1

Atomic structure and atomic spectraI. The structure and spectra of hydrogenic atoms

In the forthcoming lectures, we shall see how to use quantum mechanics to describe theelectronic structure of an atom, the arrangement of electrons around its nucleus. The conceptswe are going to study are of central importance for understanding the structures and reactionsof atoms and molecules, and hence have extensive chemical applications. We need todistinguish between hydrogenic atoms and many-electron atoms. Hydrogenic atoms areone-electron atoms and ions of general atomic number Z, they include H, He+, Li2+, U91+, etc.Many-electron atoms are atoms and ions with more than one electron, and include all neutralatoms other than H. Hydrogenic atoms, and H in particular, are important because theirstructures can be discussed exactly. They also provide a set of concepts that are used todescribe the structures of many-electron atoms and the structures of molecules too.

An important concept is the wavenumber ˜ ν of the radiation emitted or absorbed by anatom. The wavenumber is related to the wavelength λ and frequency ν by

˜ ν = 1/ λ = ν / cwhere c is the speed of light. It is common in spectroscopy to express wavenumbers in theunits of cm-1; a wavenumber of 1000 cm-1 signifies, for example, that there are 1000complete wavelengths per centimeter. The record of wavenumbers emitted or absorbed by anatom (or a molecule) is called its spectrum (from the Greek word for appearance).

2

When an electric discharge ispassed through gaseous hydrogen, theH2 molecules are dissociated and theenergetically excited H atoms that areproduced emit light of discretefrequencies. The first importantcontribution to understanding thisspectrum by the Swiss schoolteacherJoahann Balmer, who pointed out in1885 that the wavenumbers of the

light in the visible region fit the expression ˜ ν ∝122 −

1n2 n = 3, 4,...

The lines this formula describes are called the Balmer series. When further lines werediscovered in the ultaviolet, giving the Lyman series, and in the infrared, the Paschenseries, the Swedish spectroscopist Johannes Rydberg noted (in 1890) that all could be fitted

to ˜ ν = ℜH

1n1

2 −1n2

2

ℜH = 109677cm

−1

with n1 = 1 (Lyman), 2 (Balmer), and 3 (Paschen), and that in each case n2 = n1 + 1, n1 + 2,... The constant ℜH is now called the Rydberg constant for the hydrogen atom.

3

The form of the above equation suggests that the wavenumber of each spectral line canbe written as the difference of two terms, each of the form T =ℜH / n

2

The concept of a spectroscopic term is the content of the Ritz combination principle:The wavenumber of any spectral line is the difference of two terms.Two terms combine to produce a spectral line of wavenumber

˜ ν = T1 − T2For us, it is easy to explain the Ritz combination principle, unlike the early

spectroscopists, we know about the existence of photons of energy hν. The conservation ofenergy implies the Bohr frequency condition:

When an atom changes its energy by ΔE, the difference is carried awayas a photon of frequency ν, where ΔE = hν

Thus, if each term represents an energy hcT, the difference in energywhen the atom undergoes a transition between two terms is

ΔE = hcT1 − hcT2and the frequency of the light emitted is ν = cT1 − cT2This expression rearranges into the Ritz formula when expressed in termsof wavenumbers. Since only certain wavenumbers are observed in aspectrum, only certain energy states of atoms are permitted.

4

The structure of hydrogenic atomsWe can find a description of the structure of the hydrogen atom by adopting

Rutherford’s nuclear model, in which an electron travels around a nucleus of charge Ze, andthen solving the Schrödinger equation for that electron’s wavefunction. This seems to becomplicated because it involves two particles, the electron and the nucleus, and hence sixcoordinates.

However, the full equation can be separated into two equations, one for the motion ofthe atom as whole through space, and the other for the motion of the electron around thenucleus. We have already solved the first of these equations – it corresponds to the freemotion of a particle in space. The equation for the motion of the electron around the nucleusinvolvels three coordinates of the electron relative to the nucleus. We shall see that, becausethe potential is spherically symmetrical, that equation is separable. One factor in thewavefunction is for the angular motion of the electron around the fixed point (particle on asphere). This suggests that the only new work concerns the radial dependence of thewavefunctions.

The strategy is to separate the relative motion of the electron and the nucleus from themotion of the atom as a whole. Then we separate the relative motion of the electron and thenucleus into angular and radial parts, and solve the latter.

5

The separation of internal motionWe write the Schrödinger equation for the nucleus and electron:

Hψ = Eψ with

H = −

h 2

2me

∇e2 −

h 2

2mN

∇ N2 +V

where me is the mass of the electron, mN is the mass of the nucleus (a proton if the atom is H),and V is the Coulomb potential energy of an electron (of charge –e) at a distance r from anucleus of charge Ze (with Z = 1 for H).

V =−Ze 2

4πε 0×1r

, ε0 = 8.854x10-12 J-1 C2 m-1

The laplasians ∇e2 and ∇N

2 differentiate with respect to the coordinates of the electron and

nucleus (so ∇e2 has the terms ∂ 2 /∂xe

2, etc., where xe is the x-coordinate of the electron.

The equation can be expressed in terms of the separation of theparticles, r, and the location of the center of mass, R. We shall do thetransformation of the derivatives with respect to x-coordinate: theothers follow the same way.

The location of the center of mass is at X = me

mxe +

mN

mxN

with m the total mass. The separation of the particles is x = xe − xN

6

∂∂xe

=∂X∂xe

×∂∂X

+∂x∂xe

×∂∂x

=me

m×∂∂X

+∂∂x

;∂∂xN

=mN

m×∂∂X

−∂∂x

Therefore, the x-part of the sum of the two laplasians becomes

€

1me

∂2

∂xe2 +

1mN

∂2

∂xN2 =

1m

∂2

∂X 2 +1me

+1mN

∂∂x2

The y- and z-components can be dealt with similarly, and so1me

∇e2 +

1mN

∇ N2 =

1m∇cm

2 +1µ∇2

with1µ=1me

+1mN

, where µ is called reduced mass.

Now we write the overall wavefunction as ψ =ψ R( )ψ r( )and carry through the separation of variables procedure. The total equation separates into anequation for the free motion of a particle of mass m and the motion of a particle of mass µaround the center of mass:

7

€

−h2

2me∇e2ψ −

h2

2mN∇N2 ψ −

Ze2ψ4πε0r

=

−h2ψ r( )2m

∇cm2 ψ R( )−

h2ψ R( )2µ

∇2ψ r( )−Ze2ψ r( )ψ R( )

4πε0r= Eψ r( )ψ R( )

Division by ψ r( )ψ R( ) gives

−

h2

2m∇cm

2 ψ R( )ψ R( )

−h2

2µ∇2ψ r( )ψ r( )

−Ze2

4πε 0r= E

The first term is independent of r and the second and third are independent of R; hencethe equation is separable. The separation leaves the Schrödinger equation for the internalmotion (structure) of the atom:

−

h2

2µ∇2ψ +Vψ = Eψ V = −

Ze2

4πε0rwhich is the equation for hydrogen atom we must solve.

8

Separation of the internal coordinatesThe next step is to separate the angular variation of the wavefunction from its radial

dependence. We write ψ r,θ ,φ( ) = R r( )Y θ,φ( )This Schrödinger equation can be also separated into two equations:

Λ2Y = −l l +1( )Y

d 2 rR( )dr 2

−2µh2 VrR =

−2µErRh2

where now

V =

−Ze 2

4πε0r+l l +1( )h 2

2µr 2The first equation is the angular wave equation, and it is the same as the one we solved forthe angular momentum of a particle in three dimensions. The cyclic boundary conditions arealso the same. Hence, the angular wavefunction Y is a spherical harmonic specified by thequantum numbers l and ml, with

ml = 0,±1,±2,...,±land l a positive integer. However, just as ml came under a new constraint (that it lies between-l and +l) when it appeared in an equation that involved l, so we shall see that l comes under anew constraint (that it cannot exceed a certain value) because it also appears in anotherequation.

9

The radial wavefunctionThe second equation is the radial wavefunction for a hydrogenic atom.We can anticipate some features of its solution by examining the form ofthe potential energy V. The first term in V is the Coulomb potentialenergy of the electron in the field of the nucleus. The second term in Vstems from the centrifugal force that arises from the angular momentumof the electron around the nucleus. When l = 0, the electron has noangular momentum, and the potential is pure coulombic and attractive atall radii. When l ≠ 0, the centrifugal term gives a positive contribution tothe effective potential energy. When the electron is close to the nucleus(r is small), this repulsive term dominates the attractive coulombiccomponent and the net effect is a repulsion of the electron from thenucleus. The potential energies, the one for l = 0 and the one for l ≠ 0,

are qualitatively very different close to the nucleus, but they are similar at large distancesbecause the centrifugal contribution tends to zero more rapidly than the coulombiccontribution. Therefore, we can expect the solutions with l = 0 and l ≠ 0 to be quite differentnear the nucleus but similar far away from it.

Acceptable solutions can be found only for the following energies

En =

−Z 2µe4

32π 2ε02h2 ×

1n 2 n =1,2,...

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The corresponding wavefunctions are

Rn ,l = ρlLn ,l ρ( )e− ρ / 2

with n =1,2,... and l = 0,1,2,...,n −1

The variable ρ that occurs in R is proportional to the radius r:

ρ =2Zn×µme

×ra0

a0 =4πε 0h

2

mee2 = 52.9pm

where a0 is the Bohr radius (it first appeared in Bohr’s model of the hydrogen atom as theradius of its lowest energy orbit):

For most atoms it is good enough to make the approximation µ/me = 1 (the actual value for H,the worst case, is 0.99946), when the relations is simply

ρ =2Zn×ra0

The factor L(ρ) that occurs in R is a polynomial in ρ, called an associated Laguerrepolynomial. An example is the one with n = 3 and l = 0:

L3,0 = 6 − 6ρ + ρ2

11

Orbital n l Rn,l

1s 1 0

€

2 Za0

3/2

e−ρ /2

2s 2 0

€

12 2

Za0

3/2

2− 12ρ

e−ρ / 4

2p 2 1

€

14 6

Za0

3/2

ρe−ρ / 4

3s 3 0

€

19 3

Za0

3/2

6− 2ρ +19ρ2

e−ρ /6

3p 3 1

€

127 6

Za0

3/2

4 − 13ρ

ρe−ρ /6

3d 3 2

€

181 30

Za0

3/2

ρ2e−ρ /6

The full wavefunction is

€

ψ = RY , where Y are spherical harmonics. In the Table,

€

ρ = 2Zr a0 .

12

13

Locating the radial nodesLet us calculate the radii at which the radial wavefunction with n = 3 and l = 0 is zero in

the hydrogen atom. The radial wavefunction vanishes as r → ∞ on account of the exponentialfactor in R. This is true whatever the value of n and l. However, it also vanishes where the

polynomial factor L = 0: 6− 6ρ + ρ2 = 0 ρ = 3± 3 = 4.732 or 1.268Since for n = 3 and Z = 1 r = 3/ 2( )a0ρ , the radial nodes 1.90a0 and 7.10a0 (101 and 375pm). The number of radial nodes in general is n – l – 1, not counting the one at infinity.

Atomic orbitals and their energiesThe wavefunctions of hydrogenic atoms are called orbitals. All the orbitals have the

form ψ n ,l ,ml= Rn ,lYl ,ml

where R is one of the hydrogenic radial wavefunctions (Laguerre polynomials) and Y is oneof the spherical harmonics. Each orbital is defined by three quantum numbers, n, l, and ml.When an electron is described by one of these wavefunctions, we say that it occupies thatorbital. Thus, an electron described by the wavefunction ψ1,0,0 is said to occupy the orbitalwith n = 1, l = 0, and ml =0. Two of the quantum numbers, l and ml, come from the angularsolutions, and relate to the angular momentum of the electron around the nucleus:

14

An electron in an orbital with quantum number l has an angular momentum of magnitude

l l +1( ){ }1/ 2 h with l = 0,1,2,...n −1.An electron in an orbital with quantum number ml has a z-component of the angular

momentum

€

mlh with

€

ml = 0,±1,±2,...,±l .The third quantum number, n, is called the principal quantum number. It can take

values n =1,2,3,... and determines the energy through

En =−hcℜn2

hcℜ =Z 2µe4

32π 2ε02h 2

It is a peculiarity of hydrogenic atoms that the energy is independent of l and ml,and depends only on n. Therefore, all orbitals of a given n have the same energy– are degenerate – whatever values of l and ml.

The energy levelsAll the energies in the above equation are negative. This means that we are

dealing with bound states of the atom, in which the energy of the electron is lowerthan when it is infinitely far away and at rest (which corresponds to the zero ofenergy). There are also solutions of the Schrödinger equation with positiveenergies. These correspond to unbound states of the electron, the states to whichan electron is raised when it ejected from the atom by a high-energy collision or

photon. The energies of the unbound electron are not quantized, and form thecontinuum states of an atom.

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In the case of hydrogen, Z = 1 and the bound state energies are:

En =−hcℜH

n2 hcℜH =

µe4

32π 2ε02h 2

The spectrum of atomic hydrogen can be now explained by supposing that it undergoes atransition from a state with principal quantum number n2 (energy −hcℜH / n2

2) to a state

with principal quantum number n1 (energy −hcℜH / n12) and discards the energy

difference as a photon of energy hν and frequency ν.

Since energy is conserved overall, the frequency is given by

hν = hcℜH

n12 −

hcℜH

n22 or ν = cℜH

1n12 −

1n22

That is, since ˜ ν =νc

, ˜ ν = ℜH

1n1

2 −1n2

2

which is in the form required by experiment. When we insert the values of the fundamentalconstants into the expression for ℜH we get almost exact agreement with experiment. Theonly discrepancies arise from the neglect of relativistic corrections, which the Schrödingerequation ignores.

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Ionization energiesThe ionization energy I of an atom is the minimum energy required to ionize the atom

in its ground state, its lowest energy state. The ground state of hydrogen is the state with n =1, which has the energy E1 = −hcℜHThe atom is ionized when the electron has been excited to the level E = 0 to n = ∞. Therefore,the energy that must be supplied is

I = hcℜH = 2.179 ×10−18 J =13.60eV

The experimental determination of ionization energies of atoms depends on thedetection of the series limit, the wavenumber at which the series terminates and becomes acontinuum. Suppose the upper state lies at an energy Eupper, then when the atom makes atransition to Elower a photon is emitted of wavenumber ˜ ν , where hc ˜ ν = Eupper − Elower

Since the upper energy levels are given by an expression of the formEupper = −hcℜ / n

2, the emission lines will occur at

˜ ν = −−ℜn2 −

Elower

hcThe separation between successive lines decreases as n increases and so thelines converge. Moreover, a plot of the wavenumbers against 1/n2 should

give a straight line of slope −ℜ and intercept −Elower / hc. The value of–Elower is the ionzation energy of the atom in that state (and is the ground stateionization energy when the lower state is the ground state).

17

Shells and subshellsAll the orbitals of a given value of n form a single shell of the atom. In a

hydrogenic atom, all orbitals of given n, and therefore belonging to thesame shell, have the same energy. It is common to refer to successive shellsby the letters: n = 1 2 3 4 ...

K L M N ...Thus, all the orbitals of the shell with n = 2 from the L shell of the atom.

The orbitals with the same value of n but different values of l form thesubshells of a given shell. These subshells are generally referred to by theletters s, p, ... using

l = 0 1 2 3 4 ...s p d f g ...

(the letters then run alphabettically). Thus, the subshell with l = 1 of theshell with n = 3 is called 3p subshell, and its three orbitals are called the 3porbitals.

An electron that occupies one of the 3p orbitals is called a 3p electron. Since lcan range from 0 to n –1, there are n subshells of a shell with principalquantum number n. Thus, when n = 1, there is only one subshell with l = 1.When n = 2, there are two subshells, the 2s subshell (with l = 0) and the 2psubshell (with l = 1).

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When n = 1 there is only one orbital, that with l = 0 and ml = 0 (the only value of ml

permitted). When n = 2, there are four orbitals, one in the s subshell with l = 0 and ml = 0,and three in the l = 1 subshell with ml = 0, +1. When n = 3, there are nine orbitals (one with l= 0 and ml = 0, three with l = 1 subshell, and five with l = 2):

Number of orbitalsn l: 0 1 2 3 4 Total number

s p d f g1 1 12 1 3 43 1 3 5 94 1 3 5 7 165 1 3 5 7 9 25

In general, the number of orbitals in a shell of principal quantum number n is n2, so in ahydrogenic atom each shell is n 2-degenerate.s-orbitals

The orbital corresponding to the ground state (for which the electron is most tightly boundto the nucleus), is the one with n = 1 (and therefore necessarily with l = 0 and ml = 0, the onlypossible values of these quantum numbers when n = 1). We can write

for Z = 1

€

ψ =1πa0

3

1/2

e−r /a0

19

As this wavefunction is independent of angle it has the same value at all pointsof constant radius; that is, the 1s orbital is spherically symmetrical. However,ψ does depend on distance from the nucleus, and its amplitude decays

exponentially from its maximum value

€

1/πa03( )1/2

at the nucleus (at r = 0). It

follows that the most probable point at which the electron will be found is atthe nucleus itself.We can understand why the lowest energy wavefunction decays with distanceat a particular rate by thinking about the potential and kinetic energy

contributions to the total energy of the atom. The closer the electron is to the nucleus onaverage, the lower its average potential energy. This suggests that the lowest potential energyshould be obtained with a sharply peaked wavefunction that has a large amplitude at thenucleus and is zero everywhere else. However, this implies a high kinetic energy, becausesuch a wavefunction has a very sharp curvature close to the nucleus, and high curvaturecorresponds to high kinetic energy. The electron would have very low kinetic energy if itswavefunction had only a very slight curvature. However, such a wavefunction spreads togreat distances from the nucleus and the average potential energy of the electron would beless negative. The actual ground state is a compromise between these two extremes: itspreads some way from the nucleus (so the potential energy is not as low as in the firstexample, but nor is it very high) and has a reasonably low curvature (so the kinetic energy isnot very low, but nor is it as high as in the first example).

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One way of depicting the probability of finding the electron in any volume element is torepresent ψ2 by the density of shading in a diagram. A simpler procedure is to show only theboundary surface, the shape that captures 90% of the electron probability and shows thelocation of any angular nodes. For the 1s orbital, the boundary surface is a sphere.

All s orbitals are spherically symmetric, but differ in the number of radial nodes (thevalues of r at which ψ = 0). For instance, the 2s orbital has a radial node where L2,0 = 0:

L2,0 = 2 − ρ = 0 and r = ρa0 implies r = 2a0Hence, the 2s orbital has a radial node at 2a0. Similarly, the 3s orbital has two nodes at 1.90a0

and 7.10a0. The energies of the s orbitals increase (the electron becomes less tightly bound)as n increases because the average distance of the electron from the nucleus increases.

21

Calculating the mean radius of an orbitalThe mean value is the expectation value

r = rψ 2∫ dτ dτ = r 2drsinθdθdφThe integration over the angular part gives unity in each case (because the sphericalharmonic are normalized), and so the only work relates to the integration over r:

r = rR2

0

∞

∫ × r 2dr

Integrals of this kind are evaluated using xne− axdx0

∞

∫ =n!an+1

For a 1s orbital,

R = 2 1a03

e− r /a0 r =

4a03 r 3e−2r / a0

0

∞

∫ dr = 32a0

For a 2s orbital, R =18a0

3

1/ 2

2 − ra0

e− r / 2 a0

22

r =18a0

3 r 3 2 − ra0

0

∞

∫2

e− r /a0dr = 6a0The general expression for the mean radius is

r n ,l ,ml= n 2 1+

121− l l +1( )

n 2

×a0Z

Radial distribution functions The wavefunction tells us,through ψ2, the probability offinding an electron in any region.We can imagine a probe with avolume dτ and sensitive toelectrons, and which we canmove around near the nucleus ofa hydrogen atom. Since theprobability density in the groundsta te of the a tom is

ψ 2 ∝ e−2r /a0 the reading fromthe detector decreasesexponentially as the probe ismoved out along any radius.

23

Now consider the probability of finding the electron anywhere on a spherical shell ofthickness dr at a radius r. The sensitive volume of the probe is now the volume of the shell,which is 4πr2dr. The probability that the electron will be found in this shell is the probabilitydensity at the radius r multiplied by the volume of the probe:

ψ 2 × 4πr2dr = Pdr , with P = 4πr 2ψ 2

The radial distribution function P is a probability density in the sense that when it ismultiplied by dr, it give the probability of finding the electron anywhere in a shell of

thickness dr at the radius r. For a 1s orbital, P∝ r 2e−2 r /a0

Since r2 increases with radius and ψ2 decreases, P is zero at the nucleus, zero at infinity(because the exponential is zero there), and passes through a maximum at an intermediateradius. The maximum of P marks the most probable radius (as distinct from point) at whichthe electron will be found, and for a 1s orbital occurs at r = a0, the Bohr radius. When wecarry through the same calculation for the radial distribution function of the 2s orbital (byforming 4πr2ψ2), we find that the most probable radius at which the electron will be found is5.2a0 = 275 pm, which reflects the expansion of the atom as its energy increases.Calculating the most probable radius

Let us calculate the most probable radius at which an electron will be found when itoccupies a 1s orbital of a hydrogenic atom of atomic number Z.We need to find the radius at which the radial distribution function of the hydrogenic 1sorbital has a maximum value by solving dP/dr = 0.

24

ψ =Z 3

πa03

1/ 2

e− Zr / a0 P = 4πr 2ψ 2 =4Z 3

a02 × e−2Zr /a0

dPdr

=4Z 3

a03 2r − 2Zr

2

a0

e−2Zr /a0 = 0 at r= r* r* = a0

ZH He Li Be B C N O F Ne

r*/pm 52.9 25.6 17.6 13.2 10.6 8.82 7.56 6.61 5.88 5.29The 1s orbital is drawn towards the nucleus as the nuclear charge increases. At uranium themost probable radius is only 0.58 pm, almost 100 time closer than for hydrogen. The electronthen experiences strong accelerations, and relativistic effects are important.

p orbitalsA p electron has an angular momentum (of

magnitude 2h), and this momentum hasa profound effect on the shape of thewavefunction close to the nucleus: porbitals have zero amplitude at r = 0. This

can be understood classically in terms of the centrifugal effect of the angular momentum,which flings the electrons away from the nucleus. It is also what we could expect from the

form of the effective potential energy, which rises to infinity as r → 0 and excludes thewavefunction from the nucleus. The same centrifugal effect appears in all orbitals with l > 0.

25

All orbitals with l > 0 (such as p, d, f, etc., orbitals) have zero amplitude at the nucleus, andconsequently zero probability of finding the electron there. We can see from the form of theradial wavefunction that R∝ ρ l

at ρ = 0 for any l > 0.The three 2p orbitals are distinguished by the three different values that ml can take

when l = 1. Since the quantum number ml tells us the angular momentum around an axis,these different values of ml denote orbitals in which the electron has different angularmomenta around the arbitrary z-axis but the same magnitude of momentum (because l is thesame for all three). The orbital with ml = 0, for instance, has zero angular momentum aroundthe z-axis. It has the form f(r)cosθ , and the electron density, which is proportional to cos2θ,has its maximum value on either side of the nucleus along the z-axis (for θ = 0 and 180º). Forthis reason, the orbital is also called a pz orbital. The orbital amplitude is zero when θ = 90º,so the xy-plane is a nodal plane of the orbital. On this plane there is zero probability offinding an electron that occupies this orbital.

The orbitals with ml = +1 (which are proprtional to sinθe+iφ) do have angular momentumabout the z-axis. The orbital with the factor e+iφ corresponds to rotation in one direction andthat with the factor e-iφ corresponds to motion in the opposite direction. They have zeroamplitude where θ = 0 and 180º (along the z-axis) and maximum amplitude where θ = 90º,which is in the xy-plane. To draw the functions it is usual to take the real linear combinations

f sinθ eiφ + e− iφ( )∝ f sinθ cosφ ∝ xff sinθ eiφ − e− iφ( )∝ f sinθ sinφ ∝ yf

26

When normalized, they are called the px and py orbitals respectively. These combinations arestanding waves with no net angular momentum around the z-axis. They are composed ofequal and opposite values of ml. The px orbital has the same shape as a pz orbital, but isdirected along the x-axis; the py orbital is similarly directed along the y-axis.

d orbitalsWhen n = 3, l can be 0, 1, or 2. This givesone 3s orbital, three 3p orbitals, and five3d orbitals. The five d orbitals have ml = 2,1, 0, -1, -2 and correspond to five differentangular momenta around the z-axis (butthe same magnitude of angularmomentum, since l =2 in each case). Asfor the p orbitals, d orbitals with oppositevalues of ml (and hence opposite senses ofmotion around the z -axis) may becombined in pairs to give standing waves,and their boundary surfaces are shown onthe figure.

27

Spectral transitions and selection rules

We know the energies of all the states of hydrogenic atoms: En ,l,ml=−hcℜn 2

When the electron undergoes a transition, a change of state, from an orbital with quantumnumbers n1, l1, ml1 to another (lower-energy) orbital with quantum numbers n2, l2, ml2, itundergoes a change of energy ΔE and discards the exess energy as a photon ofelectromagnetic radiation with a frequency ν given by the Bohr frequency condition. Are allpossible transitions permissible and does the spectrum of the atom arises from the transitionof an electron from any initial orbital to any other orbital? No!!!

This is not so because a photon has unit intrinsic spin angular momentum. If a photon isgenerated by an electron undergoing a transition, the angular momentum of the electron mustchange to compensate for the angular momentum carried away by the photon as its spin.Thus, an electron in a d orbital with l = 2 cannot make a transition into an s orbital with l = 0because the photon cannot carry away enough angular momentum. Similarly, an s electroncannot make a transition to another s orbital, because then there is no change in the electron’sangular momentum to make up for the angular momentum carried away by the photon. Itfollows that some spectral transitions are allowed, meaning that they can occur, while othersare forbidden, meaning that they cannot occur.

28

A selection rule is a statement about whichtransitions are allowed. They are derived (for atoms) byidentifying the transitions that conserve angularmomentum when a photon is emitted or absorbed. Theselection rules for hydrogenic atoms are

Δl = ±1 Δml = 0,±1The principal quantum number n can change by anyamount consistent with the Δl for the transition because itdoes not relate directly to the angular momentum.

The selection rule enables us to construct a Grotriandiagram, which is a diagram that summarizes the energiesof the states and the allowed transitions between them. Thedensities of the transition lines in the diagram denote theirrelative intensities in the spectrum. The intensities may becalculated from the wavefunctions of the two states.

29

Hydrogenic atoms summary

The wavefunctions of hydrogenic atoms depend on three quantum numbers:Principal quantum number: n =1,2,3,...Angular momentum quantum number l = 0,1,2,...,n −1Magnetic quantum number: ml = l, l −1, l − 2,...,−lThe energy is related to n by En =

−hcℜn2

hcℜ =

Z 2µe4

32π 2ε02h 2

The magnitude of the angular momentum of the electron is l l +1( ){ }1/ 2 h and its

component on an arbitrary axis is mlh. Each energy level is n2 degenerate.The wavefunctions are products of radial and angular components:

ψ = R r( )Y θ,φ( )The angular wavefunctions Y are the spherical harmonics and the radial wavefunctions R are theassociated Laguerre functions.

The selection rules for spectroscopic transitions areΔn unrestricted Δl = ±1