Electronic and Nuclear States

C H A P T E R

10 Electronic and Nuclear States 10.0 Introduction Since history of science can be lost to current generations, it is of importance to note that in 1900 many prominent scientists did not believe atoms existed and there were only two known forces: electromagnetism and gravity. Planck started science down the road to discreteness in 1900 when he quantized the oscillators of metals to explain the heat capacity of solids. It was the paper in 1905 on Brownian motion by Einstein that convinced mainstream scientists that atoms existed. Einstein’s Theory of Special Relativity dealt with electrodynamics and his Theory of General Relativity dealt only with gravity. It was not until the 1930s that the strong and weak nuclear forces were a center of theoretical study. The strong nuclear force was discussed in Section 9.12. The weak nuclear force governs nuclear decay. Although textbooks in chemistry may discuss the four forces of Nature and the structure and properties of matter, it is safe to say at this point in time that none of these books provide an account of the origin of forces and matter that we call atoms. After Einstein developed his Theory of General Relativity that was published in 1915 with a successful explanation of the perihelion motion of Mercury, he soon discovered that when the field equations were applied to the Universe as a whole, the Universe was unstable. Acting on the belief of physicists at that time that the Universe was static and stable, in 1917 Einstein introduced the "cosmological constant" to "counteract" the attraction of the gravitational interactions between celestial objects. But even with the cosmological constant, the Einstein equations were unstable with three solutions for the description of the dynamics of the Universe: stable, collapsing, and expanding. One person who took the expanding Universe quite seriously was Georges Henri Joseph E´douard Lemaıˆtre (1894e1966), a Belgium priest with an interest in cosmology. If the Universe was expanding, it had to have a start sometime. The 1927 cosmic model of Lemaıˆtre assumed that the initial state of the Universe was the static Einstein Universe, with the radius R0. Borrowing from the fission process of radioactive elements, this Primeval Atom broke up into parts that became the expanding Universe. It was George Gamow (1904e68) who suggested in the 1940s that the recession of the galaxies implied a common point of origin. He thus suggested an "explosion" which started all of space and matter. Sir Fred Hoyle, who did not like this scenario christened the Gamow model as the "Big Bang," and this title is still what it is called today. All that came into existence at the Big Bang was energy, for this is the only thing that could possibly fit into a volume of space equal to an assumed Planck volume, (a))3, where a) ¼ (Gh/2pc3)1/2 ¼ 1.614  1035 m. Needless to say, the temperature at the time of the Big Bang was exceedingly hot to account for all of the mass and radiation in the Universe in accordance with Einstein’s expression E ¼ mc2. As the Universe expanding, it also cooled in

Physical Chemistry http://dx.doi.org/10.1016/B978-0-12-800514-9.00010-9

461

Copyright © 2017 Elsevier Inc. All rights reserved.

462

10. ELECTRONIC AND NUCLEAR STATES

accordance with the expressions for adiabatic expansion of an ideal gas. To better understand what happens in the expansion/cooling process, consider the metaphor of air composed of nitrogen, oxygen, water, and carbon dioxide. At 500 K all of the components are in the gas state, and all obey the ideal gas equation. As the temperature is lowered to 373 K, the water condenses to the liquid state. When the temperature reaches 194.7 K, the carbon dioxide leaves the gas phase. Oxygen liquefies at 90.19 K and finally nitrogen undergoes a phase transition at 77 K. In a similar way, the four forces of Nature come into existence as well as matter as the Universe continues to cool to its present background temperature of about 2.7 K. It is important to know that the free neutron is unstable and results in the decay to a proton and the creation of an electron. 1 0n

/ 11 p þ e

[10.0.1]

Equation [10.0.1] provides all of the components to make up the atoms. However, the Big Bang accounts for the existence of only the lightest of elements, such as hydrogen, helium, and lithium. Where did all of the other elements come from? The first thing to consider is the energy needed to bring two protons to the point where they touch. Assume a Coulomb potential and a separation distance of 8.7  1016 m, which is the diameter of the proton. The characteristic temperature Tc ¼ ε/kB is approximately 2  1010 K, which means that extremely high temperatures are required for the fusion reaction. (Note: due to quantum mechanical tunneling the temperature for the fusion reaction can be considerably less than 1010.) Where might this energy come from? The first candidate is the interior of stars. A star is formed by the collapse of a gas cloud, mainly of hydrogen since hydrogen is the most abundant element formed from the Big Bang. As the gaseous cloud collapses heat is given off, which in turn increases the kinetic energy of the gas particles. At sufficiently high temperatures, the kinetic energy of the protons is sufficient for the fusion process. What happens after the initial synthesis of atoms depends upon whether the fusion process is exothermic or endothermic. To answer this question, we turn to a plot of the binding energy per nucleon (DEbind/A) as a function of the number of nucleons in the atom (A) as shown in Figure 10.0.1. For elements in which A is less than 56 (iron, Fe), energy is released in the nucleosynthesis process. The released energy in the form of radiation provides a pressure that opposes the gravitational collapse and the star attains an equilibrium size. However, once the element iron is synthesized, the situation swiftly changes. Since there is no longer radiation pressure to maintain the size of the star, there is an immediate collapse. For stars with mass above a critical mass, the star explodes into a supernova. The energy of the explosion forces the fusion of nuclei to the heavier elements for which A is greater than 56. The fact that elements with A > 56 are naturally occurring on Earth means that we are made of star-stuff from supernova explosions. Humans tend to honor those who excel in a species in events that have captured the interests of humans. Sports have always been at the forefront of such accolades. Although his home run record of 714 has been surpassed by Barry Bonds (762) and Hank Aaron (755) in the United States and by Sadahara Oh (868) in Japan, Babe Ruth still remains as the King of Swat in the eyes of baseball aficionados. Horse racing has been around for about 5000 years and became a professional sport on January 1, 1702. Among the greatest horses of all time is Man o’ War. Man o’ War’s record was almost perfect: 20 wins, 1 second place, and 0 losses. The time he came in second was during the period when there were no starting gates, and he was oriented in the wrong direction when the race started. Man o’ War did not run in the Kentucky Derby because his owner, August Belmont,

463

10.0 INTRODUCTION

Δ E bind (MeV) A 9 8 7 6

12 4

Fe

C

235

U

He

238

U

6

Li

5 4 3 2 1

56

3

H He

4 2 1

H 0

H 30

60

90

120 150 180 210 240

A FIGURE 10.0.1 Nucleon binding energies of the elements. The binding energy per nucleon, DEbind/A, versus the number of nucleons, A, shows a maximum value at iron as indicated by the dashed red line. Elements to the left of the dashed red line are synthesized by fusion reactions within stars. Synthesis of elements to the right of the red line requires an input of energy only available through supernova explosion of stars. Since elements heavier than iron exist on Earth, we are made from supernova debris.

Jr., did not like to race in Kentucky. (The Belmont Stakes race was named after August Belmont, Sr., who financed the first race. The Belmont Stakes is the third leg of the Triple Crown in horse racing.) Man o’ War sired 64 stakes winners, including Kentucky Derby winners Clyde Van Dusen (1929) and War Admiral (won the 1937 Triple Crowndthe Kentucky Derby, Preakness, and Belmont Stakes). Kentucky Derby winners in the lineage of Man o’ War and the year they won the Kentucky Derby include Venetian Way (1960), Lucky Pulpit (2014), and the 2015 Triple Crown winner American Pharoah. It can be safely stated that the elucidation of the structure of the atom is the intellectual offspring and lineage of one man: Sir Ernest Rutherford. In 1898, Rutherford discovered alpha and beta radiation emitted by uranium. Rutherford also named the radiation discovered by Paul Villard in 1900 as "gamma" radiation. With Frederick Soddy, Rutherford studied the transmutation of the elements. Soddy wrote about radium in 1909 in a book titled The Interpretation of Radium (Soddy, third edition, 1912). H.G. Wells (1914) was so inspired by this publication by Soddy that he wrote The World Set Free in which the phrase "atomic bomb" was first used. In experiments performed in 1912 with Hans Geiger (of Geiger counter fame) and Ernest Marsden, alpha particle scattering by gold foil indicated that some of the alpha particles were scattered back towards the source. Rutherford likened this result as firing the guns on a battleship at a piece of tissue paper and having the shells bouncing back from the tissue. These experiments indicated that most of the mass in the atom was in a very small region of the atom. The name "proton" was given to the nucleus of the hydrogen atom by Rutherford in 1920. The word "proton" is Greek for "first" as hydrogen is the lightest known element. On the basis of his studies on nuclear reactions, Rutherford proposed

464

10. ELECTRONIC AND NUCLEAR STATES

that the proton was a fundamental particle of Nature. Rutherford also proposed that a neutral particle must be present in the nucleus to keep the positive charges together in the nucleus. He named this hypothetical particle the "neutron" in 1920, which was experimentally verified in 1932 by James Chadwick. While working with Rutherford, Niels Bohr proposed his shell model of the electronic states of the atom based on the quantization of angular momentum. Many who worked with Rutherford also received the Nobel Prize, a partial list with the year they won the Nobel Prize being: Frederick Soddy (1921), Niels Bohr (1922), James Chadwick (1935), Otto Hahn (1944), Edward Appleton (1947), Cecil Powell (1959), Ernest Walton, and John Cockcroft (shared 1951). A building is composed of fundamental materials and units arranged in accordance with a blueprint that identifies the functionality of the building, such as the location of bathrooms, elevators, and offices. Likewise, atoms and nuclei have internal structure, which means they are constructed from more fundamental particles arranged according to a plan that defines their properties. The configuration of electrons about a nucleus is governed by electric forces that are arranged in accordance with the Schro¨dinger and Dirac equations. The resulting Periodic Table of the Elements presents an orderly arrangement of the structural properties of the building blocks from which molecules can be built. Nuclei are also composed of fundamental particles that are governed by the strong and weak nuclear forces. Hadrons are subject to the strong nuclear force that holds the nuclei together, and leptons are subject only to the weak nuclear force. Hadrons are considered to be composite particles which eventually decay to the proton. Murray Gell-Mann proposed the "Eightfold Way" that is similar to a Periodic Table of the Hadrons based on the properties of baryons and mesons. The neutrino was postulated in 1930 by Wolfgang Pauli to account for "missing energy" in beta decay. The name "neutrino" was given by Enrico Fermi during a conference in 1932 and a Solvay Conference in 1933. The present chapter explores the conditions under which chemical bonds are formed to construct molecules and the internal states of the nucleus that may lead to nuclear fission. The chemical and nuclear energies are compared, and sources of energy for the future are discussed.

10.1 Quantum Mechanics, Special Relativity, and Description of the Electron A paper in the Proceedings of the Royal Society, 1929, A 123, 714e733 begins with the statement regarding the necessity of introducing the spin of the electron: "Already before the arrival of quantum mechanics there existed a theory of atomic structure, based on Bohr’s ideas of quantize orbits, which was fairly successful in a wide field. To get agreement with experiment it was found necessary to introduce the spin of the electron, giving a doubling in the number of orbits of an electron in an atom. With the help of this spin and Pauli’s exclusion principle, a satisfactory theory of multiplet terms was obtained when one made the additional assumption that the electrons in an atom all set themselves with their spins parallel or antiparallel." The author then points out: "The fact that one had to make this additional assumption was, however, a serious disadvantage, as no theoretical reasons to support it could be given." Just as Newton wrote that he had no hypothesis for the force of gravity, but justified its existence because the 1/r dependence of gravity worked, this author noted that there was no justification to

¨ DINGER EQUATION 10.1.1 DIRAC’S APPROACH TO SOLVE THE RELATIVISTIC SCHRO

465

impose spin on an electron in the wave theory of Schro¨dinger other than it worked to explain certain properties of the electron. What impressed this author was the 1905 paper by Albert Einstein that appeared in a 1905 issue of Annalen der Physik. The title of the paper was "Zur Electrodynamik bewegter Ko¨per" (on the electrodynamics of moving bodies). Contained in the Einstein paper was the equation of the following form,  
 E2 ¼ c 2 p2 þ m2 c 2 ¼ c 2 p2x þ p2y þ p2z þ m2 c 2

[10.1.1]

This equation is the basis of Einstein’s Theory of Special Relativity. In his attempt to reconcile two descriptions of the electron, quantum mechanics and Einstein’s Theory of Special Relativity, the equations of Paul Adrien Maurice Dirac (1902e84) showed that the "spin" of the electron "naturally fell out" in his equations. There were also some more profound unexpected results. The title of the paper is "Quantum Mechanics of Many-Electron Systems." And what is it about his equations? Frank Wilczek wrote of his equations (Farmelo, 2002, pp. 102e103): "Of all the equations in physics, perhaps the most ‘magical’ is the Dirac equation. It is the most freely invented, the least conditioned by experiment, the one with the strangest and most startling consequences."

10.1.1 Dirac’s Approach to Solve the Relativistic Schro¨dinger Equation In regard to his goal to connect quantum theory with relativity theory, Dirac’s approach to unite quantum mechanics with special relativity was to list what the equation must have from what it ought to have. The equation must conform to Einstein’s Theory of Special Relativity that put space and time on an equal footing. The equation must be consistent with transformation theory. The equation must resemble ordinary quantum theory of an electron moving slowly compared to the speed of light. The first requirement gives the beginning equation of his quest, which is the relativistic energy of Einstein, given in Equation [10.1.1]. How Dirac approached this problem is outlined next, following Sherwin’s book on quantum mechanics (Sherwin, 1960, pp. 281e332). The first step is to substitute the quantum mechanical operators for the momentum in Equation [10.1.1]. It is here that Dirac pulled the rabbit out of the hat. Dirac did not plug in directly the operators, but rather "invented" a function that expressed the momentum part as a squared function involving the operators for the electron,  2 v v v 2 E ¼ c ax þ ay þ az þ bme c vx vy vz 2

2

[10.1.2]

where ax, ay, az, and b are unknown parameters to be determined. The Reader may object to this formulation because the nonrelativistic Schro¨dinger equation does not have cross terms whereas Equation [10.1.2] has obvious cross terms. The cross terms are illustrated in the following equation involving matrices, A2 þ B2 ¼ ðA þ BÞ2 ¼ A2 þ AB þ BA þ B2

[10.1.3]

466

10. ELECTRONIC AND NUCLEAR STATES

where the left side represents the Schro¨dinger equation (sum of squared terms) and the two expressions in the right represent the Dirac formulation (square of summed terms). Note that for matrices the cross terms do not commute, viz, AB s BA. However, Dirac required that these matrices anticommute for them to cancel in accordance with the "¼" sign, AB þ BA ¼ 0

[10.1.4]

and that their square results in the identity matrix, A2 ¼ B 2 ¼ I

[10.1.5]

It was clear to Dirac that the matrices ax, ay, az, and b must be 4  4 matrices. If Equations [10.1.4] and [10.1.5] held for the matrices a and b, then the following identities result, 0

1 0

0

B B0 1 B b¼B B B0 0 @

0

C 0 I 0 C C C¼@ C 0 C 0 A

0 1

0 0

1 1

0

A

I

[10.1.6]

1

0

where I and 0 are 2  2 identity matrices, and 0

0

0

0

0

1

1

0

1

0

0

0

0

B B0 B ax ¼ B B B0 @ 0

1

1

C 0 0 0C C C¼@ C 0C sx A

sx

1 A

[10.1.7]

0

0 0 i

B B0 0 i B ay ¼ B B B 0 i 0 @

1

C 0 0 0 C C C¼@ C 0 C sy A

i

0

0

0

0

0

1

0

sy

1 A

[10.1.8]

0

and 0

B B0 B az ¼ B B B1 @ 0

0 0 1

1

C 0 0 0 1 C C C¼@ C 0 0 C sz A 0

0

sz 0

1 A

[10.1.9]

10.1.3 PREDICTION OF ANTIMATTER

467

where the blocks in the matrices are sx ¼

sy ¼

0 1

! [10.1.10]

1 0 0 i

! [10.1.11]

i

0

1

0

0

1

and az ¼

! [10.1.12]

The s matrices have the properties sksk ¼ 1 and sjsk ¼ 0 if j s k.

10.1.2 The Electron "Spin" The block form of Equations [10.1.6]e[10.1.9] is the identity matrix I, the null matrix 0, and the Pauli matrices s. The 2  2 Pauli spin matrices were developed by Wolfgang Pauli (1900e58) as an ad hoc means of describing the electron. Pauli proposed a model of a spinning electron to account for the physical observations, such as the splitting of spectral lines in a magnetic field. In the Dirac solution the Pauli "spin" matrices were not introduced as a spin property of the system. The s matrices were introduced because of the unconventional approach of Dirac to solve the Schro¨dinger equation by means of Equation [10.1.2]. Rather than spin the quantum mechanical property is a probability flux or probability current, as given by Equation [8.1.12]. The probability current is clockwise or counterclockwise. The fact that the "spin" naturally fell out in the relativistic Schro¨dinger equation means that the "spin" is a quantum mechanical property. The nucleus likewise has an associated spin. The statistical properties of matter depend upon their spin. The quantum states of half-integer spin (FermieDirac particles) can only be singly occupied whereas the quantum states for particles with integer spin (BoseeEinstein particles) have no restrictions on the population of quantum states.

10.1.3 Prediction of Antimatter The time-dependent part of the Schro¨dinger equation for a free particle is 

h v4ðtÞ ¼ W4ðtÞ i2p vt

[10.1.13]

468

10. ELECTRONIC AND NUCLEAR STATES

where W is the total energy. The complete wave function is a vector of four components, 0

ψ1

1

0

ψ1

1

B C B C B C B C  B ψ2 C B ψ2 C i2p B C B C Wt J ¼ ψ 4ðtÞ ¼ B C4ðtÞ ¼ B Cexp  Bψ C Bψ C h B 3C B 3C @ A @ A ψ4 ψ4

[10.1.14]

The equation to solve is of the form (Sherwin, 1960, p. 282) 
 hc v v v  ax þ ay þ az ψ  b me c 2 þ W ψ ¼ 0 i vx vy vz

[10.1.15]

Expressing Equation [10.1.15] in terms of the matrices given by Equations [10.1.6]e[10.1.9] one has the set of equations, where the retention of zeros is to emphasize the matrix character of the set of equations,


 hc v hc v v ψ þ  ψ ¼0 W þ me c ψ1 þ 0 þ i2p vz 3 i2p vx vy 4 2



 
hc v v hc v 2  ψ3  ψ ¼0 0 þ W þ me c ψ2 þ i2p vx vy i2p vz 4 






 2

hc v hc v v ψ1 þ  ψ þ W  me c ψ3 þ 0 ¼ 0 i2p vz i2p vx vy 2

[10.1.16]

 
hc v v hc v  ψ2 þ 0 þ W  me c 2 ψ4 ¼ 0 ψ1  i2p vx vy i2p vz If only the x-direction is considered then Equation [10.1.16] becomes


 hc v ψ ¼0 W þ me c 2 ψ1 þ i2p vx 4




 hc v ψ ¼0 W þ me c 2 ψ2 þ i2p vx 3


hc v ψ2 þ W  me c 2 ψ3 ¼ 0 i2p vx 
hc v ψ1 þ W  me c 2 ψ4 ¼ 0 i2p vx

[10.1.17]

10.1.4 SEA OF ELECTRONS

469

The equation for the amplitudes is obtained with the substitution 0

ψ1

1

0

A1

1

0

1

A B 1C B C B C B B C B C BA C   B ψ2 C B A2 C C 2p i2p B 2C B C B C x ¼B px x B C¼B Cexp i Cexp B A3 C Bψ C BA C l h B B 3C B 3C C @ @ A @ A A A A 4 ψ4 4

[10.1.18]

The set of equations for the amplitudes is
 hc W þ me c 2 A 1 þ A 4 ¼ 0 l 
hc W þ me c 2 A 2 þ A 3 ¼ 0 l 
hc A 2 þ W  me c 2 A 3 ¼ 0 l

[10.1.19]


hc A 1 þ W  me c 2 A 4 ¼ 0 l Nontrivial solutions to a set of equations exist only if the determinant of the coefficients is zero. Therefore from Equation [10.1.19] one has from this criterion, W2  m2e c 4  p2 c 2 ¼ 0

[10.1.20]

where p ¼ h/l. The energy, W, therefore is W¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m2e c 4 þ p2 c 2

[10.1.21]

The positive energy is associated with matter, while the negative energy is associated with antimatter. This led Dirac to propose a "sea of electrons" to explain antimatter.

10.1.4 Sea of Electrons The electron has a negative charge, represented by e. The positron has the same mass as the electron but with a positive charge, represented by eþ. To explain the antielectron, Dirac proposed that there was a "sea of electrons" that cannot be observed directly. When an electron came out of that "sea" it left behind a "positive hole." This positive hole is the antiparticle of the electron, now called the positron. The relativistic Dirac equation not only "explained" the spin of particles, but also predicted the existence of antiparticles. The "sea of electrons" is illustrated in Figure 10.1.1.

470

10. ELECTRONIC AND NUCLEAR STATES

FIGURE 10.1.1 Sea of electron. To account for the electronepositron pair, Dirac proposed a "sea of electrons" composed of electrons in lowest filled energy states. This sea (yellow box) is not noticed in everyday events. As an electron is promoted to a higher state, it leaves behind a positive "hole" in the sea of electrons. This positive hole is the positron.

There is more implied in Dirac’s equation than simply the electronepositron pair production. Einstein’s most famous equation, E ¼ mc2, allows mass to be created out of energy, and charge conservation requires that electrons and positrons be created in pairs. A vacuum is not empty of particles, but rather is swarming with virtual particles being created and destroyed during a time span given by the Indeterminacy Principle, Dt < h/2pDE ¼ hc2/2pm where m ¼ 2mpair is twice the mass mpair for each pair of particles. The "sea of virtual particles" is currently associated with "dark energy" in cosmological models. As the fabric of space expands (Theory of General Relativity), the amount of dark energy increases.

10.1.5 Time-Dependent Dirac Equation for a Free Particle in One Dimension The grouping of wave functions in Equation [10.1.19] allows one to write for the time-dependent wave function along the x-axis, 0

0

1

C B BA C   B 2C p W C B ψI ðx; tÞ ¼ B Cexp i2p h x  h t B A3 C A @

[10.1.22]

0 and 0

A1

1

C B B 0 C   C B p W Cexp i2p ψII ðx; tÞ ¼ B x t C B h h B 0 C A @ A4

[10.1.23]

10.1.5 TIME-DEPENDENT DIRAC EQUATION FOR A FREE PARTICLE IN ONE DIMENSION

471

where it is understood that p ¼ px. From Equation [10.1.19], a little algebra gives a relationship between the coefficients within a group, with the results, 0

0

1

C B B pc C C B   B W þ m c2 C p W C B e t ψI ðx; tÞ ¼ A3 B Cexp i2p x  C B h h C B 1 C B A @ 0

[10.1.24]

and 0

1 pc B C B W þ me c 2 C B C B C   B C 0 p W B C ψII ðx; tÞ ¼ A4 B x t Cexp i2p B C h h B C 0 B C B C @ A 1

[10.1.25]

The Dirac wave function has two components; the amplitude of one is more than the amplitude of the other. The two components oscillate in orthogonal planes, as illustrated in Figure 10.1.2.

FIGURE 10.1.2 Components of the wave function for the Dirac equation for a free particle. The large vertical component is the real part of the wave function (green) with a small amplitude orthogonal component 180 degree out of phase (red). Note: The amplitude of the small component is highly exaggerated in the figure.

472

10. ELECTRONIC AND NUCLEAR STATES

10.1.6 The Nodes and Bound Particles The nonrelativistic Schro¨dinger equation for a bound particle has identifiable nodes. The question that comes to mind about the probability interpretation of the square of the wave function is how does the particle move from one side of the box to the other if the probability is zero at a node? The Dirac equation for the particle in a box with W ¼ E  V provides the answer. The wave function for a particle in a one-dimensional box is (Sherwin, 1960, p. 327); 0

0

1

C B B C  C B B
ihckp  cos kp x C C B 2p W þ m c 2 L  L C B e W C B ψI ðx; tÞ ¼ A3 B Cexp i2p t  C B h kp C B C B sin x C B L C B A @ 0 and

[10.1.26]

0

1  ihckp kp x C  cos B
B 2p W þ me c 2 L C L B C B C B C  B C 0 W B C ψII ðx; tÞ ¼ A4 B Cexp i2p t B C h B C 0 B C B C  B C kp @ A sin x L

[10.1.27]

with the energies sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  chkp 2 þ m2e c 4 Wn ¼ þ L

[10.1.28]

Equations [10.1.26] and [10.1.27] answer the questions about how a particle gets from one side to the other in the nonrelativistic Schro¨dinger equation which has nodes at specific locations. There are no nodes in the Dirac solutions, so the question is moot.

10.1.7 The Electron in an Electric FielddThe Beginning of Quantum Electrodynamics Dirac required of his theory that it must conform to Einstein’s Theory of Special Relativity that space and time be placed on equal footing. The notation of the Dirac equation can be generalized by expressing the variables under one mantle. The objective is to express the variables in terms of the Einstein summation notation. This involves superscripts as well as subscripts, so the Reader is

473

10.1.9 BEAUTY IN THE EQUATIONS

forewarned that the superscripts are not power laws. The change in notation is: x ¼ x1, y ¼ x2, z ¼ x3, and t ¼ x0. The Dirac equation for these parameters is   0 v 1 v 2 v 3 v i g 0 þ g 1 þ g 2 þ g 3 ψ ¼ me cψ vx vx vx vx

[10.1.29]

In the presence of an electric field A and employing the Einstein summation convention, the equation for the electron becomes.   m
g ivm  qe Am  me c ψ ¼ 0

[10.1.30]

where the subscript denotes a summation and the superscript indicates the component.

10.1.8 The Relativistic de Broglie Wavelength One result of the Theory of Special Relativity is that the mass of an object increases with velocity v of the particle. The corresponding expression for the momentum of the particle is, mv p pr ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffi v2 v2 1 2 1 2 c c

[10.1.31]

The corresponding de Broglie wavelength is (r and n denote relativistic and nonrelativistic, respectively) lr dB

h h ¼ ¼ pr p

rffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi v2 v2 1  2 ¼ ln dB 1  2 c c

[10.1.32]

The importance of Equation [10.1.32] is transparent in the Bohr model of the atom. For heavy nuclei such as uranium, the speed of the inner shell electrons is very fast because of the proximity to the nucleus. With the shorter de Broglie wavelength at relativistic speeds, the radius of the orbit is reduced. As a consequence the energy of excitation is correspondingly altered.

10.1.9 Beauty in the Equations Dirac believed that equations must be beautiful. Just as a sour note spoils a Mozart piano concerto, or a punctuation change ruins a poem, the beautiful equation cannot be altered in any way for then its importance is diminished by magnitudes. In Dirac’s own words, "This result is too beautiful to be false; it is more important to have beauty in one’s equations than to have then fit experiment." From "The Evolution of the Physicist’s Picture of Nature," Scientific American, 208 (5) (1963) What is it about an equation that makes it beautiful? One criterion is that it must be concise. Another criterion is that it should contain a wealth of information. The equations of Maxwell

474

10. ELECTRONIC AND NUCLEAR STATES

and Newton listed in Table 7.0.1 are beautiful because they concisely summarize all of electromagnetism and mechanics of classical physics. Two equations of the modern era are those of Dirac and Einstein. The Dirac equation can be written in very concise form, i

h m g vm ψ  me cψ ¼ 0 2p

[10.1.33]

The Einstein field equations are also described as beautiful, Gmn ¼

8pG Tmn c4

[10.1.34]

where G is the gravitation constant. The Einstein tensor Gmn describes the geometry of space-time and the stresseenergy tensor Tmn describes how the flux of density and energy move through space-time. But conciseness alone is not what makes an equation beautiful. A beautiful equation not only accounts for what is known about the Universe but what has yet to be known, even to the surprise of the inventors of the equations. Dirac set out to bring together the quantum mechanics of Schro¨dinger and Special Relativity of Einstein and up popped an explanation for the ad hoc addition of spin as a quantum mechanical property of particles and also doubled the number of particles in the Universe with the prediction of antiparticles. Einstein was in search for a theory that extended his Theory of Special Relativity to include acceleration reference frames. His basic conclusion from a thought experiment that gravity and acceleration are the same animal wearing different coats led to the Theory of General Relativity. Hidden within the belly of the beast was an expanding Universe and black holes, both of which Einstein refused to believe at first because it went against his philosophical views of how the Universe worked. There is one other aspect of the beauty of Equations [10.1.31] and [10.1.32]. The equations that successfully, without exception, explain the beginning, the evolution, and possible end of the Universe can both be written on the same T-shirt.

10.2 Paul DiracdA Concise Picture History has many stories about people who made discoveries with "startling consequences" while thinking a different discovery was made which had much less startling consequence. One example is Christopher Columbus. As history books record, Columbus set sail to discover a new route to India but unknowingly discovered a new continent. Dirac set sail to discover an equation to bring together special relativity and quantum mechanics, but, unknown to him at the time, he discovered a new set of particles. Graham Farmleo has written an excellent biography of Paul Adrien Maurice Dirac (Farmelo, 2009). Dirac thought the two solutions to his equation described both the electron and the proton. When it was brought to his attention that the proton weighed considerably more than the electron, it then became clear that his equation predicted antiparticles. Dirac said of his equation that it was smarter than he was. The "most startling consequences" was the prediction of antimatter, which is the basis of the Standard Model in particle physics and central to the Big Bang model of the origin of the Universe.

10.2 PAUL DIRACdA CONCISE PICTURE

475

Paul Dirac was very shy and a man of very few words. These characteristics were instilled in him as a young boy. His father, Charles, forced his children to speak only in French in order that they learn the language. If the children made mistakes, they were "punished" by having to eat supper with their mother in the kitchen. Because Paul had problems in expressing in French what he wanted to say, he chose to remain silent. The biographer Sheilla Jones wrote of him: "He didn’t talk; he did equations. Even when he gave a lecture, he tended to carefully write his equations on the blackboard, perhaps offer a statement or two, and that was it." From The Quantum Ten: A Story of Passion, Tragedy, Ambition and Science, by Sheilla Jones, 2008, p. 224. P.A.M. Dirac began his career as electrical engineer, but converted to physics and became a superb theoretical physicist. Paul Dirac is said to be the "father of particle physics" because of his equation which predicted antiparticles. Long before string theory became a fashion, Dirac proposed that particles were discrete lines of force rather than point particles. Dirac made the comment (Farmelo, 2009, p. 370): "One can look upon the muon as an electron excited by radial oscillations."

FIGURE 10.2.1 Bust of P.A.M. Dirac. Bust of Paul Dirac on the outside balcony of Linda Hall Library of Science and Technology in Kansas City, MO.

476

10. ELECTRONIC AND NUCLEAR STATES

Paul Dirac introduced the terms "boson," "fermion," and "graviton" into the literature. In the third edition of his book, Dirac introduced the "bra-ket" notation (Dirac, 1947, p. 19). He also invented the Dirac delta function. And the list goes ondfor a very long way! When P.A.M. Dirac was informed that he would share the 1933 Nobel Prize in physics with Erwin Schro¨dinger, his initial response was to refuse the prize. Not because he was to share it with Schro¨dinger, but because he did not want the publicity associated with the Nobel Prize in physics. However, he did eventually consent to accept the Nobel Prize when he was told he would get even more public notoriety if he refused the Prize. A bust of P.A.M. Dirac located on an outer balcony of Linda Hall Library of Science and Technology, located in Kansas City, is shown in Figure 10.2.1.

10.3 The Schro¨dinger Atoms and the Periodic Table of the Elements The quest for a Periodic Table of the Elements has a long history. It might be said that it began with the ancient Greeks, Democritus with the concept of the "atom" and the view that all things were made of earth, air, fire, and water. Keeping in mind that nothing was known about the structure of the atom, the early attempts ordered the chemical properties according to the weights of the elements. In 1817, Johann Dobereiner proposed the Law of the Triads, having noted that a group of three elements had very similar properties. The group of three elements was based on weights: two of the elements in the group were virtually equidistant from the weight of the third element. An example of a triad is calcium (weight 40), strontium (weight 87), and barium (weight 137). Another law is the Law of Octaves proposed by John Newlands in 1863. He arranged 56 elements into 11 groups based on similar properties. He noticed that every eighth element had similar properties. The periodicity of chemical properties is like an octave in music, hence the name. The person most associated with the development of the Periodic Table of the Elements is Dmitri Ivanovich Mendeleev (1834e1907). Writing the chemical properties and the weights of each element on cards, he arranged the cards in order of increasing weight. This was like a game of chemical solitaire. There were gaps, and Mendeleev predicted the existence of elements to fill these gaps. Missing in Mendeleev’s table, of course, were the noble gases. This is understandable because, as the name suggests, the noble gases were chemically inert.

Ordering of Elements in the Periodic Table of the Elements It was in 1895 that Rutherford reported the discovery of argon. It was the discovery of the nucleus and the nuclear charge by Rutherford in 1911 that changed the ordering system. Unknown at this time was the existence of neutrons. The neutrons could alter the order of elements by weight. The discovery of the nucleus led to the determination of nuclear charge, known also as atomic number. In 1913, Henry Moseley reported that X-ray spectral lines correlated well with the atomic number. A final change in the Periodic Table of the Elements occurred in the 1940s. Glenn Seaborg, discoverer of all the transuranic elements from 94 to 102, placed the actinide series below the lanthanide series. One of the tests for any model of the atom is to "explain" the Periodic Table of the Elements. The Bohr model of the atom was developed for a single electron in the presence of the potential field set up by a nucleus of charge magnitude Z. As indicated in Figure 9.1.1 the stability of the Bohr orbitals has an explanation that resides solely in quantum mechanics. The circumference of the

¨ DINGER ATOMS AND THE PERIODIC TABLE OF THE ELEMENTS 10.3 THE SCHRO

477

nth orbit must be an integer number of the fundamental de Broglie wave for the first orbit, ie, ln ¼ nl1. The Bohr model of the atom with its "planet-like" orbitals cannot explain the structure of the Periodic Table of the Elements. Attempts to improve the Bohr model by extension to elliptical orbitals failed. As with the Bohr model, the Schro¨dinger model of the atom addresses the interaction of a single electron with the nucleus. The solutions to the Schro¨dinger equation are therefore "hydrogenlike" wave functions. The complete wave function for the electron states in an atom now has a theoretical foundation, ψn;‘;m‘ ;ms ðr; q; fÞ ¼ Rn;‘ ðrÞQ‘;m‘ ðqÞFm‘ ðfÞSms

[10.3.1]

where Sms represents the "spin portion" of the wave function. Since the spatial and spin wave function are uniquely defined by the particular set of quantum numbers, it is frugal to use the more concise "bra-ket" notation. The wave function is thus defined by ψn;‘;m‘ ;ms ðr; q; fÞ ¼ jn; ‘; m‘ ; ms i ¼ jn; ‘; m‘ ijms i

[10.3.2]

and its complex conjugate by, ψn;‘;m‘ ;ms ðr; q; fÞ ¼ hn; ‘; m‘ ; ms j ¼ hn; ‘; m‘ jhms j

[10.3.3]

Unlike the Bohr model the radial part of the hydrogen-like atoms, Rn‘(r), does not impose any orbital shape. Selected expressions of Rn‘(r) are given in Table 9.6.2. Clearly all of the functions of Rn‘(r) have their greatest value at the origin of the coordinate system. If the square of the wave function, R2n;‘ , is a measure of the probability of finding the electron, then the first place to look for the electron would be at the nucleus. But this approach ignores the fact that the volume increases with distance. What is important in describing the probability of the electron in an atom is the volume probability, which for the radial wave function is PV ¼ 4pr2 R2n;‘ ðrÞ. Shown in Figure 10.3.1 are plots of Rn‘(r) and r2 R2n;‘ ðrÞ versus r for the 1s orbital. The maximum in the

R1,0(r)

r 2 R 2 (r ) 1,0

1.0

0.5

0.8

0.4

0.6

0.3

0.4

0.2

0.2

0.1 2

4

r

6

8

10

2

4

6

r

8

10

12

14

FIGURE 10.3.1 Radial wave function and volume probability for 1s orbital. The radial wave function for the 1s orbital is on the left and the volume probability distribution on the right, both in arbitrary units. The maximum in the function r2 R21;0 ðrÞ is at the location of the Bohr radius, ao.

478

10. ELECTRONIC AND NUCLEAR STATES

volume probability function is at the Bohr radius ao, in agreement with the Bohr model of the atom. Shown in Figure 10.3.2 are plots of Rn‘(r) and r2 R2n;‘ ðrÞ versus r for the 2s, 3s, and 4s orbitals. Their amplitudes were adjusted for comparison purposes. These plots indicate that the electrons in these spherical orbitals spend sometime inside the inner shells of the atom.

r 2 R 2n,0 (r)

Rn,0(r)

3.0

3

2.5

2

2.0

1 1

5

10

15

2 3

20

r

1.5 1.0 0.5 5

10

15

r

20

FIGURE 10.3.2 Radial distribution and volume probability for 2s, 3s, and 4s orbitals. The radial wave function and volume probability functions for the 2s (blue), 3s (green), and 4s (red) are shown above in arbitrary units.

But the radial wave function itself does not explain the structure of the Periodic Table of the Elements. Recall that the spherical wave functions QðqÞFðfÞ ¼ Y‘;m‘ ðq; fÞ ¼ j‘ijm‘ i are part of the complete solution to the hydrogen atom. In the conventional notation the angular momentum quantum number is represented by the symbol ‘ and the orientation of the angular momentum vector to the z-axis is called the azimuthal quantum number denoted by m‘. Neither of these quantum numbers relate to the energy of the system but rather to the degeneracy of the energy levels. Because of the nesting of the equations there is a specific relationship between the quantum numbers. The restrictions are: n ¼ 1; 2; ...: ‘ ¼ 0; 3; .::; n  1

[10.3.4]

m‘ ¼ ‘; ‘ þ 1; ..; 0; 1; 2; .::; ‘ It was the melding of quantum theory with special relativity by Dirac that gave the full theoretical explanation of the structure of the Periodic Table of the Elements. The additional quantum number was the spin state of the electron, ms ¼ þ1=2; 1=2

[10.3.5]

Since the shape of the orbitals depends solely on the spherical harmonic Y‘;m‘ ðq; fÞ, the term symbols reflect the quantum numbers ‘ and m‘ . Let us write the spatial functions in the bra-ket notation for a few atomic orbitals,

¨ DINGER ATOMS AND THE PERIODIC TABLE OF THE ELEMENTS 10.3 THE SCHRO

479

1s ¼ j1; 0; 0i

[10.3.6]

2s ¼ j2; 0; 0i

[10.3.7]

2pm‘ ¼ j2; 1; m‘ i ðm‘ ¼ 0; 1Þ

[10.3.8]

3dm‘ ¼ j3; 2; m‘ i ðm‘ ¼ 0; 1; 2Þ

[10.3.9]

Aufbau Principle and the Periodic Table The relativistic form of the Schro¨dinger equation of Dirac provided a theoretical basis for the complete set of quantum numbers: n, ‘, m‘, and ms. The nested set of equations solved by Schro¨dinger gave the degeneracy with respect to angular moment and the degeneracy factor for spin give the degeneracy for the energy level of quantum number n, D ¼ 2ð2j‘j þ 1Þ

[10.3.10]

where the first factor of "2" accounts for the spin degeneracy. The degeneracy factor of Equation [10.3.10] tells how many electrons can be accommodated in the ‘ state. The structure of the Periodic Table of the Elements is thus parsed into regions associated with the orbitals s, p, d, and f. If we represent the set of degeneracies associated with the quantum number n, viz, [n, {{jlj, D}}], we have for the first four values of n: [{1, {0, 2}]; [2, {{0, 2},{1, 6}}]; [3, {{0, 2}, {1, 6}, {2, 10}}]; and [4, {{0,2}, {1, 6}, {2, 10}, {3, 14}}]. Since the energy depends only on the principle quantum number the energy degeneracies are the sums of the angular momentum degeneracies: [1, 2]; [2, 8]; [3, 18]; and [4, 32]. The structure of the Periodic Table of the Elements is therefore a reflection of the orbital degeneracies as shown in Figure 10.3.3. The two columns bordered by red are the s electrons, the green boxed region are the p electrons, the central blue boxed region the d electrons, and the lower yellow boxed region the f electrons. A simple mnemonic for the order of filling in the orbitals to reconstruct the structure of the Periodic Table of the Elements is shown in Figure 10.3.4. There are several ways to represent the orbitals in triangular form, but this particular orientation looks like a Christmas tree, and hence its reference name. One simply follows the yellow brick road represented by the dashed arrow to determine the electron configuration of the atom. For example, the electron configuration for krypton with 36 electrons is 1s22s22p63s23p64s23d104p6. Sometimes it is convenient to express the electron configurations of the high atomic number atoms in terms of the "nearest" inert gas configuration. Consider potassium with the atomic number 19. The electronic configuration for this atom is 1s22s22p63s23p64s1. The configuration for argon, atomic number 18, is 1s22s22p63s23p6. Hence potassium could also be represented by the nearest inert gas configuration as [Ar]4s1. The basis of the mnemonic in Figure 10.3.4 is simply the degeneracy factors regarding the numerical order of the quantum numbers in relationship of an atom to its position on the Periodic Table of the Elements. The mnemonic is not, however, a good barometer for the order of filling successive energy states within the atom, which is the aufbau principle was proposed by Bohr and taught in many chemistry classes. In this regard the mnemonic is successful about 80% of the ground state electronic configurations. The reason lies in the penetration of the orbitals

480

10. ELECTRONIC AND NUCLEAR STATES

s

p

n 1

2

H

He

3

4

5

6

7

8

9

10

Li

Be

B

C

N

O

F

Ne

11

12

13

14

15

16

17

18

Na

Mg

Al

Si

P

S

Cl

Ar

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

K

Ca

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

Rb

Sr

Y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

6

55

56

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

Cs

Ba

Lu

Hf

Ta

W

Re

Os

Ir

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Rn

7

87

88

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

Fr

Ra

Lr

Rf

Db

Sg

Bh

Hs

Mt

Ds

Rg

Cn

Uut

Fl

Uup

Lv

Uus

Uuo

1 2 3 4 5

d

f 57

58

59

60

61

62

63

64

65

66

67

68

69

70

La

Ce

Pr

Nd

Om

Sm

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

89

90

91

92

93

94

95

96

97

98

99

100

101

102

Ac

Th

Pa

U

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

FIGURE 10.3.3 Periodical Table of the Elements and the orbitals. The solutions to the nested Schro¨dinger equation provide a natural explanation of the structure of the current version of the Periodic Table of the Elements. The table is arranged by the atomic orbitals being filled. The rows indicate the value of the principal quantum number n and the columns indicate the value of the quantum number ‘: s (‘ ¼ 0, red), p (‘ ¼ 1, green)), d (‘ ¼ 2, orange), and f (‘ ¼ 3, yellow). Representative shapes are shown above each section.

into the interior of the atom and the electrostatic interactions between the electrons. The discrepancies are with the d and f orbitals in the outer orbitals and to some extent the incompleteness of the orbitals. One example is copper, Cu, with 29 electrons. According to the mnemonic the electron configuration of copper is [Ar] 3d94s2. However, the energy state of copper has the electron configuration [Ar] 3d10 4s1, where the 3d orbital is completely full and the 4s orbital is half full.

481

10.4 MULTIELECTRON ATOMS

1s 2s 3s 4s

7s

3p 4p

5s 6s

2p

5p 6p

4d 5d

6d

7p

3d

7d

4f 5f

5g

6f 7f

6g 7g

6h 7h

7i

FIGURE 10.3.4 Filling orbitals in the Periodic Table of the Elements. A convenient mnemonic for remembering the shape of the Periodic table of the Elements is given by the fir tree arrangement above. It is emphasized that this mnemonic is a poor gauge of the actual ordering of energy states within the atom since there are some variations due to electroneelectron interactions.

10.4 Multielectron Atoms The Hamiltonian for a multielectron atom of Z ¼ ne electrons is n " X e

2

2

2

!

#

n 1 X n X e

e

h v v v q2e  2  H ¼ þ þ þ rj rj;i 8p me vx2j vy2j vz2j j¼1 j¼1 i¼jþ1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} 2

H ð0Þ

Zq2e

[10.4.1]

H0

where the presence of the electron mass me indicates the assumption of an infinitely massive nucleus. The set of terms within the single summation operation involve only the set of coordinates {rj} for the individual electrons relative to the nucleus, and is denoted as H ð0Þ . If this were the only contribution to the Hamiltonian then the single expression involving the ne electrons could be separated into ne equations, each involving only one electron. It is the second term, the one with the double summation and denoted by H 0, which spoils this separation of variables because it requires knowledge of the set of electroneelectron relative coordinates {rj,i}. The second term therefore is the "perturbation" term which prohibits exact separation of variables. If we follow the perturbation theory outlined in Chapter 8 ("Quantum Systems with Constant Potential"), the energy for an electron in the nth quantum level for the atom of nuclear charge Z is       E D E D E D ð0Þ   ð0Þ ð0Þ  ð0Þ  ð0Þ ð0Þ  0  ð0Þ ð0Þ ð1Þ ψn H ψn ¼ ψn H ψn þ ψn H ψn ¼ En þ En

[10.4.2]

482

10. ELECTRONIC AND NUCLEAR STATES

We may write for the unperturbed hydrogen-like energy ! 2p2 me q4e ð0Þ 2 ¼ Z2 EH;n En ¼ Z  2 2 h n

[10.4.3]

where me is the reduced mass of the electron and EH,n is the hydrogen atom energy for the nth level.

Example 10.4.1

The Electronic Energy for the Ground State of the Helium Atom

The electrical interactions in the helium atom are the two electrons with the nucleus and the interaction between the two electrons. If it were not for the electroneelectron interactions, the ionization energy would be that of the Bohr atom for both electrons. But the electroneelectron interaction is repulsive, which means the ionization energy for the first electron is less than that of the Bohr atom. Assume the electroneelectron interaction is a perturbation on the system. Use perturbation theory to calculate the ground state energy for the helium atom. Discuss the consequences of electroneelectron interactions on the energy levels and the ionization energies of the multielectron atom relative to the unperturbed hydrogen atom analogs.

Solution Let us take as the simplest example the helium atom for which Z ¼ 2. The Hamiltonian for the electrons in the helium atom is given by ð0Þ 0 H ¼ H ð0Þ He;1 ð1Þ þ H He;1 ð2Þ þ H ð1; 2Þ

[10.4.4]

where the electrons 1 and 2 are identified by the parentheses and the perturbation term is dependent upon the relative electroneelectron separation distance,

H 0 ð1; 2Þ ¼

q2e r1;2

[10.4.5]

The electrons are "spinning" charged particles and therefore interact with each other through their respective spin states as well as Coulombic interactions. To a first approximation we neglect these spin coupling terms. The solution to the unperturbed part of the Hamiltonian is thus written as the product of hydrogen-like wave functions, which for the ground state is given by the normalized function, ð0Þ

4He ¼ 1sð1Þ1sð2Þ

[10.4.6]

The energy for the helium atom is therefore given by ðoÞ

ðoÞ

EHe ¼ h1sð1Þ1sð2ÞjH He;1 þ H He;2 þ H 0 ð1; 2Þj1sð1Þ1sð2Þi ðoÞ

ðoÞ

¼ h1sð1Þ1sð2ÞjH He;1 j1sð1Þ1sð2Þi þ h1sð1Þ1sð2ÞjH He;2 j1sð1Þ1sð2Þi þ h1sð1Þ1sð2ÞjH 0 ð1; 2Þj1sð1Þ1sð2Þi ð0Þ

ð0Þ

¼ Z2 EH;1 þ Z2 EH;2 þ h1sð1Þ1sð2ÞjH 0 ð1; 2Þj1sð1Þ1sð2Þi ¼ 8EH;1 þ h1sð1Þ1sð2ÞjH 0 ð1; 2Þj1sð1Þ1sð2Þi [10.4.7]

483

10.4 MULTIELECTRON ATOMS

In this notation, EH,1 is the unperturbed ground state energy of the hydrogen atom. The integral involving the perturbation term is evaluated by first transforming to relative coordinates expressed in terms of the locations r1 and r2 from the nucleus. The first-order correction term to the energy is found to be 0

h1sð1Þ1sð2ÞjH ð1; 2Þj1sð1Þ1sð2Þi ¼

ð1Þ E0

5ð2pÞ2 Zme q4e 5 5 ¼ ¼  ZEH;1 ¼  EH;1 2 4 2 8h

[10.4.8]

We now combine Equations [10.4.7] and [10.4.8] to obtain the ground state for the helium atom, 5 11 EHe ¼ 8EH;1  EH;1 ¼ EH;1 2 2

[10.4.9]

Discussion If the electrons in the helium atom did not interact with each other, then the ground state energy of the helium atom would simply be the sum of energies for the ground state hydrogen atom as modified by the nuclear charge and the number of electrons. In this case the total electron energy would be 2EHe,1 ¼ 8EH,1. The energy to ionize He by removal of one electron, ie, to form Heþ, would in this case simply be EHe,1 ¼ 4EH,1. However, the effect of the repulsive interaction between the two electrons is to raise the ground state energy level of the helium atom by an amount (5/2)EH,1. Since the hydrogen-like energies are rigorous for the one-electron atom, the ionization energy is different for the two electrons. We therefore have for helium the first (I1(He)) and second (I2(He)) ionization energies, with I1(H) ¼ 13.598 eV, I1 ðHeÞ ¼ 4EH;1 þ ð5=2ÞEH;1 ¼ ð3=2ÞI1 ðHÞ ¼ ð3=2Þð13:598Þ ¼ 20:397 eV

[10.4.10]

I2 ðHeÞ ¼ 4EH;1 ¼ 4I1 ðHÞ ¼ 4ð13:598Þ ¼ 4ð13:598Þ ¼ 54:392 eV

[10.4.11]

The experimental values for the first and second ionization potentials are 24.587 and 54.4178 eV, respectively. The agreement between the theoretical and experimental values of the second ionization potential is quite good. This result is no surprise since the one-electron solutions to the Schro¨dinger equation are exact. Inclusion of the first-order correction term to the first ionization energy improves the agreement between theory and experiment, but with significant error remaining between the two values.

The calculation given in Example 10.4.1 indicates that the effect of the repulsive electrone electron interactions on the electronic energies of the one-electron atom acts in a direction to raise the energy states (make more positive) based on the one-electron atom. For multielectron atoms this adjustment in energy states is dependent upon the energy level (the n quantum number) and the number of other electrons in the atomic system.

484

10. ELECTRONIC AND NUCLEAR STATES

Example 10.4.2

The ElectroneElectron Interactions for the Lithium Atom

The theoretical expression for the electroneelectron interaction for the helium atom leaves a margin of error. One can reverse the calculation to use the experimental values of the ionization energy to estimate the electroneelectron interactions. Use the ionization energies of lithium in the ground state to estimate the perturbation energies.

Solution The lithium atom has the electron configuration 1s22s. There are three electroneelectron interactions, but in two different quantum states. The first ionization energy is the removal of the 2s electron, the second ionization energy is the removal of one of the electrons in the 1s orbital, and the third ionization energy is the removal of the last electron in the 1s orbital. If electroneelectron interaction energies are additive, then the first ionization energy is not only the energy of interaction of the 2s electron with the nucleus but also with the two electrons in the 1s orbital. The notation for the electroneelectron interaction energy is E(n, m), where n and m are the shell quantum numbers. The ionization energy for the first electron, Li / Liþ ¼ e, is therefore I1 ðLiÞ ¼ ðZ=nÞ2 I1 ðHÞ  2Eð2; 1Þ ¼ ð3=2Þ 2 I1 ðHÞ  2Eð2; 1Þ ¼ ð9=4ÞI1 ðHÞ  2Eð2; 1Þ

[10.4.12]

where I1(H) is the ionization of the ground state of hydrogen. Note that the minus sign in front of E(2, 1) is because the ionization energies are the energies added to the system to remove the electrons. Using similar reasoning the second ionization energy for lithium, Liþ / Li2þ ¼ e, is
 I2 Liþ ¼ 9I1 ðHÞ  Eð1; 1Þ

[10.4.13]

and the third ionization energy is
 I2 Li2þ ¼ 9I1 ðHÞ

[10.4.14]

The experimental values of the ionization energies are I1(Li) ¼ 5.3917 eV; I2(Liþ) ¼ 75.6402 eV; and I3(Li2þ) ¼ 122.4543 eV. The average interaction of the electron in the 2s orbital with an electron in the 1s orbital is found by rearrangement of Equation [10.2.12] and substitution of the appropriate numbers, Eð2; 1Þ ¼

ð9=4ÞI1 ðHÞ  I1 ðLiÞ ð9=4Þð13:598Þ  5:3917 ¼ ¼ 12:602 eV 2 2

In a similar way the interaction between two electrons in the 1s orbit is
 Eð1; 1Þ ¼ 9I1 ðHÞ  I2 Liþ ¼ 9ð13:598Þ  75:6402 ¼ 46:742 eV It is noted that the ratio of the electroneelectron repulsion terms is 46.742/12.602 ¼ 3.71. The Bohr energies for the 1s and 2s orbitals are the reciprocal of the square of the respective quantum numbers. An interpretation of this slight difference is due to the fact that the electroneelectron repulsion term in the 1s orbital slightly expands the radius of the 1s orbital.

10.5 ELECTRON SPIN: SINGLET AND TRIPLET STATES OF EXCITED HELIUM

485

10.5 Electron Spin: Singlet and Triplet States of Excited Helium Consider first the simplest multielectron atom, the helium atom in its ground state. Since the electrons are identical particles one cannot "label" each electron but must take into consideration our ignorance and allow them to be interchanged. This time we include the spin quantum number. The following notation is introduced to identify the spin states of the electron. Let the up spin state be represented by a ¼ [ and the down spin state by b ¼ Y. As long as the two electrons are in the same shell the spin states of the electrons must be either a(1)b(2) or a(2)b(1). Since one cannot put labels on the electrons, these two spin configurations are considered to be equally probable. Therefore the spin function is represented by the superposition of the two states. However, there are two possibilities with respect to interchange of the two electrons: the symmetric case a(1)b(2) þ a(2)b(1) or the antisymmetric case a(1)b(2)  a(2)b(1). The decision as to which to use is determined by the spatial part of the wave function. In this case with both electron in the ground state, so the spatial part is simply 1s(1)1s(2). An interchange of electrons makes no change in the spatial wave function. Thus for electron exchange the overall wave function for helium in the ground state is 4He ð1; 2Þ ¼ 1sð1Þ1sð2Þ

½að1Þbð2Þ að2Þbð1Þ pffiffiffi 2

[10.5.1]

Notice that for the exchange of electrons, 4He(1, 2) ¼ 4He(2, 1), which is a condition for Fermi particles. Now consider the situation in which one of the electrons is promoted to the 2s orbital. Since the 1s and 2s orbitals are different spatial orbitals there is no restriction on the relative values of the spin states. Because the two states have different values of the principle quantum number, there are two spatial wave functions to consider, 1s and 2s. Again, the inability to label the electrons means that the spatial parts of the wave functions must also be superimposed. The restriction on the spin states of the electrons is also lifted with the two spatial wave functions. Let us first assume that the spin states do not change when one electron is promoted to the 2s orbital. Since the spin state remains antisymmetric, the spatial part must be symmetric with respect to electron interchange. Therefore the first excited state of the helium atoms can be 1 z 4He ð1; 2Þ

¼ 1 SHe ¼

½1sð1Þ2sð2Þþ 1sð2Þ2sð1Þ ½að1Þbð2Þ að2Þbð1Þ pffiffiffi pffiffiffi 2 2

[10.5.2]

where "z" denotes the excited state. The superscript "1" indicates that there is only one way to write this wave function, and thus is referred to as the singlet state. The notation is changed to emphasize that the excited state involves the 2s orbital and not one of the 2p orbitals for the z n ¼ 2 quantum state, 1 4He ð1; 2Þ/1 SHe . Note that the spin and spatial parts of the wave function are individually normalized to unity, leading to an overall normalization factor of 1/2. The second possibility is that the spatial part is antisymmetric. This means that the spin part must be symmetric. If the spin state of the promoted electron is unchanged, then the "" sign for the spin function in Equation [10.3.2] is replaced by a "þ" sign. Since the electrons cannot be labeled, the promotion of an electron can have two electron spin states that are the same, ie, ab / bb or ab / aa. The three ways to write the wave function for a promoted electron in the excited helium atom with an antisymmetric spatial wave function are

486

10. ELECTRONIC AND NUCLEAR STATES

3 z 4He ð1; 2Þ

¼ 3 SHe

8 > > 1 > > pffiffiffi ½1sð1Þ2sð2Þ 1sð2Þ2sð1Það1Það2Þ > > > 2 > > > > > < 1 ¼ ½1sð1Þ2sð2Þ 1sð2Þ2sð1Þ½að1Þbð2Þþ bð1Það2Þ > 2 > > > > > > > 1 > > pffiffiffi ½1sð1Þ2sð2Þ 1sð2Þ2sð1Þbð1Þbð2Þ > > : 2

[10.5.3]

The wave functions in Equation [10.5.3] are referred to as the triplet states.

Example 10.5.1 Calculation of the Unperturbed Electronic Energy for the First Excited State of the Helium Atom Use the wave functions in Equations [10.5.2] and [10.5.3] to calculate the energy of the first excited state of the helium atom. Assume that there are no electroneelectron interactions, and that the Bohr energies apply.

Solution The electron spin functions do not contribute to the energy of the helium atom in the absence of an applied electric or magnetic field. Since the question asks for the unperturbed energies of the excited helium atom, the electroneelectron pair interaction energy also does not enter into the calculation. The calculation of the unperturbed energy for helium in the excited state is therefore a straightforward application of the Bohr atom results. Using the values Z ¼ 2 for the n ¼ 1 and 2 states, we have ð0Þ

EHez ¼

Z2 Z2 ð2Þ2 ð2Þ2 E þ E ¼ E þ EH;1 ¼ 5EH;1 ¼ 5ð13:598Þ ¼ 67:99 eV H;1 H;1 H;1 n21 n22 ð1Þ2 ð2Þ2

Recall that the electronic energy of the atom Bohr atom depends only on the principal quantum number n. For the n ¼ 2 shell there are p orbitals as well as the s orbitals. The excited electron could therefore be in the p0, pþ1, or p1 orbitals, where the subscript denotes the allowed values of the azimuthal quantum number m‘ m‘. This means that there are four pairings with the 1s orbital and eight such pairings when one also allows for interchange of the electron identity, viz, 2s(1)2p1(2) and 2s(2)2p1(1). At the level of the zero-order approximation of the energy of the first excited state of helium, all of these combinations of wave functions give exactly the same energy. We may therefore extend the notation to include the p orbitals, 1

PHe ¼

½1sð1Þ2pð2Þþ 1sð2Þ2pð1Þ ½að1Þbð2Þ að2Þbð1Þ pffiffiffi pffiffiffi 2 2

[10.5.4]

10.6 SPATIALLY DIRECTED ATOMIC ORBITALS: 2PX, 2PY, AND 2PZ

487

and

3

PHe

8 > > 1 > > pffiffiffi ½1sð1Þ2pð2Þ 1sð2Þ2pð1Það1Það2Þ > > > 2 > > > > > < 1 ¼ ½1sð1Þ2pð2Þ 1sð2Þ2pð1Þ½að1Þbð2Þþ bð1Það2Þ > 2 > > > > > > > 1 > > pffiffiffi ½1sð1Þ2pð2Þ 1sð2Þ2pð1Þbð1Þbð2Þ > > : 2

[10.5.5]

Keep in mind that 2p in Equations [10.5.4] and [10.5.5] symbolically represents the three orbitals 2p0, 2pþ1, and 2p1. These orbitals, however, are equivalent when it comes to the calculation of the perturbation energies. The perturbation energies involve integrals of two types: the Coulomb integrals J,
  Js ¼ q2e h1sð1Þ2sð2Þj 1 r1;2 j1sð1Þ2sð2Þi

[10.5.6]


  Jp ¼ q2e h1sð1Þ2pð2Þj 1 r1;2 j1sð1Þ2pð2Þi

[10.5.7]

and the exchange integrals K,
  Ks ¼ q2e h1sð1Þ2sð2Þj 1 r1;2 j1sð2Þ2sð1Þi

[10.5.8]


  Kp ¼ q2e h1sð1Þ2pð2Þj 1 r1;2 j1sð2Þ2pð1Þi

[10.5.9]

The order of magnitude of these integrals is Jp > Js > Ks > Kp > 0 (Davis, 1965). As we saw in Example 10.4.1 the unperturbed energy for the excited state of helium is ð0Þ

EHez ¼ 5EH;1 . Recall that the symmetric spatial parts of the wave functions give a "þ" correction factor whereas the asymmetric spatial parts give a "" correction to the energy. There are, thereð0Þ

fore, four possibilities of energies with their quantum states: E1 ¼ EHez þ Jp þ Kp for 1 PHe ; E2 ¼ ð0Þ

ð0Þ

ð0Þ

EHez þ Jp  Kp for 3 PHe ; E3 ¼ EHez þ Js þ Ks for 1 SHe ; and E4 ¼ EHez þ Js  Ks for 3 SHe. The order of the energies is E1 > E2 > E3 > E4.

10.6 Spatially Directed Atomic Orbitals: 2px, 2py, and 2pz The p orbital is defined by the value ‘ ¼ 1, which is designated as a p and m‘ ¼ 1, 0, or þ1. The notations p0, p1, and pþ1 identify the f dependence of the p orbitals, viz, F0 ðfÞ, F1 ðfÞ, and Fþ1 ðfÞ, respectively. By choice, the coordinate system used in Section 9.2 defined the z-axis as the axis of rotation with the angle f as a measure of that rotation from the x-axis. The z-axis was thus defined by the azimuthal quantum number m‘ ¼ 0. Therefore

488

10. ELECTRONIC AND NUCLEAR STATES

1 F0 ðfÞ ¼ pffiffiffiffiffiffi ¼ p0 2p

[10.6.1]

The f dependence of this orbital is given by the square of the function, F0 F0 ðfÞ ¼ p0 p0 ¼ p20 ¼

1 2p

[10.6.2]

The corresponding functions for m‘ ¼ 1 and for m‘ ¼ þ1 are 1 F1 ðfÞ ¼ pffiffiffiffiffiffi expðifÞ ¼ p1 2p

[10.6.3]

and 1 Fþ1 ðfÞ ¼ pffiffiffiffiffiffi expðþifÞ ¼ pþ1 2p

[10.6.4]

Notice that each wave function is the complex conjugate of the other, pþ pþ ¼ p pþ ¼

1 ¼ pþ p ¼ p p ¼ p0 p0 2p

[10.6.5]

This means that the wave functions pþ1 and p1 have no special direction! In fact, Equation [10.6.5] is valid for all values of m. The solutions pþ1 and p1 are therefore unacceptable if one wants to construct molecules which have geometric structures, ie, spatial extensions. It is necessary, therefore, to find a way to generate orbitals with direction. The superposition principle comes to the rescue to introduce a dependence on the angle f for values of m‘. The addition of the f-dependent functions in the present exercise gives Fþ1 ðfÞ þ F1 ðfÞ pþ1 þ p1 pffiffiffi pffiffiffi ¼ 2 2

[10.6.6]

1 1 ¼ pffiffiffipffiffiffiffiffiffi f½cosðfÞþ i sinðfÞ þ ½cosðfÞ i sinðfÞg ¼ pffiffiffiffi cosðfÞ p 2 2p and the subtraction of the f-dependent functions gives Fþ1 ðfÞ  F1 ðfÞ pþ1  p1 1 pffiffiffi pffiffiffi ¼ ¼ pffiffiffiffi sinðfÞ p 2i 2i

[10.6.7]

This linear combination of these wave functions establishes a direction along the x- and y-axes. Since the angle f is measured from the x-axis, the wave function given by Equation [10.6.6] has its

10.6 SPATIALLY DIRECTED ATOMIC ORBITALS: 2PX, 2PY, AND 2PZ

489

maximum value along the x-axis, whereas Equation [10.6.7] has its maximum along the y-axis. The angle dependence of 2px and 2py orbitals is therefore p þp 1 2px ðq; fÞ ¼ Q1;þ1 ðqÞ þ pffiffiffi 1 ¼ 2 2

rffiffiffiffi 3 sinðqÞcosðfÞ p

[10.6.8]

and p p 1 2py ðq; fÞ ¼ Q1;þ1 ðqÞ þ pffiffiffi 1 ¼ 2 2

rffiffiffiffi 3 sinðqÞsinðfÞ p

[10.6.9]

The 2pz wave function is 1 2pz ðq; fÞ ¼ Q0;0 ðqÞp0 ¼ 2

rffiffiffiffi 3 cosðqÞ p

[10.6.10]

From Table 9.4.1 and Equation [10.6.1] the angle dependence of the 2s orbital is 1 2sðq; fÞ ¼ pffiffiffiffi 2 p

[10.6.11]

The shape of an orbital is a measure of the probability of finding an electron in certain regions of space. The angular shape of the orbitals are given by the square of the functions [2s(q, f)]2, [2px(q, f)]2, [2py(q, f)]2, and [2pz(q, f)]2 and is shown in Figure 10.6.1, as they would appear in the atom.

FIGURE 10.6.1

The shapes of the 2s and 2p orbitals in the atom. The angle dependences of the 2px, 2py, 2pz, and 2s orbitals are defined by Equations [10.6.8]e[10.6.11], respectively. Shown in the above figure are the angular probabilities [2s(q, f)]2 (cyan) and the [2px(q, f)]2, [2py(q, f)]2 and [2pz(q, f)]]2 (green) to show the relationship of all the orbitals in the atom. The three green lobes are mutually perpendicular.

490

10. ELECTRONIC AND NUCLEAR STATES

The wave functions for the p orbitals include the radial dependence, which in the present notation is 2pj ¼ Rn;1 ðrÞ2pj ðq; fÞ

[10.6.12]

The angle dependence of the d and f orbitals is determined in a similar manner. The mathematical expressions for selected hydrogen-like wave functions are given in Table 9.6.1.

10.7 Spatially Directed Bonding OrbitalsdHybridization The familiar px and py orbitals in general chemistry textbooks are hybridized orbitals. Any atomic orbital that involve the x and y coordinates are also hybridized orbitals. Chemical bonds are directional, forming the bridge between atoms in a compound. In the formation of these bonds, orbitals that lie in the desired direction may be superimposed to form a stronger union between the bonding atoms. Likewise, orbitals can be combined to form "hybrid orbitals" with a direction that is intermediate to those engaged in the hybridization. Because the energy of the atomic states is dependent upon only on the principle quantum number, the energy levels of the atom are degenerated. Hence hybridization is more easily accomplished by combining orbitals with the same energy. Thus hybridization can occur with the s, p, d, and f orbitals in the same atomic shell. There is one basic principle in constructing hybrid orbitals. Symmetry plays an important role in determining the relative compositions of the hybridized orbitals. The hybridized orbitals must be identical in all respects except for their directions. There is a whole field of chemistry devoted to the chemistry of the carbon atom. This field is called organic chemistry. Another important field in chemistry studies the carbon molecules associated with life: biochemistry. A more specialized field that deals with life molecules is medicinal chemistry. All of these fields are a result of one simple fact. Carbon is present in about 9 million compounds of the 10 million known compounds (http://www.nwscc.edu/nsfdc/ organicchemistry/Organic%20Web,htm). The wide variety of carbon compounds is a result of the unique position of the carbon atom in the Periodic Table of the Elements. Because carbon is in the second row, the outer shell electrons are sufficiently close to the nucleus to form relatively strong bonds with other elements. Having the atomic number 6 allows carbon the dual role of both an optimist and a pessimist. With four electrons in its outer shell, the optimist sees the outer shell as being half full and the pessimist sees the outer shell as being half empty. The carbon atom with an oxidation state of 0 can go the whole gambit of possibilities, from the value of þ4 in carbon dioxide to the value of 4 in methane. Carbon can also bind with other carbon atoms to form large molecules with ring structures. The electron configuration of carbon is 1s22s22p2. Hybridization of bonds is possible because of the wave description of the electrons. Waves can be added, subtracted, or left alone. Because the 2p orbitals are orthogonal to each other (see Figure 10.6.1), they cannot on their own form hybrid orbitals. It takes the 2s orbital to allow hybridization, and the 2p orbitals to give hybridization a direction. The four electrons of carbon in the n ¼ 2 shell can combine to form 2sp, 2sp2, and 2sp3 hybrid orbitals. The formation of hybrid bonds gives geometry to the molecules because the px, py, and pz orbitals point in orthogonal directions. There are two general rules in forming hybrid orbitals. The first is the conservation of orbitals: the number of hybrid orbitals generated is equal to the orbitals needed to make the hybrid

10.7.1 THE SP HYBRID ORBITAL

491

orbitals. The second general rule is that the s orbital must be equally shared in the piecemeal construction of the hybrid orbital. This is because all hybrid orbitals must be rotationally equivalent. The simplest way to determine how much of the s and p orbitals are to be included in a hybrid orbital is to think in terms of conservation of probabilities. Since each of the composite s and p orbitals are individually normalized to unity, the sum of the squares of their respective coefficients must also be unity. The methods used to create the hybrid orbitals for the carbon atom also apply to the formation of all hybrid orbitals, such as the use of d and f orbitals in the formation of coordination bonds associated with chemical complexes. Furthermore, the notation of the hybrid orbitals is not given with the atomic shell number "2" even though the 2s and 2p orbitals of carbon are used in the construction of hybrid orbitals. All atoms in the Periodic Table of the Elements can, in principle, form these hybrid orbitals. This is why the chemistry of elements in the same column has similar chemical properties.

10.7.1 The sp hybrid orbital As the name implies, the sp hybrid is composed of a superposition of the 2s and one of the 2p orbitals. Assume the hybrid orbital is to be constructed to lie along the x-axis. Since the 2px orbital already resides on the x-axis, the 2py and 2pz are not involved in the construction of the sp hybrid. Since two atomic orbitals are used, there must be two hybrid orbitals that are constructed. For symmetry reasons the two hybrid orbitals must be 180 degree apart, heading in opposite directions along an axis. Keep in mind that a node in the orbital means that there is a change in sign of the wave function, from positive to negative. The addition of orbitals means that the overlapping parts of the constituent lobes of the same sign reinforce each other whereas those regions of opposite sign result cancellation. By convention the s orbital, which has no nodes, is given a positive sign. The direction of the sp hybrid is the sole responsibility of the 2px orbital. A change in direction of the sp orbital is achieved simply by changing the signs of the lobes of the 2px orbital. Since the hybrid orbitals are equivalent, the s orbital is equally shared with each of the hybrid orbitals. The conservation of probability thus dictates that the contribution of 2s character to each hybrid is 1/2. This means the 2px contribution to the probability must also be 1/2 to conserve the probability of 1. The two sp hybrids are therefore, ψsp; ¼

2s  2px pffiffiffi 2

[10.7.1]

pffiffiffi The 2s and 2px orbitals are individually normalized and the factor of 2 accounts for the linear combination of the two orbitals, in which case the probability is normalized, E h2sj2si þ h2p j2p i 1 þ 1 D  x x ¼1 ¼ ψsp; ψsp; ¼ 2 2

[10.7.2]

Recall that the number of orbitals used to construct the hybrid orbitals must equal the number of hybrid orbitals. This also applies to electrons. The two sp hybrid orbitals can each have two electrons.

492

10. ELECTRONIC AND NUCLEAR STATES

10.7.2 The sp2 Hybrid Orbitals In the combination of the 2s orbital with two of the 2p orbitals one obtains three identical sp2 hybrid orbitals. This means that the 2s orbital must be shared equally with the three hybrid orbitals. Therefore "1/3" the probability is due to the 2s orbital used to construct the hybrid orbitals. This "equal sharing" is not true, however, for the 2px and 2py orbitals. Think in terms of three vectors in the following construction. For convenience, let us define one of the sp2 hybrid orbitals to lie along the x-axis. Because of the orthogonality of the 2px and 2py orbitals the hybrid orbital along the x-axis does not have any "2py character" in its construction. Since 1/3 of the hybrid orbital probability is the 2s orbital, the remaining 2/3 of the hybrid orbital probabilities must come from the unhybridized 2px orbital. Hence we write for this hybrid orbital, rffiffiffi rffiffiffi 1 2 2s þ 2p 41 ¼ 3 3 x

[10.7.3]

Because of the orthonormality of the 2s and 2px wave functions, it is a simple matter to show that h41 j41 i ¼ 1. The remaining two hybrids are likewise equal. They both must therefore have 1/3 of the 2s orbital probability and equally share the 2py orbital and the remaining part of the 2px orbital. Furthermore, since the three hybrid orbitals are equal they must have symmetry through a rotation of an angle of 120 degree. This means that the remaining two hybrid orbitals must therefore have their positive lobe along the x-axis in a direction opposite of the 41 orbital. By inspection one may write for the remaining two hybrid sp2 orbitals, rffiffiffi rffiffiffi rffiffiffi 1 1 1 2s  2px þ 42 ¼ 2p 3 6 2 y

[10.7.4]

and rffiffiffi rffiffiffi rffiffiffi 1 1 1 2s  2px  2p 43 ¼ 3 6 2 y

[10.7.5]

There are two important observations in regard to Equations [10.7.3]e[10.7.5]. The first observation is that the sum of the squares of the coefficients for each orbital is unity: (1/3) þ (2/3) ¼ 1 for 41, (1/3) þ (1/6) þ (1/2) ¼ 1 for 42, and (1/3) þ (1/6) þ (1/2) ¼ 1 for 43. The second observation is that the sum of the squares of the coefficients for each composite carbon orbital is also unity: (1/3) þ (1/3) þ (1/3) ¼ 1 for the 2s orbital, (2/3) þ (1/6) þ (1/6) ¼ 1 for the 2px orbital, and (1/2) þ (1/2) ¼ 1 for the 2py orbital.

10.7.3 The sp3 Hybrid The four equivalent 2sp3 hybrid orbitals can easily be constructed from their symmetry properties in a manner employed for the 2sp and 2sp2 hybrid orbitals. The simplest approach is to again define one of the hybrid orbitals as lying along one of the axes. Let the first hybrid orbital lie along the positive x-axis. Since only the 2px and 2s orbitals are involved, it is tempting to share equally the two orbitals to form the sp3 hybrid along this axis. But the 2s orbital must be equally shared among the four sp3 hybrid orbitals. Therefore to conserve probability the sp3 hybrid orbital is

10.7.4 THE GEOMETRY OF THE HYBRID ORBITALS

1 41 ¼ pffiffiffi 2s þ 4

rffiffiffi 3 2p 4 x

493

[10.7.6]

pffiffiffiffiffiffiffiffi Similarly the coefficients for the 2s orbital must be 1=4 for the remaining sp3 hybrid orbitals. To construct the remaining three sp3 hybrid orbitals, advantage is taken that these orbitals are collectively a tetrahedron structure. This means that the remaining part of the 2px orbital must be equally shared between the three, but in opposite direction of the 41 hybrid orbital. Therefore pffiffiffiffiffiffiffiffiffiffi the coefficients for the contribution of the 2px orbital must be  1=12. The simplest way to deal with the 2py orbitals is to place the second hybrid orbital in the xez plane. This means that the 2py orbital does not contribute. The coefficient for the 2pz orbital is thus determined by the square of the 2s and 2px coefficients, viz, (1/4) þ (1/12) ¼ 4/12. Hence the second hybrid orbital is 1 42 ¼ pffiffiffi 2s  4

rffiffiffiffiffi rffiffiffi 1 2 2px þ 2p 12 3 z

[10.7.7]

It is left to the Reader to reason through the arguments for the remaining two hybrid orbitals 1 43 ¼ pffiffiffi 2s  4

rffiffiffiffiffi rffiffiffi rffiffiffi 1 1 1 2px þ 2py  2p 12 2 6 z

[10.7.8]

and 1 44 ¼ pffiffiffi 2s  4

rffiffiffiffiffi rffiffiffi rffiffiffi 1 1 1 2p 2px  2py  12 2 6 z

[10.7.9]

10.7.4 The Geometry of the Hybrid Orbitals The geometry of a molecule is dependent upon the geometry of the hybrid orbitals. Threedimensional contours of the sp, sp2, and sp3 hybrid orbitals as they appear in relationship to the unhybridized 2p orbitals are shown in Figure 10.7.1. In all cases the green orbitals are the unhybridized 2p orbitals. The two sp hybrid orbitals are shown as orange and blue. Both sp hybrid orbitals can each accommodate two electrons. With the 180 degree between the hybrid orbitals, the sp hybrid bond is a linear unit in the construction of the geometry of the molecule. The three sp2 hybrid orbitals are shown as orange, blue, and purple. All three sp2 hybrid orbitals can each accommodate two electrons. The three sp3 orbitals define a plane in which the angle between adjacent orbitals is 120 degree. The sp2 orbitals are a planar unit with 120 degree in the construction of the molecular geometry. The four sp3 hybrid orbitals are distinguished by their color: orange, blue, purple, and red. All four sp3 hybrid orbitals can each accommodate two electrons. The sp3 hybrid orbitals are separated by 109 degree and are the tetrahedral unit in the geometry of a molecule. To illustrate the effect of hybridization, shown in Figure 10.7.2 are the constant contour plots of the unhybridized 2px orbital and the hybridized sp orbital along the x-axis. It is clear that positive

494

10. ELECTRONIC AND NUCLEAR STATES

2sp

2pz

sp3

sp2

sp

FIGURE 10.7.1 The shapes of hybrid orbitals on the atom. Top: The unhybridized 2pz orbital is on the left, and one of the hybridized 2sp orbitals on the right, where the negative lobe of one of the sp hybrid orbital is indicated by the arrow. Bottom: The hybrid orbitals in the carbon atom. sp hybrid: The two sp hybrid orbitals, the red and the blue lobes, are 180 degrees apart. There are two unhybridized p orbitals (green). The sp hybrid orbital is a basic unit for construction of a linear molecule. sp2 hybrid: The three sp2 hybrid orbitals lie in a plane separated by 120 degrees. The sp2 hybrid orbitals are red, blue, and purple lobes. The green orbital is the unhybridized 2pz orbital. The sp2 hybrid is a basic unit for construction of a nonlinear planar molecule. sp3 hybrid: The four sp3 hybrid orbitals red, blue, purple, and orange represent the tetrahedral unit for construction of molecular geometries.

2px

sp

FIGURE 10.7.2 Contour plots for the 2px orbital and the sp hybrid orbital. Left: A contour plot is the unhybridized 2px orbital. Note the symmetry in shape of the contours through both perpendicular planes to the plane of the image. Right: The contour plot of the sp hybrid clearly shows the effect of sign on the sp hybrid orbital. The positive lobe of the 2px orbital reinforces the positive nature of the 2s orbital (left side) whereas the superposition of lobes of opposite destructively interfere.

10.8 THE CHEMICAL BOND: VALENCE BOND AND MOLECULAR ORBITAL APPROACHES

495

2s orbital reinforces the electron density in one lobe of the 2px orbital at the expense of the electron density of the other lobe.

10.8 The Chemical Bond: Valence Bond and Molecular Orbital Approaches The chemical bond is the sharing of electrons between two or more atoms. This is achieved by overlapping of orbitals. Unlike orbitals that are located at one central coordinate system of an atom, the orbitals participating in a chemical bond are located on different atoms. Therefore orthogonality of orbitals does not apply. The "overlap" integral is not zero. There are two fundamental descriptions of the chemical bonddthe valence bond (VB, as developed by Heitler, London, Slater, and Pauling) and the molecular orbital (MO, as developed by Mulliken) theory. Both theoretical approaches start with the same Hamiltonian but differ in the manner in which the terms are grouped. The VB and MO theories are illustrated in the present section for the diatomic molecule. Now bear with the notation for if one looks very closely at what is to be done, one will discover that all we are doing is algebraic manipulations of the symbols. The coordinate system is located at the center-of-mass of a diatomic molecule, composed of the nuclei A and B. The Hamiltonian in this relative coordinate system is given in general by

X

1 h2 1 h2 1 h2 ZA ZB q2e 2 2 2 H ¼ V  V  Vj þ 2 8p2 mA A 2 8p2 mB B 2 8p2 me rAB |fflfflfflfflffl{zfflfflfflffl ffl} j |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} repulsive kinetic energy operators nucleus  nucleus 0 X Z q2 X Z q2 1 X X q2 B A e e e Aþ þ @ rAj rBj rji j i j i>j |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} repulsive attractive electron  electron nucleus  electron

[10.8.1]

where the derivatives in the kinetic energy operators are expressed in terms of the coordinates of the appropriate particles relative to the origin of the coordinate system, and the distances rij are relative distances between the appropriate particles. Let us rewrite this operator in the following manner,

H ¼

h2 h2 2 V  V2 þ H elec 8p2 mA A 8p2 mB B

[10.8.2]

where the electronic Hamiltonian is defined as,

H elec

X

! XX 2 2 h2 q2e ZA ZB q2e 2 ZA q e ZB q e  2 Vj  þ þ ¼  rAj rBj rji rAB 8p me j j i>j |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} electron coordinates

[10.8.3]

496

10. ELECTRONIC AND NUCLEAR STATES

Since the electrons are much lighter than the nuclei, it is assumed that the electrons move around the nuclei many times before there is significant movement by the nuclei. The nuclei of the diatomic molecules are assumed to be fixed in space in the calculation of the electronic states of diatomic molecules. Therefore the kinetic energy of the relative motion of the nuclei may be equated to zero in Equation [10.8.2] and the focus directed to the solution of Equation [10.8.3]. Notice, however, that the electronic Hamiltonian is a function of the coordinates of all of the electrons with the interaction potential for the nuclei at a fixed separation distance rAB as an additive constant. Hence the procedure is to calculate the electronic energy for several distances rAB and generate a potential energy surface on which the nuclei move. The molecular motions associated with the nuclei are discussed in the chapter on molecular spectroscopy. The hydrogen molecule is the simplest homonuclear diatomic to illustrate the use of symmetry in the wave function and to estimate the electronic energy. The explicit form of the Hamiltonian for the electrons in the hydrogen molecule is,

H e;H2 ¼ 

h2 h2 q2e q2e q2e q2e q2e q2e 2 2 V  V     þ þ 8p2 me 1 8p2 me 2 rA;1 rB;1 rA;2 rB;2 r1;2 rA;B

[10.8.4]

where the subscripts 1 and 2 refer to the electrons and A and B to the hydrogen atoms. This Hamiltonian can be partitioned in at least two different ways to provide an estimate for the electronic energy states.

10.8.1 The Valence Bond Approach The VB approach is to partition the Hamiltonian into two separated atomic Hamiltonians and a perturbation term, viz, h2 q2e h2 q2e q2e q2e q2e q2e 2 2 V   V    þ þ rA;2 rB;1 r1;2 rA;B 8p2 me 1 rI;1 8p2 me 2 rII;2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

H e;H2 ¼ 

H A;1

H B;2

[10.8.5]

H0

If we neglect the perturbation term H 0 then the problem reduces to two hydrogen atom problems with the ground state wave functions given by 1s for each atom. As in previous cases, the system wave function is the product of these two wave functions. However, we have identified electron 1 with atom A and electron 2 with atom B. Without specific knowledge that this assignment is correct, one must allow for interchange of the electrons. Using the postulates of quantum mechanics the four normalized valence bond wave functions for the hydrogen molecule are,

4u;VB

8 > að1Það2Þ > > > > > > < ½að1Þbð2Þþ að2Þbð1Þ 1 pffiffiffi ½1sðA; 1Þ1sðB; 2Þ 1sðA; 2Þ1sðB; 1Þ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 > 2 > 2 1 SA;B > > > > > : bð1Þbð2Þ

[10.8.6]

497

10.8.2 THE MOLECULAR ORBITAL APPROACH

4g;VB ¼

½1sðA; 1Þ1sðB; 2Þþ 1sðA; 2Þ1sðB; 1Þ ½að1Þbð2Þ að2Þbð1Þ pffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 2 2 1þ SA;B

[10.8.7]

where the subscript g refers to the spatial even wave function and the subscript u to the spatial odd wave function. The normalization factor has the "cross integral" S, defined as, SA;B ¼ h1sðA; 1Þj1sðB; 2Þi ¼ h1sðA; 2Þj1sðB; 1Þi s 0

[10.8.8]

The wave functions 1s(A, 1) and 1s(B, 1) are not orthogonal because they are centered on different atoms, denoted by A and B. The expression for the electronic energy of the hydrogen molecule in the ground state is written in terms of the energies of the isolated hydrogen atoms. That is,         ð0Þ ð0Þ EH2 ¼ 4g;VB H H2 4g;VB ¼ 2EH þ 4g;VB H 0 4g;VB ¼

ð0Þ 2EH

q2 1 1 þ e  2q2e h1sðA; 1Þj j1sðA; 1Þi þ 2q2e h1sðA; 1Þj j1sðA; 2Þi rB;1 r1;2 rA;B

[10.8.9]

ð0Þ

where EH is the ground state energy for the hydrogen atom and the factors of 2 result from the fact that interchange of A and B and of 1 and 2 results in the same numerical values of the inte1 1 grals, ie, h1sðA; 1Þj j1sðA; 1Þi ¼ h1sðB; 2Þj j1sðB; 2Þi. rB;1 rA;2

10.8.2 The Molecular Orbital Approach The second way in which the hydrogen molecule Hamiltonian can be partitioned, the molecular orbital (MO) approach, is to associate each electron with both nuclei, ie, h2 q2e q2e h2 q2e q2e q2e q2e 2 2 V    V   þ þ 8p2 me 1 rA;1 rB;1 8p2 me 2 rA;2 rB;2 r1;2 rA;B |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}

H e;H2 ¼ 

H1

H2

[10.8.10]

H0

Notice that the Coulombic interactions in H 1 and H 2 contain only one electron with both of the nuclei. Hence the electron "belongs" to both nuclei comprising the "molecule" and its molecular orbital can be expressed in terms of a linear combination of atomic orbitals (LCAOeMO). The spatial component of the ground state molecular orbital is denoted by 1ssg ð1Þ and is defined as, 1 1ssg ð1Þ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 ½1sðA; 1Þþ 1sðB; 1Þ 2 1þ SA;B

[10.8.11]

498

10. ELECTRONIC AND NUCLEAR STATES

In this notation the spatially symmetric molecular orbital is comprised of a linear combination of two 1s atomic orbitals. A similar expression is obtained for the second electron. The LCAOeMO wave function for the ground and first excited states are therefore,

4u;MO

8 > að1Það2Þ > > > > > > < ½að1Þbð2Þþ að2Þbð1Þ pffiffiffi ¼ ½1ssu ð1Þ1ssu ð2Þ > 2 > > > > > > : bð1Þbð2Þ

4g;MO ¼

½1ssg ð1Þ1ssg ð2Þ½að1Þbð2Þ að2Þbð1Þ pffiffiffi 2

[10.8.12]

[10.8.13]

The electronic energy for the hydrogen molecule in the LCAOeMO approach is therefore         ð0Þ EH2 ¼ 4g;MO H H2 4g;MO ¼ 2EHþ þ 4g;MO H 0 4g;MO 2

¼

ð0Þ 2EHþ 2

 1   q2 q2  4g;MO þ e þ 2 e 4g;MO  r1;2 rA;B rA;B

[10.8.14]

ð0Þ

where EHþ is the ground electronic energy for the hydrogen molecule ion. 2

10.8.3 Comparison of the VB and LCAOeMO Approaches The differences between the VB and LCAOeMO approaches are most easily demonstrated by comparing directly the wave functions expressed in terms of the constituent atomic orbitals. The spatial function for the hydrogen molecule in the LCAOeMO theory is, 1  ½1sðA; 1Þþ 1sðB; 1Þ½1sðA; 2Þþ 1sðB; 2Þ 4g;MO ¼
2 1 þ SA;B [10.8.15] 1  f1sðA; 1Þ1sðA; 2Þ þ 1sðB; 1Þ1sðB; 2Þ þ 2½1sðA; 1Þ1sðB; 2Þg ¼
2 1 þ SA;B Note that the VB theory dictates that only one electron at a time is associated with either atom, whereas the LCAOeMO approach allows two electrons to be on the same atom (polarization) as well as on one atom at a time. Therefore the LCAOeMO approach has a higher electron density between the two atoms than does the VB approach. The difference is illustrated in Figure 10.8.1.

10.9 LCAOeMO DESCRIPTION OF DOUBLE AND TRIPLE BONDS

499

FIGURE 10.8.1 Electron densities of the valence bond and molecular orbital theories. Top: Visualization of orbital overlap in the valence bond approach. The region of overlap is the darkened area, which represents the localized bond. Bottom: Visualization of the overlap in the LCAOeMO theory of bonding. The electron can roam the volume surrounding both atoms, in which the electron density is greatest between the atoms as indicated by the increase in density of color.

10.9 LCAOeMO Description of Double and Triple Bonds Double and triplet bonds occur quite often in organic compounds. Double and triple bonds are composed of two types of bonds, the sigma bond (s) and the pi bond (p). The p bonds are composed of unhybridized p orbital on neighboring atoms that overlap. The s bond is a hybridized orbital that lies on the axis connecting the two atoms. Shown in Figure 10.9.1 are threedimensional contour plots of the double bond and the triple bond. The p bonds are in green and the s bond is red. In the case of the double bond, the s bond is the overlap of sp2 hybrid orbitals from each atom. An example is the planar molecule ethylene, H2C]CH2. In the case of the triple bond, the accompanying s bond is the overlap of sp hybrid orbitals from each atom. An example is the linear molecule acetylene, HC^CH. Since two p hybrid orbitals are involved in the triple bond, the s bond must be an sp hybrid orbital. Consider the simplest case of the ethylene molecule. The LCAOeMO description gives two spatial wave functions, 2p1  2p2 2p ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ð1  SÞ

[10.9.1]

500

10. ELECTRONIC AND NUCLEAR STATES

π

π σ π

π

πσ π

FIGURE 10.9.1 Bonding in the double and triple bonds. Left: The double bond involves the overlapping of two unhybridized p orbitals and two hybridized sp orbitals. The overlap of the two p orbitals, referred to as a p bond, is shown in green in which case each lobe is identified with the symbol p. The overlapping sp orbitals, one from each atom, form a sigma bond is indicated by the blue envelop and identified with the symbol s. Right: The triple bond involves two p bonds, one being the green lobes and the symbol p and the other by red lobes and the symbol p, and a sigma bond indicated by the blue envelop and identified with the symbol s.

where the subscripts "1" and "2" refer to the two carbon atoms, and the overlap integral S in the normalization factor is, S ¼ hp1 jp2 i ¼ hp2 jp1 i

[10.9.2]

The reason that S is not zero is because the two orbitals are on different centers, 1 and 2. What does the overlap integral signify? Consider the relative locations of the "þ" and "" lobes of the two p orbitals. If both p orbitals are aligned in the same direction then S is a positive number. A positive value of S means that the electron density between the two atoms in reinforced. This class of wave functions promotes binding between the atoms, so 2pþ is a bonding orbital. Similarly a negative value of S results if the adjacent lobes are aligned in opposite directions. A negative value of S means that the electron density between the atoms is diminished. Therefore 2p is an antibonding orbital. The molecular orbitals are illustrated in Figure 10.9.2 along with the associated wave functions designed to represent the positive lobes of the orbitals. The energy of the p bonds are calculated to be E ¼ h2p jH j2p i ¼

h2p1  2p2 jH j2p1  2p2 i ¼ab 2ð1  SÞ

[10.9.3]

   where a ¼ hpj H pj i=ð2  2SÞ and b ¼ hpj H jpi i=ð2  2SÞ. (Do not confuse the integrals a and b in Equation [10.9.3] with the spin up and spin down functions. Both notations are standard in the literature. The Reader can distinguish between the two notations by the context in which they are used.) Since Eþ ¼ a þ b is the energy for 2pþ the value for b is a negative number. A schematic of the energy level diagram for the p bonds in the ethylene molecule is given in Figure 10.9.3. Notice that the total energy is conserved, (a þ b) þ (a  b) ¼ 2a. The d and f orbitals also form hybrid bonds that are important in transition metal coordination complexes. The crystal field theory developed in the 1930s by Bethe and van Vleck was combined

501

10.10 LCAOeMO FOR BENZENE

FIGURE 10.9.2 Comparison of p bond wave functions with particle in a one-dimensional box wave functions. The positive lobes are represented as filled areas and the negative lobes are the open areas. The left side shows the relative orientations of the lobes of the participating p orbitals and the right side is the corresponding wave. The bottom row is the bonding pi orbital (pþ) and the top row is the antibonding pi orbital (p).

α−β α

α

α+β Energy level diagram of the p bonds in ethylene. The effect of the inclusion of the nearestneighbor interaction energy b is to "split" the energy. The 2pþ orbital is of the lower energy and is referred to as a "bonding orbital" because of the increased electron density between the atoms. The higher energy state, 2p, is the "antibonding orbital."

FIGURE 10.9.3

with molecular orbital theory to form ligand field theory. The splitting of d orbitals to form hybrid orbitals provides an explanation of the color of complexes.

10.10 LCAOeMO for Benzene Benzene is a cyclic molecule composed of six carbon atoms. The sigma bonds between adjacent carbon atoms are sp2 hybrid orbitals, which "leaves behind" one unhybridized p orbital on each atom. These remaining p orbitals, however, can overlap with each other in the formation of p bonds.

502

10. ELECTRONIC AND NUCLEAR STATES

Principal of Conservation of Probability It is assumed that the p orbitals are individually normalized so that 5 X 

  2pj 2pj ¼ 6

[10.10.1]

j¼0

where the subscript "j" identifies the atom and the 2p orbitals are the unhybridized 2pz orbitals on each atom. Let us now make the assumption that the orbitals do not overlap, which is manifested in the integral 

  2pk 2pj ¼ 0 ðj s kÞ

[10.10.2]

It is to be emphasized that Equation [10.10.2] does not mean that the wave functions are orthogonal since they are centered on different nuclei, but rather that they do not overlap. Thus Equation [10.10.2] is introduced strictly for the purpose of simplicity in the construction of the molecular orbitals from the atomic orbitals. It is a simple matter to show that the same results obtain if the pffiffiffiffiffiffiffiffiffiffiffi normalization factor includes the overlap integral in the denominator, ie, 1 þ S.

Linear Combination of Atomic Orbitals for the Benzene Ring We are going to systematically determine the six molecular orbitals that can be obtained from a linear combination of the six unhybridized p orbitals in the benzene ring. In this process the power of symmetry arguments will be exploited. As in the case of the particle in a box, the molecular orbitals will be identified with the number of nodes. One must therefore determine the number of distinguishable ways that one can represent these nodes in relationship to the positions of the carbon atoms. To do this, each node is represented as a straight line on the ring. Any intersection of a line with a carbon atom means that that particular carbon atom does not contribute to the molecular orbital. The lowest energy state does not have a node. Hence all of the p orbitals contribute to this state. Because the labeling of the carbon atoms is arbitrary, any rotation of the benzene ring through multiples of the angle 60 degree (¼360 degree/n ¼ 360 degree/6) will result in the same picture. The ground state molecular orbital of benzene is therefore  1  ψ0 ¼ pffiffiffi p0 þ p1 þ p2 þ p3 þ p4 þ p5 6

[10.10.3]

where the numerical value of the quantum number n is omitted, viz, 2p0 is given as p0. The prefactor 1/O6 is the normalization factor since each of the p orbitals is individually normalized. Visualization of the LCOA orbitals for the ψ0 state is shown in Figure 10.10.1. The next simplest molecular orbital to visualize is the highest energy state, ie, the most nodes. Again all six carbon atoms contribute to the molecular orbital. Since a line passing through a carbon atom eliminates that carbon atom, the nodes for the benzene ring must therefore lie between the carbon atoms. Hence the highest energy molecular orbital is  1  ψ5 ¼ pffiffiffi p0  p1 þ p2  p3 þ p4  p5 6

[10.10.4]

503

10.10 LCAOeMO FOR BENZENE

0 5

1

4

2 3

FIGURE 10.10.1 Lowest energy orbital in benzene. The positive lobes of the p orbitals are shaded. The red and yellow connecting lines indicate the p orbital contribution to the double bonds. The carbon atoms are numbered 0 to 5.

This wave function is visualized in Figure 10.10.2. The "þ" and "" signs accentuate the notion of the standing wave, and that the p orbitals shown in the benzene illustration alternate sign.

0

0 1

5

2

4 3

5

1

4

2 3

FIGURE 10.10.2 Highest energy state in benzene. Left: The benzene structure showing the locations of the nodes and alternating signs for the orbitals. Right: The signs of the p orbitals, where the shaded lobe denotes a positive sign.

In accordance with Equations [10.10.3] and [10.10.4], we see that the molecular orbitals ψ0 and ψ5 have "used up" 1/3 of the probabilities of the orbitals. We must now distribute the remaining 2/3 probability density of each p orbital. The first energy state above the ground state has only one node. The line representing this node can be drawn in two ways, as shown in Figure 10.10.3. In the picture on the left the orbitals 1, 2, 4, and 5 contribute the same magnitude with 1 and 5 having the opposite sign as 2 and 4. The

504

10. ELECTRONIC AND NUCLEAR STATES

0

0

5

1

5

1

4

2

4

2

3

3

FIGURE 10.10.3 Double degenerate states with one node for benzene. The two figures show the two locations of nodes and the corresponding signs for the benzene ring.

orbitals 0 and 3 contribute the same amount but of opposite sign to each other, where their contribution is greater than the 1, 2, 4, and 5 orbitals. This symmetry information is manifested in the expression,  1  ψ1 ¼ pffiffiffi Ap0 þ Bp1  Bp2  Ap3  Bp4 þ Bp5 6

[10.10.5]

where the coefficients A and B represent the relative contributions of each p orbital to the wave function. The coefficients A and B are determined below. The wave function on the right is simpler because of symmetry all four orbitals contribute equally to the total probability. Therefore the wave function for the second illustration is  1  ψ2 ¼ pffiffiffi p1 þ p2  p4  p5 4

[10.10.6]

Notice that in this case all of the p orbitals contribute the same amount but differ in sign. The last two orbitals are those with two nodes and are represented in Figure 10.10.4. It is again noted that the p orbitals in the figure on the left contribute equally but with different signs, hence  1  ψ3 ¼ pffiffiffi p1  p2 þ p4  p5 4

[10.10.7]

Similar arguments as applied to Equation [10.10.5] give for the final wave function,  1  ψ4 ¼ pffiffiffi Cp0  Dp1  Dp2 þ Cp3  Cp4  Dp5 6

[10.10.8]

where again for symmetry reasons C s D. The final step in the determination of the molecular orbitals is the evaluation of the coefficients A, B, C, and D. To do this we form a table using the principal of conservation of probabilities. The table

505

10.10 LCAOeMO FOR BENZENE

0

0

5

1

5

1

4

2

4

2

3

3

FIGURE 10.10.4 Double degenerate states with two nodes for benzene. The two figures show the two locations of two nodes and the corresponding signs for the benzene ring.

 is constructed by placing hψj ψk i in the rows with the corresponding coefficients of the atomic orbitals c2j in the columns. Hence from Equations [10.10.3]e[10.10.8], Table 10.10.1 is formed. TABLE 10.10.1 Normalized Probability Coefficients of Benzene c20

c21

c22

c23

c24

c25

ψ20

1 6

1 6

1 6

1 6

1 6

1 6

ψ21

A2 6

B2 6

B2 6

A2 6

B2 6

B2 6

ψ22

0 0

1 4 1 4

0

ψ23

1 4 1 4

1 4 1 4

1 4 1 4

ψ24

C2 6 1 6

D2 6

D2 6

D2 6

D2 6

1 6

1 6

C2 6 1 6

1 6

1 6

ψ25

0

Because of the normalization of the individual wave functions the sum in both the columns and the rows must be unity. This leads to the equations for the rows, A2 þ 2B2 ¼ 3

[10.10.9]

C2 þ 2D2 ¼ 3

[10.10.10]

and for the columns, A2 þ C2 ¼ 4

[10.10.11]

B2 þ D2 ¼ 1

[10.10.12]

506

10. ELECTRONIC AND NUCLEAR STATES

Solving the simultaneous equations one has A ¼ C ¼

pffiffiffi pffiffiffi 2 and B ¼ D ¼ 1= 2 or

 1
ψ1 ¼ pffiffiffiffiffi 2p0 þ p1  p2  2p3  p4 þ p5 12

[10.10.13]

and  1
ψ4 ¼ pffiffiffiffiffi 2p0  p1  p2 þ 2p3  p4  p5 12

[10.10.14]

An illustration of the wave function ψ1 is given in Figure 10.10.5. Inclusion of orbital overlap between adjacent atoms is achieved by dividing each of the Equations [10.10.3,] [10.10.4], [10.10.6], [10.10.7], [10.10.13], and [10.10.14] by the appropriate normalization factor.

5

0 1

4 3

2

FIGURE 10.10.5 Visualization of the ψ1 wave function for benzene. The de Broglie wave represents the relative magnitude and sign of the contributions of the various pz orbitals to the ψ1 hybrid orbital.

10.11 Hu¨ckel Molecular Orbital Description of Benzene The Hu¨ckel model for molecular orbitals makes the assumption that only the nearest-neighbor atomic orbitals have significant overlap in the formation of the p bond. In this case the orthogonality condition of Equation [10.10.2] does not hold, but rather is replaced by the integral expression,    pj pjþ1 ¼ S

[10.11.1]

where modulus 6 is assumed, ie, 5 þ 1 ¼ 0. For those atoms further removed than the nearestneighbor the integral in Equation [10.11.1] is zero. The normalization factor is now dependent

¨ CKEL MOLECULAR ORBITAL DESCRIPTION OF BENZENE 10.11 HU

507

upon the specific form of the wave function. For example, using the ground state function defined by Equation [10.10.3] one has, 5 X

hpk jpk i þ

hψ0 jψ0 i ¼

5 X

k¼0

hpk jpkþ1 i

k¼0

6

¼ 1 þ 2S

[10.11.2]

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Therefore the term 1= 1 þ 2S must enter as a product in Equation [10.10.3] to obtain the normalized Hu¨ckel wave function for the ground state. It is left as an exercise to determine the corrections to the other wave functions to obtain the corresponding Hu¨ckel wave functions from those given above. Let us further define the energy integrals for the normalized Hu¨ckel wave functions, hpk jH jpk i ¼a 1 þ 2S

[10.11.3]

and fore the nearest-neighbor interactions hpk jH jpkþ1 i ¼b 1 þ 2S

[10.11.4]

where a < 0 and b < 0. It is a simple matter to determine that the energies of the six molecular orbitals of benzene are hψ0 jH jψ0 i ¼ a þ 2b

[10.11.5]

hψ1 jH jψ1 i ¼ a þ b

[10.11.6]

hψ2 jH jψ2 i ¼ a þ b

[10.11.7]

hψ3 jH jψ3 i ¼ a  b

[10.11.8]

hψ4 jH jψ4 i ¼ a  b

[10.11.9]

and hψ5 jH jψ5 i ¼ a  2b

[10.11.10]

The states 1 and 2 represent doubly degenerated levels as do the states 3 and 4. An energy level diagram for the six energy levels of the Hu¨ckel orbitals is given in Figure 10.11.1. The ground state of the p orbitals is when the 0, 1, and 2 energy levels are fully occupied, each having two electrons. The "delocalization energy" is defined as the difference between the "normal" p orbitals and the ground state with the p orbitals. Thus the delocalization energy for the benzene ring is 8b.

508

10. ELECTRONIC AND NUCLEAR STATES

E5 E4

E3

p0 p1 p2 p3 p4 p 5

_ _ _ _ _ _ _ _ _ _ _ _ _ β 2β E1

E2 E0

FIGURE 10.11.1 Energy level diagram of the Hu¨ckel Energy for the benzene molecule. The "splitting" energies of b and 2b are a result of the interactions between p orbitals with nearest-neighbor atoms. As each orbital can accommodate two electrons the ground state energy of the benzene ring p bonds is the fully occupied energies E0, E1, and E2.

10.12 Other Effects That Alter the Electronic Energy States of Atoms and Molecules One-electron hydrogen-like orbitals give rise to energy states for noninteracting electrons. The presence of repulsive interactions between the electrons within an atom results in higher energy values for the internal states, as indicated by the application of the first-order perturbation theory to the helium atom. Introduction of an "effective" nuclear charge through "shielding" also alters the energy of the atom. We have not discussed the effect of electron spin and the coupling of the spin with other spins and the orbital angular momentum on the energy states of the atoms. These latter effects are important for atoms with unpaired electrons. The application of an external electromagnetic field also alters the energy states of the atoms. A schematic diagram of these effects is given in Figure 10.12.1.

hydrogen-like

shielding of nuclear charge

electron electron repulsion

spin-orbit interaction

external EM

FIGURE 10.12.1 Schematic diagram of different effects on the internal energy states of multielectron atoms (Davis, 1965).

The electronic energy states of molecules are also altered through electroneelectron interactions and spin-orbital coupling. In addition to these interactions the molecular system also permits delocalization of the electrons through the formation of hybrid orbitals that may involve several atoms. The nature of the chemical bond may be altered through the substitution of different groups on nearby atoms. Electron withdrawing of donating groups may affect the chemical reaction by weakening the target bond.

10.13 EINSTEIN MODEL FOR STEADY-STATE ELECTRONIC TRANSITIONS

509

10.13 Einstein Model for Steady-State Electronic Transitions No doubt the Reader is familiar with "laser pointers" that can be purchased for a few dollars. It is a "pointer" because the red beam of light is very narrow and intense. The red beam of light can be aimed at a point on a slide in a public presentation to draw attention to that spot. The word "laser" is an acronym for light amplification through stimulated emission of radiation. The stimulated emission results in a beam of light that is highly coherent, like soldiers being "in step" marching in a parade. The heliumeneon (HeNe) laser serves as an example. The laser tube consists of 90% helium and 10% neon at pressure of approximately 1 Torr. There are two mirrors that serve to pass the beam from one end to the other of the laser cavity. The beam stimulates emission of light from the excited neon atom to be "in step" with the stimulating light. The underlying mechanism for the HeNe laser exemplifies many properties of electronic transitions in atoms and molecules. The electric discharge ionizes the helium atom from its ground state 1s2. When an electron recombines with Heþ it has other orbitals to chose from than the 1s orbital. Some of the electrons go to the empty 2s orbital in the 1s2s singlet state ð1 S0 Þ. But now the 2s electron is "stuck" because the selection rules do not allow it to return to the 1s orbital by radiationless transition. However, it can pass this excited energy to the Ne atom by means of a collision because the energy diagram of the Ne atom has a state very near the transition energy 1s / 2s for the He atom. It is from the excited state of Ne that the light of wavelength 632.8 nm originates. A schematic diagram of the process is illustrated in Figure 10.13.1. For a laser to work there must be a population inversion (see Chapter 13, "Quantum Statistical Mechanics").

1s2s

single triplet

1s2s

1s22s22p55s 2

2

5

1s 2s 2p 4s

1s22s22p54p 632.8 nm 1s22s22p53p

1s22s22p53s

1s2

ground

He

1s22s22p53p

Ne

FIGURE 10.13.1 Electronic transitions for the heliumeneon laser. This diagram illustrates the electronic transitions in helium and neon. The initial collision raises the ground state of helium to the singlet and triplet excited states. As return to the ground state is forbidden, the energy is transferred to the neon atom at a state that has equal energy of transition with the helium atom (dashed arrows). The red lines represent the transitions that result in monochromatic and coherent lines in the laser. The 632.8-nm line is the most intense. The dotted arrows represent spontaneous transmissions. Adaptation of Figure 2, http://web.physics.uscb.edu/wphys128/ xperiments/laser/LaserFall06.pdf.

510

10. ELECTRONIC AND NUCLEAR STATES

There are tunable lasers that provide more than one wavelength. Argon ion lasers, for example, can be tuned to a wide range of wavelengths in the visible and ultraviolet regions. They are (in nanometers) 351.1, 363.8, 454.6, 457.9, 465.8, 476.5, 488.0, 496.5, 501.7, 514.5, 528.7, and 1092.3. The basis of the laser is the 1917 paper by Einstein on transition probabilities in atoms (Einstein, 1917). For simplicity of derivation, consider a simple two-state system for which electrons can move between the two states m and n. According to the Boltzmann distribution (see Chapter 12, "Classical Statistical Mechanics"), the probability of being in a particular state of a system is proportional to the "statistical weight" pn times the "weighting factor" exp(εn/kBT). The trann sition between states m and n is illustrated in Figure 10.13.2. The parameters Bm n and Bm are the transition probabilities for absorption and emission, respectively, in the presence of electromagnetic radiation. The parameter Anm is the spontaneous transition probability for the spontaneous transition from the upper to the lower state. Under steady state conditions there is no net change. The number of electrons being promoted from state n to the m state is equal to the number in the m state returning to the n state by stimulated emission plus spontaneous emission of radiation. In mathematical form, the steady state condition is 

  εn εm  n m n B rðlÞ ¼ pm exp  Bm rðlÞþ Am pn exp  k BT n k BT |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} emission absorption

[10.13.1]

m Bm n

Bnm

A nm

n FIGURE 10.13.2 Stimulated and spontaneous electronic transitions between two states.

where r(l) is the radiation density at wavelength l. Rearrangement of Equation [10.13.1] to solve for r(l), one has 

Anm εm pm exp  Anm Bnm k BT    ¼ rðlÞ ¼ pn Bm εn εm εn  εm n m n pn exp  exp  B  pm exp  B 1 k BT n k B T m pm Bnm k BT

[10.13.2]

511

10.14 NUCLEAR STRUCTURE

Since light is absorbed only if the energy of the photon is equal to the energy difference within the 3 n n atom, εm  εn ¼ hy. Furthermore, setting pm Bnm ¼ pn Bm n and Am =Bm ¼ 8pðy=cÞ results in Planck’s Radiation Law (see Equation [7.2.9]) rðlÞ ¼ ¼

8pðy=cÞ3  hynm exp 1 k BT

[10.13.3]

It is to be noted that the derivation of Planck’s Radiation Law from transition probabilities was only a sidebar of the Einstein paper. The main focus was on the motion of a molecule in a radiation field.

10.14 Nuclear Structure Just as a high-speed collision between two automobile leaves its inner parts of gears, wheels, and other parts scattered all over the street, the collision of two nuclei in a particle accelerator leaves tracks of subatomic particles. By 1961 there was a particle zoo. Shown in Figure 10.14.1 is an artist’s rendition of colliding particles in a painting titled "Atom Smash."

FIGURE 10.14.1 Atom smash. This painting, titled "Atom Smash" is an artist’s impression of the collision between two nuclei producing subatomic particles and nuclear fragments. (Painting by Lowell Smithson.)

The quest of particle physics is to understand the relationships between the fundamental particles that give rise to the composite particles, not too unlike Mendeleev’s Periodic Table of the Elements showed the relationships between the properties of atoms and the composition of the atoms.

512

10. ELECTRONIC AND NUCLEAR STATES

The principle components of the Universe fall in three categories: particles, forces, and fields. All three are related in the quantum field equations, which bring together quantum mechanics and special relativity. The first field equation for particles was derived by Dirac in 1928, which predicted the existence of antiparticles. The concept is straightforward. A particle has an associated field, and the particles interact through a force field. Associated with the force field is a particle that carries the force. For example, an electron is a particle that has an electric field, the force of which is carried by the photon. Particles and forces are paired. The strong and weak nuclear forces subdivide particles into two main classes: hadrons and leptons. Hadrons are the heaviest particles and are subject to the strong nuclear force. Hadrons are considered to be composite particles, which through decay eventually reach the proton stage. Leptons are the light particles, subject to the weak nuclear force but not the strong nuclear force. Leptons are considered to be fundamental particles.

10.14.1 The Particles: Hadrons and Leptons Hadrons are further subclassified according to quantum statistics: baryons and mesons. Baryons are fermions and mesons are bosons. A list of hadrons is given in Table 10.14.1. TABLE 10.14.1 List of Hadrons Particle

Symbol

Mass GeV/c2

kg

BARYONS Proton

pþ

0.9381

1.6725  1027

Neutron

n

0.9465

1.6875  1027

Lambda

L

Pþ P0 P

1.1156

1.9889  1027

1.1894

2.1205  1027

1.1926

2.1262  1027

1.1974

2.1348  1027

Xi neutral

X0

1.315

2.3444  1027

Xi minus

X

1.321

2.3551  1027

Pion

P0

0.13497

2.4063  1028

Pion

P

0.139568

2.4883  1028

Kaon

K

0.49365

8.8008  1028

Kaon

K0L;S

0.4977

8.8730  1028

Eta

h

0.547

9.7522  1028

Sigma plus Sigma neutral Sigma minus

MESONS

Murray Gell-Mann set out to devise a "Periodic Table" of a select set of hadrons, and in 1961 came out with the "Eightfold Way." The Eightfold Way consists of a diagram based on the properties of the baryons and mesons. The baryon octet of Gell-Mann’s Eightfold way is shown in Figure 10.14.2.

513

10.14.2 THE FORCE CARRYING PARTICLES: QUARKS AND BOSONS

+1

q

0

-1

0

n Σ-

0

p+ 0

Σ

Σ+

Λ0

Ξ0

Ξ-1 -2

s

FIGURE 10.14.2 The Eightfold Way. Murray Gell-Mann set out to classify hadrons in the form of a table, similar to the way the Periodic Table of the Elements reveals the substructure of the electronic states of the atoms. The "s" is the spin and the "c" is the charge of the particle.

The leptons are considered to be fundamental particles. They have a spin 1/2 and do not partake in strong interactions. As fundamental particles, some leptons are negatively charged. A list of leptons is given in Table 10.14.2. TABLE 10.14.2 Leptons Particle

Symbol

Antiparticle

Charge

Mass (MeV/c2)

Electron

e

eþ

1

0.511

Electron neutrino

ne

ne

0

<0.0000022

Muon

m

mþ

1

105.7

Muon neutrino

nm

nm

0

<1.70

Tau

s

sþ

1

1777

Tau neutrino

ns

ns

0

<15.5

10.14.2 The Force Carrying Particles: Quarks and Bosons In 1964, Gell-Mann, and independently George Zweig, suggested that the hadrons were not elementary particles but rather composed of other particles. To make things work the most economical scheme was the introduction of new particles with fraction charges, 2/3 and 1/3. But there were no such particles in Nature, so he referred to them as "mathematical." The process by which these new particles were name is somewhat unusual (Gell-Mann, 2012). He had chosen the "sound" for the name of the new particles, but he did not know how the sound

514

10. ELECTRONIC AND NUCLEAR STATES

should be spelled. One possible spelling was "kwork." The spelling that was eventually chosen was "quark," which came from his reading of James Joyce’s book Finnegan’s Wake: Three quarks for Muster Mark! Sure he has not got much of a bark An sure any he has it’s all beside the mark. Gell-Mann noted in the interview that names of other particles based on Greek words became obsolete as new information was gained. The word "proton," for example, meant "first thing." But of course the proton is not the first thing. On the other hand "quark" did not mean anything at all, so it could never become obsolete. The selection of literature was appropriate because three "flavors" of quarks were proposed: up (u), down (d), and strange (s). The forces between the quarks were given the names of colors: red, green, and blue. Hence the given name of the new dynamical theory was quantum chromodynamics. Each quark was assigned a spin and a charge. The charges, however, were fractions of the electron charge. Both d and s quarks have a charge of 1/3 and the u quark has a charge of þ2/3. The proton has a þ1 charge, which is obtained from 2/3 þ 2/3  1/3. Therefore the proton composition is (u, u, d). The neutron is neutral, which is obtained from þ2/3  1/3  1/3 ¼ 0. Thus the neutron has the composition (u, d, d). The names of the quarks are not arbitrary. The up (u) and down (d) quarks are named for their components of spin. The strange (s) quark was named for the strange particles found in cosmic rays in 1947, the kaon (K) and pion (p). The charm (c) quark was named by Glashow because he was "pleased" by the symmetry it brought to the subatomic world (complemented the s quark). If one had a "down" and "up" quark, then Harari believed that there should logically be a top (t) and bottom (b) quark. A list of quarks is given in Table 10.14.3. TABLE 10.14.3 Quarks Name

Symbol

Antiparticle

Charge q

Mass (MeV/c2)

Up

u

u

þ2/3

1.5e3.3

Down

d

d

1/3

3.5e6.0

Charm

c

c

þ2/3

1160e1340

Strange

s

s

1/3

70e130

Top

t

t

þ2/3

169,100e173,300

Bottom

b

b

1/3

4130e4370

Bosons are also force carrying particles. The photon is the most familiar. The strong nuclear force is carried by "gluons." The theory of the weak nuclear force required three new bosons as mediators of the weak interaction: Wþ, W, and Z. The "W" stands for the "weak" nuclear force. Neutrino absorption and emission are mediated by Wþ and W, where the electrical charge is associated with electron and positron emission. The W bosons are associated with transmutation of the elements. The "Z" boson was named by Steven Weinberg as being the last particle needed for the Standard Model and it has "zero" charge. The Z boson is not therefore involved with transmutation. The Z boson is the particle that mediates momentum, spin, and energy transfer. The bosons are listed in Table 10.14.4.

515

10.14.3 THE STANDARD MODEL

TABLE 10.14.4 Bosons Name

Symbol

Antiparticle

Charge (q)

Spin

Mass (GeV/c2)

Interaction

Photon

g

Self

0

1

0

Electromagnetic

W boson

W

Wþ

1

1

80.4

Weak

Z boson

Z

Self

0

1

91.2

Weak

Gluon

g

Self

0

1

0

Strong

Self

0

0

125.3

Mass

Self

0

2

0

Gravitation

Higgs boson

H

Gravitona

G

0

aNot confirmed.

10.14.3 The Standard Model The Standard Model is a theory in particle physics which addresses three of the four known forces in Nature: electromagnetic force, weak nuclear force, and the strong nuclear force. The current formulation was finalized in the mid-1970s. The Standard Model is based on symmetry principles, such as rotation. Consider, for example, the table setting for a dinner party of eight. The setting of the table is not changed if rotated by angles of 45, 90, 135, 180, 225, 270, or 315 degree. The symmetry is broken if one member of the party drinks from the water glass on the right. Similarly symmetry in the Standard Model obtains if all of the particles are massless. Presumably this is the condition when the Universe first came into existencedsymmetry prevails in a manner similar to all particles in the gas phase. Now enter the Higgs particle, first proposed in the 1960s. It is the interaction with the Higgs particle that all other particles gained mass. It is this reason the Higgs boson is referred to as the "God particle" first introduced by Leon Lederman and Dick Teresi in the title of their 1993 book The God Particle: If the Universe is the Answer, What is the Question? The phrase "God Particle" originated as a joke, as explained by Teresi in an interview with Melissa Block (2013) on the radio show "All Things Considered." When Teresi asked Lederman what was the ultimate particle in the Universe, of which we can make anything else. Lederman drew a "funny looking picture" and jokingly referred to it as "The God Particle." Teresi suggested that they use it as a working title for their book. Teresi told Lederman that publishers never used any title he suggested, so it became the working title. The publisher did accept it as an appeal to potential Readers, and the rest is history. The Higgs boson was included even though the experimental confirmation had to wait 30 years. Confidence in its eventual confirmation may be a consequence of the "beauty" of the Standard Model. There are 12 elementary fermions in the Standard Model, along with their antiparticles. There are six quarks and six leptons. Each classification is grouped in pairs, which form three generations which are given in order of increasing mass. Another characteristic of the first generation is that the particles do not decay. None of the neutrinos decay. The gauge bosons all have spin of 1, and therefore do not have to obey the Pauli Exclusion Principle. Along with the other components of the standard model, the three generations of quarks (indicated by colors to explain how some quarks could coexist in what should be identical quantum states) and leptons are shown in Figure 10.14.3. Shown in Figure 10.14.4 are the components of the proton and the neutron as well as how the transfer of a gluon between quarks can change the color of the individual quark by not the overall identity of the particles. The existence of all of the components of the Standard Model has been verified with the announcement on July 4, 2012 that evidence for the Higgs boson was obtained at the LHC.

516

10. ELECTRONIC AND NUCLEAR STATES

quarks

up

charm

H

g gluon

top

s

b

down

strange

bottom

photon

e

μ

τ

Z

electron

muon

tau

Z boson

υe

υμ

υτ

electron neutrino

muon neutrino

tau neutrino

d

leptons

t

c

I

II

Higgs boson

γ gauge gaugebosons bosons

u

W W boson

III

FIGURE 10.14.3 The standard model.

d

u d

u

u

proton

A

d

neutron

B

C

FIGURE 10.14.4 Proton, neutron, and color change. Top: The proton and neutron constructed of quarks with a color charge. Bottoma: (A) hadron composed of three quarks of different colors before color change. (B) The blue quark emits a blue-antigreen gluon, where the removal of the blue is the dominant center color and the antigreen in the gluon leaves behind the color green for the quark. (C) The antigreen of the gluon removes the green of the lower left quark and the blue of the gluon colors the quark. a Modification of the images in "Color charge" (https://en.wikipedia.org/wiki/Color_charge).

517

10.15 ENERGY FROM THE SUN: FUSION

The Standard Model is self-consistent. The Standard Model is far from being a complete theory of fundamental interactions. It does not include the gravitational field as described in the Theory of General Relativity, dark matter, or dark energy. The Standard Model is most likely one facet of the jewel that is a bigger picture of the Universe.

10.15 Energy From the Sun: Fusion As indicated in Figure 10.0.1, the binding energy per nucleon for A < 56 increases with Z. This means that in the nucleosynthesis to atoms of more tightly bound nucleons energy is released. Hans Bethe attended a conference organized by George Gamow in 1938 in Washington, DC. The astrophysicists told the physicists what they new about stars, but without a specific source of energy. The simplest model was 1 1 2 þ 1H þ 1H / 1H þ e

þ ne

[10.15.1]

Stimulated by this conference and another Washington Conference of April 1938, Hans Bethe (1967) examined reactions between protons and other nuclei as sources of energy from the Sun. Bethe wrote two papers on the source of energy in the Sun: a protoneproton (p-p) paper with Charles Critchfield and a carbonenitrogeneoxygen (CNO) cycle. Since the CNO cycle takes place in stars heavier than our Sun, only the p-p cycle is shown in Figure 10.15.1. There are two parallel reactions at the start that result in 32 He, which then combine to form 42 He and two protons. The net reaction is 411 H / 42 He þ 2bþ þ 4ne

[10.15.2]

p

p

+

p

+

β ν

β+

p

+

ν

n p

p n

ν p

ν

p

p

n

p n

p

p

p n n p

+

p

+

p

FIGURE 10.15.1 The protoneproton cycle for powering the Sun. Two protons (p) collide and give off a

positron (bþ) and a neutrino ðnÞ to form deuterium (pn), where "n" denotes the neutron. The deuterium collides with another proton with the release of a positron to form helium-3 (ppn). Two helium-3 nuclei collide to give helium-4 and two protons, bringing the reaction back to the stating material of the cycle. Note: pn and ppn refer to the "p" and "n" in the figure.

518

10. ELECTRONIC AND NUCLEAR STATES

From Appendix B the proton and electron masses are mp ¼ 1.672621637  1027 kg and me ¼ 9.10938  1031 kg, and from the Internet the mass of helium-4 is 6.6464764  1027 kg, the change in mass for the reaction in Equation [10.15.2] is Dm ¼ 4.0366396  1029 kg. The synthesis pathway thus has an energy difference  2 DE ¼ ðDmÞc 2 ¼ 4:0366396  1029 2:99792458  108 m=s ¼ 3:62795  1012 J

[10.15.3]

The frequency associated with this energy, if released at once, is approximately 5.48  1021 cps. This energy is associated with the gamma ray region in accordance with the information in Appendix V. Due to random motion resulting from inelastic collisions with nuclei, the photon may take millions of years to emerge at the surface as a component of the spectrum of light measured for the Sun. Meanwhile the radiation pressure offsets the gravitational collapse.

10.16 The Liquid Drop Model of the Nucleus and Fission The shell model of the nucleus was presented in Section 9.13 in which the nucleons were assumed to be like an ideal gas confined in a spherical container. The shell structure within the nucleus was identical to that of the electrons about the nucleus because both were assumed to be described by spherical harmonics. The liquid drop model is the other extremedinteractions occur between all of the nucleons. To a first approximation, it is assumed that the strength of interaction is independent of the distance between the nucleons, and that these interactions are pairwise additive. There are, therefore, Z(Z  1)/2 pairwise repulsive Coulomb interactions (þEC) and the strong nuclear force provides A(A  1)/2 attractive interactions (ESN). The total interaction energy for the nucleons is EðZ; AÞ ¼

ZðZ  1Þ AðA  1Þ EC  ESN < 0 2 2

[10.16.1]

which is always less than zero since A > Z and the magnitude of ENS is greater than the magnitude of EC. Clearly, if the only consideration to nuclear stability is Equation [10.15.1], then all nuclei should be stable since E(Z, A) is always less than zero. What is the source of the instability?

10.16.1 A Wafer Thin Mint A very graphic example of instability is found in Monty Python’s film titled "The Meaning of Life" (Monty Python is a British comedy group). There is a scene in which a very large man, Mr. Creosote, has finished a meal that consisted of virtually everything on the menu (https:// www.youtube.com/watch?v¼HJZPzQESq_0). He was asked by his waiter if he wanted a "wafer thin mint" after his meal. At first Mr. Creosote refused, but then gave in. The result was an explosion of food throughout the restaurant. Why Mr. Creosote exploded is clear in the video. He expanded to such an extent that his skin could not withstand the internal pressure brought on by that very tiny mint. The key is the strength of the skinda surface effect.

10.16.2 The Liquid Drop Nucleus and Nuclear Fission The "wafer thin mint" of a nucleus is the captured neutron that causes the nucleus to split into two parts. The liquid drop model first proposed by George Gamow and later modified by Niels

10.16.2 THE LIQUID DROP NUCLEUS AND NUCLEAR FISSION

519

Bohr, and John Wheeler. The semiempirical formula of Carl Friedrich von Weiza¨cker provides a simple model to explain nuclear fission. A major feature of the liquid drop model is the distinction between surface nucleons and bulk nucleons. Implicit in this distinction is that the interactions are distant-dependent. There are more nearest-neighbors for a nucleon in the bulk of the nucleus than for those nucleons on the surface. It is for this reason that vapor pressure on solutions is a surface concentration phenomenon (see Figure 5.5.1). Also associated with the energy of a nucleus is the filling of the shells and the Pauli Exclusion Principle. Since the masses are nearly equal, the energy states of the proton are very close to the energy states of the neutron. One can then "pair up" a proton with a neutron. But this pairing is limited because of the asymmetry in numbers: there are more neutrons than protons, thus there are more neutrons in higher energy states than protons. Because of spin, the nucleons tend to "pair up" in the nucleus. This effect also contributes to the binding energy. The binding energy (in MeV) for the liquid drop model is thus of the empirical form known as the semiempirical mass formulation, Ebind ðMeVÞ ¼ C1 A  C2 A

3=2

ðA 2ZÞ2 C5  C3 1=3  C4 þ dZN 3=4 A A A Z2

[10.16.2]

where the first term is the volume term (associated with the packing fraction of the nucleonsdsee Section 6.3 for the packing fraction) with C1 ¼ 15.75 MeV, the second term is the surface term with C2 ¼ 17.8 MeV, the third term is the Coulomb term with C3 ¼ 0.711 MeV, the fourth term is the Pauli "asymmetry" term with C4 ¼ 23.7 MeV, and the last term accounts for pairing of the nucleons with C5 ¼ 34 MeV (http://www.nuclear-power.net/nuclear-power/fission/liquiddrop-model/). The function dZN is dependent upon whether the protons (Z) and neutrons (N) are even or odd in number where the position in the subscript identifies the particle. For example, dEO means that the number of protons is even (Z ¼ E) and the number of neutrons is odd (N ¼ O). The values of dZN are dEE ¼ 1, dOO ¼ þ1, and dEO ¼ 0 ¼ dOE. The distribution of even/odd stable nuclei is given in Table 10.16.1. Shown in Figure 10.16.1 is a graphic representation of the semiempirical binding energy formula given by Equation [10.16.2]. TABLE 10.16.1 The Distribution of Stable Nucleia A

N

Z

Number of Stable Nuclei

Even

Even

Even

166

Odd

Odd

8

Even

Odd

57

Odd

Even

53

Odd

aEisberg, Robert, 1961. Fundamentals of Modern Physics, John Wiley & Sons, Inc., New York, p. 589.

The liquid drop model explains the spherical shape of the nucleus but does not explain all of the properties of the nucleus. What if some additional energy was added to the liquid dropdin the form of an energetic neutron? Just as Jell-O shakes when the plate it is on is shaken, the extra energy provided by the neutron caused the spherical drop to undergo interval vibrations with a change in the shape of the drop. The vibration modes that are set up can be envisioned as the vibration of a diatomic molecule with nearly equal masses. A cartoon for the spilling of a liquid drop is given in Figure 10.16.2.

200

-8 ΔE bind -7 -6

Z

A

(MeV )

-5

100

-4 -3 -2 -1 -0

0 100

0

200

N FIGURE 10.16.1 Graphic representation of the semiempirical mass expression. The binding energy is shown as a plot of the number of protons (Z) versus the number of neutrons (N), where the binding energy is color-coded in accordance with the scale on the right of the image. The dark red area represents iron. Modification of image found in article "Nuclear binding energy" (https://en.wikipedia.org/wiki/Nuclear_binding_energy).

n

p n n p p n p n n p n p n p n p p n n p n

n

n

p n n p p p n p n n n p p n p n n p n p n n p p p n n n n p p n n p n n p n p

p n n p p n n p n p n p p n n p n n n p p pn n p n n p n p n p n p p n n n

n

n p n p

p

p n p n

p

p p p n n n p p n p n n p n p n p n p p n n n pn

E

A

B

C

FIGURE 10.16.2 Cartoon of nuclear fission. The protons are in red and the neutrons are in blue. (A) The binding energy of the nucleus to absorb the incoming neutron. (B) The absorbed neutron increases the energy within the nucleus that causes the nucleus to be unstable, and eventually a mode results in the splitting of the nucleus into nearly two equal halves. (C) The total energy of the two daughter elements is less than that of the parent, which means that energy is released to the surroundings.

10.17 NUCLEAR POWER UNLEASHED

521

10.17 Nuclear Power Unleashed As stated in the introduction to this chapter, nucleosynthesis of elements beyond iron requires supernova explosions. The energy of a rapidly collapsing star is infused in the synthesized nuclei, which makes them unstable. An article in The Times (London) on September 12, 1933, there was an interview with Rutherford that touched upon using nuclear energy. Rutherford stated that transmutation of the elements can occur by bombarding the elements with high energy charge particles, notably protons. However, Rutherford noted that the high repulsive Coulomb energy would require the charge particles to be accelerated to high energies, and the resulting released energy would be too small to be of any value.

630,726 Leo Szlard read that newspaper article and realized that neutrons could achieve the transmutation process: neutrons were not subject to Coulomb interactions. During the brief time of waiting for a traffic light to change, Szlard had an epiphany. Knowing that chain reactions occur in chemical reactions, Szlard realized that if the fission process released two or more neutrons, each neutron could engage in a fission reaction: one became 2, 2 became 4, 4 became 16, and so forth. This means that a tremendous amount of energy could be released in a very short period of time to create an explosion. On March 12, 1934, Szlard applied for a patent and was awarded British Patent Number 630,726 entitled Improvements in or Relating to the Transformation of Chemical Elements. In the patent Szlard even discussed the concept of a critical mass. Szilard was ahead of his time because the fissile materials required to sustain a neutron-induced chain reaction were not known at that time. (This background material is from L’Annunziata, 2007, pp. 239e240.) (Note: the term "fission" was coined by Frisch in 1939 but used in this paragraph for consistency of discussion).

1938e45 On December 17, 1938, Otto Hahn and Fritz Strassmann bombarded 235U with neutrons and discovered barium atoms in the products. Not knowing what to make of this result, they turned to Lisa Meitner and Otto Frisch. The explanation was that the neutron converted 235U to the unstable 236U which subsequently decayed to 141Ba and 92Kr plus three neutrons. Now knowing that a fissile material exists, and the discovery was made in Germany with Hitler in power, Szlard became fearful that Germany might invent an atomic bomb. Along with Eugene Wigner, Szlard wrote a letter to President Franklin D. Roosevelt regarding this possibility and asked Albert Einstein, the most famous scientist of the day (as well as now) to sign it. President Roosevelt acted, and thus began the humble beginnings of what was to be known as the Manhattan Project. On December 2, 1942, Enrico Fermi and his team achieved the first self-sustaining nuclear chain reaction. On July 16, 1945, at 5:29 and 45 seconds am, the Trinity test of a plutonium device, called "The Gadget," was successful. It was then that J. Robert Oppenheimer, utter the prophetic words from the Bhagavad Gita: "Now I am become Death, the destroyer of worlds."

1954e2016 On June 26, 1954, at Obninsk, Russia, the first nuclear power plant APS-1 was connected to the power grid. The APS-1 generated 5 MW of power for commercial use. By the year 2016 there were

522

10. ELECTRONIC AND NUCLEAR STATES

442 nuclear power plants in 31 countries (https://www.euronuclear.org/info/encyclopedia/n/ nuclear-power-plant-world-wide.htm). However, major disasters at nuclear power plants and the effect on the environment have raised caution about the future of nuclear power as a source of energy in the future (Caldicott, 2011; Weart, 2012; Mahaffey, 2014).

10.18 Fossil Fuels and Nuclear Fuels: A Comparison The energy content of fossil fuels lies in the chemical bonds in combustion reactions that result in carbon dioxide and water as major products. Closely associated with the energy output is the change in the oxidation state of carbon, a race to the þ4 oxidation state in CO2. The fossil fuel energy is therefore the heat of combustion, DHcomb. The energy content in nuclear fuels is associated with the change in mass from reactants to products. The released energy is thus calculated from the Einstein relationship DE ¼ (Dm)c2.

Heat of Combustion of Methane Methane is the major composition of what is called "natural gas." The chemical reaction of interest is CH4 ðgÞ þ 2O2 / 2H2 O þ CO2

[10.18.1]

The heat of combustion for methane given in Appendix H is 212.79 kcal/mol. Using the conversion factors in Appendix A for 1 J ¼ 0.239 cal and from Appendix B that 1 mole ¼ NA ¼ 6.0221  1023 particles, the chemical energy for a molecule of methane is   cal J mol J 18 212:79  10 ¼ 1:478  10 mol 0:239 cal 6:0221  1023 particle particle 3

[10.18.2]

Nuclear Energy A typical nuclear reaction of 236 92 U

/

90 143 38 Sr þ 54 Xe

236

U is

þ 310 n

[10.18.3]

The change in mass for this reaction is (in atomic mass units u) Dm ¼ ðmSr90 þ mXe143  mn Þ  mU236 ¼ ð89:907738u þ 142:9352u þ 3ð1:00866uÞ  236:045561u ¼ 0:17665u

[10.18.4]

10.19 ENERGY FOR THE FUTURE

523

Using the information in Appendix B to convert atomic mass units to kilograms, 1 u ¼ 1.6605  1027 kg, the energy lost by the system to the surroundings is DEsurr

 2  27 kg 8 m ¼ ðDmÞc ¼ ð0:17665uÞ 1:6605 10 ¼ 2:64  1011 J 3 10 u s 2

[10.18.5]

Conclusion The energy available in the nuclear reaction is approximately 10 million times more than the available energy for burning methane.

10.19 Energy for the Future The primary topic of this chapter is energy: chemical bonding energy and nuclear fission energy. When it comes to the future and the Ecosphere in which we live, the only difference between fossil fuel power plants and nuclear power plants is that the burning of fossil fuels releases carbon dioxide (and other gases such as sulfur dioxide for some coals) into the atmosphere. However, even nuclear power relies on the use of fossil fuels for (Caldicott, 2011) uranium mining and milling, treatment of mill tailings, conversion of uranium to the useful form of uranium hexafluoride, enrichment of uranium, fabrication of fuel elements, the construction and dismantlement of nuclear reactors, cleanup after its useful lifetime, the disposal of nuclear waste, and safe storage of nuclear waste for 240,000 years. Muller (2015) mentions a lower limit of at least 100,000 yearsd still a very long time. Both fossil and nuclear sources are extractive, which means that both will eventually vanish from the scene. There is a reserve of approximately 250 years for fossil fuel and approximately 50 years for uranium. Breeder reactors, that is, reactors that create more fuel than they use up, are not the solution because the radioactive waste is much more lethal than at present. If the rate of population growth from 1950 to 2012 continues unabated through the year 2100, then the estimated population in 2100 is 26 billion people. But this number will not obtain because, as anyone who tried to maintain a self-sustaining aquarium or terrarium knows, a population cannot grow beyond its natural resources. The natural resources on Earth, which is a paludarium, are being used up or made toxic to maintain, and expand upon, a way of life. Humans can survive without air for 3 min, without water for 3 days, and without food for 3 weeks. Yet with this knowledge, the air is being polluted, the water is being used up in industrial practices, and the land is eroding at an increasing ratedand all three are being poisoned with toxic chemicals. The continued path towards increased creature comfort is the means of checking the population growth. In regard to global warming and climate change, we are assured by the proponents of fossil fuels that the human species will adapt to the hostile world. But if humans are required to adapt, then why not adapt to sustainable energy sources? Sustainable energy sources are those naturally occurring sources that existed before humans became the dominant species. All of life depends upon solar energy, so why not use solar energy to run our mechanical machines? Solar energy also provides wind power, which is currently being developed. The EartheMoon system provides gravitational energy through the regular rhythm of the tides.

524

10. ELECTRONIC AND NUCLEAR STATES

This is not to say we should eliminate nuclear energy and fossil fuels. Nuclear energy produces good, through nuclear medicine and the treatment of cancer. Fossil fuels will also find use as long as there is portable equipment. But what must be changed is the source of energy on the grand scale, such as to produce electricity. Since the human species will need electricity as long as the species exists, why not plan for longevity of the species through the development of sustainable energy sources? As of this writing (2016), 140% of Denmark’s electrical demand was met by wind power (https://www.theguardian.com/environment/2015/jul/10/denmark-wind-windfarmpower-exceed-electricity-demand). France and Germany have integrated solar power into its power grid despite cloudy weather (https://www.google.com/amp/www.ibtimes.com/solar-power-europe-can-germany-francelead-renewable-energy-despite-cloudy-climates-2204040%3famp¼1), and Portugal ran on the combination of solar, wind, and hydropower alone for four days (https://www.theguardian. com/environment/2016/may/18/Portugal-runs-for-four-days-straight-on-renewqable-energyalone). Pakistan is looking to the future in its development of a solar park (http://www.dawn. com/news/1205484). However, efforts to advance sustainable sources of energy in the United States are up against the energy industries of fossil and nuclear fuels (https://cleantechnica. com/2013/02/09/germany-solar-power-lessons/). Regardless of this opposition, wind and solar power are being developed at the local level, such as the wind turbines in Smokey Hills, Kansas. According to a 2015 report the production at three Kansas wind farms tops the national average (http://www.khi.org/news/article/production-at-three-kansas-wind-farmstops-national-average). The three farms have a combined maximum capacity of about 450 megawatts annually. Solar panels are becoming popular for private homes, businesses, and public locations, as shown in Figure 10.19.1 along with a photo of wind turbines in the Smokey Hills project.

FIGURE 10.19.1 Sustainable sources of energy. Left top: An array of solar panels in a yard of a private residence. Left bottom: A rooftop array of solar panels on a building at the Kansas City Zoo. Right: Part of the wind turbines in the Wind Farm Project of Smokey Hills Kansas. The Smokey Hills Wind Farm composed of volunteer land owners in Lincoln and Ellsworth Counties. It is operated by Enel Green Power.

525

PROBLEMS

Problems 10.1 The first-order perturbation term h1sð1Þ1sð2ÞjH 0 j1sð1Þ1sð2Þi for the helium atom was shown in Example 10.4.1 to be (5/4)ZEH,1, where Z ¼ 2 for helium and EH,1 is the ionization energy of the hydrogen atom. This value is calculated from the "pure" hydrogen-like wave functions 1s. One can use the experimental values of the ionization energies to calculate a more realistic value of the perturbation energy. Given the experimental values of the first and second ionization energies, I1(He) ¼ 24.58741 eV and I2(He) ¼ 54.41778 eV, respectively, obtain the experimental value of the perturbation energy hψ1 ð1Þψ1 ð2ÞjH 0 jψ1 ð1Þψ1 ð2Þi, where jψ1 ð1Þψ1 ð2Þi is the symbolic representation of the "true" wave function for electrons 1 and 2 in the lowest orbital in the helium atom. 10.2 Given that the lithium atom is in its ground electronic state, identify all of the perturbation terms that represent the interactions between the electrons. 10.3 The third ionization energy for lithium is I3(Li) ¼ 122.45429 eV. This corresponds to the ionization Li2þ / Li3þ þ e. Compare this experimental value with the predicted ionization energy of the Bohr atom. 10.4 The three ionization energies for the lithium atom in the ground state are I1(Li) ¼ 5.39172 eV; I2(Li) ¼ 75.64018 eV; and I3(Li) ¼ 122.45429 eV. Following Problem 10.1, evaluate the perturbation energies hψ1 ð1Þψ1 ð2ÞjH 0 jψ1 ð1Þψ1 ð2Þi and hψ1 ð1Þψ2 ð3Þ jH 0 jψ1 ð1Þψ2 ð3Þi ¼ hψ1 ð2Þψ2 ð3ÞjH 0 jψ1 ð2Þψ2 ð3Þi. 10.5 The six ionization energies for the carbon atom are given in Appendix R. Is there anything unusual about the trend in the ionization energies? Compare the reported values of the ionization energies with the Bohr values. Discuss the comparisons. 10.6 The cyclobutadiene molecule has the following conjugated structure,

0

1

0

1

3

2

3

2

where the carbons are numbered 0, 1, 2, and 3. Determine the normalized LCAOeMO wave functions for cyclobutadiene. Express the Hu¨ckel energies Ej of each wave function in terms of the energy integrals a and b. Construct an energy diagram similar to that for benzene in Figure 10.10.1. Note: The sum of all of the energies of the LCAOeMO states must conserve the total energy of the system at the start, which in this system is 4a. Start with the highest and lowest energy wave functions to provide some insight about the other wave functions such as the total energy is conserved. 10.7 The 1,3-butadiene molecule has the double bond structure C]CeC]C. What are the molecular orbitals associated with this structure and the corresponding energy expressions?

526

10. ELECTRONIC AND NUCLEAR STATES

10.8 The formula for acetylene is C2H2. Identify the bond types in this molecule. What are the hybrid orbitals? 10.9 Let the benzene molecule "open up" to form the linear 1,3,5 -hexatriene, C]CeC¼]CeC]C. What are the molecular orbitals for this molecule? Determine the energy diagram for this molecule in a manner shown in Section 10.10 for benzene. Discuss any differences and/or similarities between the two systems? 10.10 Consider the molecule cyclooctatetraene.

Following the example of benzene, determine the molecular orbitals and the corresponding energy diagram. What is different about the energy diagram for cyclooctatetraene in comparison to that of benzene? What is the energy for the transition from the ground state to the first excited state? 10.11 The formula for acetylene is C2H2. Identify the bond types in this molecule. What are the hybrid orbitals? 10.12 Balance the equation for the combustion of acetylene, C2H2(g) þ O2(g) / CO2 þ O2. From the bond energies in Appendix K, estimate the enthalpy of combustion of acetylene. Compare this value with the enthalpy of combustion for acetylene in Appendix H. 10.13 From the heat of combustion of ethylene in Appendix H and the double bond energy C]C in Appendix K, estimate the CeH bond energy. How does this compare with the combustion of methane from data in Appendix H? 10.14 Consider the uranium atom, atomic number 92. What is the electron configuration of uranium estimated from the aufbau principle? If an inner shell electron at the n ¼ 2 level is knocked out of the atom and an electron in its outermost shell drops to replace it, what is the frequency of the emitted photon? What region of the electromagnetic spectrum would one look for that radiation? Assume Bohr energy levels and use the information in Appendix V. 10.15 What is the wavelength for the photon emitted if the electron in Problem 10.14 was in the L shell? What region of the electromagnetic spectrum would one find this line? 10.16 Calculate the radius of the n ¼ 1 shell (K shell) for uranium using the Bohr expression for the radius. What is the nonrelativistic de Broglie wavelength for this electron? Using the de Broglie wavelength, estimate the speed of the 1s electron. What conclusions can be drawn from the speed of the electron?

PROBLEMS

527

10.17 Use the Bohr expression for the energy of the atom and calculate the ionization energy for the outermost electron in nitrogen. How does this value compare with that in Appendix R? One interpretation of differences of this nature is that the other electrons "screen" the outermost electrons from the "true" atomic charge. What would be the "screened Coulomb charge" on the nitrogen nucleus if this interpretation is valid? 10.18 Sodium has one more electron than neon, which means the nuclear charge for sodium is greater. Why then are the ionization energies for the first electron greater for neon? Support your argument by numerical calculations based on the Bohr atom energies. See Appendix Q. 10.19 The most intense line for the argon ion laser is 488 nm. What is the corresponding energy difference in the levels in the argon atom? 10.20 The halogens fluorine (9), chlorine (17), bromine (35), and iodine (53) are in the fluorine family of the Periodic Table of the Elements. Consider the enthalpies of formation of methane and the halogenated series fluoromethane, chlorimethane, bromomethane, and iodomethane given in Appendix D. If the average contribution of the CeH bond for methane is the same for the halogenated compounds, estimate the bond energy for CeX. Explain the differences in the strength of these bonds. Take into consideration the ground state electron configuration, mass, and size of each atom. 10.21 Consider the C2HX series C2H6 (ethane), C2H4 (ethylene), and C2H2 (acetylene). What is the hybridization of carbon for each molecule? From the data in Appendix D on heats of formation, estimate the energy of the double bond C]C and the triple bond C^C. Compare these values with those given in Appendix K. Is there a correlation of carbonecarbon bond types with the heat of combustion given in Appendix H? 10.22 The atomic mass of 12C is 12.0107 and of 13C is 13.00335. What is the mass, in kilograms, of each atom? Using the information in Appendix B, what is the total mass of an equivalent number of protons and neutrons for each atom? Account for any discrepancies. What is the difference in mass per nucleon (total of the number of protons and neutrons) for each atom? Account for the discrepancy. 10.23 The atomic mass of 238U is 238.02891 and of 235U is 235.043924. Repeat the calculations in Problem 10.22. Compare the results for 12C, 13C, 238U, and 235U. Is there a trend? 10.24 The collision of a neutron with a nucleus to give the helium nucleus (the alpha particle) has the general representation, A 1 A4 4 Z X þ 0 n / Z2 X þ 2 He

where Z is the number of protons (nuclear charge), A is the mass number (total number of neutrons plus protons), and X is the chemical symbol for the element. If the atom that decays is 235 92 U, what is the product atom? What is the discrepancy in mass on the reactant side of the ledger and the product side of the ledger? Convert the amount of missing mass into energy. If this energy is used to impart kinetic energy on the alpha particle, what is the corresponding speed of the alpha particle?

528

10. ELECTRONIC AND NUCLEAR STATES

10.25 According to the Big Bang theory, only hydrogen, helium, and lithium atoms were formed in the early stages of the Universe. All subsequent elements were synthesized inside of stars through sequences of nuclear collisions that fused the nuclei to form a new element. Consider the fusion of 32 He with 42 He to form 74 Be and a gamma ray, 3 4 7 2 He þ 2 He / 4 Be þ

g

If the diameter of the helium nucleus is taken to be 5  1015 m, calculate the Coulomb potential energy, ε, in Joules, when the nuclei touch. Calculate the characteristic temperature T that corresponds to that energy, ie, Q ¼ ε/kB, where kB is the Boltzmann constant. Calculate the relative speed crel of the two helium nuclei at this temperature, viz, mc2rel 3 ¼ k BQ 2 2 What conclusions can be drawn from these calculations? 10.26 Consider the three atoms and their relative atomic masses: 40 20 Ca

ð39:962590863Þ;

56 26 Fe

ð55:93493633Þ;

64 30 Zn

ð63:92914201Þ

Calculate the missing mass used for binding the nucleons, using the information for the protons and neutrons in Appendix B. Calculate the binding energy per nucleon for each isotope. What conclusions can be drawn from these calculations? 10.27 Use the semiempirical mass formula, Equation [10.16.2], to calculate the binding energies for the isotopes in Problem 10.26. How do these calculated values compare with the values determined in Problem 10.26? 10.28 Consider the total binding energies (in MeV) for the following isotopes: 126 C (92.16); 58 26 Fe 60 52 (509.944366); 26 Fe (525.345093); and 26 Fe(447.696747). Use Equation [10.16.2] and calculate the total binding energies for these isotopes. What conclusions can be drawn? Take into consideration odd and even numbers of nucleons. 10.29 One of the steps for lithium burning (lithium plus a proton yields two helium-4 atoms) is 3 4 7 2 He þ 2 He / 4 Be

Given the relative masses of 3.0160293201 u for helium-3, 4.00260325413 u for helium-4, and 7.01692983 u for beryllium-7, what is the energy difference in MeV for this reaction? 10.30 A breeder reactor is a reactor that uses nuclear fuel to generate more of the same nuclear fuel. One example is the use of 239 94 Pu to provide neutrons that react with the more 238 abundant isotope of uranium, 92 U, which eventually generates another 239 94 Pu. If the change takes place in two steps, identify the components.

References Bethe, H., 1967. "Energy production in stars", Nobel Lecture. https://www.nobelprize.org/nobel_prizes/physics/ laureates/1967/bethe-lcture.pdf. Block, M., March 15, 2013. All Things Considered, the Man Who Coined ‘The God Particle’ Explains: It Was a Joke. http:// www.npr.org/sections/the two-way/2013/03/15/174440162/the-man-who-coined-the-god-partice-explains-itwas-a-joke.

REFERENCES

529

Caldicott, H., 2011. Nuclear Power is Not the Answer. The New Press, New York. Davis Jr., J.C., 1965. Advanced Physical Chemistry: Molecules, Structure, and Spectra. The Ronald Press Company, New York. Dirac, P.A.M., 1947. The Principles of Quantum Mechanics, third ed. Clarendon Press, Oxford. Einstein, A., 1917. "On the quantum theory of radiation," Physikalische Zeitschrift 18, 121e128 (an English translation can be viewed at http://cua.mit.eduu/8.421/Papers/Einstein%201917.pdf). Farmelo, G., editor, 2002. It Must Be Beautiful: Great Equations of Modern Science. Granta Books, London. Farmelo, G., 2009. The Strangest Man: The Hidden Life of Paul Dirac, Mystic of the Atom. Basic Books, A Member of the Perseus Books Group, New York. Gell-Mann, M., January 12, 2012. Interview on Quarks, YouTube, http://www.youtube.com/watch? v¼KDkaMuN0DA0. L’Annunziata, M.F., 2007. Radioactivity: Introduction and History. Elsevier, Boston. Mahaffey, J., 2014. Atomic Accidents: A History of Nuclear Meltdowns and Disasters from the Ozark Mountains to Fukushima. Pegasus Books, New York. Muller, D., 2015. "Uranium: Twisting the Dragons Tale", Genepool Productions Film Victoria and Screen Australia. Sherwin, C.W., 1960. Introduction to Quantum Mechanics. Holt, Rinehart and Winston, New York. Soddy, F., 1912. The Interpretation of Radium, third ed., reprinted as Scholar’s Choice. Wells, H.G., 1914. The World Set Free, reprinted 2007 by The Book Tree, San Diego. Weart, S.R., 2012. The Rise of Nuclear Fear. Harvard University Press, Cambridge.