physical chemistry 2 nd edition thomas engel, philip reid chapter 22 quantum states for...

Physical Chemistry 2Physical Chemistry 2ndnd Edition EditionThomas Engel, Philip Reid

Chapter 22 Chapter 22 Quantum States for Many-Electron Atoms

and Atomic Spectroscopy

Chapter 22: Quantum States for Many-Electron Atoms

© 2010 Pearson Education South Asia Pte Ltd

Physical Chemistry 2nd Edition

OutlineOutline

1. Good Quantum Numbers, Terms, Levels, and States

2. The Energy of a Configuration Depends on Both Orbital and Spin Angular Momentum

3. Spin-Orbit Coupling Breaks Up a Term into Levels

4. The Essentials of Atomic Spectroscopy5. Analytical Techniques Based on Atomic

Spectroscopy




OutlineOutline

6. The Doppler Effect7. The Helium-Neon Laser8. Laser Isotope Separation9. Auger Electron and X-Ray Photoelectron

Spectroscopies10.Selective Chemistry of Excited States:

O(3P) and O(1D)




22.1 Good Quantum Numbers, Terms, Levels, 22.1 Good Quantum Numbers, Terms, Levels, and Statesand States

• The H atom quantum numbers are good quantum numbers as the set of operators commutes with the total energy operator .

• A useful model is to generate the total orbital and spin momentum vectors L and S:

zz ssll ˆ and ˆ,ˆ,ˆ 22

H

i

ii

i sSlL





• The capitalized form of the operators refers to the resultant for all electrons in unfilled subshells of the atom.

2

2,

2

2,

ˆˆ and ˆˆ

ˆˆ and ˆˆ

ii

iizz

ii

iizz

iLiL

sSsS




Example 22.1Example 22.1

Is an eigenfunction of the operator ? If so, what is its eigenvalue MS?

1112212211112,1 ssss

S




SolutionSolution

This result shows that the wave function is an eigenfunction of 0M with ˆ

s zS

2,1011122122111122

1112212

2211112

1112212

2211112

1112212211112ˆ1112212211111ˆ

1112212211112ˆ1ˆ

2,12ˆ1ˆ2,1ˆ

electron on only acts ˆ where2ˆ1ˆˆ

sssshh

ssh

ssh

ssh

ssh

ssssssssss

ssssss

ssS

iisssS

zz

zz

zzz

zzzz





• Another interaction need to be added to form good quantum numbers for many-electron atoms.

• This is defined by total angular momentum, J

• A term is a group of states that has the same L and S values.

SLJ




22.2 The Energy of a Configuration Depends on 22.2 The Energy of a Configuration Depends on Both Orbital and Spin Angular MomentumBoth Orbital and Spin Angular Momentum

• The vector model of angular momentum can be used to depict singlet and triplet states.




22.2 The Energy of a Configuration 22.2 The Energy of a Configuration Depends on Both Orbital and Spin Angular Depends on Both Orbital and Spin Angular MomentumMomentum

• To extract the possible terms that are consistent with a p2 configuration,





What terms result from the configuration ns1d1? How many quantum states are associated with each term?




SolutionSolution

Because the electrons are not in the same subshell, the Pauli principle does not limit the combinations of ml and ms. Using the guidelines formulated earlier,

Therefore, the terms that arise from the configuration ns1d1 are 3D and 1D.

202 ,12/12/1 ,02/12/1 minmaxmin LSS




SolutionSolution

Table 21.4 shows how these terms arise from the individual quantum numbers. In setting up the table, we have relied only on the z components of the vectors msi and mli. Using these components, can be easily calculated because no vector addition is involved. Because each term has(2S+1)(2L+1) states, the 3D term consists of 15 states, and the 1D term consists of 5 states as shown in Table 21.3.

and i

liLi

siS mMmM




22.2 The Energy of a Configuration 22.2 The Energy of a Configuration Depends on Both Orbital and Spin Angular Depends on Both Orbital and Spin Angular MomentumMomentum

Hund’s 1st rules: The lowest energy term is that which has the greatest spin multiplicity. For example, the 3P term of an np2 configuration is lower in energy than the 1D and 1S terms.

Hund’s 2nd rules: For terms that have the same spin multiplicity, the term with the greatest orbital angular momentum lies lowest in energy. For example, the 1D term of an np2 configuration is lower in energy than the 1S term.





Determine the lowest energy term for the p2 and d6 configurations.




SolutionSolution

The placement of the electrons is as shown here:

For the p2 configuration, ML,max=1 and MS,max=1.

Therefore, the lowest energy term is 3P. For the d6 configuration, ML,max=2 and MS,max=2.




SolutionSolution

Therefore, the lowest energy term is 5D. It is important to realize that this procedure only provides a recipe for finding the lowest energy term. The picture used in therecipe has no basis in reality, because no association of a term with particular values of ms and ml can be made.




22.3 Spin-Orbit Coupling Breaks Up a Term into 22.3 Spin-Orbit Coupling Breaks Up a Term into LevelsLevels

Hund’s 3rd rules: The order in energy of levels in a term is given by the following:• If the unfilled subshell is exactly or more than half full, the level with the highest J value has the lowest energy.• If the unfilled subshell is less than half full, the level with the lowest J value has the lowest energy.





What values of J are consistent with the terms 2P and 3D? How many states with different values of MJ correspond to each?




SolutionSolution

The quantum number J can take on all values given by

For the 2P term, L=1 and S=1/2. Therefore, J can have the values 3/2 and 1/2. There are 2J+1 values of MJ , or 4 and 2 states, respectively.

For the 3D term, L=2 and S=1. Therefore, J can have the values 3, 2, and 1. There are 2J+1 values of MJ or 7, 5, and 3 states, respectively.

SLSLSLSLJ ,...,2,1,




22.4 The Essentials of Atomic Spectroscopy22.4 The Essentials of Atomic Spectroscopy

•All spectroscopies involve the absorption that induces transitions between states of a quantum mechanical system.•The frequency for absorption lines in the hydrogen spectrum is given by

where n = principal quantum number RH = Rydberg constant

2222320

4 1111

8~

finalinitialH

finalinitial

e

nnR

nnch

emv





The absorption spectrum of the hydrogen atom shows lines at 82,258; 97,491; 102,823; 105,290; and 106,631 cm-1. There are no lower frequency lines in the spectrum. Use graphical methods to determine ninitial and the ionization energy of the hydrogen atom in this state.




SolutionSolution

The following equation allows ninitial and the ionization energy to be determined from a limited number of transitions between bound states.

The plot of versus assumed values of has a slope of -RH and an intercept with the frequency axis of . However, both ninitial and nfinal are unknown, so

that in plotting the data, nfinal values have to be assigned to the observed frequencies.

2222320

4 1111

8~

finalinitialH

finalinitial

e

nnR

nnch

emv

v~




SolutionSolution

Because there are no lower frequency lines in the spectrum, the lowest value for nfinal is ninitial=1. We try different combinations of nfinal and ninitial values to see if the slope and intercept are consistent with the expected values of -RH and –RH/ninitial2. In this case, the sequence of spectral lines is assumed to correspond to nfinal=2, 3, 4, 5, and 6 for an assumed value of ninitial=1; nfinal=3, 4, 5, 6, and 7 for an assumed value of ninitial=2; and nfinal=4, 5, 6, 7, and 8 for an assumed value of ninitial=3.

v~




SolutionSolution

The plots are shown in the following figure:




SolutionSolution

The slopes and intercepts calculated for these assumed values of ninitial are:

By examining the consistency of these values with the expected values, we can conclude that ninitial=1. The ionization energy of the hydrogen atom in thisstate is hcRH.




22.4 The Essentials of Atomic 22.4 The Essentials of Atomic SpectroscopySpectroscopy

• Information from atomic spectra is generally displayed in a standard format called a Grotrian diagram.




22.5 Analytical Techniques Based 22.5 Analytical Techniques Based on Atomic Spectroscopy on Atomic Spectroscopy

• The concentration of lead in human blood and the presence of toxic metals in drinking water are determined using atomic emission and atomic absorption spectroscopy.





The 2S1/2 → 2P3/2 transition in sodium has a

wavelength of 589.0 nm. This is one of the lines characteristic of the sodium vapor lamps used for lighting streets, and it gives the lamps their yellow-orange color. Calculate the ratio of the number of atoms in these two states at 1500, 2500, and 3500 K. The following figure is a Grotrian diagram for Na (not to scale) in which the transition of interest is shown as a blue line.




SolutionSolution




SolutionSolution

The ratio of atoms in the upper and lower levels is given by the Boltzmann distribution:

The degeneracies, g, are given by 2J+1, which is the number of states in each level:

kThc

lower

upperkT

lower

upper

lower

upper transitionlowerupper eg

ge

g

g

n

n //

212

2 and 412

32

qgg lowerupper




SolutionSolution

From the Boltzmann distribution,

3500Kat 10 1.865 and

2500K,at 101.144 1500K,at 10699.16.24421

exp2

10381.110589

10998.210626.6exp

2

4

3-

4-7

239

834

/

T

eg

g

n

n kThc

lower

upper

lower

upper transition




22.6 The Doppler Effect22.6 The Doppler Effect

• The Doppler shift is used to measure the speed at which stars and other radiating astronomical objects are moving relative to the Earth.

where vz = velocity component c = speed of light

v0 = light frequency which the source is stationary

cvcv zz 110





A line in the Lyman emission series for atomic hydrogen (nfinal=1), for which the wavelength is at 121.6 nm for an atom at rest, is seen for a particular quasar at 445.1 nm. Is the source approaching toward or receding from the observer? What is the magnitude of the velocity?




SolutionSolution

Because the frequency observed is less than that which would be observed for an atom at rest, the object is receding. The relative velocity is given by

1-82

2

20

20

20

20

2

0

s m 10582.2 ;8611.01.445/6.1211

1.445/6.1211

/1

/1

/1

/1

or ,1/1

z

z

zz

v

c

v

c

v

c

v




22.7 The Helium-Neon Laser22.7 The Helium-Neon Laser

• To understand the He-Ne laser, we need to use the concepts of absorption, spontaneous emission, and stimulated emission.

• Spontaneous emission is a random process where photons phases and propagation directions are random.

• In stimulated emission, the phase and direction of propagation are the same as that of the incident photon.





• There are 4 levels of energy involved in a lasing transition.





• In a He-Ne laser operating as an optical resonator, there are coherent stimulated emission (parallel lines) and incoherent spontaneous emission (wave lines).





• The linewidth of a transition in a He-Ne laser is Doppler broadened through the Maxwell-Boltzmann velocity distribution.

Resonator transmission decreases the linewidth to less than the Doppler limit.

Amplification threshold reduces the number of frequencies supported by the resonator.





The distribution function that describes the probability of finding a particular value of magnitude of the velocity along one dimension, v, in a gas at temperature T is given by

This velocity distribution leads to the broadening of a laser line in frequency given by

dvekT

mdvvf kTmv 2/2

2

dvv

vv

kT

mc

kT

mdvvI

2

0

02

2exp

2





The symbol c stands for the speed of light, and k is the Boltzmann constant. We next calculate the broadening of the 632.8 nm line in the He-Ne laser at a function of T.a. Plot I(v) for T=100, 300, and 1000 K, using the mass appropriate for a Ne atom, and determine the width in frequency at half the maximum amplitude of I(v) for each of the three temperatures.b. Assuming that the amplification threshold is 50% of the maximum amplitude, how many modes could lead to amplification in a cavity of length 100 cm?




SolutionSolution

a. This function is of the form of a normal or Gaussian distribution given by

The full width at half height is 2.35 , or for this case, . This gives half widths of 7.554 × 108 s-1,

1.308 × 109 s-1, and 2.388 × 109 s-1 at temperatures of 100, 300, and 1000 K, respectively.

dvedvvf vvv

20

0

2/,

2

1

220 /35.2 mckTv




SolutionSolution

The functions I(v) are plotted here:




SolutionSolution

b. The frequency spacing between two modes is given by

The width of the velocity distribution will support 5 modes at 100 K, 8 modes at 300 K, and 15 modes at 1000 K. The smaller Doppler broadening at lowtemperatures reduces the number of possible modes considerably.

188

105.112

10998.2

2

sd

cv




22.8 Laser Isotope Separation22.8 Laser Isotope Separation

• Laser isotope separation is to separate atoms and molecules into their different isotopes.

• A tunable copper vapor laser with narrow line width is used to excite ground-state uranium atoms to an excited state involving the 7s electrons.




22.9 Laser Isotope Separation22.9 Laser Isotope Separation

• In Auger electron spectroscopy (AES), A core level hole is formed by energy transfer from an incident photon or electron.

• The core hole is filled through relaxation from a higher level and a third electron is emitted to conserve energy.





Upon impingement of X-rays from a laboratory source, titanium atoms near the surface of bulk TiO2 emit electrons into a vacuum with energy of 790 eV. Thefinite mean free path of these electrons leads to an attenuation of the signal for Ti atoms beneath the surface according to . In this equation, d is the distance to the surface and is the mean free path. If is 2.0 nm, what is the sensitivity of Ti atoms 10.0 nm below the surface relative to those at the surface?

/0 deIdI




SolutionSolution

Substituting in the equation , we obtain

This result illustrates the surface sensitivity of the technique.

30.2/0.10 107.60

100 eI

I

/0 deIdI




22.10 Selective Chemistry of Excited States: 22.10 Selective Chemistry of Excited States: O(O(33P) and O(P) and O(11D)D)

• The interaction of sunlight with molecules in atmosphere leads interconnected set of chemical reactions.

• For wavelengths less than 315 nm, the dissociation of molecular oxygen occurs as

physical chemistry 2 nd edition thomas engel, philip reid chapter 22 quantum states for...

Documents

quantum states

electron atoms

physical chemistry

good quantum numbers

group of states

triplet states

nd edition example

nd edition solution