Physics2203,Fall2012ModernPhysics

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 Monday,Oct.15th,2012.‐‐Ch.10,Wearegoingtocreatetheperiodictable.Supplementalreading:“ModernPhysics”Serway/Moses/Moyer

 Announcements‐‐‐NextMidtermonOct.31st—Chapters8‐12.‐‐‐Midtermgradespostedtoday.‐‐SubmitadetailedoutlineofyourpaperbyMonday,Oct.29th

3d

4p

Red Δml = −1Blue Δml = 0Green Δml = +1

Threelines

Quiz#3:average7.5/10

We now have four quantum numbersn =1,2,3,....l = 0,1,2,3,....,(n −1)ml = l,l −1,.,0,...,−l

ms = ±12

En = −1n2

mec2 ke2( )2

c( )2

Next Energy n=2l = 0,1l =1 :ml = 1,0,-1l = 0: ml = 0

ms = ±12

Lowest energy n=1l = 0ml = 0

ms = ±12

Next Energy n=3l = 0 and, 1 and 2 l = 2 :ml = 2,1,0,-1,-2l =1 :ml = 1,0,-1l = 0 :ml = 0

ms = ±12

Howdowefillthesequantumstates?Howmanyelectronsineachquantumstate?Withinagivenshell(n)whichstatesfillfirst?

We need to learn how to fill different levels ⇒ Zeff : s, p, d, fPauli Exclusion Principle ⇒ How to occupy quantum states

Hund's Rule ⇒ Lowest Energy maximizes Spin Sz , consistent with Pauli ExclusionHund's Rule (2)⇒ Maximizes L z whenever possible

ThePauliPrincipleisaconsequenceofQuantumMechanics:IdenXcalparXclesareindis;nguishableinQM.

(a)and(b)showtwodifferentclassicalpicturesofatwo‐electronsca\eringprocess.Itiseasytokeeptrackofelectron1andelectron2becauseweknowthepaths.

(c)Depictsaquantumprocesswherethepathsareblurredbythewavenatureoftheelectrons.Oncetheyinteractyoucan’tknowwhoiswho—indis;nguishable!

Let ψ x1, x2( ) describe the wave function of two identical particles in 1D

ψ x1, x2( )2dx1dx2 ⇒ probability of finding

#1 between x1 ⇒ x1 + dx1

#2 between x2 ⇒ x2 + dx2

What is required is that

ψ x1, x2( )2= ψ x2 , x1( )

2 or

ψ r

1,r

2( )2= ψ r

2 ,r

1( )2

There are two obvious options

ψ r

1,r

2( ) = ±ψ r

2 ,r

1( )

+ sign are Bosons (photons): spin 0− sign are Fermions (electrons, protons, ): spin 1/2

ψ r

1,r

2( ) = ±ψ r

2 ,r

1( ) + sign are Bosons (photons)− sign are Fermions (electrons, protons, )

Look at product wave Functions

ψab r

1,r

2( ) =ψa (r1)ψb (r2 )

These product wave functions are solution, butψa (r1)ψb (r2 ) is not proper for indistinguishabity.

ψab r

1,r

2( ) =ψa (r1)ψb (r2 ) ±ψa (r2 )ψb (r1)

+ sign are Bosons (photons)− sign are Fermions (electrons, protons, )

ψab r

2 ,r

1( ) =ψa (r2 )ψb (r1) ±ψa (r1)ψb (r2 )

ψab r

2 ,r

1( ) = −ψ r

1,r

2( )

Fermions: ψab r

1,r

2( ) =ψa (r1)ψb (r2 ) -ψa (r2 )ψb (r1)

Lets try to put two electrons into the same state, a=b

ψab r

2 ,r

1( ) =ψa (r2 )ψa (r1) -ψa (r1)ψa (r2 )

ψab r

2 ,r

1( ) =ψa (r2 )ψa (r1) -ψa (r2 )ψa (r1) ≡ 0

Bosons: ψab r

1,r

2( ) =ψa (r1)ψb (r2 ) +ψa (r2 )ψb (r1)

Lets try to put two bosons into the same state, a=b

ψab r

2 ,r

1( ) =ψa (r2 )ψa (r1) +ψa (r1)ψa (r2 )

ψab r

2 ,r

1( ) =ψa (r2 )ψa (r1) +ψa (r2 )ψa (r1) ≡ 2ψa (r2 )ψa (r1)

Bose Condensation

1945NobelPrizefordiscoveryoftheExclusionPrinciple.”

n=1n=2

n=3

n=4

n=5

En,l = −Zeff

2 (n,l)n2

mec2 ke2( )2

c( )2 Foragivenn,howwillZeffdependuponl?

Letusconstructthe2electrongroundstate(Xmeindependent)oftheHeatom

Assump;on:independent‐parXcleapproximaXon.Eachelectronseestheprotonbutnoteachother.

According to the Pauli exclusion principle the two electons go inton=1, l=0, ml = 0,ms =1 / 2n=1, l=0, ml = 0,ms = −1 / 2Spectroscopic notation: 1s2

ψ100 r( ) = 1π1/2

2a0

3/2

e−2r /a0 : Z=2

ψa = (1,0,0,+) and ψb = (1,0,0,−)

Ea = Eb = −21

2

(13.6eV ) = −54.4eV

ψ(r1,r2 ) =ψ100+(r1)ψ100−(r2 )−ψ100−(r1)ψ100+(r2 )

Ea = Eb = −21

2

(13.6eV ) = −54.4eV

ψ(r1,r2 ) =ψ100+(r1)ψ100−(r2 )−ψ100−(r1)ψ100+(r2 )

The spatial dependence is the same only spin part different

ψ(r1,r2 ) = 1π

2a0

3

e−2(r1+r2 )/a0 +− − −+

Veryimportant:secondspinisalwaysoppositefirst–Pauli!

Total Energy is E=Ea + Eb = −108.8eVExperimental number is -79.0 eVNeed to include screening of one electron by the other.

En = −mZ

eff

2 ke2( )2

2n2( )2

ΔEn→n ' = −mZ

eff

2 ke2( )2

2 ( )21n2 −

1n '2

Eionization =mZ 2 ke2( )2

2n2( )2 =13.6eV Z

eff

2( )

rn =n2

2

ke2Zm

ForHethe1sstatesarelowerthanforHbecauseZ=2,butoneelectronwillscreentheothersoweshouldhave1>Zeff<2.

The measured ionization is 24.6eV

24.6eV =mZ 2 ke2( )2

2n2( )2 = 13.6eV Z

eff

2( )Zeff = 1.35e

The first excited state is 19.8eV

19.8eV = −mZ

eff

2 ke2( )2

2 ( )2112 −

122

Zeff = 1.39e

The measured radius is 0.05 nm

0.05 = n2

2

ke2ZmZeff = 1.6e

Pauli Exclusion Principle ⇒ How to occupy quantum states

Li has the configuration 1s2 2s1 :⇒ Because of Pauli Exclusion Principle.

LetsStartoffbymakingsureweknowhowtousethePauliExclusionPrinciple

HerewewilljustassignstatesUsingthePauliExclusionPrinciple H

1s 2s

He

Li

Be

Sz

1/2

0

1/2

0

En = −mZ

eff

2 ke2( )2

2n2( )2

ΔEn→n ' = −mZ

eff

2 ke2( )2

2 ( )21n2 −

1n '2

Eionization =mZ 2 ke2( )2

2n2( )2 =13.6eV Z

eff

2( )

rn =n2

2

ke2Zm

ForLithe1sshellisfilled.ThefirstguessisthatZ=1.Sincethethirdelectronisinthen=2statetheionizaXonenergywouldbe3.4eV.

The measured ionization is 5.4eV

5.4eV =mZ 2 ke2( )2

2n2( )2 =

13.6eV4

Zeff

2( )Zeff = 1.26e⇒Why??

The measured radius is 0.20 nm

0.20 = 42

ke2ZmZeff ≈ 1.5e

This simple picture breaks down for too large Z and n.

En = −mZ

eff

2 ke2( )2

2n2( )2

ΔEn→n ' = −mZ

eff

2 ke2( )2

2 ( )21n2 −

1n '2

Eionization =mZ 2 ke2( )2

2n2( )2 =13.6eV Z

eff

2( )

rn =n2

2

ke2Zm

ForBeboththe1sand2sshellsarefilled.Can’tdomuchhereusingthesimplehydrogenmodel.YouwouldthinkthatBewouldbelikeHewithahighexcitaXonenergybutitisnot.ThelowestexcitaXonenergyisonly2.7eV.YoufindtheeffecXveZ!

Next comes B, C, N, O, F,---Need to learn about Pauli and Hund!

2p

Be

1s 2s

Firstexcitedstate

Be

1s 2s

Groundstate

We need to learn how to fill different levels ⇒ Zeff : s, p, d, f

6Atoms:2pelectrons

Quantum Numbers still the same.Principal n= 1, 2, 3, ...Orbital l = 0,1,2,...(n-1) Magnetic ml ≤ l

But with Zeff (n,l) the l statesfor a fixed n will have different En,l En.0 < En,1 < En,2

Lookatthen=2states.Themaximum(mostprobableposiXon)is4‐5Xmeslargerthat1s.Butthe2shasasecondmaximummuchclosertor=0than2p.Bo\omlineisthatZeffforthe2slargerthanfor2p.Forn=3theZeff(3s)>Zeff(3p)>Zeff(3d).

4s<3d

2s<2p

Hund's Rule ⇒ Lowest Energy maximizes Spin Sz , consistent with Pauli ExclusionHund's Rule (2)⇒ Maximizes L z whenever possibleNowwearereadytokeepgoingwiththeperiodicchart

1s 2s

2p

ml=1 ml=0 ml=‐1

B

C

N

0

F

Ne

SZ LZ

1/2 1

1 1

3/2 0

1 1

1/2 1

0 0

Na‐‐‐3s(Ne)—1s22s22p63s1

Na‐‐‐3s(Ne)—1s22s22p63s1:IonizaXonenergyis5.14eV

E = −Zeff2

32(13.6eV ) = 5.14eV

Zeff = 35.1413.6

=1.84e

YoulookupradiusandcalculateaZefffromtheradius?

2:(25points)ThisisaproblemaboutusingHund’sRuleandthePauliPrincipleinamanyelectronatom,Oxygen(Z=8).WritedownthegroundstateconfiguraXonofOusingthePauliPrinciple.

ElectronicconfiguraXon1s2

AnswerElectronicconfiguraXon1s22s22p4

b) UsetheHund’sruleandthePauliprincipletodescribethegroundstate,usingtheboxesbelowandarrowsforthespinstates.

0

1s 2s

2p

ml=1 ml=0 ml=‐1 SZ LZ

1 1

2:(25points)ThisisaproblemaboutusingHund’sRuleandthePauliPrincipleinamanyelectronatom,Oxygen(Z=8).WritedownthegroundstateconfiguraXonofOusingthePauliPrinciple.

c)Theenergyneededtoremoveoneofthe1selectronsin540eV.FindZeff.

Zeff= Whyisn’tZeff=8?

The measured ionization is 540eV

540eV =mZ 2 ke2( )2

2n2( )2 = 13.6eV Z

eff

2( )Zeff = 6.3e

We need to learn how to fill different levels ⇒ Zeff : s, p, d, f

Example of complexity ⇒ Filling 3 d shell⇒10 electrons

11Atoms

Ni configuration is Ar closed shell plus 3d 8 4s2 : Cu is 3d10 4s1

n=3

Hund's Rule ⇒ Lowest Energy maximizes Spin Sz , consistent with Pauli Exclusion

Nitrogen is 1s2 2s2 2p3 with Sz = 3 / 2 :Why???

InPrinciplewejustusetheSchödingerequaXontosolvefortheenergyofamulX‐electronatom.InpracXcewecan’tsolvetheproblemfor2electrons,i.e.He!

−2

2m∇2Ψ r1,r2,.....rn( ) +U r1,r2,.....rn( )Ψ r1,r2,.....rn( ) = EΨ r1,r2,.....rn( )

First and essential approximation is assume we can represent the potential as some effective potential only dependent upon the electron density n(r).

Second assumption is that the many particle wave function can be written as a product function.

Ψ r1,r2,.....rn( ) =ψ 1 r1( )ψ 2 r2( ).......ψ n rn( )

This gives us n differential equations to solve.

−

2

2m∇2 +Un,effψ n = Enψ n

This gives us n differential equations to solve.

−

2

2m∇2 +Un,effψ n = Enψ n

Example HeUse two 1s functions with some Zeff

Use Poissions Equation to find Ueff

Use Ueff to calculate wavefuctionsover and over again until you converge.

What will this effective potential look likeIf the electron is very close to the nucleus Zeff = ZIf the electron is far from the nucleauZeff = 1

Guess the wave functions.Use Poissions Equation to find Ueff

Use Ueff to calculate a new set of wavefuctionsover and over again until you converge.Self Consistent CalculationHartree theoryHartree Fock TheoryDensity Functional Theory.

Guess the wave functions.Use Poissions Equation to find Ueff

Use Ueff to calculate a new set of wavefuctionsover and over again until you converge.Self Consistent CalculationHartree theoryHartree Fock TheoryDensity Functional Theory.

Kohn and Sham proved that there is Functional of the density F(ρ) that isan exact solution to the many electronSchördinger EquationsBut only God knows what this Function is!!

ViolatesHund’s1stRule

ViolatesPauliPrinciple

ViolatesPauliPrinciple

ViolatesHund’s2ndRule