Content check — University Physics 3, Chapter 9: Condensed Matter Physics

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 9: Condensed Matter Physics

57 problems · 1 figure

Check Your Understanding 9.1 §9.1 source ↗

Why is the potential energy associated with the exclusion principle positive in Example 9.1?

Answer

It corresponds to a repulsive force between core electrons in the ions.

Check Your Understanding 9.2 §9.2 source ↗

What does the energy separation between absorption lines in a rotational spectrum of a diatomic molecule tell you?

Answer

the moment of inertia

Check Your Understanding 9.3 §9.3 source ↗

If the dissociation energy were larger, would that make it easier or more difficult to break the solid apart?

Answer

more difficult

Check Your Understanding 9.4 §9.4 source ↗

What happens to the ground state energy of an electron if the dimensions of the solid increase?

Answer

It decreases.

Check Your Understanding 9.5 §9.7 source ↗

How does the magnitude of the forward bias current compare with the reverse bias current?

Answer

The forward bias current is much larger. To a good approximation, diodes permit current flow in only one direction.

Check Your Understanding 9.6 §9.8 source ↗

What conditions are necessary for superconductivity?

Answer

a low temperature and low magnetic field

9-1 §9.1 source ↗

What is the main difference between an ionic bond, a covalent bond, and a van der Waals bond?

Answer

An ionic bond is formed by the attraction of a positive and negative ion. A covalent bond is formed by the sharing of one or more electrons between atoms. A van der Waals bond is formed by the attraction of two electrically polarized molecules.

9-3 §9.1 source ↗

Describe three steps to ionic bonding.

Answer

1. An electron is removed from one atom. The resulting atom is a positive ion. 2. An electron is absorbed from another atom. The result atom is a negative ion. 3. The positive and negative ions are attracted together until an equilibrium separation is reached.

9-5 §9.1 source ↗

For the

H_{2}

molecule, why must the spins the electron spins be antiparallel?

Answer

Bonding is associated with a spatial function that is symmetric under exchange of the two electrons. In this state, the electron density is largest between the atoms. The total function must be antisymmetric (since electrons are fermions), so the spin function must be antisymmetric. In this state, the spins of the electrons are antiparallel.

9-7 §9.2 source ↗

Rank the energy spacing

(Δ E)

of the following transitions from least to greatest: an electron energy transition in an atom (atomic energy), the rotational energy of a molecule, or the vibrational energy of a molecule?

Answer

rotational energy, vibrational energy, and atomic energy

9-9 §9.3 source ↗

Why is the equilibrium separation distance between

K^{+} and {Cl}^{−}

different for a diatomic molecule than for solid KCl?

Answer

Each ion is in the field of multiple ions of the other opposite charge.

9-11 §9.3 source ↗

In sodium chloride, how many

{Cl}^{–}

atoms are “nearest neighbors” of

{Na}^{+}

? How many

{Na}^{+}

atoms are “nearest neighbors” of

{Cl}^{−}

Answer

6, 6

9-13 §9.3 source ↗

The NaCl crystal structure is FCC. The equilibrium spacing is

r_{0} = 0.282 nm

. If each ion occupies a cubic volume of

r_{0}^{3}

, estimate the distance between “nearest neighbor”

{Na}^{+}

ions (center-to-center)?

Answer

0.399 nm

9-15 §9.4 source ↗

If the electron number density (N/V) of a metal increases by a factor 8, what happens to the Fermi energy

(E_{F}) ?

Answer

increases by a factor of

\sqrt[3]{8^{2}} = 4

9-17 §9.4 source ↗

Why does the graph in Figure 9.12 increase gradually from the origin?

Figure a is a graph of g in parentheses E versus E. The curve starts at zero and goes up and right. It is labeled g in parentheses E is proportional to E raised to half. Figure b is a graph of f in parentheses E versus E. There is a horizontal line at y value I and a vertical line at x value E subscript F. These, along with the axes form a rectangle in the first quadrant. Figure c is a graph of g in parentheses E, f in parentheses E versus E. The curves from figure a and b are superimposed here. The point on the curve with an x value of E subscript F has a y value of g in parentheses E subscript F. — Figure 9.12 in the book ↗

Answer

For larger energies, the number of accessible states increases.

9-19 §9.5 source ↗

What are the two main approaches used to determine the energy levels of electrons in a crystal?

Answer

(1) Solve Schrödinger’s equation for the allowed states and energies. (2) Determine energy levels for the case of a very large lattice spacing and then determine the energy levels as this spacing is reduced.

9-21 §9.5 source ↗

How does the number of energy levels in a band correspond to the number, N, of atoms.

Answer

For N atoms spaced far apart, there are N different wave functions, all with the same energy (similar to the case of an electron in the double well of

H_{2}) .

As the atoms are pushed together, the energies of these N different wave functions are split. By the exclusion principle, each electron must each have a unique set of quantum numbers, so the N atoms bringing N electrons together must have at least N states.

9-23 §9.5 source ↗

Why are some materials semiconductors?

Answer

For a semiconductor, there is a relatively large energy gap between the lowest completely filled band and the next available unfilled band. Typically, a number of electrons traverse the gap and therefore the electrical conductivity is small. The properties of a semiconductor are sensitivity to temperature: As the temperature is increased, thermal excitations promote charge carriers from the valence band across the gap and into the conduction band.

9-25 §9.6 source ↗

What kind of semiconductor is produced if germanium is doped with (a) arsenic, and (b) gallium?

Answer

a. Germanium has four valence electrons. If germanium doped with arsenic (five valence electrons), four are used in bonding and one electron will be left for conduction. This produces an n-type material. b. If germanium is doped with gallium (three valence electrons), all three electrons are used in bonding, leaving one hole for conduction. This results in a p-type material.

9-27 §9.6 source ↗

What is the Hall effect and what is it used for?

Answer

The Hall effect is the production of a potential difference due to motion of a conductor through an external magnetic field. This effect can be used to determine the drift velocity of the charge carriers (electrons or hole). If the current density is measured, this effect can also determine the number of charge carriers per unit volume.

9-29 §9.6 source ↗

For a p-type semiconductor, how do impurity atoms alter the energy structure of the solid?

Answer

It produces new unfilled energy levels just above the filled valence band. These levels accept electrons from the valence band.

9-31 §9.7 source ↗

When p- and n-type materials are joined, why does the depletion layer not grow indefinitely?

Answer

The electric field produced by the uncovered ions reduces further diffusion. In equilibrium, the diffusion and drift currents cancel so the net current is zero. Therefore, the resistance of the depletion region is large.

9-33 §9.7 source ↗

Why does the reverse bias configuration lead to a very small current?

Answer

The positive terminal is applied to the n-side, which uncovers more ions near the junction (widens the depletion layer), increases the junction voltage difference, and therefore reduces the diffusion of holes across the junction.

9-35 §9.7 source ↗

Explain how an audio amplifier works, using the transistor concept.

Answer

Sound moves a diaphragm in and out, which varies the input or base current of the transistor circuit. The transistor amplifies this signal (p-n-p semiconductor). The output or collector current drives a speaker.

9-37 §9.8 source ↗

How does BCS theory explain superconductivity?

Answer

BSC theory explains superconductivity in terms of the interactions between electron pairs (Cooper pairs). One electron in a pair interacts with the lattice, which interacts with the second electron. The combine electron-lattice-electron interaction binds the electron pair together in a way that overcomes their mutual repulsion.

9-39 §9.8 source ↗

What impact does an increasing magnetic field have on the critical temperature of a semiconductor?

Answer

As the magnitude of the magnetic field is increased, the critical temperature decreases.

9-41 §9.1 source ↗

Potassium chloride (KCl) is a molecule formed by an ionic bond. At equilibrium separation the atoms are

r_{0} = 0.279 nm

apart. Determine the electrostatic potential energy of the atoms.

Answer

U = −5.16 eV

9-43 §9.1 source ↗

The measured energy dissociated energy of KCl is 4.43 eV. Use the results of the preceding problem to determine the energy of repulsion of the ions due to the exclusion principle.

Answer

−4.43 eV =

−4.69 eV +

U_{ex},

U_{ex} =

0.26 eV

9-45 §9.2 source ↗

For the preceding problem, find the equilibrium separation of the H and Cl atoms. Compare this with the actual value.

Answer

The measured value is 0.484 nm, and the actual value is close to 0.127 nm. The laboratory results are the same order of magnitude, but a factor 4 high.

9-47 §9.2 source ↗

The characteristic energy of the

N_{2}

molecule is

2.48 \times 10^{−4} eV

. Determine the separation distance between the nitrogen atoms

Answer

0.110 nm

9-49 §9.2 source ↗

A diatomic

F_{2}

molecule is in the

l = 1

state. (a) What is the energy of the molecule? (b) How much energy is radiated in a transition from a

l = 2

to a

l = 1

state?

Answer

E =

2.2 \times 10^{−4} eV

; b.

Δ E =

4.4 \times 10^{−4} eV

9-51 §9.3 source ↗

The CsI crystal structure is BCC. The equilibrium spacing is approximately

r_{0} = 0.46 nm

. If

{Cs}^{+}

ion occupies a cubic volume of

r_{0}^{3}

, what is the distance of this ion to its “nearest neighbor”

I^{+}

ion?

Answer

0.65 nm

9-53 §9.3 source ↗

The measured density of a NaF crystal is

2.558 {g/cm}^{3}

. What is the equilibrium separate distance of

{Na}^{+}

and

{Fl}^{−}

ions?

Answer

r_{0} = 0.240 nm

9-55 §9.3 source ↗

Determine the dissociation energy of 12 moles of sodium chloride (NaCl). (Hint: the repulsion constant n is approximately 8.)

Answer

2196 kcal

9-57 §9.3 source ↗

What value of the repulsion constant, n, gives the measured dissociation energy of 171 kcal/mol for KCl?

Answer

11.5

9-59 §9.4 source ↗

What is the difference in energy between the

n_{x} = n_{y} = n_{z} = 4

state and the state with the next higher energy? What is the percentage change in the energy between the

n_{x} = n_{y} = n_{z} = 4

state and the state with the next higher energy? (b) Compare these with the difference in energy and the percentage change in the energy between the

n_{x} =

n_{y} =

n_{z} =

400

state and the state with the next higher energy.

Answer

4 %

; b.

4.2 \times 10^{−4} %

; for very large values of the quantum numbers, the spacing between adjacent energy levels is very small (“in the continuum”). This is consistent with the expectation that for large quantum numbers, quantum and classical mechanics give approximately the same predictions.

9-61 §9.4 source ↗

What value of energy corresponds to a density of states of

1.10 \times 10^{24} {eV}^{−1}

Answer

10.0 eV

9-63 §9.4 source ↗

Consider a cube of copper with edges 1.50 mm long. Estimate the number of electron quantum states in this cube whose energies are in the range 3.75 to 3.77 eV.

Answer

4.55 \times 10^{9}

9-65 §9.4 source ↗

Determine the Fermi energy and temperature for copper at

T = 0 K

Answer

Fermi energy,

E_{F} = 7.03 eV,

Temperature,

T_{F} = 8.2 \times 10^{4} K

9-67 §9.5 source ↗

What is the main difference between an insulator and a semiconductor?

Answer

For an insulator, the energy gap between the valence band and the conduction band is larger than for a semiconductor.

9-69 §9.5 source ↗

A valence electron in a crystal absorbs a photon of wavelength,

λ = 0.300 nm

. This is just enough energy to allow the electron to jump from the valence band to the conduction band. What is the size of the energy gap?

Answer

4.13 keV

9-71 §9.6 source ↗

Suppose that the cross-sectional area of the strip (the area of the face perpendicular to the electric current) presented to the in the preceding problem is

1 {mm}^{2}

and the current is independently measured to be 2 mA. What is the number density of the charge carriers?

Answer

n =

1.56 \times 10^{19} {holes/m}^{3}

9-73 §9.6 source ↗

The Hall effect is demonstrated in the laboratory. A thin rectangular strip of semiconductor with width 5 cm and cross-sectional area

2 {mm}^{2}

is attached to a battery and immersed in a field perpendicular to its surface. The Hall voltage reads 12.5 V and the measured drift velocity is 50 m/s. What is the magnetic field?

Answer

5 T

9-75 §9.7 source ↗

A p-n diode has a reverse saturation current

1.44 \times 10^{−8} A

. It is forward biased so that it has a current of

6.78 \times 10^{−1} A

moving through it. What bias voltage is being applied if the temperature is 300 K?

Answer

V_{b} = 0.458 V

9-77 §9.7 source ↗

Applying the positive end of a battery to the p-side and the negative end to the n-side of a p-n junction, the measured current is

8.76 \times 10^{−1} A

. Reversing this polarity give a reverse saturation current of

4.41 \times 10^{−8} A

. What is the temperature if the bias voltage is 1.2 V?

Answer

T = 829 K

9-79 §9.8 source ↗

At what temperature, in terms of

T_{C}

, is the critical field of a superconductor one-half its value at

T = 0 K

Answer

T = 0.707 T_{c}

9-81 §9.8 source ↗

A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a

50 - Ω

resistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?

Answer

61 kV

9-83 §9.AP source ↗

Potassium fluoride (KF) is a molecule formed by an ionic bond. At equilibrium separation the atoms are

r_{0} = 0.255 nm

apart. Determine the electrostatic potential energy of the atoms. The electron affinity of F is 3.40 eV and the ionization energy of K is 4.34 eV. Determine dissociation energy. (Neglect the energy of repulsion.)

Answer

\begin{matrix} U_{coul} & = & −5.65 eV \\ E_{form} & = & −4.71 eV, E_{diss} = 4.71 eV \end{matrix}

9-85 §9.AP source ↗

The separation between hydrogen atoms in a

H_{2}

molecule is about 0.075 nm. Determine the characteristic energy of rotation in eV.

Answer

E_{0 r} =

7.43 \times 10^{−3} eV

9-87 §9.AP source ↗

Determine the lowest three rotational energy levels of

H_{2} .

Answer

E_{0 r} =

7.43 \times 10^{−3} eV

;

l = 0; E_{r} = 0 eV

(no rotation);

l =

1;

E_{r} =

1.49 \times 10^{−2} eV

;

l =

2;

E_{r} =

4.46 \times 10^{−2} eV

9-89 §9.AP source ↗

List five main characteristics of ionic crystals that result from their high dissociation energy.

Answer

i. They are fairly hard and stable.

ii. They vaporize at relatively high temperatures (1000 to 2000 K).

iii. They are transparent to visible radiation, because photons in the visible portion of the spectrum are not energetic enough to excite an electron from its ground state to an excited state.

iv. They are poor electrical conductors because they contain effectively no free electrons.

v. They are usually soluble in water, because the water molecule has a large dipole moment whose electric field is strong enough to break the electrostatic bonds between the ions.

9-91 §9.AP source ↗

Astronomers claim to find evidence of

{He}_{2}

from light spectra of a distant star. Do you believe them?

Answer

No, He atoms do not contain valence electrons that can be shared in the formation of a chemical bond.

9-93 §9.AP source ↗

Show that the average energy of an electron in a one-dimensional metal is related to the Fermi energy by

\overset{−}{E} = \frac{1}{2} E_{F} .

Answer

\sum_{1}^{N / 2} n^{2} =

\frac{1}{3} {(\frac{N}{2})}^{3},

\overset{−}{E} = \frac{1}{3} E_{F}

9-95 §9.AP source ↗

Estimate the fraction of Si atoms that must be replaced by As atoms in order to form an impurity band.

Answer

An impurity band will be formed when the density of the donor atoms is high enough that the orbits of the extra electrons overlap. We saw earlier that the orbital radius is about 50 Angstroms, so the maximum distance between the impurities for a band to form is 100 Angstroms. Thus if we use 1 Angstrom as the interatomic distance between the Si atoms, we find that 1 out of 100 atoms along a linear chain must be a donor atom. And in a three-dimensional crystal, roughly 1 out of

10^{6}

atoms must be replaced by a donor atom in order for an impurity band to form.

9-97 §9.AP source ↗

Determine the Fermi energies for (a) Mg, (b) Na, and (c) Zn.

Answer

E_{F} = 7.11 eV

; b.

E_{F} = 3.24 eV

; c.

E_{F} = 9.46 eV

9-99 §9.AP source ↗

What value of the repulsion constant, n, gives the measured dissociation energy of 158 kcal/mol for CsCl?

Answer

9.15 \approx 9

9-101 §9.CP source ↗

For an electron in a three-dimensional metal, show that the average energy is given by $\overset{−}{E} =$ $\frac{1}{N} \int_{0}^{E_{F}} E g (E) d E =$ $\frac{3}{5} E_{F},$

Where N is the total number electrons in the metal.

Answer

In three dimensions, the energy of an electron is given by:

E = R^{2} E_{1},

where

R^{2} =

n_{1}^{2} +

n_{2}^{2} +

n_{3}^{2}

. Each allowed energy state corresponds to node in N space

(n_{1}, n_{2}, n_{3})

. The number of particles corresponds to the number of states (nodes) in the first octant, within a sphere of radius, R. This number is given by:

N =

2 (\frac{1}{8}) (\frac{4}{3}) π R^{3},

where the factor 2 accounts for two states of spin. The density of states is found by differentiating this expression by energy:

g (E) =

\frac{π V}{2} {(\frac{8 m_{e}}{h^{2}})}^{3 / 2} E^{1 / 2}

. Integrating gives:

\overset{−}{E} = \frac{3}{5} E_{F} .

Loading chapter 10…

Problems and answers are adapted from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-3/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.