Chapter 5 Many Electron Atoms

5.1 Schrödinger Equation

SEE THE HANDOUT

5.2 The Inclusion of Exchange

The electrons are identical and so the product of the wave function (equation 11 on the handout) is not acceptable. We need to find the acceptable wave functions which include the spin of the electrons from the beginning by starting with spin-orbitals.

nlm_lm_s

nlm_l

sm_s

everything is lower case because the equation deals with only a single electron spinor.

nlm_lm_s

=R_nl(r)Y_lm(q ,f )c

sm_s

the spin orbitals satisfy

é
ê
ê
ë

Ñ²

+U(r)

ù
ú
ú
û

nlm_lm_s

=E_nlu

nlm_lm_s

we let the set of quantum numbers

nlm_lm_sºa

if one of the quantum numbers change, giving say nlm_lm_s' , then we call the whole set b . We then have spin-orbitals ( u_a , u_b , u_g , ...) then we can use a simple product wavefunction

(1)u

(2)u

(3)...

however this is also not acceptable. We need to form a combination of these. Slater showed that an overall anti-symmetric (both with space and spin inclusive) wave function can only be constructed as

Y =

½
½
½
½
½
½
½
½
½
½
½

(1)

...

(1)

(2)

...

(2)

·
·
·

(N)

...

(N)

½
½
½
½
½
½
½
½
½
½
½

this is the Slater Determinant. We note that the exchange of two electrons is simply done by exchanging two rows. Therefore we can use -Y_C as required, for example in Helium, (1s)² is the ground state so

=Y₁₀₀c

=Y₁₀₀

=Y₁₀₀c

=Y₁₀₀¯

Y_C=

½
½
½
½

Y₁₀₀(1) (1)	Y₁₀₀(1)¯ (1)
Y₁₀₀(2) (2)	Y₁₀₀(2)¯ (2)

½
½
½
½

Y_C=Y₁₀₀(1)Y₁₀₀(2)

[ (1)¯ (2)-¯ (1) (2)]=Y_Sc

this is the same result as we got before. If any two columns of a determinant are equal then the determinant vanishes. The conculsion for this is if any two electrons are in the same spin-orbital then Y_C vanishes, so an overall anti-symmetric wave function cannot be constructed. The can also be rephased to state that no two electrons can have the same set of four quantum numbers nlm_lm_s . Again this can be rewritten to state that not more than two electrons can exist in a state labelled by n , l and m_l .

This is the Pauli Exclusion Principle.

5.3 Self Consistent Field Calculation (Hartree Method)

We need a method for finding R_nl(r) , see the handout.

The result of is is where we would normally have E=E_n for hydrogen we instead get Eº E_n,l for a many electron atom. However the Hartree method for the Self Consistent Field uses the Slater Determinants, which were devised by both Hartree and Fock, is more accurate since it takes into account the exchange from the start of the technique.

5.4 Periodic Table

Notes about the periodic table:

the elements are arranged according to their chemical properties
the basis of physical chemistry comes from the study of how the ground state of atoms form molecules
the starting point comes from the knowledge of the binding energy of the electrons for the ground state of all atoms, ie. a weakly bound electron may attach itself to another atom if this is energetically favourable. This leads to form a chemical bond

We can calculate the binding energy for hydrogen and helium using Quantum Mechanics and for many electron atoms we use the Hartree-Fock to solve this problem numerically. However this does not give much physical insight into the processes going on. So instead we will consider the general features of the potential ( U(r) ) to gain the physical insight we want about the internal workings of molecules.

5.4.1 Core Penetration and Shielding

We we be considering the one electron in the many electron atom. The electron samples all the positions weighted by |Y |² . For a small radius ( r ) the electron is inside the charge cloud of the other electrons, so it sees the full nuclear charge

r® 0Þ U(r)®

-Z

For a large r the electron is outside the charge cloud of the other electrons so it sees a Z_eff=+1 for the core instead. The core consisting of the nucleus and all the other electrons. So we say

when r® 0 that the electron penetrates the core
when r®¥ that the electron is shielded by the other electrons, shielding increase for increasing values of n and l

Example One - Lithium:: so if we start with a bare nucleus with Z=3 and add the first electron, we put it into U₁₀₀_1/2 . We now choose m_S=+1/2 which has been choosen arbitrary. f_nlm_{_l}ºY_nlm_{_l} is exact however Z=3 . The binding energy is therefore
binding energy=3²E_1s(H) (for Li²⁺)
so the single electron is very tightly bound and so very close to the nucleus. If we now add the second electron so that U_100-_1/2 (this is required by the Pauli Exclusion Principle) we now have Li⁺ . We note that the electrons with the same n and l are called equivalent electrons. Both the equivalent electrons in Li⁺ are now tightly bound but are not exactly 3²E_1s(H) each since we have electron-electron repulsion which lifts the energy level. The 1s shell is now full and so the Pauli Exclusion Principle says that the next electron must go into the 2s or 2p energy level. Both the electrons have more energy that the 1s ( n² dependence) which is even higher due to shielding, however the 2s state is somewhat lower that the 2p due to the extra core penetration. So the third electron goes to U₂₀₀_1/2 which provides us with the neutral Lithium atom.

Therefore the ground state of Lithium is (1s)²2s , and the binding energy of the outer electrons is the ionisation energy (for Lithium this is 5.4eV ). We note that the Hartree Self Consistent Field and an application of the Pauli Exclusion Principle give us a reasonable set of results. The Hartree-Fock technique though gives better results as it has the Pauli Exclusion Principle built in.
Example Two - Beryllium:: so if start with a base nucleus with Z=4 and add the first electron. This electron goes to U₁₀₀_1/2 , but as Z=4 we get Be³⁺ where the binding energy is much larger (due to the extra proton) compared to Li²⁺ and so the electron is much more tightly bound
binding energy=4²E_1s(H)
the second electron goes into U_100-_1/2 , it is still very tightly bound however it is not exactly 4²E_1s(H) due to electron-electron repulsion.

the third electron goes to U₂₀₀_1/2 which is much lower than the corresponding state in Lithium since even if the shielding was complete then the electron would still see a 2+ core.

the forth electron goes to U_200-_1/2 , the 2s electrons are now not as tightly bound as they would be in the absence of each other. However they are still more tightly bound than the 2s state for Litium (since Z is bigger).

From this we can conclude that the ground state of Be is 1s²2s² (or (1s)²(2s)² ). The ionisation energy is 9.3eV .
Example Three - Other Elements:: we can continue for the other elements in the periodic table as we have done for example one and two.

If we continue with developing this technique up the periodic table we come across some interesting features

Boron (B) where Z=5 we find a new shell so the structure is
1s²2s²2p
the 2p shell can take up to six electrons, in accordence to the Pauli Exclusion Principle

n=2 , l=1 m_l m_s

1 ±1/2

0 ±1/2

-1 ±1/2

In general the shells have 2(2l+1) states available. The 2p shell fills progressively until Neon
Neon (Ne) where Z=10 where the structure is
1s²2s²2p⁶
the ionisation potential increases as the 2p shell fills. There is now a competition between the increasing attractive charge on the core and also the electron-electron repulsion, this occurs as
- the electrons become more tightly bound as Z increases
- the electrons are more weakly bound as the electron-electron repulsion increases
the ionisation potential of Neon is 21.56eV
Sodium (Na) where Z=11 . The next electron fills the 3s level however the Pauli Exclusion Principle states that 3s is a lower energy state than the 3p and 3d states. So the configuration of Sodium is
1s²2s²2p⁶3s (or [Ne]_3s)
Due to this configuration the single electron outside the closed shells are weakly bound due to the effective shielding (but are more strongly bound than the 3s state of hydrogen due to penetration). So the ionisation potential is much lower than expected, 5.14eV , when compared to the 3s state of hydrogen which is -1.5eV
the 3s shell fills up at Magnesium where Z=12
the 3p shell opens up at Aluminium where Z=13
the 3p shell fills up at Argon where Z=18

we notice a rule in the features we discover throughout the periodic table, known as the aufbau process. If you take an electron then it goes into the lowest avaliable n level. For a given n the electron goes into the lowest avaliable level. This is not a very good rule as is breaks down at Potassium where Z=19 where the configuration is [Ar]_4s and not [Ar]_3d . This occurs since core penetration brings the 4s state below the 3d state.

So if we return to the features that occur further into the periodic table

the 4s shell begins to fill at Calcium where Z=20 ( [Ar]_4s^₂ )
when we get to Scandium (Sc) where Z=21 is when the 3d shell starts to fill ( [Ar]_4s^₂_3d )
the filling becomes irregular up to Zinc (Zn) where Z=30 since the 3d state competes with the 4s state, for example Cromium (Cr) has a configuration [Ar]_4s3d^₅ and not as expected [Ar]_4s^₂_3d^₄ .

from this we call the first transistion group the elements Scandium (Sc) to Zinc (Zn).

The chemical properties are:

Alkalis (Li, Na, K, ...):: are loosely bound and have a single valence electron, this results in this group being very reactive
Halogens (F, Cl, Br, ...):: have a hole in the outer shells, this makes this group also very reactive however for a different reason
Noble Gasses (He, Ne, Ar, ...):: have closed shells and so a large ionisation potential, this makes them inert

5.5 The Gross Structure of the Alkalis

The Alkalis group acts similarly to Hydrogen as they both have one electron on the outside of a closed shell. This makes the energy levels well represented by an emperical fudged formula

E_nl=-

(n-d_l)²

atomic units

where d_l is called a quantum defect, this is only a function of l to an approximation. This means we do not need a huge table were d is different for each situation, we can get by with only a few values for d as it only changes with l . This quantum defect allows us to mathematically represent the effects of shielding and penetration.

	d_s	d_p	d_d
Lithium (Li)	0.40	0.04	0.00
Sodium (Na)	1.35	0.85	0.01
Potassium (K)	2.19	1.71	0.25

these are the selection rule remains the same as it is for hydrogen. We note that D n=anything including zero.

In alkalis the confguration is the same but they have different l values and so have different energies. This results in the resonance line (a stronger transistion), for example in lithium it is a 2s-2p transistion, this is similar to the case with hydrodren where the strongest transistion is the 1s-2p.

5.6 Rydberg Atoms

For a high n and l configuration, all atoms look like hydrogen, ie. one electron outside a 1+ positive core. Any atom in a configuration that has a high n and l is known as a Rydberg Atom.

For n~ 100 the atom is of the size µ m (micrometres).