Infinite Potential Well … bottom line

Equationr Schrodinge

0V)ψ(E2m

dx

ψd22

2

0 ax

V(x)

0

¥

V = 0

Electron

V = ¥ V = ¥

Energies are quantized, defined by one single quantum number, n = 1, 2, 3, 4 …

8ma

hnE

22

n

Tunneling

“ … If an electron comes up a potential barrier greater than its energy … there is a finite probability that it will “pass” through the barrier…”

A

B

C

D

E

… for an electron !We place an electron in region I… with energy E less than VO

(E<VO)… what is the probability the electron will be in I … II … III ??

I II III

V(x)Vo

x = 0 x = a x

How do we calculate the probability ??… we need to solve Schrodinger’s equation … apply boundary

conditions etc.

Tunneling

I II III

V(x)Vo

x = 0 x = a x

-jkx2

jkx1I eAeA(x)ψ

0V)ψ(E2m

dx

ψd22

2

-jkx2

jkx1III eCeC(x)ψ

x-2

x1II eBeB(x)ψ

2O2

22 E)-2m(V

α and 2mE

k with

We now need to apply BC’s at x=0 and x=a …The properties of ψ require that it be continuous and single

valued

Tunneling

I II III

V(x)Vo

x = 0 x = a x

(x)ψ ... (x)ψ ... (x)ψ IIIIII

yIIIyII

I

Incident

Reflected

A1

A2

Therefore … the solution suggests that the electron can be found beyond the barrier VO … EVEN THOUGH its energy E is less than VO!

Tunneling

(x)ψ

(x)ψT 2

I

2

III

I II III

V(x)Vo

x = 0 x = a x

yIIIyII

I

Incident

Reflected

A1

A2

E)-4E(V

VD where

a) (αDsinh1

1T

O

2O

2

2O2 E)-2m(V

α

What are the important factors that influence the tunneling probability ??

… the energy of the electron… the width and height of the barrier

For a wide or high barrier …

2O

OO

a) α 2(O

V

E)16E(VTwith

eTT and 1a α

Application of Tunneling(a)

Materialsurface

Probe ScanItunne

x

V(x)

Metal

y(x) Second Metal

Vacuum

Vo

(b)

lx

Itunnel

Tunnelingcurrent grayscale value(nA)

x (Å)

y (Å)

The Potential “Box”

y

z

x

0

a

c

bV = 0

V = ¥

V = ¥

V = ¥

V = ¥

If you confine an electron in a box … what would you expect the wave-function to be?

Think of it as a combination of 3 one-dimensional infinite potential wells… and therefore

the general solution will have the form of:

cπn

k ,bπn

k ,aπn

k where

z)y)sin(kx)sin(kAsin(kz)y,ψ(x,

3z

2y

1x

zyx

The Potential “Box”

The solution to the electron in a “box” problem results in 3 quantum numbers

A specific solution or eigenfunction i.e.

is called a state …

Note that the electron energy is quantized and depends on 3 quantum numbers

)n n (n8

E 23

22

212

2

n,n,n 321

ma

h

etc. ψor ψor ψor ψ 2,1,31,4,22,1,21,1,2

The H-Atom … An OverviewDescribe the H-atom …

i.e. what does the nucleus look like?how much charge is there at the nucleus?

i.e. Z=1how many electrons?

The H-atom represents the simplest system we can use to have a look at a real quantum physics example

The H-Atom … Force & PEObviously the electron is being attracted to the nucleus

because of the …… Coulombic attraction between two opposite charges!

The force between two charges is:

and the potential energy is given by:

rε4πe-

V(r)o

2

2o

21

rε4πQQ

F

The H-Atom … Spherical Coordinates

Due to the spherical symmetry of the H-atom … it makes sense to work in the spherical coordinate system instead of the cartesian one … i.e. x, y, z r,θ,φ

rε4πe-

V(r)o

2

x

y

z

r

q

Nucleus

f

P(r,q,f)

+Ze

e

rsinq V(r)

r

4peor Ze2

V(r) =

+Ze

The H-Atom … Wavefunction

No need to go through the solution in detail … … we do however need to understand the origin of certain parameters and functions!

Obtaining the wavefunction for the H-atom electron can be done by solving …

… in 3-dimensions

i.e. one would expect to get … 3 quantum numbers!

And the general wave function looks like:

φ) , (θY(r)Rφ)θ,(r,ψll ml,ln,ml,n,

The H-Atom … Quantum Numbers

Two functions … R a function of r and Y a function of θ and φ

Three quantum numbers! … n, l, ml

The spherical part i.e. R depends on n and l … while the angular or spherical one, Y depends on l and ml

n=1,2,3,4,…… is the Principal Quantum Numberl=0,1,2,3 ……(n-1) is the Orbital Angular Momentum

Quantum Numberml=-l, -(l-1), -(l-2), ……-2, -1, 0, 1, 2, …+l is the Magnetic

Quantum Number… for now MEMORIZE these!

φ), (θY(r)Rφ)θ,(r,ψll ml,ln,ml,n,

The H-Atom … Quantum Numbers

l

n 0-> s

1-> p

2-> d

3-> f

4-> g

1 1s

2 2s 2p

3 3s 3p 3d

4 4s 4p 4d 4f

5 5s 5p 5d 5f 5g

Let’s check the validity of these Energy States

3,2,22,3,12,3,01,2,42,3,-12,0,11,1,01,2,34,1,21,0,0

Energy !

Electron energies depend on n only … given by:

What does this energy represent ?… the energy required to remove the electron from the n=1 state (i.e. to free the electron)

also known as the ionization energy …

Electrons prefer to minimize their energy … therefore most likely to be found in n=1 state known as the ground state!

2

2

222o

24

n n

(13.6eV)Z

nh8ε

ZmeE

eV 13.6E1

An electron with velocity 2.1E6 m/s strikes a H atom. Find the n th energy level the electron will excite to. Calculate the wavelength of the light as the electron returns to ground

state.

K.E. = 12.5eV; n = 3.51 -> 3; ΔE = 12.09 eV; λ = 102.6 nm

Energy !Electron energy, En.

0E = KE

15

5

10

1 13.6 eV Ground state

2 3.40

3 1.514 0.855 0.54

n

n = ¥

Ionizationenergy, EI

n = 1

Continuum of energy. Electron is free

Exc

it ed

s tat

es

Orbital Angular Momentum

Just like energy (En) … angular momentum L is also quantized… by ‘l’

… what happens when l=0 ?

1/2l 1)][l(lL

x

z

Orbiting electron

qLL z

Bexternal

y

Orbital Angular Momentum

0

1

2

1

2

ml

l = 2

z

L = h 2(2+1)

Bexternal

x

z

Orbiting electron

qLL z

Bexternal

y

For l=2 … ml would be ……-2, -1, 0, 1, 2

lZ mL

Selection Rules …

Electron has momentum … also photons have intrinsic momentum

When photons are absorbed … in addition to the energy conservation momentum must also be conserved …

Selection rules: Δl=±1 … Δml=0, ±1

i.e. if electron is in ground state 1,0,0 … (n,l,ml)

If enough energy is gained to move up to n=2 then what are l and ml?

l …0, 1 … and ml … -1, 0, 1

Therefore … n=2, l=1, and ml=-1,0,1

Selection Rules …

Energy

0

1s1

2s 2p2

3s 3p3 3d

4s 4p4 4d 4f

5s 5p5 5d 5f

l = 0 l = 1 l = 2 l = 3

13.6eV

nl

Photon

Selection Rule Example

An electron in State (3,2,-2). What are the energy states in Shell N, this electron can jump to?

Spin … (intrinsic angular momentum)

1/2s with 1)][s(sS 1/2

Spin: last quantum number required to fully describe an electron!

The component of the spin along a magnetic field is also quantized (i.e. if B-field is in Z-direction)

1/2m with mS ssZ

The Quantum Numbers

Radial Probability

Bohr Radius … the radial distance where the radial probability is maximum

0 0 .2 0.4

r (nm)

1s

00 0.2 0.4 0 .6 0 .8

0

2p

2s

r (nm)

n = 2n = 1

r2|R2,0|2

r2|R2,1|2

r2|R1,0|2

0 0.2 0 .4 0.6 0.8

r (nm)

0

2p

0 0.2 0 .4

1s

0

n = 1

0

2s

n = 2

r (nm)

R1,0 R20

R21

φ), (θY(r)Rφ)θ,(r,ψll ml,ln,ml,n,

“Angular” Probability

s-states symmetricalp-states directional

x

y

z

x

y

z

|Y|2 for a 2px orbital

|Y|2 for a 2pz orbital(ml = 0)

x

y

z

x

y

z

|Y|2 for a 1s orbital

|Y|2 for a 2py orbital

+ Z e

N u c le u s

r 1

r 2

r 1 2

E le c tro n 1

E le c tro n 2

- e

- e

A helium-like atom. The nucleus has a charge of +Ze, where for He Z = 2.If one electron is removed, we have the He+ ion which is equivalent to thehydrogenic atom with Z = 2.

Multi electron atom : He

Energy

1sn

1 2 3 4 5 6

2s2p

3s3p

3d 4s4p

4d

4f

5s5p

5d

5f

6s

6p

5g

K

L

M

N

O

Energy states for multi electron atom

Energy depends on both n and l

Pauli Exclusion Principle & Hund’s Rule

NO two electrons can have the same set of quantum numbers …

i.e. if one electron in ψ1,0,0,1/2 then a second electron in the same system will have … ψ1,0,0,-1/2

Electrons in the same n, l orbital “like” to have “parallel" spins …

L i

B e B

K

L p

(n = 1 )

(n = 2 )s

-1 0 1 = m

H

s

H e

K

L p

(n = 1 )

(n = 2 )

Electronic configurations for the first five elements. Each box representsan orbital (n, , m ).

F N e

s

p

sK

L

s

p

sK

L

C N O

Electronic configurations for C, N, O, F and Ne atoms. Notice that Hund'srule forces electrons to align their spins in C, N and O. The Ne atom hasall the K and L orbitals full.