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8. Atoms with more than one electron

Links to specific sections in the text:
8.a. Orbital angular momentum and magnetic moment
8.b. Spin
8.c. Spin-orbit effects
8.d. Pauli exclusion principle
8.e. The periodic system
8.f. X-ray emission

Studying heavier atoms is much more complicated than studying the hydrgen atom. no exact solutions are known. The problem is that even if there are only two electrons presont, like in the He atom, one must take into account the intercaction of the two electrons. The electrons repel each other. A lucky circumstance allows for the non-interacting model to work better (the non-interacting model is one in which the electron-electron interaction is neglected). The lucky circumstance is the exclusion principle, which disallows two electrons to be in the same state. Being in different states the wavefunction is different and the overlap is small. If the probability of finding the electrons close to each other is small then in the first, rough approximation the Coulomb interaction between electrons can be neglected.

### 8.a. Orbital angular momentum and magnetic moment

Everyone knows that currents in a wire loop create a magnetic dipole field perpendicular to the plane of the loop. Of course the same is true for a charged particle in a circular orbit. The magnetic dipole moment of such a particle is m = IA, where I is the current and A is the area of the loop. For a single particle the current is I = q v/2pr. Here v/2pr is frequency of rotation. Now the area is r2p, So we obtain for the magnetic moment: m = qvr / 2. Now the angular momentum of the charged particle is | L| =mvr, so combining the two equations we obtain th erelation betwen the magnetic moment and the angular momentum as

m = q|L|/ 2m.

The ratio of the magnetic moment and of the angular momentum is the gyromagnetic ratio, g, which is always g=q/2m for a rotating charged particle. The fixed gyromagnetic ratio tells one that rotation is always accompanied with angular momentum and magnetic moment at the same ratio. We learned that quantum mechanics the angular momentum is quantized. The quanta of angular momentum (at least of the z-component) are integer multiples of hbar. So the combination that always appears in quantum physics is the product of the gyromagnetic ratio and hbar, called the Bohr magneton:

mB = e hbar / 2m.

This could be figured out from dimensional considerations (up to the multiplier 1/2) if we try to find quantities of the dimension of a magnetic diplole moment constructed from constants characterizing the electron e and m, and universal constants, hbar, c.

The presence of a magnetic moment for nonzero angular momentum has important consequences on the behavior of atoms in magnetic field. First of all a dipole has a nonzero energy in magnetic field. The magnetic energy of a dipole is

U = - m B

We can always choose the z-axis to point in the direction of the magnetic field. Then

U = - B m z = B e Lz / 2m = B e hbar ml / 2 m = hbar wL ml,

where ml is the magnetic quantum number and wL is the Larmor frequency,

wL = e B / 2 m.

Consequently, in an external magnetic field the energy of levels of hydrogen atom will be

E = En + hbar wL ml.

The hitherto degenerate levels, having the same principal quantum numbers, but different magnetic quantum numbers split. The split levels are equidistant, split by "magnetic quanta," hbar wL. This pheneomenon of line splitting is called the Zeeman effect. It can be used, e.g., to measure the magnetic field in stars. The Zeeman effect is the reason why ml is called the magnetic quantum number. In fact, after the invention of quantum mechanics it became soon evident that the observed splitting of lines, though similar to the ones predicted, freqently differed from the prediction considerably. This discrepancy lead to the discovery of electron spin by Wolfgang Pauli.

The Larmor frequency has an additional significance. A dipole in a magnetic field is acted upon by the torque

G = m x B

According to the laws of dynamics the angular momentum will change as

dL/dt = G.

Since m is parallel to L the dL is perpendicular to L and B. In other words the magnitude of L will be unchanged but its direction will be keep changing describing a rotating motion around the direction of B. This is called precession. The frequency of this precession is exactly the Larmor frequency as

| dL/dt| = | m| B sin q = eB|L| sin q / 2m

Now |L| sin q = |Lperp | is just the rotating, perpendicular (to the B axis) component of L. Since Lz is unchanged we can write

|d Lperp/dt |= (e B / 2 m) |Lperp|.

Now in a rotation motion |d Lperp| = d f |Lperp|, thus we obtain

d f / dt = w = e B/ 2 L = wL.

The frequency of procession is equal to the Larmor frequency.

> Ezample: Suppose that we find that the lines corresponding to the transition from the n=2 to the n=1 level of He of a star are split into three levels. The splitting between subsequent levels is DE = 10-5eV. What is the magnetic field at the star?

> Solution: Using DE = hbar wL= hbar e B / 2m we obtain

> B= 2 m DE / e hbar = 2 x 9.1x10-31kg x 10-5V / 1.06x 10-34J s=0.182 Tesla.

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### 8.b. Spin

Discrepancy with the observed and predicted Zeeman effect lead to the discovery of electron spin by Pauli. Pauli discovered that observations and theory can be brought to an agreement if we assume that electrons have an internal angular momentum. It is not tied to any rotational motion but rather it is an intrinsic property. This angular momentum is called the spin. Many other particles, like protons, photons, etc. have spin, as well. The spin is quantized just like any other angular momentum. We learned that the orbital angular momentum is quantized such that |L| = hbar [l(l+1)]1/2, where l is an integer. The quantization of spin is similar, except in addition to integer, half-integer values of the quantum number s (that corresponds to l) are allowed. The value of s depends on the type of the particle. In particular, for electrons, or protons s=1/2. For photons s=1. Thus the magnitude of the spin carried by an electron is |s| = hbar [s(s+1)]1/2 = hbar [1/2(1/2+1)]1/2 = hbar (3/4)1/2. The same way the z component of the spin angular momentum is sz = hbar ms , where ms = s, s-1, ..., -s. Since s=1/2 this leaves only ms = 1/2 and -1/2. That is to say the electron spin has two orientations. Just like for orbital angular momentum, the spin cannot be exactly aligned with the z axis |sz| = hbar/2 < |s| = hbar (3/4)1/2.The x and y components of the spin are fuzzy, they cannot be measured simultaneously with the z component. Again, just like for orbital angular momenta the endpoint of the spin vector is on a circle that has the radius of (s2-sz2)1/2=hbar (3/4 -1/4)1/2= hbar/ 21/2. Semiclassically one would say the enpoint of the spin vector precesses on a circle of this radius around the z axis. Quantum mechanically we can say that one can find the endpoint of the spin vector anywhere on the circle with equal probability.

The magnetic moment attached to the electron spin is m = -(e/2m) g s. The gyromagnetic ratio e/2m is valid for orbital motion, but there is no real reason to assume that the gyromagnetic ratio is the same for electron spin. Indeed it turns out that g is almost exactly 2. The relativistic theory of the electron, the Dirac eqeution predicts that g=2. Theories that go beyond the Dirac equation (Quantum Electrodynamics) predicts exactly the existing tiny deviations (less than a percent)of g from 2. Its value is 2.0232... This is called the anomalous magnetic moment of the electron. The deviation of g from two can be measured to a precision of 11 significant digits. It is one of the greatest triumphs of Quantum Electrodynamics, the quantized theory of electrons and photons to be able to calculate the anomalous magnetic moment to the same precision and come to a complete agreement with experiments.

So what is the total magnetic moment of the electron?

> µ = µl +µs = (-e/2m) [ L + g s ].

Since g=2 the total magnetic moment and the total angular momentum J = L + s are not parallel to each other.

Spin was discovered in the Stern-Gerlach experiment. A beam of silver atoms went through a region of inhomogeneous magnetic field. Inhomogeneous magnetic field acts with a force on magnetic dipoles.The force is F = dB/dz µ. If the magnetic field and its derivative point into the direction of the z axis then the force is proportional to the z component of the magnetic moment. In silver the orbital angular momentum of the outer electron is 0, it is in an s-state. The expectation of the outcome of the experiment is that since µz is quantized, that one gets bright spots on the screen wheer the silver atoms hit the screen, corresponding tothe appropriate values of µz, i.e at spots calculated from the vertical force Fz = dB/dz (e/2m)g hbar m, where m is the magnetic quantum number. Stern and Gerlach saw two bright spots on the screen indicating that they observe a magnetic moment which has only two allowed orientation. This is impossible for obital angular momentum, that has always 2l+1 possible orientations, with l=0,1,2,... The explanation is that though silver has a lot of electrons they are in closed shells, in which all the magnetic moments cancel because for a closed shell S ml=0. Nuclear magnetic moments are all much smaller than electron magnetic moments, because the gyromagnetic ratio is much smaller for nucleons, it is inversely proportional to mass. The single valence electron is in as l=0 state, so it does not contribute to the total magnetic moment. The only significant contribution comes form the spin magentic moment of the valence electron, accounting for the two bright spots on the screen.

Let us investigate now the Zeeman effect for states of the hydrogen atom taking into account spin. The interaction energy is UB = - m B = -Bµz = B(e/2m) hbar ( ml + g ms ) Note that not only ml is an integer, gms is also 2x1/2=1 or 2x(-1/2) = -1. The l=0, s states then split into two with UB = hbar wL and - hbar wL. The p-states (l=1) can have m=-1,0, and 1, so adding the splitting due to spin on top of that we get the following 3x2=6 magnetic energy levels UB = 2 hbar wL , UB = hbar wL , UB = 0 (doubly degenerate), UB = - hbar wL , UB = -2 hbar wL .

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### 8.c. Spin-orbit effects

The relativistic theory of electrons resolves some of the degeneracies predicted by the Schrodinger equation. The basic reason is that angular momentum conservation is valid only for the total angular momentum, J = L + s. The intricate reason for the accidental degeneracy of the Bohr energy levels is not valid any more and the energy levels become dependent of the values of j, the quantum number characterizing the total angular momentum of the electron, J. As every angular momentum, the relation between the magnitude of the total angular momentum J, and of the quantum number j is

|J| = hbar [j(j+1)]1/2

The general theory of angular momentum is fairly complicated. It suffices to say that for any angular momentum operator, J, in quantum mechanics the allowed values for the quantum number j are 0, 1/2, 1, 3/2,... (integers and half-integers). One can prove that for a given orbital quantum number, l, and spin quantum number, s, the possible vaules of the total angular momentum quantum member j are:

j= l+s, l+s-1, ..., l-s. (except for l=0, when j=s)

Then, obviously, if s is half integer (like s=1/2 for electrons) then j is half integer as well. If s is integer (s=0 for p mesons, s=1 for photons) then j is integer as well, because l is always integer. The addition of angular momentum is more or less in agreement with classical ideas that two vectors can be added in many different ways, starting from parallel and ending up with antiparallel (where the magnitudes are subtracted). Unlike in classical physics, in quantum mechanics there is only a discrete number of choices as shown in the above list. If s=1/2, as for a single electron, then j= l +1/2 or l-1/2 when l0, or j=1/2 for l=0.

> Example: Show by comparing the magnitudes of j, l, and s that quantum mechanically one cannot add vectors L and s either completely parallel or completely antiparallel.

> Solution: We have to show that |J| <|L| + |s| and the equality is never true. Also |L| - |s| < |J| and the equality is never true, except in the case when l=0. The magnitudes of the total angular momentum for the two possible states are |J| = hbar [(l+1/2)(l+3/2)]1/2 and |J| = hbar [(l-1/2)(l+1/2)]1/2. If we add the magnitudes of the orbital angular momentum and spin, then we get |L| + |s| = hbar {[l(l+1)]1/2+ (3/4)1/2}> |J| = hbar [(l+1/2)(l+3/2)]1/2 . This is so because squaring this equation we obtain that

> (l+1/2)(l+3/2) < l(l+1) +3/4 +2 [l(l+1)]1/2 (3/4)1/2

> or equivalently

> l < [l(l+1)]1/2 (3)1/2,

> a relation that is obviously true, because l< [l(l+1)]1/2 and 1< (3)1/2.

> In a similar manner one can show that if j= l-/12 (i.e. the orbital momentum and spin are quantum mechanically "antiparallel") then |J|= hbar [(l-1/2)(l+1/2)]1/2 > |L| - |s| = hbar {[l(l+1)]1/2- (3/4)1/2}.

So let us list now the lowest states of the Hydrogen atom.

For n=1, l=0, j=1/2. There are two states now: Jz = hbar 1/2 and -hbar 1/2. These states are, of course, degenerate, due to rotation invariance. The spectroscopic notation for them is 1S1/2. The general notation is n lj where for l the notation S, P, D, F, ... used if l=0,1,2,3,...

For n=2, one has 8 states (in general according to nonrelativistic quantum mechanics with spin there are 2n2 states in the n th shell, the duplication of states due to the spin up or down option). They would be all degenerate in nonrelativistic quantum mechanics. If l=0, then j=1/2 and one has the doubly degenerate states of 2S1/2. If l=1, then one can have j= 1+1/2 = 3/2 and j=1-1/2 =1/2. These states are 4 and two fold degenerate because of the possible choices of mj. The notation for these states is 2P3/2 and 2P1/2. In fact, according to relativistic quantum mechanics the states 2S1/2 and 2P1/2 are still degenerate because the energy levels depend only on th e quantum numbers n and j. So the relativistic picture resolves the eightfold degeneracy into two fourfold degenerate systems of state: on one hand 2P3/2 (four states: mj= -3/2, -1/2,1/2 and 3/2) and on the other 2S1/2 (2 states) and 2P1/2 (2 states). While the degeneracy with respect to Jz is the consequence of rotation invariance and as such must be exact, the degeneracy between the 2S1/2 and 2P1/2 levels is still "accidental."

The resolution of the degeneracy among states with different j is called a spin-orbit effect. The spin-orbit effect name comes from the fact that J2 = (L + s)2 = L2 + s2 + 2 L s At fixed L2 and s2 the term 2 L s (i.e. spin-orbit interaction) determines the value of J2 and with that the fine structure of the spectrum. The splits between states of the same n and differentr j are relativistic effects (predicted by the relativistic variant of the Schrodinger equation, the Dirac equation) and as such they are small. Fine structure splittings are typically 10-3 eV.

The degeneracy between the levels 2S1/2 and 2P1/2 gets resolved if the so-called superfine structure is taken into account. This is caused by effects beyond relativistic quantum mechanics. The energy difference, excited by radio waves was first measured by Lamb. It is called the Lamb shift. It is in the radio frequency domain. It is of O(10-6eV). Its value measured in MHz ( DE = h f) is 1057.9MHz. The agreement between the prediction of quantum electrodynamics and the experimental value is better than 0.1 MHz, within experimental errors.

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### 8.d. Pauli exclusion principle

To explain the Mendeleev table of elements Wolfgang Pauli postulated in 1925 that no two electrons in an atom can be in the same state. In other words, one can approximately classify electrons in heavier atoms using hydrogen-like levels but then the complete set of quantum numbers characterizing the state of the electron should not be the same. the complete set includes the quantum numbers n, l, m, and ms. This means that in every shell characterized by the principal quantum number n, can be only 2n2 electrons. The set can be equivalently characterized by the set n,l,j,mj.

A deeper explanation of the exclusion principle lies in the indistinguishability of identical particles. Classically, two balls can always be distinguished from each other, provided one takes a close enough look. Particles do not have any other properties than their quantum numbers and if those are identical they are truly indistinguishable. In two electron scattering there is no way to distinguish which of the electrons came from the left which from the right. The mathematical expression of this lack of distinguishability is given by the symmetry of the probability for the exchange of identical particles.

The wavefunction for two particles must depend on the coordinates of both particles. Suppose that the two particle wavefunction is y(r1,r2). This is the solution of the Schrodinger equation written down for a two particle system in external field. Then

dP = |y(r1,r2)|2dx1dy1dz1dx2dy2dz2

is the probability that we find particle 1 in volume dx1dy1dz1 around point r1 and particle 2 in dx2dy2dz2 around point r2. But if the particles are indistinguishable then we really do not know whether the first particle is at r1 and the second particle is at r2 or vica versa. Consequently, this probability must be the same as |y(r2,r1)|2dx1dy1dz1dx2dy2dz2. In other words the probability density must be symmetric in the coordinates

|y(r1,r2)|2=|y(r2,r1)|2.

Now there are two simple ways we can solve this for equation for the wavefunction itself. If the wavefunction is symmetric,

y(r1,r2)=y(r2,r1),

or antisymmetric

y(r1,r2)=-y(r2,r1),

in both cases the relation for magnitude squares is satisfied. One can prove based on very deep theroetical reasons that the wavefunction for all particles that have half integer spins, s=1/2, 3/2,... must obey the latter equation, must be antisymmetric. These particles are called fermions. Examples are electrons, protons, neutrons, mu mesons, neutrinos, etc. Two particle wavefunctions for identical particles that have integer spin like s=0,1,... must be symmetric in the exchange of coordinates. These particles are called bosons. Examples are pions, photons, d and He nuclei, etc.

Note that the two symmetries have very different physical consequences. Due to the antisymmetry, y(r,r)=0. Then the probability that two fermions are at the same point is zero (This conclusion should be modified if we also consider spin, as we will observe later). There is no similar constraint on identical bosons.

In the approximation of neglecting the interaction between electrons it is easy to write down solutions of the Schrodinger equation that have the appropriate symmetry. Assume e.g. that the time independent Schrodinger equation has the form

Ey(r1,r2)= - (hbar2/2m)[ D1 + D2 ]y(r1,r2) + [V(r1) + V(r2)]y(r1,r2)

As we discussed it earlier such an equation has a factorized solution

y(r1,r2)=y1(r1)y2(r2),

where y1(r1) and y2(r2) are solutions of the single particle equations

E1y1(r1)= - (hbar2/2m) D1 y1 (r1) + V(r1) y1 (r1), and

E2y2(r2)= - (hbar2/2m) D2 y2 (r2) + V(r2) y2 (r2),

with E1 + E2 = E. The solution y(r1,r2)=y1(r1)y2(r2) is not antisymmetric, as it should be for electrons. But then there is another solution of the same energy

y'(r1,r2)=y1(r2)y2(r1),

because having the same schrodinger equation the solution y2(r1) is certainly allowed for particle 1, and the solution y1(r2) is allowed for particle two. The energy is unchanged and a linear combination of the two solutions is a solution as well. Taking

y(r1,r2)=y1(r1)y2(r2)-y1(r2)y2(r1)

we end up with a solution of the schrodinger equation which is antisymmetric in the coordinate at the same time. This form immediately implies that the two wavefunctions y1 and y2 must be different other wise y would vanish identically. The conclusion is that the antisymmetry of the two particle wavefunction implies that the Pauli exclusion principle is satisfied. One has to keep in mind that the wavefunction depends on quantum numbers n, l, m, and ms. Thus, the spacial part of the wavefunction maybe identical for y1 and y2 provided the spin, or ms dependent parts are different. We will define the spin wavefunctions, |+> and |-> corresponding to states in which ms=+1/2 and ms=-1/2, respectively. Then the complete hydrogenlike wavefunction wouldhave the form of

y(r) = Rn,l(r) Ylm(q,f) |+> ( or |-> )

> Example: Ground state of the He. atom. The He nucleus has charge 2e, Z=2, so the neutral atom contains two electrons. Neglecting Coulomb repulsion between the two electrons the Schroding equation has separable solutions. The solutions are hydrogen like wavefunctions, where E = E1 + E2, the energies of the individual electrons. The ground state is obtained if we choose n1=n2=1. Then antisymetry requires that the wavefunctions of the two electrons are different. One can choose ms=1/2 for the first particle, ms=-1/2 for the second. The usual notation for such a two particle spin wavefunction is |+->. Then antisymmetrizing we get the ground state wavefunction of the he atom as

> y(r1,r2)= R1,0(r1) R1,0(r2) (1/4p )(1/21/2)[ |+-> - |-+> ].

> The multiplier (1/4p ) is just [ Y00(q,f) ]2 while (1/21/2) is needed to normalize the combination of the two spin wavefunctions. The wavefunctions R1,0(r) are hydrogenlike, with Z=2:

> R1,0(r) = (1/p1/2)(2/a)3/2 e-2r/a

> The ground state energy predicted by this non-interacting model is E = E1 + E2 = - 2 (Z2/n2) 13.6 eV = - 2 (4) 13.6 eV = -108.8 eV = - 8 Ry the (where 1 Ry = 13.6 eV). The true, experimentally observed, energy is E = -5.807 Ry. This is something we would expect, since the repulsion of the electrons adds a positive contribution to the energy.

There are many ways to take into account (approximately) the effect of the repulsion of electrons. The simplest way would be the following. Remember that the expectation value of the hamiltonian in an eiegenstate of energy E of the Hamiltonian is just E. This is so because

<H> = Ú dx dy dz y*(r) H y(r) = Ú dx dy dz y*(r) E y(r) = E Ú dx dy dz y*(r) y(r) = E.

Using the ground state determined from the Schrodinger equation without interaction term (i.e. the product of hydrogen-like wave functions), denoted by y0, we can calculate the expectation value of the complete Hamiltonian. If we denote the hamiltonian without the interaction term as H0, and the interaction term as Hint then we obtain

Ú dx dy dz y0*(r) H y0 (r) = Ú dx dy dz y0*(r) ( H0+ Hint) y0 (r) = E0 + Ú dx dy dz y0*(r) Hint y0 (r)

Knowing the form of Hint and y0 we can calculate the last integral and get 2.5Ry. Adding that to -8Ry we get an estimate for the total energy that is -5.5Ry, pretty close to the experimental value of E = -5.807 Ry.

One can show that in any other state but in the true ground state the expectation value <H> = Ú dx dy dz y*(r) H y(r) is larger then E1, the ground state energy. In other words take any normalized function y(r) then

<H> = Ú dx dy dz y*(r) H y(r) > E1,

expect in the case when y(r) is the ground state wavefunction, when the equality sign would result. Then one could experiment with various trial wavefunctions to lower the<H>. the lower <H> is the closer we are to the true ground state energy and true ground state wavefunction. The advantage is that we do not need to solve the Schrodinger equation, only integrate (possibly numerically) a function. Of course, if one has a good physiscal lintuition what function to choose one gets better results. In the particular case of the He atom twe guess that the shape of the wavefunctions is correct, except there is a screening effect, whereby to some extent one of the electrons screens the charge of nucleus for the other one, so that it sees an effective charge that is really smaller then 2e. This would be partially taken care of if we chose the wavefunction insted of R(r) µ e-2r/a something like R(r) µ e-Zr/a where Z is an effective value for the nuclear charge presumably smaller then 2. Using this trial wavefunction we can calculate the expectation value of the Hamiltonian, including the coulomb repulsion term between electrons. One obtains the result as a function of Z. Now one could use any Z. Since we know that no matter what z we use the energy we obtainis higher then the ground state energy it makes sense to minimize the obtained expression in th eeffective charge Z. The minimum is obtained at Z=27/16 < 2, as expected. The value of the minimum is -5.7Ry, atremendous improvement compared to the value -8Ry, even compared to our first, better estimate of -5.5RY.

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### 8.e. The periodic system

The Pauli exclusion principle, or alternatively, the antisymmetry of wavefunctions provides a tool to build up heavier atoms. We have seen that electron wavefunctions for the He atom can be constructed. Before discussing heavier atoms I would like to make only one more comment concerning the He atom: The orientation of spins of the two electrons is opposite.The possible total angular momenta of the two electron system, both being in s state, could be either s= s1+s2, s1+s2-1,..., |s1-s2|. In our case, since s1=s2=1/2, the only possible choices are s=1 and 0. The choice for He is s=0 because the Pauli exlusion principle implies that the spins are antiparralel, rather then parallel. Consequently, the total angular momentum of the He atom is zero.The He atom forms a completely spherically symmetric state.

Building of heavier atoms is based on a few simple principles. The first one is of course the exclusion principle. The second is that the atom as any other physical system that is able to dissipate energy tries to occupy the lowest possible energy state. These two principles do most of the trick, but are not sufficient to build up the periodic table using simple minded hydrogen like energy levels. The problem is that while the accidental degeneracy is cetainly valid for hydrogen like ions, the interaction of electrons changes the hamiltonian and completely destroys it. In fact we know that higher angular mumentum implies orbits that are larger. Wavefuncions vanish at the origin as rl, faster and faster for larger and larger l. Wavefunction for larger l experience much more pronounced screening effects, because more of the electron cloud is inside their orbits. The net result is that that the energy levels depend strongly on orbital angular momentum: the coulomb attraction is less and less for higher and higher l. The higher the l is the higher the energy level will be at. as a result we will have a hierarchy of states as

E1s < E2s < E2p < E3s < E3p < E3d µ E4s < E4p < E5s < E4d < E5p < E6s < E4f µ E5d < E6p < E7s < E5f µ E6d

In spite of the destroying of the accidental symmetry for heavier atoms it is possible to use the same quentum numbers to characterize energy levels and the notion that energy levels are filled one by one if we add more electrons to the system. Chemical properties are determined by the outermost subshell. So we get the following systematics

- Hydrogen: 1s1 configuration
- Helium: 1s2 configuration (2 electrons with opposite spins on the n=1, l=0 level). Closed shell, chemically inert.
- Lithium: 1s22 s1 configuration (2s level is lower than 2p level due to interaction of electrons).
- Beryllium: 1s22 s2 configuration (2s sublevel is closed, but small difference from 2p sublevel, therefore it is not inert).
- Boron: 1s22 s22p1 configuration. In the p level (l=1) we need to worry about the total angular momentum of electrons as well. The 2P1/2 is lower than the 2P3/2 i.e. j=1/2. Consequently, the total angular momentum of the boron electron system is j=1/2.
- Carbon: 1s22 s22p2 configuration. Here there is a question concerning the spins and ml values of the two 2p electrons. It is advantageous to have parallel spins (total spin, S=1) and different ml values. This is called Hund's rule. Remember that wavefunctions y2,1,x, y2,1,y and y2,1,z are large in different regions of space (along the x,y, and z axes respectively) and so electrons in two different states have a small overlap and Coulomb repulsion energy. In other words the energy is minimized if their ml values are different allowing parallel spins, which have slightly lower energies.
- Nitrogen: 1s22 s22p3 configuration. According to Hund's rule all three p electrons have parallel spins (S=3/2) and the orbital momenta cancel each other ( ml=-1,0, and 1).
- Oxygen: 1s22 s22p4configuration. Two of the p electrons must be have the same ml value. They must have antiparallel spins, so the total spin is S=1.
- Fluor: 1s22 s22p5 configuration. All but one of the p shell states are filled. There is one unpaired electron, spin is S=1/2.
- Neon: 1s22 s22p6 configuration. The n=1 and n=2 shells are filled. S=0, inert gas. Energy gap from 2p to 3s level is large, hard to excite Ne.

Subsequently, from sodium (Z=11) to potassium (Z=19) the 3s and 3p subshells are filled up in a way, very similar to those of 2s and 2p (Lithium - Neon). Then instead of the 3d shell, 4s shell gets filled. The 3d shells are however, lower than the 4p shells so after filling the 4s subshell, the 3d subshell gets filled. This leads to the phenomenon of transition elements. These subsequent elements from scandium (Z=21) to zinc (Z=30) have closed 4s valence shells and similar chemical properties. For lower Z values, all subsequent elements had different valence shell configurations and vastly different chemical properties.

Further transition series have the 4d, 5d, 4f (lanthanides, rare earth elements), 5f (actinides). The valence shells for transition series are the same.

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### 8.f. X-ray emission

The nuclear Coulomb potential for inner electrons in heavy elements is not very much screened. Therefore, these electrons have a very large Coulomb energy: Taking the K shell (n=1) the Coulomb energy is

EK = - ke2Z2/ 2 a0 = - 13.6 eV Z2. Taking a heavy element as lead, we obtain EK = 91.2 keV. If an energetic electron (E> 92keV) collides with a Pb atom then it may knock out one of the K electrons. In that case, following from simple energy considerations, one of the outer electrons will drop down into the n=1 level emittting a very energetic photon. These photons have very definite energies, i.e. a line spectrum. Ka corresponds to L-> K transition (63.4 keV = 91.2keV - 27.8keV), Kb corresponds to M->K transition (n=3 to n=1) 81.1keV photons, etc. Taking into account of screening, Moseley established the following rule for the energy of Ka transitions:

E( Ka) = - ke2(Z-1)2/ (2 22a0) + ke2(Z-1)2/ (2 a0) = 3 ke2(Z-1)2/ (2 22a0)

This gives a good agreement with experiments and can be used to identify elements.

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