Physics 2203, Fall 2012 Modern Physics

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1 Physics 2203, Fall 2012 Modern Physics. Monday, Oct. 15th, Ch. 10, We are going to create the periodic table. Supplemental reading: Modern Physics Serway/Moses/Moyer Announcements Next Midterm on Oct. 31 st Chapters Midterm grades posted today. Submit a detailed outline of your paper by Monday, Oct. 29th

2 Quiz #3: average 7.5/10 4p Red Δm l = 1 Blue Δm l = 0 Green Δm l = +1 Three lines 3d

4 We now have four quantum numbers n = 1,2, 3,... l = 0,1,2, 3,...,(n 1) m l = l,l 1,.,0,..., l m s = ± 1 2 Next Energy n=2 l = 0,1 l = 1 : m l = 1,0,-1 l = 0: m l = 0 m s = ± 1 2 E n = 1 m e c 2 ( ke 2 ) 2 n 2 ( c) 2 Next Energy n=3 l = 0 and, 1 and 2 l = 2 : m l = 2,1,0,-1,-2 l = 1: m l = 1,0,-1 l = 0 : m l = 0 m s = ± 1 2 Lowest energy n=1 l = 0 m l = 0 m s = ± 1 2 How do we fill these quantum states? How many electrons in each quantum state? Within a given shell (n) which states fill first?

5 Pauli Exclusion Principle How to occupy quantum states We need to learn how to fill different levels Z eff : s, p, d, f Hund's Rule Lowest Energy maximizes Spin S z, consistent with Pauli Exclusion Hund's Rule (2) Maximizes L z whenever possible

6 The Pauli Principle is a consequence of Quantum Mechanics: IdenXcal parxcles are indis;nguishable in QM. (a) and (b) show two different classical pictures of a two electron sca\ering process. It is easy to keep track of electron 1 and electron 2 because we know the paths. (c) Depicts a quantum process where the paths are blurred by the wave nature of the electrons. Once they interact you can t know who is who indis;nguishable!

7 Let ψ ( x 1, x 2 ) describe the wave function of two identical particles in 1D ψ ( x 1, x 2 ) 2 dx 1 dx 2 probability of finding What is required is that #1 between x 1 x 1 + dx 1 #2 between x 2 x 2 + dx 2 ψ ( x 1, x 2 ) 2 = ψ ( x 2, x 1 ) 2 or ψ ( r 1,r ) 2 2 = ψ ( r 2,r ) 2 1 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

8 ψ ( r 1,r ) 2 = ±ψ ( r 2,r ) 1 + sign are Bosons (photons) sign are Fermions (electrons, protons, ) Look at product wave Functions ψ ( ab r 1,r ) 2 = ψ a )ψ b ) These product wave functions are solution, but ψ a )ψ b ) is not proper for indistinguishabity. ψ ( ab r 1,r ) 2 = ψ a )ψ b ) ±ψ a )ψ b ) + sign are Bosons (photons) sign are Fermions (electrons, protons, ) ψ ( ab r 2,r ) 1 = ψ a )ψ b ) ±ψ a )ψ b ) ψ ( ab r 2,r ) 1 = ψ ( r 1,r ) 2

9 Fermions: ψ ( r ab 1,r ) 2 = ψ a )ψ b ) -ψ a )ψ b ) Lets try to put two electrons into the same state, a=b ψ ( ab r 2,r ) 1 = ψ a )ψ a ) -ψ a )ψ a ) ψ ( ab r 2,r ) 1 = ψ a )ψ a ) -ψ a )ψ a ) 0 Bosons: ψ ( ab r 1,r ) 2 = ψ a )ψ b ) +ψ a )ψ b ) Lets try to put two bosons into the same state, a=b 1945 Nobel Prize for discovery of the Exclusion Principle. ψ ( ab r 2,r ) 1 = ψ a )ψ a ) +ψ a )ψ a ) ψ ( ab r 2,r ) 1 = ψ a )ψ a ) +ψ a )ψ a ) 2ψ a )ψ a ) Bose Condensation

10 n=1 n=2 n=3 n=4 n=5 E n,l = Z 2 eff (n,l) n 2 2 m e c ( ke 2 ) 2 ( c) 2 For a given n, how will Z eff depend upon l?

12 Let us construct the 2 electron ground state (Xme independent) of the He atom Assump;on: independent parxcle approximaxon. Each electron sees the proton but not each other. According to the Pauli exclusion principle the two electons go into n=1, l=0, m l = 0, m s = 1 / 2 n=1, l=0, m l = 0, m s = 1 / 2 Spectroscopic notation: 1s 2 ψ 100 ( r) = 1 2 π 1/2 a 0 3/2 e 2r /a 0 : Z=2 ψ a = (1,0,0,+) and ψ b = (1,0,0, ) E a = E b = (13.6eV ) = 54.4eV ψ,r 2 ) = ψ 100+ )ψ 100 ) ψ 100 )ψ 100+ )

13 E a = E b = (13.6eV ) = 54.4eV ψ,r 2 ) = ψ 100+ )ψ 100 ) ψ 100 )ψ 100+ ) The spatial dependence is the same only spin part different ψ,r 2 ) = 1 2 π a 0 3 e 2 +r 2 )/a Very important: second spin is always opposite first Pauli! Total Energy is E=E a + E b = 108.8eV Experimental number is ev Need to include screening of one electron by the other.

14 For He the 1s states are lower than for H because Z=2, but one electron will screen the other so we should have 1>Z eff <2. The measured ionization is 24.6eV ( ) eV = mz 2 ke 2 2 = 13.6eV Z 2n 2 ( ) 2 eff Z eff = 1.35e The first excited state is 19.8eV 19.8eV = mz 2 ke 2 eff 2 Z eff = 1.39e ( ) 2 1 ( ) ( ) E n = mz 2 ( ke 2 ) 2 eff 2n 2 ( ) 2 ( ) 2 ΔE n n ' = mz 2 ke 2 eff 2 1 ( ) 2 E ionization = mz 2 ( ke 2 ) 2 r n = n2 2 ke 2 Zm n 1 2 n' 2 2 ( ) = 13.6eV Z 2n 2 ( ) 2 eff The measured radius is 0.05 nm 0.05 = n2 2 ke 2 Zm Z eff = 1.6e

15 Pauli Exclusion Principle How to occupy quantum states Li has the configuration 1s 2 2s 1 : Because of Pauli Exclusion Principle.

16 Lets Start off by making sure we know how to use the Pauli Exclusion Principle Here we will just assign states Using the Pauli Exclusion Principle H 1s 2s S z 1/2 He 0 Li Be 1/2 0

For Li the 1s shell is filled. The first guess is that Z=1. Since the third electron is in the n=2 state the ionizaxon energy would be 3.4eV. The measured ionization is 5.4eV ( ) 2 5.

17 For Li the 1s shell is filled. The first guess is that Z=1. Since the third electron is in the n=2 state the ionizaxon energy would be 3.4eV. The measured ionization is 5.4eV ( ) 2 5.4eV = mz 2 ke 2 = 13.6eV 2n 2 ( ) 2 4 Z eff = 1.26e Why?? 2 ( Z eff ) E n = mz 2 ( ke 2 ) 2 eff 2n 2 ( ) 2 ( ) 2 ΔE n n ' = mz 2 ke 2 eff 2 1 ( ) 2 E ionization = mz 2 ( ke 2 ) 2 r n = n2 2 ke 2 Zm n 1 2 n' 2 2 ( ) = 13.6eV Z 2n 2 ( ) 2 eff The measured radius is 0.20 nm 0.20 = 4 2 ke 2 Zm Z eff 1.5e This simple picture breaks down for too large Z and n.

For Be both the 1s and 2s shells are filled. Can t do much here using the simple hydrogen model. You would think that Be would be like He with a high excitaxon energy but it is not.

18 For Be both the 1s and 2s shells are filled. Can t do much here using the simple hydrogen model. You would think that Be would be like He with a high excitaxon energy but it is not. The lowest excitaxon energy is only 2.7 ev. You find the effecxve Z! Be 1s 2s Ground state E n = mz 2 ( ke 2 ) 2 eff 2n 2 ( ) 2 ( ) 2 ΔE n n ' = mz 2 ke 2 eff 2 1 ( ) 2 E ionization = mz 2 ( ke 2 ) 2 r n = n2 2 ke 2 Zm n 1 2 n' 2 2 ( ) = 13.6eV Z 2n 2 ( ) 2 eff 1s 2s 2p Be First excited state Next comes B, C, N, O, F,---Need to learn about Pauli and Hund!

19 We need to learn how to fill different levels Z eff : s, p, d, f 6 Atoms: 2p electrons

20 Quantum Numbers still the same. Principal n= 1, 2, 3,... Orbital l = 0,1,2,...(n-1) Magnetic m l l But with Z eff (n,l) the l states for a fixed n will have different E n,l E n.0 < E n,1 < E n,2 Look at the n=2 states. The maximum (most probable posixon) is 4 5 Xmes larger that 1s. But the 2s has a second maximum much closer to r=0 than 2p. Bo\om line is that Z eff for the 2s larger than for 2p. For n=3 the Z eff (3s)>Z eff (3p)>Z eff (3d). 4s<3d 2s<2p

21 Hund's Rule Lowest Energy maximizes Spin S z, consistent with Pauli Exclusion Hund's Rule (2) Maximizes L z whenever possible Now we are ready to keep going with the periodic chart 2p 1s 2s m l =1 m l =0 m l = 1 S Z L Z B 1/2 1 C 1 1 N 3/2 0 0 F Ne 1 1 1/

22 Na 3s(Ne) 1s22s22p63s1

23 Na 3s(Ne) 1s 2 2s 2 2p 6 3s 1 : IonizaXon energy is 5.14 ev E = Z 2 eff (13.6eV ) = 5.14eV 2 3 Z eff = = 1.84e You look up radius and calculate a Z eff from the radius?

24 2: (25 points) This is a problem about using Hund s Rule and the Pauli Principle in a many electron atom, Oxygen (Z=8). Write down the ground state configuraxon of O using the Pauli Principle. Electronic configuraxon 1s 2 Answer Electronic configuraxon 1s 2 2s 2 2p 4 b) Use the Hund s rule and the Pauli principle to describe the ground state, using the boxes below and arrows for the spin states. 2p 1s 2s m l =1 m l =0 m l = 1 S Z L Z 0 1 1

25 2: (25 points) This is a problem about using Hund s Rule and the Pauli Principle in a many electron atom, Oxygen (Z=8). Write down the ground state configuraxon of O using the Pauli Principle. c) The energy needed to remove one of the 1s electrons in 540eV. Find Z eff. Z eff = Why isn t Z eff =8? The measured ionization is 540eV ( ) 2 ( ) 540eV = mz 2 ke 2 2 = 13.6eV Z 2n 2 ( ) 2 eff Z eff = 6.3e

26 11 Atoms n=3 We need to learn how to fill different levels Z eff : s, p, d, f Example of complexity Filling 3 d shell 10 electrons Ni configuration is Ar closed shell plus 3d 8 4s 2 : Cu is 3d 10 4s 1

27 Hund's Rule Lowest Energy maximizes Spin S z, consistent with Pauli Exclusion Nitrogen is 1s 2 2s 2 2 p 3 with S z = 3 / 2 :Why???

28 In Principle we just use the Schödinger equaxon to solve for the energy of a mulxelectron atom. In pracxce we can t solve the problem for 2 electrons, i.e. He! 2 2m 2 Ψ( r 1,r2,...rn ) + U ( r 1,r2,...rn )Ψ( r 1,r2,...rn ) = EΨ( r 1,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 r 1,r2,...rn ( )ψ 2 ( r 2 )...ψ n r n ( ) = ψ 1 r 1 ( ) This gives us n differential equations to solve. 2 2m 2 + U n,eff ψ n = E n ψ n

29 This gives us n differential equations to solve. 2 2m 2 + U n,eff ψ n = E n ψ n Example He Use two 1s functions with some Z eff Use Poissions Equation to find U eff Use U eff to calculate wavefuctions over and over again until you converge. Guess the wave functions. Use Poissions Equation to find U eff Use U eff to calculate a new set of wavefuctions over and over again until you converge. Self Consistent Calculation Hartree theory Hartree Fock Theory Density Functional Theory. What will this effective potential look like If the electron is very close to the nucleus Z eff = Z If the electron is far from the nucleau Z eff = 1

Guess the wave functions. Use Poissions Equation to find U eff Use U eff to calculate a new set of wavefuctions over and over again until you converge.

30 Guess the wave functions. Use Poissions Equation to find U eff Use U eff to calculate a new set of wavefuctions over and over again until you converge. Self Consistent Calculation Hartree theory Hartree Fock Theory Density Functional Theory. Kohn and Sham proved that there is Functional of the density F(ρ) that is an exact solution to the many electron Schördinger Equations But only God knows what this Function is!!

31 Violates Hund s 1 st Rule Violates Pauli Principle Violates Pauli Principle Violates Hund s 2 nd Rule

Use the Venn Diagram to compare and contrast the Bohr Model of the atom with the Quantum Mechanical Model of atom

Use the Venn Diagram to compare and contrast the Bohr Model of the atom with the Quantum Mechanical Model of atom Bohr Model Quantum Model Energy level Atomic orbital Quantum Atomic number Quantum mechanical