# Infinite Potential Well … bottom line

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Infinite Potential Well … bottom line
a x V ( ) = E l e c t r o n Energies are quantized, defined by one single quantum number, n = 1, 2, 3, 4 …

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 V ( x ) o = a How do we calculate the probability ?? … we need to solve Schrodinger’s equation … apply boundary conditions etc.

Tunneling We now need to apply BC’s at x=0 and x=a …
V ( x ) o = a 2nd order linear differential equation. … (E-V) negative for zone II, hence no wave 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 V ( x ) o = a I n c i d e t R f l A 1 2 y I y I Therefore … the solution suggests that the electron can be found beyond the barrier VO … EVEN THOUGH its energy E is less than VO!

Tunneling 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 … I V ( x ) o = a y n c i d e t R f l A 1 2

Application of Tunneling
x V ( ) M e t a l y S c o n d u m b I ( a ) M t e r i l s u f c P o b S n I T u n e l i g c r t a y s v ( A ) x Å

The Potential “Box” ¥ z c V y b a x =
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: y z x a c b V =

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

The H-Atom … An Overview
Describe the H-atom … i.e. what does the nucleus look like? how much charge is there at the nucleus? i.e. Z=1 how 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 & PE Obviously 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:

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 x y z r q N u c l e s f P ( , ) + Z i n V ( r ) 4 p e o Z 2 = +

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:

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 Number l=0,1,2,3 ……(n-1) is the Orbital Angular Momentum Quantum Number ml=-l, -(l-1), -(l-2), ……-2, -1, 0, 1, 2, …+l is the Magnetic Quantum Number … for now MEMORIZE these!

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
3,2,2 2,3,1 2,3,0 1,2,4 2,3,-1 2,0,1 1,1,0 1,2,3 4,1,2 1,0,0

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! The value of Psi can be substituted back in

An electron with velocity 2. 1E6 m/s strikes a H atom
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; ΔE = eV; λ = nm

Energy ! ¥ K E l e c t r o n g y , = ­ 1 5 I i z . 3 6 V G u d s a C m
= K 1 5 3 6 V G u d s a 2 4 8 I i z C m f x

Orbital Angular Momentum
Just like energy (En) … angular momentum L is also quantized… by ‘l’ … what happens when l=0 ? x z O r b i t n g e l c o q L B a y

Orbital Angular Momentum
For l=2 … ml would be … …-2, -1, 0, 1, 2 1 2 m l = z L h ( + ) B e x t r n a x z O r b i t n g e l c o q L B a y

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 … E n e r g y 1 s 2 p 3 d 4 f 5 l = . 6 V P h o t

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)
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)

The Quantum Numbers

Radial Probability . 2 4 6 8 r ( n m ) p 1 s = R , . 2 4 r ( n m ) 1 s 6 8 p = | R , Bohr Radius … the radial distance where the radial probability is maximum

“Angular” Probability
x y z | Y 2 f o r a p b i t l ( m = ) 1 s s-states symmetrical p-states directional

Multi electron atom : He

Energy states for multi electron atom
1 s 2 3 4 5 6 p d f K L M N O 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 …