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PY3P05

Lectures 10-11: Multi-electron atomsLectures 10-11: Multi-electron atoms

o Schrödinger equation for

o Two-electron atoms.

o Multi-electron atoms.

o Helium-like atoms.

o Singlet and triplet states.

o Exchange energy.

PY3P05

System of non-interacting particlesSystem of non-interacting particles

o What is probability of simultaneously finding a particle 1 at (x1,y1,z1), particle 2 at (x2,y2,z2), etc. => need joint probability distribution.

o N-particle system is therefore a function of 3N coordinates:

(x1,y1,z1; x2,y2,z2; … xN,yN,zN)

o Must solve (1)

o First consider two particles which do not interact with one

another, but move in potentials V1 and V2. The Hamiltonian is

(2)

o The electron-nucleus potential for helium is

o The eigenfunctions of H1 and H2 can be written as the product:

€

ˆ H Ψ(ˆ r 1, ˆ r 2,..., ˆ r N ) = EΨ(ˆ r 1, ˆ r 2,..., ˆ r N )

€

ˆ H = ˆ H 1 + ˆ H 2

= −h2

2m1

∇12 + V1( ˆ r 1)

⎛

⎝ ⎜

⎞

⎠ ⎟+ −

h2

2m2

∇ 22 + V2( ˆ r 2)

⎛

⎝ ⎜

⎞

⎠ ⎟

€

Vi = −2e2

4πε0ri

€

i =1,2

€

( ˆ r 1, ˆ r 2) =ψ1(ˆ r 1)ψ 2(ˆ r 2)

xy

z

e1e2

r1r2

PY3P05

System of non-interacting particlesSystem of non-interacting particles

o Using this and Eqns. 1 and 2,

o That is, where E = E1+E2.

o The product wavefunction is an eigenfunction of the complete Hamiltonian H, corresponding to an eigenvalue E which is the sum of the energy eigenvalues of the two separate particles.

o For N-particles,

o Eigenvalues of each particle’s Hamiltonian determine possible energies. Total energy is thus

o Can be used as a first approximation to two interacting particles. Can then use perturbation theory to include interaction.

€

ˆ H Ψ(ˆ r 1, ˆ r 2) = ( ˆ H 1 + ˆ H 2)ψ1(ˆ r 1)ψ 2( ˆ r 2)

= (E1 + E2)ψ1(ˆ r 1)ψ 2( ˆ r 2)

€

ˆ H Ψ(ˆ r 1, ˆ r 2) = EΨ(ˆ r 1, ˆ r 2)

€

( ˆ r 1, ˆ r 2,...ˆ r N ) =ψ1(ˆ r 1)ψ 2(ˆ r 2)L ψ N (ˆ r N )

€

E = E i

i=1

N

∑

PY3P05

Application to heliumApplication to helium

o Assuming each electron in helium is non-interacting, can assume each can be treated independently with hydrogenic energy levels:

o Total energy of two-electron system in ground

state (n(1) = n(2) = 1) is therefore

o For first excited state, n(1) = 1, n(2) = 2 => E =-68 eV.

€

En = −13.6Z 2

n2

€

E = E1(1) + E1(2)

= −13.6Z 2 1

n(1)2+

1

n(2)2

⎛

⎝ ⎜

⎞

⎠ ⎟

= −13.6(2)2 1

12+

1

12

⎛

⎝ ⎜

⎞

⎠ ⎟

= −109 eV

-50

-60

-70

-80

-90

-100

-110

Energy(eV)

Neglecting electron-electron interaction

Observed

PY3P05

System of interacting particlesSystem of interacting particles

o For He-like atoms can extend to include electron-electron interaction:

o The final term represents electron-electron repulsion at a distance r12.

o Or for N-electrons, the Hamiltonian is:

and the corresponding Schrödinger equation is again of the form

where

€

ˆ H = ˆ H ii=1

N

∑

=−h2

2m∇ i

2 −Ze2

4πε0ri

⎛

⎝ ⎜

⎞

⎠ ⎟

i=1

N

∑ +e2

4πε0riji> j

N

∑

xy

z

e1e2 r12

r1r2

€

ˆ H =−h2

2m∇1

2 −Ze2

4πε0r1

⎛

⎝ ⎜

⎞

⎠ ⎟+

−h2

2m∇ 2

2 −Ze2

4πε0r2

⎛

⎝ ⎜

⎞

⎠ ⎟+

e2

4πε0r12

€

ˆ H Ψ(ˆ r 1, ˆ r 2,..., ˆ r N ) = EΨ(ˆ r 1, ˆ r 2,..., ˆ r N )

€

( ˆ r 1, ˆ r 2,..., ˆ r N ) =ψ1( ˆ r 1)ψ 2( ˆ r 2)L ψ N ( ˆ r N )

€

E = E i

i=1

N

∑and

PY3P05

Wave function for system of interacting particlesWave function for system of interacting particles

o The solutions to the equation

can again be written in the form

o The radial wave functions are solutions to

and therefore have the same analytical form as for the hydrogenic one-electron atom.

o Allowable solutions again only exist for

where Zeff = Z - nl.

o Zeff is the effective nuclear charge and nl is the shielding constant. This gives rise to the shell model for multi-electron atoms.

€

d2Rn i li

dri2 +

2

r

dRn i li

dri

+2μ

h2E +

e2

4πε0ri

⎡

⎣ ⎢

⎤

⎦ ⎥Rn i li

= 0

€

ˆ H iψ i(ˆ r i) =−h2

2m∇ i

2 −Ze2

4πε0ri

+e2

4πε0rij

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ψ i(ˆ r i)

= E iψ i( ˆ r i)

€

ψi(ˆ r i,θ i,φi) = Rn i li( ˆ r i)Υli m i

(θ i,φi)

€

En = −Zeff

2μe4

(4πε0)22h2n2

PY3P05

Atoms with two valence electronsAtoms with two valence electrons

o Includes He and Group II elements (e.g., Be, Mg, Ca, etc.). Valence electrons are indistinguishable, i.e., not physically possible to assign unique positions simultaneously.

o This means that multi-electron wave functions must have exchange symmetry:

which will be satisfied if

o That is, exchanging labels of pair of electrons has no effect on wave function.

o The “+” sign applies if the particles are bosons. These are said to be symmetric with respect to particle exchange. The “-” sign applies to fermions, which are anti-symmetric with respect to particle exchange.

o As electrons are fermions (spin 1/2), the wavefunction of a multi-electron atom must be anti-symmetric with respect to particle exchange.

€

| Ψ(ˆ r 1, ˆ r 2,..., ˆ r K , ˆ r L ,..., ˆ r N ) |2=| Ψ(ˆ r 1, ˆ r 2,..., ˆ r L , ˆ r K ,..., ˆ r N ) |2

€

( ˆ r 1, ˆ r 2,..., ˆ r K , ˆ r L ,..., ˆ r N ) = ±Ψ(ˆ r 1, ˆ r 2,..., ˆ r L , ˆ r K ,..., ˆ r N )

PY3P05

Helium wave functionsHelium wave functions

o He atom consists of a nucleus with Z = 2 and two electrons.

o Must now include electron spins. Two-electron wave function is therefore written as a product spatial and a spin wave functions:

o As electrons are indistinguishable => must be anti-symmetric. See table for allowed symmetries of spatial and spin wave functions.

xy

z

e1e2

r1r2

Z=2

€

=ψspatial (ˆ r 1, ˆ r 2)ψ spin

r12

PY3P05

Helium wave functions: Helium wave functions: spatialspatial

o State of atom is specified by configuration of two electrons. In ground state, both electrons are is 1s shell, so we have a 1s2 configuration.

o In excited state, one or both electrons will be in higher shell (e.g., 1s12s1). Configuration must therefore be written in terms of particle #1 in a state defined by four quantum numbers (called ). State of particle #2 called .

o Total wave function for a excited atom can therefore be written:

o But, this does not take into account that electrons are indistinguishable. The following is therefore equally valid:

o Because both these are solutions of Schrödinger equation, linear combination also a solutions:

where is a normalisation factor.

€

=ψ (ˆ r 1)ψ β (ˆ r 2)

€

=ψ (ˆ r 1)ψα (ˆ r 2)

€

S =1

2(ψα ( ˆ r 1)ψ β ( ˆ r 2) +ψ β ( ˆ r 1)ψα ( ˆ r 2))

ΨA =1

2(ψα ( ˆ r 1)ψ β ( ˆ r 2) −ψ β ( ˆ r 1)ψα ( ˆ r 2))

€

1/ 2

Symmetric

Asymmetric

PY3P05

Helium wave functions: Helium wave functions: spinspin

o There are two electrons => S = s1+ s2 = 0 or 1. S = 0 states are called singlets because they only have one ms value. S = 1 states are called triplets as ms = +1, 0, -1.

o There are four possible ways to combine the spins of the two electrons so that the total wave function has exchange symmetry.

o Only one possible anitsymmetric spin eigenfunction:

o There are three possible symmetric spin eigenfunctions:€

1

2[(+1/2,−1/2) − (−1/2,+1/2)]

€

(+1/2,+1/2)

1

2[(+1/2,−1/2) + (−1/2,+1/2)]

(−1/2,−1/2)

singlet

triplet

PY3P05

Helium wave functions: Helium wave functions: spinspin

o Table gives spin wave functions for a two-electron system. The arrows indicate whether the spin of the individual electrons is up or down (i.e. +1/2 or -1/2).

o The + sign in the symmetry column applies if the wave function is symmetric with respect to particle exchange, while the - sign indicates that the wave function is anti-symmetric.

o The Sz value is indicated by the quantum number for ms, which is obtained by adding the ms values of the two electrons together.

PY3P05

Helium wave functionsHelium wave functions

o Singlet and triplet states therefore have different spatial wave functions.

o Surprising as spin and spatial wavefunctions are basically independent of each other.

o This has a strong effect on the energies of the allowed states.

S ms ψspin ψspatial

0 0

+1

1 0

-1€

1/ 2(↑1↓2 −↓1↑2)

€

↑1↑2

1/ 2(↑1↓2 +↓1↑2)

↓1↓2

€

1

2(ψα ( ˆ r 1)ψ β ( ˆ r 2) −ψ β ( ˆ r 1)ψα ( ˆ r 2))

€

1

2(ψα ( ˆ r 1)ψ β ( ˆ r 2) +ψ β ( ˆ r 1)ψα ( ˆ r 2))Singlet

Triplet

PY3P05

Singles and triplet statesSingles and triplet states

o Physical interpretation of singlet and triplet states can be obtained by evaluating the total spin angular momentum (S), where

is the sum of the spin angular momenta of the two electrons.

o The magnutude of the total spin and its z-component are quantised:

where ms = -s, … +s and s = 0, 1.

o If s1 = +1/2 and s2 = -1/2 => s = 0. o Therefore ms = 0 (singlet state)

o If s1 = +1/2 and s2 = +1/2 => s = 1. o Therefore ms = -1, 0, +1 (triplet states)

€

ˆ S = ˆ S 1 + ˆ S 2

€

S = s(s +1)h

Sz = msh

z

s = 1

+1

0

-1

tripletstates1=1/2

s2=1/2

ms

s1=1/2 s2=-1/2

s = 0, ms = 0

singletstate

PY3P05

Helium termsHelium terms

o Angular momenta of electrons are described by l1, l2, s1, s2.

o As Z<30 for He, use LS or Russel Saunders coupling.

o Consider ground state configuration of He: 1s2

o Orbital angular momentum: l1=l2 = 0 => L = l1 + l2 = 0

o Gives rise to an S term.

o Spin angular momentum: s1 = s2 = 1/2 => S = 0 or 1

o Multiplicity (2S+1) is therefore 2(0) + 1 = 1 (singlet) or 2(1) + 1 = 3 (triplet)

o J = L + S, …, |L-S| => J = 1, 0.

o Therefore there are two states: 11S0 and 13S1 (also using n = 1)

o But are they both allowed quantum mechanically?

PY3P05

Helium termsHelium terms

o Must consider Pauli Exclusion principle: “In a multi-electron atom, there can never be more that one electron in the same quantum state”; or equivalently, “No two electrons can have the same set of quantum numbers”.

o Consider the 11S0 state: L = 0, S = 0, J = 0

o n1 = 1, l1 = 0, ml1 = 0, s1 = 1/2, ms1 = +1/2o n2 = 1, l2 = 0, ml2 = 0, s2 = 1/2, ms2 = -1/2

o 11S0 is therefore allowed by Pauli principle as ms quantum numbers differ.

o Now consider the 13S1 state: L = 0, S = 1, J = 1

o n1 = 1, l1 = 0, ml1 = 0, s1 = 1/2, ms1 = +1/2o n2 = 1, l2 = 0, ml2 = 0, s2 = 1/2, ms2 = +1/2

o 11S1 is therefore disallowed by Pauli principle as ms quantum numbers are the same.

PY3P05

Helium termsHelium terms

o First excited state of He: 1s12p1

o Orbital angular momentum: l1= 0, l2 = 1 => L = 1o Gives rise to an P term.

o Spin angular momentum: s1 = s2 = 1/2 => S = 0 or 1o Multiplicity (2S+1) is therefore 2(0) + 1 = 1 or 2(1) + 1 = 3

o For L = 1, S = 1 => J = L + S, …, |L-S| => J = 2, 1, 0o Produces 3P3,2,1

o Therefore have, n1 = 1, l1 = 0, s1 = 1/2

and n2 = 2, l2 = 1, s1 = 1/2

o For L = 1, S = 0 => J = 1o Term is therefore 1P1

o Allowed from consideration of Pauli principle

No violation of Pauli principle => 3P3,2,1 are allowed terms

PY3P05

Helium termsHelium terms

o Now consider excitation of both electrons from ground state to first excited state: gives a 2p2 configuration.

o Orbital angular momentum: l1 = l2=1 => L = 2, 1, 0o Produces S, P and D terms

o Spin angular momentum: s1 = s2= 1/2 = > S =1, 0 and multiplicity is 3 or 1

o *Violate Pauli Exclusion Principle (See Eisberg & Resnick, Appendix P)

L S J Term

0 0 0 1S0

1 0 1 1P1

2 0 2 1D2

0 1 1 *3S1

1 1 2, 1, 0 3P2,1,0

2 1 3, 2, 1 *3D3,2,1

PY3P05

Helium Grötrian diagramHelium Grötrian diagram

o Singlet states result when S = 0.

o Parahelium.

o Triplet states result when S = 1

o Orthohelium.

PY3P05

Exchange energyExchange energy

o Need to explain why triplet states are lower in energy that singlet states. Consider

o The expectation value of the Hamiltonian is

o The energy can be split into three parts,

E = E1 + E2 + E12

where

o The expectation value of the first two terms of the Hamiltonian is just

where ER = 13.6 eV is called the Rydberg energy.

€

ˆ H =−h2

2m∇1

2 −Ze2

4πε0r1

⎛

⎝ ⎜

⎞

⎠ ⎟+

−h2

2m∇ 2

2 −Ze2

4πε0r2

⎛

⎝ ⎜

⎞

⎠ ⎟+

e2

4πε0r12

= ˆ H 1 + ˆ H 2 + ˆ H 12

€

E = ψ spatial*∫∫ ˆ H ψ spatiald

3 ˆ r 1d3 ˆ r 2

€

E = E1 + E2

= −4ER

1

n12

+1

n22

⎛

⎝ ⎜

⎞

⎠ ⎟

€

E i = ψ spatial*∫∫ ˆ H iψ spatiald

3 ˆ r 1d3 ˆ r 2

E12 = ψ spatial*∫∫ ˆ H 12ψ spatiald

3 ˆ r 1d3 ˆ r 2

PY3P05

Exchange energyExchange energy

o The third term is the electron-electron Coulomb repulsion energy:

o Evaluating this integral gives E12 = D J

where the + sign is for singlets and the - sign for triplets and D is the direct Coulomb energy and J is the exchange Coulomb energy:

o The resulting energy is E12 ~ 2.5 ER. Note that in the exchange integral, we integrate the expectation value of 1/r12 with each electron in a different shell. See McMurry, Chapter 13.

€

E12 = ψ spatial*∫∫ ˆ H 12ψ spatiald

3 ˆ r 1d3 ˆ r 2

= ψ spatial*∫∫ e2

4πε0r12

ψ spatiald3 ˆ r 1d

3 ˆ r 2

€

Dαβ =e2

4πε0

ψα*∫∫ ( ˆ r 1)ψ β

* ( ˆ r 2)1

r12

ψα ( ˆ r 1)ψ β ( ˆ r 2)d3 ˆ r 1d3 ˆ r 2

Jαβ =e2

4πε0

ψα*∫∫ ( ˆ r 1)ψ β

* ( ˆ r 2)1

r12

ψ β ( ˆ r 1)ψα ( ˆ r 2)d3 ˆ r 1d3 ˆ r 2

PY3P05

Exchange energyExchange energy

o The total energy is therefore

where the + sign applies to singlet states (S = 0) and the -sign to triplets (S = 1).

o Energies of the singlet and triplet states differ by 2J. Splitting of spin states is direct consequence of exchange symmetry.

o We now have, E1 + E2 = -8ER and E12 = 2.5ER => E = -5.5ER = -74.8 eV

o Compares to measure value of ground state energy, 78.98 eV.

o Note:o Exchange splitting is part of gross structure of He - not a small effect. The value of 2J

is ~0.8 eV.

o Exchange energy is sometimes written in the form which shows explicitly that the change of energy is related to the relative alignment of the electron spins. If aligned = > energy goes up.

€

E = −4ER

1

n12 +

1

n22

⎛

⎝ ⎜

⎞

⎠ ⎟+ Dαβ ± Jαβ

€

ΔEexchange = −2Jαβˆ S 1 ⋅ ˆ S 2

PY3P05

Helium termsHelium terms

o Orthohelium states are lower in energy than the parahelium states. Explanation for this is:

1. Parallel spins make the spin part of the wavefunction symmetric.

2. Total wavefunction for electrons must be antisymmetric since electrons are fermions.

3. This forces space part of wavefunction to be antisymmetric.

4. Antisymmetric space wavefunction implies a larger average distance between electrons than a symmetric function. Results as square of antisymmetric function must go to zero at the origin => probability for small separations of the two electrons is smaller than for a symmetric space wavefunction.

5. If electrons are on the average further apart, then there will be less shielding of the nucleus by the ground state electron, and the excited state electron will therefore be more exposed to the nucleus. This implies that it will be more tightly bound and of lower energy.

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