Physics 361Principles of Modern Physics

Lecture 23

The quantum mechanical model of the many-electron atom

This lecture• Reduced Mass• Many Electron Atoms• Screening• Filling Atomic Shells

Many Electron AtomsWe have been working out the two-bodyproblem, between an electron anda positive nucleus with Z positively-charged protons.

The system we will work out isschematically represented at right,though the classical type orbitsare not what we expect from QM.

This type of problem is much more difficult than the hydrogen atom, which can be solved exactly. That said, we will use the solutions of the hydrogen atom in order to understand many-electron atoms.

Before we try to understand the many-electron atom, let’s first go over why the hydrogen atom is exactly solvable.

What did we mean by radius?In the radial function or the radial probability.

Distance between the +nucleus and the electron.

What did we mean by radius?

+e-

Every two-body problem can be split into a relative coordinate – the one we’vebeen using

and a center of mass coordinate . This coordinate describes how the two bodies move together in space acted on by external forces.

The two coordinates are completely separable -- independent of each other.

Everything we have derived (and/or discussed) so far remains the same if we consider that both bodies move, except we must alter the mass used in the Schrödinger equation to the “reduced mass”.

Breaking up the two-body problem into center of mass and relative coordinates.

+e-

Reduced Mass

The mass changes to the reduced mass

, where the masses are for the positive andnegative charges.

+e-

In the case that one of the masses ismuch larger than the other, eg,

The reduced mass is approximately the smaller of the two.

Also, the CM is just the location of thelarger,

This is exactly the situation with theone-electron atoms since,

.So using the electron mass instead ofreduced mass gives only a very small error. To get exact results, all you needto do is to insert the reduced mass in all our previous results.

Reduced Mass Approximation, and Why “Reduced”?

We can consider first order approximation for the case where

This gives, , which is less than the electron mass,ie, it is reduced!!

+e-

We just need to alter equations with mass, that is,

Most of the time, the reduced subscript is removed, and the mass is understood to be “reduced”. But what does “reduced” mean?

Reduced Mass Approximation, and Why “Reduced”?

We can consider first order approximation for the case where

This gives, , which is less than the electron mass,ie, it is reduced!!

+e-

We just need to alter equations with mass, that is,

goes to,

Most of the time, the reduced subscript is removed, and the mass is understood to be “reduced”. But what does “reduced” mean?

More than one body – not exactly solvableWhile the two-body problem can be solved exactly, when additional bodies are included the problem can only beapproximated (at best).

How to approximate many-electron atoms? We will utilize the solutions of the one-electron atom and incorporate the effects of the other electrons via electrostatic screening

Electrostatic Screening – Forces between Spherically Symmetric Objects

-q

+q

-q

+q

In electrostatics, the force and potential energybetween two charged bodies does not dependon their distribution of charge, as long as their distribution is radially symmetric.

For example, the force on the negative charge atupper right is the same, whether +q is spread out in space -- the lower right picture.

Electrostatic Screening from Inside Charges – Gauss’ Law

-q

+q

-q

+q

The fact that the force only depends on the net charge within a sphere is a result of Gauss’ Law(from electrostatics).

Gauss’ Law is the surface integral is of electric fieldonly depends on the net charge within the surface.

Since radial symmetry implies radial fields,the fields only depend qin. Eg,

qin

No Electrostatic Screening from Radially-Symmetric Outside Charges – Gauss’ Law

-q

+q

The electrostatic forces on a charge are not affectedby a radial charge distribution on its outside.

That is, both charge configurations (with or without the outside charge) at right give the same force on the –q charge.

-q

+q

qoutside

Electrostatic Screening Related to AtomsFor atoms we have a nucleus of charge .If each electron interacted only with the nucleus,this would give the hydrogen-like wave functions and energy levels we have been discussing.

, which is the approximate radius.

However, the electrons also interact with each other!!

We can approximate this interaction by assuming that the other electrons screen the nuclear charge some effective amount. That is, we have a screening charge. This reduces the forces and the effective nuclear charge in the above results to where is some number of charges which are closer to the nucleus which screen its charge. Electrons farther away from the nucleus can be ignored!!

Many Electron Wave Function – Pauli Exclusion PrincipleWe can approximate the states of the many-electron atom as hydrogen states with different nuclear charges, .

Spin ½ particles, like electrons, are fermionswhich obey the Pauli Exclusion Principle.This can be derived in relativistic QM, but itsimplications can seen throughout the non-relativistic regime.

Exclusion Principle: Only one fermion is allowed to occupy any one state of a multi-state wave function.Each electron must have unique set of quantum numbers.Concentrating on the spatial portion of the many-bodywave function, we can write (for short hand)

and can approximate the many-body function as the product of the hydrogen-like functions

Start Adding Electrons to Atoms – Go Through Periodic Table Let’s see how to do this as we go through the periodic table.

we have already worked out the hydrogen atom. Its ground state is given by

Two electrons -- HeLet’s add an electron to a helium atom that has only one electron. The first electron will be in the hydrogen ground state – with bound state given by (-13.6 eV)x4 and with aradius approximately two times smaller than for hydrogen.

The second electron can both increase Z and decrease n, by falling to the other 1s state.

As radius decreases, so does .

This minimizes .

For large radius, .

Three electrons -- LiWe have filled the 1s sub-shell with two electrons. This is denoted as 1s2.Also note that this is the complete n=1 shell.The n=1 shell is also denoted as the K shell.

Where does the third electron go?

Must go into a n=2 state (known as the L shell).

Notice below that both n=2 angular momentum states have average radii much larger than the n=1 state.

But which of the n=2 subshells? 2s or 2p?The smaller average radius of 2p might make this seem favorable....

Three electrons -- LiBut no! The 2s level gets filled first.To understand this focus on the smallbump near zero. This little bump will contribute a lot to lowering the energy of this state, even though the average radius is larger than for the 2p state. This is because the energy depends on , not .

So 2s gets filled before 2p.

Four electrons -- BeThis completes the 2s subshell, just like with He, so it is written as 1s22s2.

Short-hand notation is [He]2s2.

Five electrons -- BWe now start to fill the 2p sub-shell.This goes to 1s22s22p1 configuration that minimizes spin-orbit coupling energy.

Six electrons -- CThere are three 2p orbitals.

Does the second 2p electron fill the same 2p orbital, or go into a different one?

It goes into a different one – this keeps the electrons farther apart and, thus, reduces electrostatic energy.

To see this we rewrite the orbitalsas linear superposition.

This is allowed because we still end up with three different orbitals. The benefit is that they now all look the same – just with their lobes directed along different axes.

Since these orbitals are spatially separated from each other, filling them each separately with one electron lowers the energy, rather than double filling them.

Six electrons -- C

Due to the anti-symmetry requirement of the many electron wave function, the electrons on separate 2p orbitals will like to have their spins aligned. This forces their spatial wave functions to be anti-symmetric, which pushes the electrons apart and loers their electrostatic energy.

This tendency for spins to align on separate orbitals is denoted as Hund’s rule.

Eight electrons andhigher – beyond Oxygen

Nitrogen just has all three 2p orbitals filled with one electron each, all having the same spin.

When we get to oxygen, we now need to fill one of the 2p orbitals with a 2nd electron.

This continues until we get to the next Noble gas neon.

Then we have 1s22s22p6

For sodium we do the same thing as with Li

Potassium

When we get to potassium, we have [Ne]3s23p6 and we need to place the next electron. One might think that this should go into the 3d orbital, but it does not.

Potassium and Calcium

When we get to potassium, we have [Ne]3s23p6 and we need to place the next electron. One might think that this should go into the 3d orbital, but it does not.

The radial probability density of the 3d, 4s, and 4p orbitals is sketched at the right. Notice that both of these have highly penetrating lobes at small radius. These potions significantly lower their energies (ie, increase their binding energies).

This results in the 4s orbital having a slightly lower energy than the 3d, so it fills first.

P(r)

Energies of least-boundelectrons

At the right is the relative order of the energy levels of the many-electron atom.

When adding up the electrons in an atom, the electron must go to the lowest-energy state available.

In general, as the nuclear charge Z increases, this order is of the levels is maintained. However, there is a famous counter-example (which we will discuss in two slides from now).

Transition Metals

The 3d orbital is now filled after the 4s. There are 5 such orbital angular momentum states, each with two spin states – which gives 10 states. Thus, there are 10 elements associated with filling the transition metals for each shell (or row in the periodic table).

Filling d-orbitals

The d-orbitals are filled before the end of the transition-metal rows. In the group (column) of Cu, the energy of the 3d and 4s levels become flipped for the least-bound electron.

This makes it more energetically favorable for an electron to move from the 4s state to the 3d for Cu, Ag, and Au.

Thus, they have a electron configurations like Cu:

[Ar]4s13d10

This provides an unbound electron that helps conduct electric current.

switched

Filling the rest of the periodic table

After a d orbital is filled, the next p orbital is filled (blue circle at right).

The right-most atoms on the periodic table have filled p orbitals. These are highly stable atoms because their outer orbitals (the s and p orbitals) are all completely filled. They are known as the Noble Gases. They are not chemically reactive (so are a type of “inert gas”).

The filling of the f orbitals is achieved in the rows at the bottom of the periodic table (circled in red).These are known as the “rare earths”. The “Lanthanides” for the filling of the 4f orbital and the “Actinides” for the filling of the 5f orbital.

Trends within the periodic table

Since the outer electrons dominate the chemical interactions between atoms, they largely influence the behavior of the elements. Thus there are periodic behaviors in reactivity and the physical properties of the elements.

For example: -The first group “Alkalis” are highly reactive.-The second to last group “Halogens” are also highly reactive, and form stable compounds with the Alkalis.-However, the last group is nearly inert (hardly reactive at all).

Size Trends

As we move from filling the s to p orbitals along a row, we are continually adding more electrons with nearly the same average radius. Thus, these additional electrons are not fully screened, and interact with a larger proportion of the nuclear charge.

Since ,

This implies that these least-bound electrons have a decreasing average radius as we go from left to right in the periodic table.

At the start of each row, we have an increment of n, which increases the radius dramatically.

Ionization-EnergyTrends

As the radius of the atoms decreases in moving along a row, the electrons are more strongly bound and require an increased amount of energy in order to liberate them from the atom.

This manifests itself as a periodic behavior for the ionization energy of atoms. This is also important for determining the bonding between the elements.