photons of red light do not have enough energy overcome
the electron binding energy, no matter how many arrive. Photons of
blue light contain more energy which is sufficient to eject electrons.
Any excess energy is KE.
The wave-particle duality…light can have wave properties
in some circumstances (diffraction, interference) and particle
properties in others (photons and quantization).
The Bohr atom…
Using Planck’s quantum
theory, Bohr postulated
that the electron in the H
atom traveled only in
discrete orbits with
rn = n2ao/2p.
ao=0.529Å
and total energy:
Z2
E n   2 Ry
n
Ry = 2.18 x 10-18J
n=3, E = -2.42x10-19 J
n=2, E = -5.45x10-19 J
e
n=1, E = -2.18x10-18 J
The Bohr H atom and the H atom spectrum.
n
n
n
n
=
=
=
=

5
4
3
n = 2
E4 = -1 .3 6 x1 0 -1 9 J
E3 = -2 .4 2 x1 0 -1 9 J
E2 = -5 .4 5 x1 0 -1 9 J
E
n = 1
E1 = -2 .1 8 x1 0 -1 8 J
For the general case of an electron going from the n1
state to the n2 state…
Eatom  E final  E initi al  E2  E1
Z2
E n   2 Ry
n
 Z 2   1 1  2
Z2
Eatom   2 Ry  
 2 Ry
 
 2 
Z Ry




2
n2
 n1
 n1 n 2 
For n1>n2, ∆Eatom is negative indicating energy lost by the
atom and released as a photon.
For n2>n1, ∆Eatom is positive indicating that energy must be
added to excite the electron to a higher energy level.
Ionization…
 1 1  2
Eatom  
n 2  n 2 
Z Ry
 1
2 
1 1  2
E phot on  E atom  
n 2  n 2 
Z Ry
 2
1 
For an electron dropping to the n2 level from the n1 level
(i.e. n1 > n2). Ephoton is always positive.
n=E/h, and for the H atom, Z = 1.
E phot on  1 1 Z 2 Ry
2.18x10 18 J
n phot on 
 
 2 



2
h
6.626x10 34 J • s
n 2 n1  h
 1 1 
15 1
 


3.29x10
s .
n 2 n 2 
 2
1 
What is n for the photon released when an electron drops
from the third to the second orbit?
If light can behave like a particle, can a particle act like light…
DeBroglie…
Standing waves of given
length (like a guitar string)
can exist with only certain
wavelengths…
n
L
2
where n is some integer.
Nodes/amplitude.
If standing waves are quantized, might it not be possible to
characterize the electron in an atom as a standing wave.
2pr=n
The DeBroglie Wave Equation
2 pr  n
nh
m e vr 
2p
h

mev
What is the wavelength of an electron with a velocity of
1.0 x 106m/s?
A 3.0 tonne Mack truck traveling at 40 m/s?
But do electrons really behave like waves?
Schrödinger and the solution to the H atom.
Wave functions describe the properties of standing waves.
Can wave functions describe the properties of electrons
around the H atom?
2
Z
E n   2 Ry
n
The energy of an electron in the H atom has an energy
that depends on an integer: n, henceforth known as
The Principal
Quantum Number
Wave functions…
For each value of n (1, 2, 3 etc.) there are n2 different wave
functions, any of which are valid for an electron of the H
atom.
First of all, what is a wavefunction? It is a mathematical
description of the wave properties of an electron in the H
atom. As a wavefunction, it has the properties of a wave:
Amplitude
Nodes
Phase
Wavefunctions can be added together just like physical
waves resulting in constructive and destructive interference.
Y has no real physical significance.
Y2, however, represents the probability of finding an
electron at a particular point around the H nucleus.
A three dimensional plot of 2 vs. distance from the nucleus, r,
represents a volume of space.
As representations of the wavefunctions, we draw that
region of space in which there is a 90% probability of
finding an electron having that wavefunction. In homage to
Bohr, these are called…
Orbitals
Let’s look at the simplest wavefunction know as 1s.
r 2 1 s
1 s
r
r
For n = 1, there is only 1 (n2) orbital. Higher values of n have
several types of orbitals. These are distinguished by a second
quantum number:
l (ell), the angular momentum quantum number.
For any value of n, l can have any value of 0, 1, 2, 3…, n-1.
Therefore, for any n, there are n types of orbitals.
l can only be 0.
For n = 2, l can be 0 or 1.
For n = 3, l can be 0, 1, or 2.
For n = 4, l can be 0, 1, 2 or 3.
For n = 1,
Etc.
l
Orbital type
0
1
2
3
4
s
p
d
f
g
Any orbital will have n-1 nodes.
l = 1: The p orbitals
2p
l
l
For any value of there turns out to be 2 +1 orbitals of that
type. These are distinguished by a third quantum
number;
the magnetic quantum number: ml.
ml can have any integer value from -l to +l
ml determines
orbital orientation.
Each of these orbitals is
associated with a set of
three quantum numbers.
l
Any orbital has angular nodes.
For n = 1 there is the 1s orbital only
For n = 2, there is the 2s and three 2p orbitals
l
For n = 3, there are now three possible values of : 0, 1 and 2.
This tells us that there are three types of orbitals: s, p and d.
l
For = 2, there are five possible values of ml: -2, -1, 0, 1, 2.
Therefore there must be five d orbitals.
For each n there are n2 orbitals each of which can be identified
l
by a set of three quantum numbers: n, , ml.
n - indicates orbital size and energy
l - indicates orbital type
ml - indicates orbital orientation
Affordable housing for the H atom electrons…
5s
4s
5p
4p
5d
4d
3s
3p
3d
2s
2p
5f
4f
E
Shell and subshell
1s
Electrons have the property of spin - also quantized.
There are two possible spin states specified by the spin
quantum number: ms which has a value of +1/2 or -1/2..
Any orbital can be identified by a set of three quantum numbers.
Any electron can be identified by a s set of four quantum numbers.
n,
l and m
l
indicates the orbital electron is in
ms - indicates spin