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

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

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

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

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

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!

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.

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

( ) { } 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!!!

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

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)

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)

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

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

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)

{ } 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

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

The solutions have the same features we have seen already: Energy is quantized En = - R Z2 / n2 = - 2.178 x 10-18 Z2 / n2 J [ n = 1,2,3 …] Wavefunctions have shapes which depend on the quantum numbers There are (n-1) nodes in the wavefunctions

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

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

En = - R Z2 / n2 = - 2.178 x 10-18 Z2 / n2 J [ n = 1,2,3 …] … degenerate

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

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

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

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