1. What’s coming up??? Nov 8,10,12 Hydrogen atom Ch. 9
Oct 25 The atmosphere, part 1 Ch. 8
Oct 27 Midterm … No lecture
Oct 29 The atmosphere, part 2 Ch. 8
Nov 1 Light, blackbodies, Bohr Ch. 9
Nov 3,5 Postulates of QM, p-in-a-box Ch. 9
Nov 8,10,12 Hydrogen atom Ch. 9
Nov 15 Multi-electron atoms Ch.10
Nov 17 Periodic properties Ch. 10
Nov 19 Periodic properties Ch. 10
Nov 22 Valence-bond; Lewis structures Ch. 11
Nov 24 Hybrid orbitals; VSEPR Ch. 11, 12
Nov 26 VSEPR Ch. 12
Nov 29 MO theory Ch. 12
Dec 1 MO theory Ch. 12
Dec 2 Review for exam

2. Y (x) A kx = sin PARTICLE IN A BOX ENERGY Y(0) = 0 Y(L) = 0 x Lx L
BOUNDARY CONDITION

3. PARTICLE IN A BOX
ENERGY
n = 3
QUANTIZED
n = 2
n =1

4. PARTICLE IN A BOX The application of the BOUNDARY CONDITIONS
Gives a series of QUANTIZED ENERGY LEVELS
ONLY CERTAIN ENERGIES ALLOWED!
DETERMINED BY THE NUMBER n
n is a QUANTUM NUMBER

5. 2 n p Y = (x) sin x L L THE WAVEFUNCTIONS A NODE ENERGY n = 3 Y
CHANGES SIGN
n = 2
THE NUMBER OF NODES IS GIVEN BY N-1 ….. INCREASING NUMBER OF NODES AS THE ENERGY INCREASES
n =1

7. What does Y Mean????? The answer lies in WAVE-PARTICLE DUALITY
Electrons have both wavelike and particle like properties.
Because of the wavelike character of electron
we CANNOT say that an electron
WILL be found at certain point in an atom!

8. THE HEISENBERG UNCERTAINTY PRINCIPLE
He postulated that if...
Dx is the uncertainty in the particle’s position
Dp is the uncertainty in the particle’s momentum
For a particle like an electron, we cannot know
both the position and velocity
to any meaningful precision simultaneously.

11. PARTICLE IN A BOX FOR ONE DIMENSION SCHRODINGER EQUATION 1-D REQUIRES
ONE QUANTUM NUMBER!

12. ( ) { } THREE DIMENSIONS E h m n L = + 8 h d d d - + + Y ( x , y , z )
SCHRODINGER EQUATION { } h 2 d 2 d 2 d 2 - + + Y ( x , y , z ) = E Y ( x , y , z ) p 2 2 2 2 n n n n n n n 8 m dx dy dz x y z x y z E h m n L x y z = + 2 8 ( )
3-D REQUIRES
THREE QUANTUM NUMBERS!!!

13. E h m n L = + 8 Note that when Lx=Ly=Lz E2,3,4 = E 3,2,4 = E4,2,3
These levels are said to be degenerate – this
means they are different wavefunctions, but
have the same energy

14. HYDROGEN ATOM The hydrogen atom is composed of a proton (+1)z
HYDROGEN ATOM y
Electron at (x,y,z)
Proton at (0,0,0) x
The hydrogen atom is
composed of a proton (+1)
and an electron (-1). If the
proton is located at the origin,
the electron is at (x, y, z)

15. We want to obtain the energy of the hydrogen atom system
We want to obtain the energy of the hydrogen atom system. We will do this the same way as we got it for the particle-in-a-box: by performing the “energy operation” on the wavefunction which describes the H atom system.
HY = EY
Remember that this equation is called the Schrodinger wave equation (SWE)

16. HY = EY { } Y = EY H = KE operator + PE operator H = + V + V2
-(h2 / 8p2m)
+ V 2 {
+ V }
Y = EY
-(h2 / 8p2m) 2
Where = {d2/dx2 + d2/dy2 +d2/dz2}

17. Now there is a difference from the particle-in-a-box problem: here there is a potential energy involved ….
Coulombic attraction between the proton and electron
V = -Ze2 / r

18. Since the potential energy depends
on the separation between the
proton and the electron, it is more
convenient to think about the problem
using a different co-ordinate system:
(x,y,z)  (r,q,j)

19. { } Y = EY + V Where now has terms in {d2/dr2 ; d2/dq2 ; d2/dj2}z
Electron at (r, q, j)
Proton at (0,0,0) x
After the transformation we still
have the Schrodinger equation 2 {
+ V }
Y = EY
-(h2 / 8p2m) 2
Where now has terms in
{d2/dr2 ; d2/dq2 ; d2/dj2}
and V = -Ze2 / r

20. Y(x,y,z)  Y(r,q,j) = R(r) x Y(q,j)
The result of solving the Schrodinger equation this way is that we can split the hydrogen wavefunction into two:
Y(x,y,z)  Y(r,q,j) = R(r) x Y(q,j)
Depends on angular variables
Depends on r only

21. The solutions have the same features we have seen already:
Energy is quantized
En = - R Z2 / n2
= x Z2 / n2 J [ n = 1,2,3 …]
Wavefunctions have shapes which depend on the quantum numbers
There are (n-1) nodes in the wavefunctions

22. Because we have 3 spatial dimensions, we end up with 3 quantum numbers:
n, l, ml
n = 1,2,3, …; l = 0,1,2 … (n-1); ml = -l, -l+1, …0…l-1, l
n is the principal quantum number – gives energy and level
l is the orbital angular momentum quantum number – it gives the shape of the wavefunction
ml is the magnetic quantum number – it distinguishes the various degenerate wavefunctions with the same n and l

23. n l ml
1 0 (s) 0
2 0 (s) 0
1 (p) -1, 0, 1
3 0 (s) 0
2 (d) -2, -1, 0, 1, 2

24. En = - R Z2 / n2
= x Z2 / n2 J [ n = 1,2,3 …]
… degenerate

25. The result (after a lot of math!)
Node at s = 2!!

27. Probability Distribution for the 1s wavefunction:
-r/a e
3/2 ÷ ø ö ç è æ = Y 1
100 a p
Maximum probability at nucleus

28. A more interesting way to look at things is by using the radial probability distribution, which gives probabilities of finding the electron within an annulus at distance r (think of onion skins)
max. away from nucleus

29. 90% boundary:
Inside this lies
90% of the probability
nodes