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Chapter 28
Quantum Mechanics of Atoms
© 2006, B.J. Lieb
Some figures electronically reproduced by permission of
Pearson Education, Inc., Upper Saddle River, New Jersey
Giancoli, PHYSICS,6/E © 2004.
Ch 28
1
Two Approaches to Quantum Mechanics
Schrödinger Wave Equation: is a “wave” equation
similar to the equations that describe other waves
Heisenberg Method: based on matrices. We will only
study one part- the Heisenberg Uncertainty Principle
•It was soon realized that the two methods gave
equivalent results
•Modern quantum mechanics includes elements of both.
Correspondence Principle: required that a new theory
must be able to produce the old classical laws when
applied to macroscopic phenomena.
Ch 28
2
Heisenberg Uncertainty Principle
•Consider the problem of trying to “see” an electron with a
photon
•We will refer to the uncertainty in x as x
•If  is the wavelength of the light then from diffraction:
x  
•Photons have momentum p=h /  and
when the photon strikes the electron it
can give some or all of its momentum
to the electron
p
Ch 28
h

the product of these two is
h
(x)(p)    h

3
Heisenberg Uncertainty Principle
A more careful analysis of this gives
h
(x)(  p ) 
2
There is also an uncertainty principle for energy and time
(E)(t ) 
Ch 28
h
2
4
Example 28-1. The strong nuclear force has a range of about 1.5x10-15 m. In 1935 Hideki
Yukawa predicted the existence of a particle named the pion (π) that somehow “carried” the
strong nuclear force. Assume this particle can be created because the uncertainty principle
allows non-conservation of energy by an amount ΔE as long as the pion can move between
two nucleons in the nucleus in time Δt so that the uncertainty principle holds. Assume that the
pion travels at approximately the speed of light and estimate the mass of the pion. (See page
896 in textbook)
h
2
If the velocity of the pion is slightly less than c, then it can travel the distance
d = 1.5x10-15 m in the time Δt where
E t 
t ~
d
c
The rest mass energy of the pion is equal to the uncertainty in energy
E  m c
Ch 28
2
5
Example 28-1 (continued). The strong nuclear force has a range of about 1.5x10-15 m. In
1935 Hideki Yukawa predicted the existence of a particle named the pion (π) that somehow
“carried” the strong nuclear force. Assume this particle can be created because the uncertainty
principle allows non-conservation of energy by an amount ΔE as long as the pion can move
between two nucleons in the nucleus in time Δt so that the uncertainty principle holds.
Assume that the pion travels at approximately the speed of light and estimate the mass of the
pion. (See page 896 in textbook)
E t 
h
2
We substitute the above expressions for ΔE and Δt:
d  h
m c   
 c  2
2
hc
m c 
2d
(6.6 1034 Js)(3 108 m / s)

2 (1.5 1015 m)
2
m c 2  2.110 11 J  130 MeV
m  130 Mev
c2
A few years later the pion was discovered and it’s actual mass is ≈ 140 MeV/c2.
Ch 28
6
Example 28-2. Estimate the lowest possible kinetic energy of a neutron in a typical
nucleus of radius 1.0x10-15m. [Hint: A particle confined to the nucleus will have
momentum at least as large as its uncertainty. The size of the nucleus is ]
(x)( P) 
h
2
 x  1.0 1015 m
P 
h
2  x
 P  m v
h
v 
2 m x
(6.626 1034 Js)
7


6
.
3

10
m/ s
2 (1.67 1027 kg)(3 1015 m)
We will assume v ≈Δv. It should be noted that v is about 20% of c, which might
require a relativistic calculation, but we just want a rough estimate. Notice that
the neutron is in rapid motion in the nucleus.
KE 
Ch 28
1 2
mv
2
1
  (1.67 10  27 kg)(6.3 107 m / s ) 2
2
 1 eV

K E  (3.3 10 12 J )
  21 MeV
19
 1.6 10 J 
7
Schrödinger Wave Equation
•De Broglie had said that electron formed standing waves in
the Hydrogen atom
•Schrödinger found a wave equation that described these
waves and provided additional information about the
electron states
•The Schrödinger Wave Equation is a better description than
the simple picture of standing circular waves that we saw at
the end of last chapter
•It is a second order differential equation
Ch 28
8
What does solution of “Wave Equation” for the
Hydrogen Atom give?
•Three new “quantum numbers” in addition to n
•Wave function  (Greek letter psi) is the amplitude of
the matter wave at any point in space and time
• 2 is the probability of finding the electron at
a given point in space.
•Same formula for energy levels as Bohr Model
Z2
En  13.6 eV 2
n
Ch 28
n  1, 2, 3
9
Probability vs. Determinism
•Newton had introduced a era of determinism—if we knew
the position and velocity of an object, we can predict its
future behavior
•Uncertainty Principle says we can’t know this
•Predictions of wave equations for the position and
momentum of electron are based only on probabilities.
•The kinetic theory of gases is based on probabilities but
this is different because gas molecules were assumed to
move in a deterministic way but there were too many
molecules to keep track of.
Ch 28
10
Principal Quantum Number n
•Can have any integer value n = 1, 2, 3,...
•Determines energy of state with the same formula as the
Bohr Model
2
En  13.6 eV
Z
n2
n  1, 2, 3
•Other quantum numbers have a small effect on energy
Ch 28
11
Orbital Quantum Number l
•Related to the angular momentum of electron
•Must be integer less than n
l = 0, 1, 2, 3, …(n-1)
•Angular Momentum is given by
L
where
 
l(l  1) 
h
2
Example: l = 2
L
Ch 28
l (l  1)  
2(2  1)   2.5 
12
Standard Notation
l = 0 is s state
l = 1 is p state
l = 2 is d state
l = 3 is f state
(then alphabetically g, h, i, j…)
Example: state with
n = 3, l = 2 is 3 d state
n = 3, l = 1 is 3 p state
n = 3, l = 0 is 3 s state
Ch 28
13
Quantum Numbers
Magnetic Quantum Number: ml
•Integer values from –l to +l
•Example: if l = 2, then
ml = 2, 1, 0, -1, -2
•Note that there are 2 l +1 possible
values
•space quantization: the angular
momentum can only have integer
projections on the z-axis ( Lz )
•Usually defined by magnetic field.
Lz  ml 
Ch 28
14
Evidence for Magnetic Quantum Number ml
•Integer values from –l to +l
•Example: Upper State
•if l=2, then
•ml = 2, 1, 0, -1, -2
•Lower State: l=1, then
•ml = 1, 0, -1
Ch 28
15
Spin Quantum Number ms
•Two possible values ms = + ½ or ms = - ½
•Was observed that spectral lines were split, even without
magnetic field, so something with two values was needed.
•Called spin quantum number as if electron was a
negatively charged sphere that was spinning
•We now know this spin picture is not correct
We still speak of spin up: ms = + 1/2
spin down: ms = - 1/2
Ch 28
16
Wave functions
•Every state has a different wave function
•2nlm is the “probability distribution”
•Sometimes referred to as electron cloud
•Probability distribution for ground state of H atom (1s)
Ch 28
17
Wave functions
n=2, l=0, ml = 0 (2s) state
n=2, l=1, ml = 0 (2p) state
n=2, l=1, ml = ± 1 (2p) state
Ch 28
18
Other Wave Functions
Ch 28
19
Photon Selection Rules
•Quantum Mechanics provides a means of calculating the
probability of electron changing state and emitting a
photon
•Photon has spin of 1 so this favors transitions with
 l  1
•transitions that obey this are called allowed
transitions
•transitions that do not obey this are forbidden
transitions and have a very low probability.
Ch 28
20
Single Electron Atoms
•this model explains Hydrogen very precisely, but does it
apply to atoms with more electrons and more protons in the
nucleus?
•It works well for all single-electron ions such as He+ (Z = 2)
and Li++ (Z = 3).
•It can be used with adjustments for atoms where there is a
single valence electron outside of a closed shell
Ch 28
21
The Pauli Exclusion Principle
Atoms with more than one electron obey Pauli Exclusion
Principle:
•Each electron occupies a particular quantum state with n,
l, ml and ms.
•No two electrons in an atom can occupy the same
quantum state
•Thus no two electrons can have the same set of quantum
numbers n, l, ml and ms.
Ch 28
22
Helium (z = 2)
Helium:
n
l
1
0
ml
0
ms
½
1s
n=1, l=0
1
0
0
-½
1s
Each electron is attracted to nucleus but repelled by other electron.
Ch 28
23
Lithium (Z = 3)
•Lithium has a single electron in the n =2,l=0 state (2s).
This single electron experiences attraction of three protons
and the repulsion of the two inner electrons which gives a
net charge of approximately e. So the 2s electron is
somewhat hydrogen-like and is not tightly bound.
•We write the electron configuration of lithium as 1s2 2s1
n
1
1
2
Ch 28
l
0
0
0
ml
0
0
0
ms
½
-½
n = 3, l = 0
n = 2, l = 0
00 00 00
0
n = 1, l = 0
½
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Energy Level diagram for He, Li, and Na
Ch 28
25
Periodic Table of Elements
Shell: Electrons with the same value of n.
•K shell: n =1
•L shell:
n=2
•M shell: n = 3
Subshell: Electrons with the same n and l
values
Examples:
•2s Subshell
•3p Subshell
Ch 28
26
X-Ray Spectra
When we derived the formula for the energy of stationary states, we did
not let Z = 1 because the equation with Z works in many (but not all)
cases.
Z2
En  13.6 eV
n2
n  1, 2, 3
•This formula can not always be used for the electrons in a nucleus
with a large Z because it requires a central positive charge and the
presence of the other electrons changes the situation.
•The nuclear charge Z is shielded by other electrons which move
rapidly
•For an innermost electron, the effect of the outer electrons largely
cancel out and thus the formula can work.
•X-Rays are produced when electrons are accelerated through a high
voltage and they then knock an electron out of the n = 1 or 2 atomic
states
Ch 28
27
X-Ray Spectra of Molybdenum
•Consider what happens when an electron is removed from
the n = 1 state of the atom molybdenum (Z=42).
•An electron from the n = 2 or n = 3 drops down to n = 1
and a photon is emitted
•That electron ‘sees’ a nuclear charge of (Z-1) because the
other innermost electron is still there.
(Z  1)2
En  13.6 eV
n2
Ch 28
n  1, 2, 3
28
Example: X-Ray Spectra of Molybdenum
(Z  1)2
En  13.6 eV
n2
n  1, 2, 3
(42  1)2
E1  (13.6 eV )
 22900 eV  22.9 keV
12
(42  1)2
E2  (13.6 eV )
 5700 eV  5.70 keV
22
(42  1)2
E3  (13.6 eV )
 2540 eV  2.540 keV
32
We thus calculate E21= 17.2 keV
and E31= 20.4 keV and
E 
hc

or
 
hc
E
and 21=0.072 nm and 31=0.061 nm.
Ch 28
29
Laser
Light Amplification by
Stimulated Emission of Radiation
•Laser uses stimulated emission to amplify a light beam
•Stimulated Emission: an atom in an excited state is stimulated
by an incoming photon to drop to a lower state with the
emission of a photon.
•Spontaneous Emission: Process we have already discussed
where atom spontaneously drops to a lower state with the
emission of a photon.
Ch 28
30
Stimulated Emission of Radiation
•Absorption of Photon: photon must
have energy equal to Eu – El
•Stimulated Emission: photon must have
energy equal to Eu – El and two coherent
photons are emitted. Stimulated photon
emitted faster.
•Coherent: all parts of beam have the
same phase
•Laser uses stimulated emission to amplify light beam
Ch 28
31
Population Inversion
•In order to have LASER action, you must have more
atoms in the excited state than in the ground state
•This is called a population inversion because normally
the atoms are in the ground state
Ch 28
32
Population Inversion in a Ruby Laser
•In a ruby laser, atoms are put in state E2 by means of a flash
bulb.
•They then decay to E1 which is a metastable (long-lived)
state where they stay until stimulated emission occurs
Ch 28
33
Ruby Laser
•Flash bulb creates a population inversion
•A few photons are emitted spontaneously
•Many photons escape
•A few photons emitted along the axis of the ruby cause
stimulated emissions which then cause additional stimulated
emissions
•Beam builds up but a few photons pass through partially
Ch 28
transparent
mirror
34