Chapter 29 MOLECULAR AND SOLID-STATE PHYSICS

Physics Including Human Applications

Chapter 29 MOLECULAR AND SOLID-STATE PHYSICS

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GOALS When you have mastered the content of this chapter, you will be able to achieve the following goals:

Definitions Define each of the following terms, and use each term in an operational definition:

ionic bonding extrinsic semiconductor covalent bonding semiconductor fluorescence superconductivity phosphorescence nuclear magnetic resonance bioluminescence

Band Theory of Solids

State the band theory of solids, and use it to explain the optical and electrical properties of different solids.

Molecular Absorption Spectra

Explain the basis of vibrational and rotational absorption spectra of molecules.

Solid-State Problems

Solve problems involving vibrational and rotational energy states of molecules and the band theory of solids.

PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 21, Electrical Properties of Matter, Chapter 27, Quantum and Relativistic Physics, and Chapter 28, Atomic Physics.



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29.1 Introduction What factors determine the structure of molecules? Molecular construction is important in understanding the chemical properties and reactions of matter. The field of organic chemistry involves the study of specific atoms and the way they are joined together to form a molecule. The structure of the molecules is closely related to the chemical behavior of materials. In this chapter we will study the physical basis for molecular bonding and emission and absorption spectra of molecules.

Have you ever wondered why some solids are opaque and mirrorlike while others are translucent or transparent? What is the physical basis for the optical properties of solids? Have you noticed the difference in the thermal conductivity between stainless steel tableware and silver tableware? Have you noticed that good electrical conductors are usually good thermal conductors? In this chapter we will develop a model that will enable us to understand the behavior of solids in a qualitative way.

29.2 Molecules and Bonding Molecules are stable configurations of atoms. As in other naturally stable systems, the stability of molecules indicates that the energy of the molecule is lower than the energy of the system of separate atoms. We can formulate the following guide: if interacting atoms can combine and thereby reduce the total energy of the system, then the atoms will form a molecule. What can you conclude about monatomic gases (such as helium, He) as compared with diatomic gases (such as hydrogen, H2)? A useful model for describing diatomic molecules is the "spring model," in which two atoms are bound by a Hooke's law force.

The electronic states of atoms determine the nature of the bonding that occurs in the formation of molecules. As atoms are brought together there are three limiting situations that can occur. These define the pure cases of chemical bonding Figure 29.1.



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1. NO BOND IS FORMED BETWEEN INTERACTING ATOMS If all of the lowest electron states of the interacting atoms are filled, the atoms repel each other when these electronic shells overlap. This is a quantum force resulting from the Pauli exclusion principle, which does not allow two electrons to occupy the same quantum state. Helium atoms behave in this way.

2. IONIC BOND IS FORMED In this case one or more electrons are stripped from one atom and captured by another atom. The first atom is left with a positive charge and the second atom becomes negatively charged. They are held together by the electrostatic attraction of these two ions. Sodium chloride, Na+Cl-, is an example of an ionic compound. The sodium atom gives up an electron to the chlorine atom, and the resulting Na+ and Cl- attract each other to form the NaCl system. Ionic compounds are usually soluble in water. Can you give a plausible physical explanation for this solubility, remembering that the dielectric constant of water is about 80?

3. COVALENT BOND IS FORMED In this case the atoms share electrons. This sharing produces a high probability electron distribution between the atoms. These shared electrons produce an attracting force between the atoms. Hydrogen, H2, is an example of nonpolar covalent bonding, and HCl is a polar covalent bond. A nonpolar bond (such as that in H2, Cl2) is electrically symmetrical producing no external electric field. In general, nonpolar molecules are usually gases or liquids at room temperature. If nonpolar, covalent molecules form solids, they are soft and easily vaporized. What physical explanation can you give for this observation?

On the other hand, polar molecules exhibit relatively strong external electric fields due to their nonsymmetrical charge separation. For example, in HCl the shared electrons are closer to the chlorine nucleus than to the hydrogen nucleus, thus making a polar molecule as shown in Figure 29.1. Polar substances usually have higher melting and boiling points than nonpolar substances. Hydrogen gas, H2, represents a typical molecule with a covalent bond, and sodium chloride, NaCl, represents a typical molecule with an ionic bond, there are many molecules that represent intermediate bonding, with some characteristics of these two pure cases.

29.3 Absorption Spectra for Molecules The way in which a substance absorbs light provides information about the molecular structure of the substance. The spectrophotometer is an instrument designed to measure and record the absorption spectra of molecular and atomic samples. The details of the spectrophotometer were discussed in Chapter 28. As with atoms, the absorption spectra of molecules are the result of photon absorption by the system (in this case the molecules of the sample). The absorption spectra of molecules are more complex than those of atoms because in addition to the electronic energy transitions available, there are additional vibrational and rotational energy transitions available in molecules. The quantum equation that applies to the photon absorption process is given by Planck's equation,

E = hƒ (29.1)



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where h is Planck's constant = 4.13 x 10- 15 eV-sec and f is the frequency of the photons absorbed.

The output of the spectrophotometer is a plot of the percent of light transmitted as a function of wavelength. The absorption of visible and ultraviolet wavelengths is the result of electronic transitions from a lower to a higher energy level in the molecules of the sample. These energy transitions are the order of electron volts in magnitude. The vibrational and rotational energy states of the molecules are quantified just as the electronic states are. These energy levels are much closer together, and the energy differences are the order of 10-2 eV for the vibrational states and 10-4 eV for the rotational states. The wavelengths absorbed by the transitions between vibrational states lie in the infrared region 1.5 to 30 microns (1m = 10-6 m), while the absorption that arises from transitions between rotational states occurs in the infrared and microwave regions (wavelengths from 30 m to 1 cm).

29.4 Energy Levels for Vibrational States The energy level differences for vibrational states depends upon the stiffness of the bond (analogous to the spring constant of a spring) and upon the effective mass of the vibrating atoms making up the molecule. For a diatomic molecule we can use a dumbbell model, Figure 29.2. The equation for the energy levels for the vibrating diatomic dumbbell is:

Evib = (n + 1/2)hf = (n + 1/2) h/2π)(k/m)1/2 (29.2)

where k is the effective spring constant of the bond, m is the effective mass of the molecule given by m = M1M2/(M1 + M2) where M1, and M2 are the atomic masses of the two vibrating atoms in kilogram and n is the vibrational quantum number (n = 0, 1, 2, ...). The expression for f in an SHM system was developed in Chapter 15.



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EXAMPLE

Find E0 (the ground state) for the vibrational level of carbon monoxide, CO. For CO molecules, k = 1870 N/m and

m = (12 x 16 / (12 + 16)) x 1.67 x 10-27 kg

m = 1.14 x 10-26 kg

Thus,

ƒ = 1 / (2π)(k/m)1/2 = 1/(2π (1870 N/m / 1.14 x 10-26 kg)1/2)

ƒ = 6.45 x 1013 Hz

E0 = (½) hƒ = 3.3 10-34 J-sec x 6.45 x 1013 Hz

= 21.29 x 10-21J

= 0.13 eV

29.5 Energy Levels for Rotational States The energy levels for rotational states depend upon the effective mass of the molecules and the separation of the atoms making up the molecule. The equation for the rotational energy level is:

Erot = J(J + 1)(h/2π)2/2mR2) = J(J + 1)(h/2π)2/2I (29.3)

where m is the effective mass, R, the separation of atoms, and J, the rotational quantum number (J = 0, 1, 2, ...). Note that mR2 is the moment of inertia, I, of the molecule about an axis through the center of mass.

EXAMPLE

Find the energy difference (J = 0 to J = 1) for rotational transitions in CO molecules.

From the previous problem we have m = 1.14 x 10 26 kg. For CO, R = 1.13 x 10-10 m. For J = 1,

E1 = 1(1 + 1)(h/2π)2/2mR2 = (1.05 x 10-34 J-sec)2/(1.14 x 10-26 kg x (1.13 x 10-10 m)2)

E1 = 7.57 x 10-23 J = 4.75 x 10-4 eV

E0 = 0

∆E = 4.73 x 10-4 eV



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29.6 Identifying Molecular Structure In practice absorption spectra give the wavelengths of the absorbed photons. This information is then used to calculate the energy differences for the vibrational and/or rotational states. The energy differences can then be used to calculate the effective spring constants for the bonds and the separation of the atoms making up the molecules. (The masses of the constituent atoms are known quantities in this case.) In complex molecules there are different bonds and different separations in different directions. These considerations lead to a complicated set of energy levels that can be used with the spectrophotometer data to analyze the molecule's structure. No two molecules have the same absorption spectra. The vibrational and rotational spectra are unique characteristics of the molecules. Certain functional groups are found to have characteristic absorption band spectra that can be used to identify new molecular structures and to help distinguish between isomers (compounds involving the same kinds of atoms, but having different structure.)

29.7 Fluorescence and Phosphorescence The phenomena of fluorescence and phosphorescence arise from the complexity of molecular energy level systems and interactions. In each case the system is excited by external radiation-usually visible or ultraviolet light. In the case of fluorescence, the system gives up some of its energy (through vibrational excitation) in radiationless collisions with other molecules. The molecule then emits a photon from a lower vibrational state, and thus a longer wavelength photon is emitted. In phosphorescence there is a radiationless relaxation from one excited state to another excited state with a long lifetime (usually called a metastable state). The subsequent decay to the ground state is relatively slow and gives a characteristic afterglow associated with phosphorescence.

29.8 Solids The solid state of matter is characterized by its relative rigidity and fixed volume. We can use as a model for a solid a system of atoms, molecules, or ions rigidly held close together. The binding forces in solids may be ionic, covalent, van der Waals, or metallic. Ionic binding results in solids where positive and negative ions are involved as in ionic molecules, and the binding energy is electrostatic energy. An example of an ionic solid is NaCl (table salt). The solid state ionic bond is strong, and ionic solids are usually hard crystals with high melting points. Would you expect ionic solids to be brittle or ductile? Ionic crystals are usually soluble in polar liquids.

Covalent solids result from the sharing of electrons just as in the molecular covalent bond. Examples of covalent crystals are the semi-conductor silicon and germanium. The solid-state covalent binding force is very strong, and these solids are very hard (diamond, for example, the hardest substance known, is a covalent solid) and have high melting points. Covalent solids are insoluble in most liquids.



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The van der Waals force in solids results when the atoms involved have dipole moments (nonsymmetric charge distribution) and display primarily dipole-dipole interactions Figure 29.3. This dipole-dipole interaction results in a weak binding force. Molecular crystals having van der Waals bonding have low melting points and boiling points. Solid methane (melting point = -259°C) is an example of such a solid.

The metallic bond results from the attraction of the free electrons (made up of the valence electrons of the atoms) that are shared by all of the positive ion cores. This gives rise to a very strong metallic bond. This metallic bond is similar to the covalent bond but it is unsaturated-that is, it could accommodate more electrons with little increase in the energy of the bond. The unsaturated nature of the metallic bond makes it possible to form mixtures of metals called alloys that vary in properties as the composition is changed. Modern technology has made good use of alloys with special electrical, magnetic, and thermal properties. Can you think of alloys that are used for specific applications?

29.9 Electron Behavior in Solids Classical physics explained the electrical conductivity of solids through the use of the electron gas model. According to this classical model good conductors have loosely bound electrons that move through the solid. The resistance of the conductor is accounted for by the collisions between the electrons and the vibrating positive ion cores. Using this model, we can explain why the resistance of a conductor should increase as the temperature is increased. In this classical model the insulators are represented as systems with tightly bound electrons. This model provides a qualitative picture of electrical phenomena in solids, but it fails to give good quantitative predictions.

Quantum physics provides us with a much more comprehensive understanding of the electron behavior in solids. The question that needs answering is what happens to the quantized energy levels of the atoms as they condense into the solid state? The exact solution of this N-atom (where N is in the order of 1023) problem is not known exactly, but quantum physics can be used to give us a model for solids. As the N atoms condense into the solid state, the individual electronic states form bands of closely spaced energy levels. The energy levels are so close within a band that for practical purposes they can be treated as a continuum. But just as with the atomic energy levels, there are certain forbidden energy gaps between allowed energy bands. According to



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the Pauli exclusion principle there is a set number of electrons in any band. The band theory model applied to a conductor, semiconductor, and an insulator is illustrated in Figure 29.4. The highest energy, completely filled band is called the valence band. The next higher energy band, which is at most partially filled, is called the conduction band.

29.10 Conductors For a conductor, such as silver, the band of energy levels that contain the conduction electrons, called the conduction band, is only partially filled. This means that there are empty energy states very near (<10-4 eV) the filled electron states in the conduction band. When energy is supplied to a conductor, an electron at the top of the distribution in the conduction band can absorb this energy and move into an empty state moving through the solid like a “free” electron. This external energy may be electrical in nature; thus a battery will produce an electrical current in the conductor. The energy may be electromagnetic (photons, in the language of quantum physics); the electron absorbs the photons and accelerates through the solid, thus producing reflected light and the opaque appearance common to conductors. If the energy is thermal in nature, the electrons contribute to the thermal conductivity of the conductor.

Question

1. Use the band model for a conductor to explain the relationship between thermal and electrical conductivity.

29.11 Insulators An insulator has a large (~ 10 eV), forbidden, energy gap between the filled valence band and an empty conduction band. Many insulators are solids made up of atoms with an even number of electrons. In this case external energy cannot move an electron to an empty state unless the energy supplied to an electron is sufficient to cause the electron to jump the forbidden energy gap. Thus we see that a battery connected to an insulator does not produce a current. Many insulators are transparent to visible light as the photon energy passes through the conductor without being absorbed by the valence band electrons. Insulators are typically poor thermal conductors.



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29.12 Intrinsic Semiconductors ]Intrinsic semiconductors, silicon and germanium, for example, are characterized by a filled valence band with a small (~1 eV) energy gap between the valence band and empty conduction band. At room temperature there is some tunneling through the forbidden gap by a few electrons. This gives rise to a small electrical conductivity by the few thermally excited electrons in the conduction band. Unlike conductors, the resistance of semiconductors decreases with increases in temperature. The reason for this behavior is that the number of carriers in the conduction band increases exponentially with temperature and dominates the increase in electron-atom collisions that occur in conductors. Since the forbidden gap is small, semiconductors are opaque to visible light, but transparent to infrared light.

Question

2. Use the band model of a semiconductor to explain photoconductivity for semiconductors.

29.13 Extrinsic Semiconductors Thus far we have considered pure materials. The basic materials in the development of solid-state electronic devices are the impure semiconductors. By adding appropriate impurities to silicon and germanium, it is possible to produce additional energy states within the forbidden energy gap.

By doping the host semiconductor with an impurity that has one more valence electron than the host has, it is possible to create donor states in the forbidden gap near the bottom of the conduction band. Electrons from these donor states are easily excited into the conduction band and thereby provide negative carriers, creating an n-type semiconductor (Figure 29.5a).

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By doping the semiconductor with an impurity that has one fewer electron than the host semiconductor, it is possible to create acceptor states in the forbidden gap near the top of the valence band. Electrons are easily excited from the valence band into these acceptor states where they are trapped. The resulting hole in the valence band serves as a positive carrier moving through the solid. Acceptor doping creates a p-type semiconductor (Figure 29.5b).



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In solid-state devices the junction of the p- and n-type materials is the basic element. In the establishment of equilibrium at the junction, the mobile charges (holes for p-type material and electrons for n-type material) diffuse across the junction leaving behind the fixed charge of the impurity state, that is, negative for acceptors with their trapped electrons and positive for donors with their lost electrons. Thus the pn junction is a rectifying junction, the basis of the semiconductor diode operation. The forward bias applied voltage lowers the barrier for charge flow, and the reverse bias voltage raises the barrier and prohibits charge flow Figure 29.6.

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29.14 Transistors Transistors consists of two pn junctions in series, that is, either a pnp sandwich or a npn sandwich. An npn transistor, set up as a simple amplifier, is illustrated in Figure 29.7.

Note that the first junction is forward biased, and the second junction is reversed biased. A small input signal (voltage) produces a change in the current flowing through the very thin middle layer (called the base and composed of p material in this example). This current flows through the large resistance second junction (reversed bias) thus producing a large output voltage signal. This voltage amplification is achieved through the use of bias voltage supplies.



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29.15 Solar Batteries The solar battery is also based on the pn junction. Electron-hole pairs are created by the incident photons absorbed near the junction. These electrons and holes move under the influence of the potential difference at the barrier. The minority carriers (electrons in the p material and holes in the n material) create an appreciable change in the minority current through the cell. For the cell to be efficient the absorbing layer must be of large area and very thin (microns thick). For the operation of a solar battery see Figure 29.8.

29.16 Luminescence and Bioluminescence Luminescence is the emission of light due to the excitation of electrons. Luminescent devices using the pn junction include the light emitting diode (LED) in which recombination of electrons and holes at the junction releases photons in the visible region of the spectrum, characteristic of the energy difference involved in the process. The solid-state laser is also a luminescent device.

Bioluminescence is the term used when referring to light production by living systems. Bioluminescence is common in a wide range of animals, insects, bacteria, and fungi. Probably the most common example of bioluminescence is that exhibited by fireflies. In this particular case there is strong evidence that bioluminescence plays a very important part in the mating process of the species. Other systems exhibiting bioluminescence are fish, marine worms, and some luminous sea bacteria.

Bioluminescent production of light involves transformation of chemical energy into the electromagnetic energy of photon emission. Most of these processes involve enzyme-catalyzed reactions. Recent research has made the chemistry and physics of bioluminescence much clearer, but the biological significance of the phenomena for many of the systems remains quite mysterious.



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29.17 Superconductivity Superconductivity was discovered by Heike K. Onnes in 1911. Now it is receiving significant technological attention. Most metals lose some of their electrical resistance as their temperatures are lowered toward absolute zero, but finally their resistances become constant. A superconductor abruptly loses all of its resistance at some critical temperature that is characteristic of the superconducting material. The critical temperatures of superconductors range from tenths of a degree above absolute zero to temperatures in the neighborhood of 20°K. Tin, lead, and mercury are three well-known superconductors. Modern technology has developed many superconducting alloys.

The physical explanation of superconductivity is provided by the Bardeen-Cooper- Schreiffer (or BCS) theory. According to this theory quantum principles make it possible for two electrons to couple in pairs of zero momentum. These pairs are coupled through vibrations of the lattice (the ion cores of the metal) in such a way that there is an actual attraction between the electrons that overcomes their electrostatic repulsion. The electron-lattice- electron interaction produces an energy gap between the superconducting state and normal state for the metal. At the critical temperature for a given superconductor the system makes an abrupt transition into the superconducting state. Since the lattice vibrations are the source of resistance at ordinary temperatures, it is not surprising that superconductors are not among the best conductors at room temperature.

In addition to their zero resistance property (which has obvious energy saving potential for electrical energy transfer), superconductors have very low thermal conductivity. A new class of alloy superconductors are now used to make solenoids that are capable of maintaining magnetic fields over 5 tesla continuously. These superconducting solenoids offer the potential for many applications of strong magnetic fields, such as magnetic levitation transportation systems, whole-body nuclear magnetic resonance, and efficient ferromagnetic ore separation.

29.18 Nuclear Magnetic Resonance Nuclear magnetic resonance, or NMR as it is commonly called, has become a very important tool in chemistry. The physical basis for the phenomenon is the resonant absorption of electromagnetic energy by protons in a magnetic field. The sample (liquid or solid) is placed in a homogeneous magnetic field and subjected to radio-frequency (in the megahertz region) radiation. The absorption takes place when the proton absorbs energy as the radio frequency equals the natural precession frequency of the proton in the field. The proton magnetic moment has a natural precession frequency about the external magnetic field given by,

ƒ = µ • B /2πs (29.4)

where s is the spin angular momentum of the proton and is equal to one half of h/2π , or 5.28 x 10-35 J/sec, µ is the magnetic moment of proton (1.41 x 10-26 J/T); and B is the external magnetic field.



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A schematic diagram of a NMR apparatus is shown in Figure 29.9. The detection system measures the radio-frequency energy absorbed by the sample.

A typical NMR spectrum for ethanol (CH3CH2O is shown in Figure 29.10. The fine structure is shown in the high resolution spectrum Figure 29.10b.

Each of the different proton groupings in the sample is in a different magnetic environment, and thus there is a slightly different resonant frequency for each grouping. The area under each absorption band is proportional to the number of protons in the absorbing group for that band. The chemist and biochemist are able to distinguish each different chemical grouping and the number of protons in each group for a particular sample. The structural information revealed in NMR spectra is highly detailed, and it has provided great insight into the nature of chemical bonding.



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SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of this summary with the number of the section where you can find the related content material.

Definitions 1. Covalent bonding is characterized by an attractive force involving

a. dipole-dipole interactions b. shared electrons c. magnetic interaction d. gravitational interaction e. none of these

2. Fluorescence and phosphorescence are distinguished from each other by the

difference in the time of emission. The radiation is present as long as sample is the irradiated. __________ radiation continues after stimulating radiation has been removed.

3. Bioluminescence involves the emission of light from biological systems. The energy source for this light is primarily a. heat b. chemical c. magnetic d. gravitational e. none of these

4. A p-type semiconductor is one that has been doped with an impurity that has valence

electrons that number ___________ (greater than/less than/equal to) host semiconductor.

5. In order to operate a superconducting solenoid magnet it is necessary to maintain a. high current b. temperature below 20° K c. zero magnetic field d. low current e. none of these

6. NMR is a valuable tool for chemists because it yields molecular data that show a. electron distribution b. proton groups c. numbers of protons d. fluorescence e. none of these



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Band Theory of Solids 7. The band theory of solids can be used to explain which of the following properties of

solids? a. mechanical b. optical c. thermal d. electrical e. none of these

Molecular Absorption Spectra 8. The vibrational spectrum of a molecular system can be used to determine the

molecular a. moment of inertia b. rotational frequency c. force constant d. shape e. vibration frequency

9. The rotational spectrum of a molecular system can be used to determine the

molecular a. moment of inertia b. vibrations frequency c. force constant d. shape e. rotational frequency

Solid-State Problems

10. The effect of the mass change that occurs when tritium, 31H, is substituted for

hydrogen in 31H2 is to multiply the vibrational frequency of 3

1H 2 by a. 3 b. SQR RT 3 c. 1/3 d. 1/SQR RT 3 e. 6

11. The effect of this substitution (problem 10) on the rotational frequency will be to multiply the rotational frequency of 3

1H 2 by a. 3 b. SQR RT 3 c. 1/3 d. 1/SQR RT 3 e. 6



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Answers 1. b (Section 29.2) 2. fluorescent, phosphorescent (Section 29.7) 3. b (Section 29.16) 4. less than (Section 29.13) 5. b (Section 29.17) 6. b, c (Section 29.18)

7. b, d (Sections 29.9, 10) 8. c, e (Section 29.4) 9. a, e (Section 29.5) 10. d (Section 29.4) 11. c (Section 29.5)

ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the

equations will help you learn to translate words into equations and to solve single-concept problems.

Equations

E = hƒ (29.1)

Evib = (n + 1/2)hf = (n + 1/2) h/2π)(k/m)1/2 (29.2)

Erot = J(J + 1)(h/2π)2/2mR2) = J(J + 1)(h/2π)2/2I (29.3)

ƒ = µ • B /2πs (29.4)

Problems

1. Find the energy in the vibrational state for n = 1 for carbon monoxide (CO) molecules. (See the example in Section 29.4 for constants.)

2. Draw a vibrational energy level diagram for n = 0, 1, 2 for CO molecules.

3. Calculate the energy of rotation for CO molecules for J = 0, 2, 3. Compare the energy of rotation with the energy of vibration.

4. For CO molecules with n = 1 place the levels for J = 1, 2, 3 on the energy level diagram drawn in problem 2. (The J = 1 rotational energy is calculated in the example of Section 29.4.)

Answers

1. 40.0 x 10-2 eV

3. 0, 14.3 x 10-4 eV, 28.5 x 10-4 eV



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EXERCISES These exercises are designed to help you apply the ideas of a section to physical

situations. When appropriate the numerical answer is given in brackets at the end of the exercise.

Section 29.2

1. Discuss the solubility of a polar and nonpolar substance in water (a polar solvent). Give a possible physical explanation for this behavior.

2. The ionization energy of a system is the energy needed to remove the most loosely bound electron in the ground state of the system in the gaseous phase. The ionization energy of the hydrogen atom is 13.6 eV, and that of the H2 molecule is 15.7 eV. Give the physical explanation for this observation.

3. Rank the four potassium halides (KBr, KCl, KI, KF) in order from most to least ionic, and give your reasons for ranking.

Section 29.4

4. The H2 molecule absorbs infrared light of 2.3 m wavelength when the vibrational state changes by Δn = 1. Find the effective spring constant for the H2 molecule. [562 N/m]

5. The frequency of vibration of the 1H19F molecule is 8.72 x 1013 Hz. a. Find the photon wavelength for a Δn = 1 vibration transition. b. Find the effective spring constant for the HF molecule. [a. 3.4 x 10-6 m; b. 476 N/m]

6. Find the ratio of the Δn = 1 transitions for H2 and D2 vibrational energy levels. (Deuterium, D, is heavy hydrogen, 2

1H.) [ΔE)D /(ΔE)H = 1/SQR RT 2]

Section 29.5

7. The separation of the H atoms in the H2 molecule is 0.074 nm. Find the smallest energy level separation for the rotational states of the H2 molecule. Find the photon wavelength for this energy. [1.06 x 10-2 eV, 1.22 x 10-6 m]

8. Find the ratio of the ΔJ = 1 transitions for H2 and D2 rotational energy levels. [ΔEH/ΔED = 2]

Section 29.8

9. Construct a model based on the deBroglie wave property of electrons in solids that will explain the existence of allowed energy bands and forbidden energy gaps.



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Section 29.13

10. Indium is used to dope germanium. Is the result a n- or p-type semiconductor? Explain. Sketch a band structure for germanium doped with indium.

11. Antimony is used to dope germanium. Is the result a n- or p-type semiconducting material? Explain. Sketch a band structure for germanium doped with antimony.

Section 29.14

12. Sketch a pnp transistor amplifier circuit that would function like the npn amplifier shown in Figure 26.8 .

13. Find the minimum energy gap in a semiconductor LED which emits visible light if the photons emitted correspond to recombination of electrons and holes. [1.77 eV]

Section 29.17 14. Make a list of elements that are well-known superconductors and locate them on the

periodic table. What conclusions can be drawn about superconductors?

PROBLEMS The following problems may involve more than one physical concept. The numerical

answer is given in brackets at the end of each problem. 15. The energy gap in germanium is 0.75 eV. Find the wavelength at which Ge begins to

absorb light. [1650 nm]

16. The energy gap of an insulator is 20 eV. What wavelength radiation would begin to excite electrons into the conduction band? [61.7 nm]

17. The energy required to break the bond holding atoms together in a molecule is called the dissociation energy. The dissociation energy for hydrogen gas at 25°C at constant pressure is 104,220 cal/mole. Find the dissociation energy in electron volts for a single molecule of H2. [4.56 eV/molecule]

18. Consider a proton that has a magnetic moment of magnitude equal to 1.41 x 10-26 J/T. a. What is the natural precession frequency of the proton in the earth's magnetic field (B = 6.00 x 10-4 T)? What kind of electromagnetic radiation has this frequency? b. What is the precession frequency of the proton in the magnetic field of a superconducting magnet (B = 5.00 T)? What kind of electromagnetic radiation has this frequency? [a. 2.55 x 104 Hz, long radio waves; b. 2.13 x 108 Hz, short radio waves]

19. Tantalum becomes a superconductor at a temperature of 4.39°K. What is the binding energy of the coupled pairs of electrons in tantalum? [3.78 x 10-4 eV]

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Chapter 29 MOLECULAR AND SOLID-STATE PHYSICS