Transcript Transparancies for Atomic Structure Section

Quantum Phenomena
II:
Matter Matters

Atomic Structure
Hydrogen atom



Quantum numbers
Electron intrinsic spin
Other atoms



More electrons!
Pauli Exclusion Principle
Periodic Table
Fundamental Physics

Particle Physics



The fundamental particles
The fundamental forces
Cosmology

The big bang
http://ppewww.ph.gla.ac.uk/~parkes/teaching/QP/QP.html
March 2005
Chris Parkes
“Curiouser and curiouser” cried Alice

Christine Davies’ first part


This lecture series




2
Some consequences of QM
Applications
Emphasis on awareness not mathematical rigour
First few lectures – Young & Freedman 41.1->41.4


Basics intro: Rutherford’s atom, blackbody radiation, photoelectric effect, wave particle duality, uncertainty principle,
schrödingers equation, intro. to H atom.
Lectures main points same, but more complex treatment
Last few lectures – Y&F 44
Understanding atoms

Key to all the elements & chemistry


What are atoms made of ?


quarks
Why are p,n clamped together in the middle?

Strong nuclear force

Second part of this course…..

How do we analyse atoms
The first part of this course…..

3
Nucleus (p,n) ,e
What are the nucleons made of


Non-relativistic QM – the schrödinger equation
Find the energy levels for a
Hydrogen atom
• Find the wavefunction for the hydrogen atom
Schrödinger Equation :
solving H atom
2  2
2
2 
 2  2  2  nlm  U nlm  En nlm

2m  x y z 
1.
This looks like p2/2m + U = E in classical mechanics
2.
n,l,m are quantum numbers
3.
E depends on n only for H (also l for multi electron atoms)
4.
BUT now we have a wavefunction (x,y,z)

BIG Difference from
classical physics.
Wavefunction

No longer know
dy where a particle is
Probability to find a particle at
dx

Just how likely it will
be at x,y,z
P(x,y,z) dx dy dz = | (x,y,z)|2 dx dy dz
  
5
dz

  x, y, z  dxdydz  1 (normalisat ion )
   
2
Spherical co-ordinates

Potential energy of one electron in orbit
around one proton

1 q1q2
1 e2
U (r )  

  ke2 r
40 r
40 r

Spherical symmetry, so use spherical
polars


rewrite schrödinger in r, , 
Rather than x,y,z
2  1 2
1


1
2 


(r )  2
(sin 
) 2 2
   U nlm  En nlm
2
2
2 m  r r
r sin  

r sin   
• Mass of electron m, Charge of proton,electron e
For single electron heavy ion would have q=Ze
•Try (r,,)
= R(r) Y(,)
•Separate out the radial parts and the angular parts,
6
LHS(r) = RHS(,)=C
Radial Equation
r d2
2mr 2
(rR )  2 [ E  U (r )]  C  l (l  1)
2
R dr

2
2
2

d

Or rearranged as

(rR )  [U (r ) 
l (l  1)]( rR )  E (rR )
2
2
2m dr
2mr
BUT this looks a lot like schrödinger eqn



With rR(r)=(r)
And with an extra term
L=mvr
What is the extra term ?



e-
Think classically
Potential U + K.E. term

1 2 (mvr)
L
 l (l  1)
mv 


2
2
2
2mr
2mr
2mr 2
 L2   2l (l  1) Total angular momentum
2
7
2
2
p
Solution

Rnl (r )  r Ln,l e
l

r
na0
n l 1
L are Laguerre functions  bi r i , where :bi 1  2
na
i 0
0
i  l 1 n
bj
(i  1)( j  2l  2)
b0  2a03 / 2



They are a series with specific solutions for n and l values
a0 is a length
known as Bohr radius
4 0 2
a0 
0.529 x10-10 m
me 2
Similiarly can solve the angular part of eqn
l |m|
Ylm ( ,  )  N lme im sin |m|   ai cos i 
Spherical harmonics involving
another series of constants ai
i 0

8
Specific solutions for l and m values
Find the H Energy levels
Reminder lmn(r,,) = Rnl(r) Ylm(,)
•
• So now we have the solution!
•
Substitute into
 

 
 2  2  2  nlm  U nlm  En nlm

2m  x y z 
2

2
2
2
-ve, relative to ionised atom
The energy only depends on n


9
n=4
n=3
And find an expression for E
me4
 13.6
E

eV
2 2 2 2
2
32  0  n
n

Ionised atom
n is Principle quantum number
Not on l,m for coulomb potential U
n=2
E0
Quantum Numbers

Atom can only be in a discrete set of states n,l,m


Principle n fixes energy - quantized


Integer in range 0 to n-1
m (or ml ) fixes z component of angular
momentum

10
Integer >=1
l fixes angular momentum L


Diff. From classical picture with any orbit
Integer in range –l to +l
If you only learn 5 things from
this…..
1.
2.
11
Solving Schrödinger  Discrete states
Quantum numbers n,l,m
1. Energy, ang. mom, z cmpt L
3.
Energy  1/n2 , scale is eV
4.
Know the ranges n,l,m can take
5.
….Hence understand how to calculate the states
Angular Momentum
• Quantum picture of Angular Momentum
12
States and their
spectroscopic notation
n
1
l
0
m
0
1s
2
0
1
0
-1,0,1,
2s
2p
0
1
2
0
-1,0,+1
-2,-1,0,1,2
3s
3p
3d
0
1
2
3
0
-1,0,+1
-2,-1,0,+1,+2
-3,-2,-1,0,+1,+2
4s
4p
4d
4f
3
4
13
Angular momentum is
QUANTIZED

We now know Energy is quantized



Familiar from seeing transition photons
E.g. Balmer series
nf=2


E
1
1
  0  2  2 

hc 
n f 
ni

L   l (l  1)
2





14
Photon
1
BUT we have also learnt

Ei
2
and l takes discrete values
s state is l=0
p state is l=1
d state is l=2
f
g
L=
L=
L=
0
2
2  3  6
Ef
Emission
TOTAL Angular momentum L
Quantum number l
m
- z component of l
- magnetic quantum
number
Cartoon of components for
l=2, p state

| L | 6
Lz  2
c.f. Classical behaviour
•state has angular mometum and this
has a component along z axis
Lz  
Lz  0
But quantum
•States are quantized
Lz  
Lz  2


15
•Ang. momentum can be zero
choice of z axis purely a convention
Important for interactions of atom with magnetic field along z
(later)
The states
•
16
Hydrogen wavefunctions
• Where is the electron ?
The first few states

Can substitute into our expressions n,l,m and find out
nlm(r,,) = Rnl(r) Ylm(,)= R(r) P() F()
Probability depend on wavefunction squared
17
Visualising the states(1)
States with zero angular momentum are isotropic


Indep.of  and 

n00(r,,)= nlm(r)
P(x,y,z) dx dy dz = | (x,y,z)|2 dx dy dz




i.e. probability in cube of vol dV is P dV
[] ?
3
Probability density fn PDF (dim. 1/length )
So P(r)dr depends on volume of shell of sphere
Integrating probability over  and 
Volume is 4r2dr
P(r )dr  4r P(r , ,  )dr
2
 4r  (r ) dr
2
r
18
dr
2
Normalised so
integral is 1
Visualising the states(2)
1s state
n=1,l=m=0
wavefunction
PDF
P(r)
in units of a

19
2s, 3s states
Hydrogen Atom PDFs
Scale increases with increasing n
l=0 spherically symmetric
m=0 no z cmpt of ang.momentum
z
x
20
Fine structure
• Energy levels given by quantum number n
• Now add a magnetic field…
21
Adding Angular
Momentum




L1 specified by l1,m1
L2 specified by l2,m2
How would we combine them ?
z

 
LTotal  L1  L2 what is ltot, mtot ?
mtot
mtot  m1  m2
L1
And obv.
mtot | ltot |
So for the total…
ltot  l1  l2 , l1  l2  1,...., l1  l2
Anti-parallel
Ltot
m1
Easy (classical like) bit, adding components
22
m2
parallel
L2
Zeeman Effect


Observe energy spectrum of H atoms
Now …add magnetic field


Nature, vol. 55
11 February 1897,
pg. 347
Atoms have moving charges, hence magnetic interaction
Spectral lines split (Pieter Zeman, 1896)
Discrete
states
as
Ang.mom.
quantized
Angular momentum has made small contribution to energy (order 10,000th )
23Fine
Structure
Zeeman effect

Potential energy contribution as classical
e
Magnetic dipole
  IA
L
2m
U     B Potential energy in magnetic field
Sodium 4p3s
Now, put magnetic field along z axis
e
e
U  zB  
Lz B   ml
B
2m
2m
U   ml  B B
Bohr magneton 
B
Orbital Magnetic Interaction energy equation
So, for example, p state l=1, with possible m=-1,0,+1, splits into 3 Energy levels
according to Zeeman effect
24
How many lines on that last photo …?
Stern-Gerlach Experiment
Experiment with silver atoms, 1921, saw some EVEN numbers of lines
Non-uniform B field, need a force not just a twist
Lz  ml 





25
l=0, ml=0
1 line
l=1, ml=-1,0,1
3 lines
l=2, ml=-2,1,0,1,2
5 lines
……
l=a,
ml has 2a+1 lines
Odd no.
“Anomalous” Zeeman Effect
EVEN numbers of splittings, something is missing……


We need another source of ang. mom.,
ml is not enough
We know we can add angular momentum

J  L S
Total
orbital
and
spin
ml=-l….+l, ms=-1/2, +1/2
J z  Lz  S z
Intrinsic property of electron
So, every previous state we can split into two
(careful though for total as 1+1/2 = 2-1/2!!)
Using +1/2 or –1/2 electron spin states
Energy splitting as before but with an extra factor of g=2
Due to relativistic effects
26
U   gms  B B
Complex Example: Sodium p state
STATES
l=1 hence j=1+1/2 or j=1-1/2
j=3/2, mj=+3/2,+1/2,-1/2,-3/2
4
j=3/2 or j=1/2, now 2 states
j=1/2, mj=+1/2,-1/2
2
27
Electron Spin
U   gms  B B
Gyromagnetic ratio g~2

Like a spinning top!


But not really…point-like particle as far as we know
Orbital and Intrinsic spin is familiar

Earth spinning on axis while orbiting the sun
Electron spin is 1/2
S and ms, just like l and ml
1 1
3 2
S   s ( s  1)   (  1)  
2 2
4
1 1
ms   , up and down spins
2 2
2
Spin
28
2
2
is just another standard characteristic of a particle like its mass or charge
Total Angular momentum: A Top 5…
1.
Orbital angular momentum L, e orbiting nucleus

2.
Quantum number l

3.


Has ms =-1/2, +1/2
So splits an l state into two
Total Angular Momentum J


29
Interacts with magnetic field, U=mlBB
Zeeman effect gives splitting of states
Spin s=1/2, intrinsic property of electron

5.
notation l=spdfg…., l=0,1,2,3,4…
l has z-component ml, (-l….+l)

4.
L2=l(l+1)h
Sum of orbital and spin
Anomalous Zeeman effect / Stern-Gerlach Expt
Multi-Electron Atoms

30
Everything that isn’t hydrogen!
Pauli Exclusion Principle

No two electrons can occupy the same quantum
mechanical state


Nothing to do with Electrostatic repulsion

Also true for neutrons

Deeply imbedded principle in QM

If all electrons were in the n=1 state all atoms
would behave like hydrogen ground state

31
Actually true for all fermions (1/2 integer spin)
No chemistry – same properties
Multi-Electron atoms

Lowest energy configuration

Energy
Start adding the electrons filling up each state
l=0
1
2
Z=18
n=3
Z=10
n=2
n=1
Z=2
•H Energy levels depend only on 1/n2
•Each state contains two electrons
32
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
1s
2s
2p
ml ms for
filling states
Full shells
But order in
shell?
Central field approximation


We have neglected any interaction of electrons
BUT we
no longer have a coulomb potential
2

1 Ze
U (r ) 
40 r
 U now depends on electrons we have already added


Approximation - Electron moving in averaged out
field due to all others
Screening Effect



Higher n, l means more screening
See less charge (Gauss’ law)
Radial solutions of schrödinger have
changed E now depends on l not just n
electrons
nucleus
p,s states extra peaks
at low r, more time
close to nucleus, less
screened
33
Energy Order of States

Screening shifts the states


f above d above p above s
but also 3d is above 4s
E Gap
number
Z=18
Z=10
Z=2
34
35
Everything you ever need to know
about Chemistry

Closed Shell











36
Z=2
Z=10
Z=18
1s2
all n=1 states full
1s22s22p6
all n=1,2 states full
1s22s22p63s23p6
Noble gases – non reactive, stable, RH column
One electron more


Helium
Neon
Argon
Krypton….
Lithium
Sodium
Potassium
Rubidium
Z=3
Z=11
Z=19
Z=37
He + 2s
Ne + 3s
Ar + 4s
Kr + 5s
Alkali metals, effective screening, weak binding,
easily get ions
Similarly Be, Mg, Ca form 2+ ions (alkaline earth
metals)
And F,Cl,Br form 1- ions to get closed shell
(halogens)