Fundamentals of University Physics. Vol III.

This is a simple process. EM radiation shines onto a sample of matter (in any particular phase). Two things can happen, the radiation is absorbed, and is re-emitted with a lower energy, or the radiation can be reemitted at a different wavelength. So this gives us information about the electronic structure of the sample. Within the sample there are quantized energy levels. Imagine there are so many atoms in the upper level and so many in the lower level. When there is incident EM radiation, the atoms are redistrobuted in terms of thier energy whenever the enrgy hv corresponds to dE (the Bohr frequeny rule), the energy gap between the distrobution of energy levels. Over different states this gives rise to absorption or emission.

The hydrogen atom is the simplest of all the atoms. There is one proton orbited by a single valence electron. They actually both orbit thier mutual centre of gravity. The potential that holds them together can be expressed as,

From the solution of the Schrödinger equation we can state the energy level values as,

(See fig 1 for information on the Lyman, Balmer and Pachen series)

The selection rule states that dn the change in energy levels can be anything.

Photon energy is therefore

The density of the electron cloud, in Hydrogen say, is proportional to the modulus of the squared wave function. The density drops off with increasing radius from the nucleus. We define a new quantity called the radial probability where we have the relation,

From the solution of the wave equation we find that for a given value of n, there are n different charge distrobutions, each associated with a different amount of angular momentum. Classically the angular momentum is expressed as,

Using quantum mechanics however we have,

Where l is the orbital angular momentum quantum number, with a value of (n-1) to zero. For example if n=3 then l can be 2,1,0 ie three electron clouds but all with the same energy.

Looking at the energy diagram in fig 1. The n=1 level is called the 1s level and so on (see diagram). By the selection rules dn can be anything and dl can be ±1. So the Lyman series can be produced by transition the p terms to the 1s level. The Balmer series is formed from any transition from 2s upto 5s across to 2p upto 5p. Simillary we get the Pachen series from transitions from the d terms to the p terms.

Space quantisation gives us restricted orbits with respect to a magnetic field direction.

As an electron orbits it has angular momentum about the atomic nuclei, but it also has additional angular momentum by spinning on it's own axis. S is the spin angular momentum, it is expressed as,

From fig 3, ms=±½

These are group one elements that have one valence electron.

We measure wave lengths in relation to corresponding energies. We can write

vbar is the wave number,

Therefore,

Also,

Now let us consider a specific series,

Sodium has eleven electrons, ten of them being in closed shells and the remaining electron in the valence shell. The ten electrons are tightly bound to the nucleus. For optical spectra they are inactive. It is the single valence electron that gives the optical spectra of the atom. Comparing to the energy levels of the hydrogen atom (see SPEC3 sheet again) because of the similarity of having one valence electron.

Then: selection rules predict the emission lines. dn=anything & dl=±1. The principal series is formed from transitions from any p level to the 3s level. These are the strongest lines. The first line of the principal series is the 3p to 3s transition. The diffuse series is formed from 3d-6d transitions to the 3p level. The sharp series is formed from 6s-4s transitions to 3p level. The first letter of each series historically determined the origon of the letters which denote energy terms.

To explain features of the energy level diagram;

We find that the sharp and the principal series are split into doublets. The diffuse series and the fundamental series are split into triplets.

This is explained in the following section.

This splits all the levels except the s levels. When we have a charge carrier traveling in a circle and we have a magnetic field which is in the same directionas the angular momentum. We can formulate the following from fig 2

But,

Therefore,

The (additional) electron magnetic moment due the electron spinning on it's own axis is,

The "2" is called the "Electron g-factor"

µL and µs interact in the following way. The electron sees a positively charged nucleus circulating it.

To determine the energy of µs in Bint,

Therefore,

Eso is the spin orbit energy and A is the "spin-orbit interaction".

For the p level, l=1, s=½, therefore j=3/2,1/2

Therefore,

For the s levels, l=0, s=½ and j=½ hence,

This means there is no energy splitting.

(Continued)

For the d state, l=2, s=½, j=3/2, 5/2

Where gJ is the Landè g-factor.

Where,

From figure 1,

We now calculate the shift of line 4 in a 1T field.

To find lambda,

Let us first consider a two electron atom. We first find all the terms, each term corresponding to a different energy level. When we say a two electron atom we mean two valence electrons.

where,

where,

where,

L=S=J=0 for closed shells.

The number of terms are restricted by the Pauli exclusion principal, however, this is unproved so far.

Hund's rule tell us that the lowest energy state has the largest value of S, then if two states have the same value of S the lowest is the one with the higher L value.

Figure one shows how this rule is applied when determining the values of sub levels.

The energy change due to spin orbit coupling is,

This is the same as for single electron atoms, but J,L and S are for the atom as a whole.

Splitting between levels is given as follows. Suppose we have three levels, J+1, J and J-1. The difference between the J+1'th and J'th level is dEso(J+1->J).

Say we have three levels 3P2 , 3P1 and 3P0 the splitting between the first two is twice as wide as the second two, for example. Also three levels, 3D3, 3D2 , 3D1 the ratio between the first and second is 3 to 2.

(see sheet)

However S=1 is forbidden by the Pauli exclusion principal.

Therefore, L=l1+l2=l2 and S=s1+s2=0 (up,down) , 1(up,up). Both singelt and triplet stages.

This causes the energy levels to be different with the same value of n.

The energy for this is expressed as

For l2=1, complete screening, ZeH=1. l2=0, incomplete screening, ZeH>1, which means the energy is decreased.

The energy associated with rotations and vibrations of the molecule is much less than that with electronic transitions. Measurement of these are made as shown in fig 4.

For higher energy absorbtions, vibrational electron states absorb at near infrared and lower energy transition we have electron rotation states at the far infrared to µ-wave.

where,

but,

Therefore,

Where J is the rotationa quantum number. Also classically,

Therefore,

Therefore,

This is called the rotational constant.

The selection rule says that J changes by one, or dJ=±1.

Therefore,

Therefore,

but,

Where,

Therefore,

First we consider the simplest molecule, two atoms. Between these two atoms there is a potential, which can be expressed as a function of the distance between them.

ie the restoring force is proportional to the square of the displacement, just as a harmonic oscillator region. To find the total energy of the molecule (KE+PE) solve the Schrödinger equation.

The solution of the equation is,

Where n is used as the vibrational quantum number and vosc is the characteristic frequency of vibration. At the distance ro, n=0 and Ev is called the zero point energy (see quantum mechanics notes).

This leads us to the field of infra red spectroscopy.

How absorption lines do we expect to see for more complicated molecules ? It is found from counting the number of degrees of freedom of the molecule. For a single atom the number of degrees of freedom is equal to 3, and these are all translational. For N atoms of the molecule, the number of degrees of freedom is simply 3N, of which, 3 are translational (unless we have a linear molecule), the rest that remain are vibrational.

Therefore, the number of modes of vibration is (3N-6) or (3N-5) if the molecule is linear).

Now let us consider the example of SO2,

The motions indicated by (1) are symmetrical stretches at a frequency, v1. (2) is an Asymmetrical stretch, at v2. (3) is a bending process at frequency v3. We would hence observe three abosorption line. For SO2 v1 = 7.35 µm, v2 = 8.69 µm and v3 = 19.23 µm.

Now let us consider the linear molecule, CO2 (O-C-O). There should be four modes of vibration (3N-5=4) but we only observe two absorption lines at 4.25 µm and 14.99 µm. An unobservable motion is a symmetrical strech, this is because there is never a dipole moment, the molecule is always symmetrical). Hoever the asymmetric stretch allows a dipole to occur, which gives us the 4.25 µm line. The bending motion gives us the 14.99 µm line, this bending is bending in the plane of the molecule. CO2 has two modes of bending, both degenerate (same energy). The other bending motion is such that the carbon (or both oxygens) vibrate in and out of the moleculat plane.

Now let us consider the HCl molecule. See Sheet.

We now consider the energy in the R branch (see sheet).

Where the first term is the degeneracy of the energy level and the second term is the Boltzmann distrobution term. The degeneracy is calculated as,

Consider sending photons at a molecule, the photons are usually reflected from the molecule. Suppose that the frequencies are changed so that we send in a frequency v and we get frequencies away from the molecule of v±dv. By energy conservation we have,

Where Ex are the initial and final energy of the molecule.

If Ef=Ei, then v=v' (this is the case for Rayleigh scattering). If Ef<>Ei then v<>v', we then have a change of frequency, this is Raman scattering. The energy path for Raman scattering is as shown in fig 2.

Fig 3 shows a plot of the Raman and Rayleigh lines.

The selection rules for Raman spectroscopy are different to usual, in this case dJ=±2. So we can determine the shift of energy lines (dE). The molecule gets excited and then ends in a final state (as in fig 2. The molecule goes to a virtual state and then rests at a final state, here the final state is the J+2 and the initial is J).

If we have rotational states also with the vibrational states we get the Rayleigh and Raman lines, but with lots of fine structure.

Now consider CO2. We have two absorption lines (see earlier), the first is for the anti-symmetrical stretch (highest energy, v1) and the second is for the molecule bending (lowest energy, v2). For the symmetrical stretch we didn't see a absorption line because there was no dipole moment. For Raman spectrocopy however, we do see a line for the symmetrical stretch (v3) this is due to the rule of mutual exclusion.

This applies for molecules with a centre of symmetry (of which linear molecule are). If such a molecule is Raman active then it is IR inactive. If it is Raman inactive then it is IR active.

Raman lines are always seen between rotational states.