1 february 2012modern physics iv lecture 31 modern physics for frommies iv the universe - small to...

1 February 2012 Modern Physics IV Lecture 3 1

Modern Physics for Frommies IVThe Universe - Small to Large

Lecture 3

Fromm Institute for Lifelong Learning University of San Francisco

Agenda• Administrative Matters• Atomic Physics• Molecules

Administrative Matters• This is Lecture 3, Lecture 8 will be Wed. 7 March

• Wikis from previous course

http://modphysicsfrommiies.wiki.usfca.edu/Note the ii in frommiis

http://modphysfromm2.wiki.usfca.edu

http://modphysfromm3.wiki.usfca.edu

http://modphysicsfrommiies.wiki.usfca.edu/



http://modphysfromm2.wiki.usfca.edu/


The current course wiki, http://modphysfromm4.wiki.usfca.edu , now includes a Glossary of Mathematical Symbols,

Other glossaries can be found on Google by searching “mathematics symbols”



Quantum Mechanical View of AtomsBohr model discarded as an accurate description of nature

Certain aspects have however been retained

e.g. Electrons in an atom exist only in discrete states of definite energy, the stationary states

Transitions between these states require the emission (or absorption of a photon.

According to wave mechanics, electrons do not travel in well defined circular orbits ala Bohr. The electron, because of its wave nature, is better thought of as spread out in space as a “cloud”.

The size and shape of the electron cloud can be found by solving the Schrödinger equation for the atom and forming the probability distribution, | |2.


Ground state of hydrogen


Schrödinger’s Equation in Spherical Coordinates

x

2

2

2 2 22

2 2 2

( , , ) , ,2

where

x y z E x y zm

x y z

Separation of variables:

Assume a solution of the form , ,x y z X x Y y Z z

y

zr

In Cartesian coordinates


Want to do the same thing with spherical symmetry


Spherical Time Independent Schrödinger Equation

2

22 2 2 2 2 2

2

0

1 1 1 2sin 0

sin sin

1where for the hydrogen atom

4

mr E V

r r r r r

eV

r

Separation of variables:

Try solution of form , ,r R r

( , , ) , l mn lr R r Y

: , cos

where the are

l lm imml l

ml

Y P e

P

spherical harmonics

associated Legendre polynomials

: lnR r associated LaGuerre functions


Quantum NumbersIf we do QM in for a particle confined in a 1-D and 3-D potential well or rigid box. (See Course II Lecture 4)

The solutions are characterized by a single quantum number (n) in the 1-D case and by three numbers (nx, ny and nz) in 3-D.

These quantum numbers arise from the imposition of boundary conditions on the solutions. We might expect that in the 3-D problem of the hydrogen atom the solutions will be characterized by numbers corresponding to Boundary conditions applied in 3-D. Restrictions on the values of these quantum numbers arise from the mathematics of the LaGuerre functions and the spherical harmonics.

Actually, we need a fourth number. There is an additional degree of freedom which I will treat in a few minutes.


202 2

0

1,2,3,

1

2

4

n

n

E m eE

n n

principle quantum number

Results from boundary conditions on solution of the R part of the separated Schrödinger eqn.

Bohr result

R part contains the potential energy

n alone determines the energy levels (actually there is a slight deviation from this)

Consequence of central inverse square force.


0,1,2,3, , 1

l n orbital angular momentum quantum number

associated with and parts of Sch. eq

n.

R r

tangential

Quantu

Classically

m boundary conditions

or

1 2

L r

hL l l

p L rmv

Note disageement with Bohr quantization where

2

in particular, the ground state has 0 0

hL n

l L

The semiclassical planetary model with electrons in orbits is not a good one


Note: All these transitions have l = 1


s p d f g h etc.

l=0 1 2 3 4 5 etc.

Notation for states: nl, e.g. 4d is n=4, l=2

, 1, ,0,1, , 1, lm l l l l magnetic quantum number

is a vector quantity, conserved in a central potential

The solution for specifies that is an integer related to

's -component.

l

L

m

L z

2z l

hL m


Aside on Angular Momentum

r

v

Particle of mass m moving with circular speed v around an axis at radius r.

L r mv

Magnitude: sin

Here, 90 sin 1

L mvr

L mvr

Direction: to plane of and with sense

determined by right hand rule.

r v


C A B

sinC AB

A

C

To the plane of AB

Right Hand Rule

Direction of advance of a right hand screw


Note that the vector product is not commutative

A B B A

Again look at Right Hand Rule

C A B

‛ A

C A × B

B × A


1 fixed

restrictedz l

L l l

L m

Space quantization

Note choice of z axis is arbitrary.


Energy is dependent solely on n.

Presence of multiple ls and ms for a given n states are degenerate

This degeneracy is removed if directional symmetry is broken by say a B or E field.

What about Lx and Ly?


2 2

z

2 2

If and are known, knowledge of 2nd component

3rd is

consequence of

also known.

x y zL

L L

L L L

If we know exactly, we know nothing of

Uncertainty pr

we know

inciple:

nothing of and

z

z

x y

L

L

L L


x

y

z

Ly

Lx

Liz

L


Magnetic effects

Normal Zeeman effect: Transition between 1s and 2p

Spectral lines broaden and split into 3 lines as B is applied and increased. 3 lines = “normal Zeeman effect”

Consider the electron “orbit” to be a current loop with

IA

2

2 2 2

e rq erv eIA A L

T r v m

Vector: 2

eL

m


Apply external magnetic field

Torque:

Potential energy: B

N B

U B

Bohr magneton

2

2

l

B

z l B

e

m

m

em

m


quantized quantizedL

Additional potential energy term:

B z B lU B m B

Each degenerate energy level, l, is split into 2l+1 separate energy levels, ml.

B has specified a direction in space (z axis) and the symmetry responsible for the degeneracy has been broken.


The Stern-Gerlach experiment:

If B is inhomogeneous there will be a net force as well as torque on the atom


For l 0 the states should separate according to ml

2 lines seen instead of the expected 3 (or 2l+1 = odd)

Haven’t seen the whole picture yet.


Electron Spin

Wolfgang Pauli:

Relativity besides n, l, ml need 4th quantum number

G. Hollenbeck and S. Goudsmit:

Propose intrinsic “spin” angular momentum for the electron

s = ½ħ

Another magnetic quantum number: ms = ½

1928, P. A. M. Dirac justifies this from relativity.

i m


Gives magnetic effects like orbital angular momentum.

Intrinsic spin → intrinsic magnetic dipole moment

New magnetic quantum numbers:

ms = ± 1/2

Doubles number of states for a given n

States are degenerate unless a spatial direction is specified, e.g. external E or B field

Quantum state now specified by {n, l, ml, ms}


Return to the Stern-Gerlach experiment

l = 0 state will give 2 lines for ms = ± 1/2

Fine structure:

Even in the absence of external fields, very high resolution spectroscopy reveals splitting of spectral lines.

Rest frame of electron: nucleus orbits and appears as a current loop. Interacts with spin magnetic moment and breaks degeneracy

Line separation is about 5 x 10-5 eV compared to the 2p 1s transition energy of 10.2 eV

Hyperfine structure:Arises from the spin angular momentum and consequent spin magnetism of the nucleon(s)


s-wave states are spherically symmetric, not so for l 0


Quantum StatisticsConsider a system of 2 particles, say electrons

1 2 1 2 1 2( , ) *P r r dv dv dv dv

1 2

1 2

Wave function for the system is ,

we observe a probability of finding the particles

in volume elements at and

r r

r r

It is easy to show that

* *

i.e. no change in observable for

1 2 1 2 2 1 2 1

identical particles no observable change if they are interchange d

* , , * , ,r r r r r r r r


So, under interchange 2 possibilities

±

If 2 identical particles interchange = +they are said to obey Bose-Einstein statistics and are called bosons.

If 2 identical particles interchange = -, they are said to obey Fermi-Dirac statistics and are called fermions.

Bosons have integral spin, e.g. photons, mesons, some atoms and nuclei, ………


Fermions have ½ integral spin, e.g. leptons, nucleons, some atoms and nuclei,………..

For fermions

Cannot have 2 identical particles with the same set of quantum numbers.

Pauli Exclusion Principle

You can stick as many bosons into a quantum state as you want.


Electrons are fermions.

Build some elements. As electrons are added the exclusion principle will have an effect.

Hydrogen: 1 e in the 1s state 1s1

He: 2 e in the 1s state, ms =1 and -1, 1s2

No more e can be added to the 1s state without violating the exclusion principle !

The K shell is filled


Li: 3rd e has to go in 2s state 2s1

Be: 4th e in the 2s state, ms =1 and -1, 2s2

2s state (subshell) is now filled

B: 5th e has to go in the 2p state 2p1

p state has 2l+1 = 3 values of ml, each with 2 values of ms, accommodating C, N, O, F and Ne as 2p2 – 2p6.

2p subshell is now filled, as is the L shell

Na: 11th e has to go in 3s state 3s1 etc.


3s and 3p each with 2l+1 ml values each having 2 values of ms.

Weirdoes:

3d, 4d and 5d subshells fill up the transition metals followed by the lanthanides and the actinides

Complicated inter electron interactions mess things up

If electrons were bosons, they would all sit in the ground state, 1s, and chemistry would be very different.

17 February 2010 Modern Physics II Lecture 6 37

Bonding in molecules

Ionic bonding:

NaCl

P for Na outer electron


Na

Na has 11 e-

10 reside in inner closed shells Last e- spends most of its time outside these shells.

The last e- feels net attraction due to +1e, not all that strong

Cl

Cl has 17 e-

12 are in closed shells 1s22s22p63s2

Others are in non spherically symmetric p states


H

ml=0, unpaired 4 states ml=±1, ms=±1

Exclusion principle allows one more e- in ml = 0 with spin oriented opposite to that of the last Cl e-

If an extra electron happens to be in the vicinity it can be in this state and could see an attraction due to Cl nucleus as much as +5e.

Stronger than the +1e attraction between Na nucleus and its outer electron charge distribution in slide #36 and an ionic bond between Na and Cl.


Covalent bonding:

H+H H2If H atoms are close together, e- clouds overlap and e- “orbit” both nuclei.

Both H s in ground state. Electron spins can be either parallel (S = 1) or antiparallel (S = 0)

1st consider S = 1:Exclusion principle 2 e- with same quantum numbers must be in different places, i.e. belong to different atoms.

(+) nuclei repel, no bond is formed.


S = 0:

e- have different values for ms, spend a lot of time in the internuclear region

(+) nuclei are attracted to the internuclear e- and a bond is formed.

In a wave picture, exclusion principle destructive interference when S=1 and constructive when S=0.


Energetics point of view:

For S = 0, e- can occupy same space, space of 2 atoms rather than 1 x is increased.

H. U. P p can be less energy is less

Molecule has lower energy than the 2 separate atoms

H2 is stable

Binding energy is 4.5 eV for H2


0.074 nm

In the vicinity of r0 we may

approximate

, constants for attractive,

repulsive parts of

, are small integers

m n

A BU

r rA B

U

m n


Activation energy often need to break earlier bonds

2 2 2

2 2

2 H + O 2H O

H and O must 1st be broken into H and O atoms

spark

UA = 0 for hypergolic materials, don’t need spark


Energy storage in biological systems

adenosine triphosphate

ATP ADP + (phosphate group) + ENERGY

1 february 2012modern physics iv lecture 31 modern physics for frommies iv the universe - small to...

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