Ch38F

Chapter 38 Molecules and Solids 1

CHAPTER 38

Molecules and Solids

1*∙Would you expect the NaCl molecule to be polar or nonpolar?

NaCl is a polar molecule.

2 ∙ Would you expect the N2 molecule to be polar or nonpolar?

N2 is a non-polar molecule.

3 ∙ Does neon occur naturally as Ne or Ne2? Why?

Neon occurs naturally as Ne, not Ne2. Neon is a rare gas atom with a closed shell electron configuration.

4 ∙ What type of bonding mechanism would you expect for (a) HF, (b) KBr, (c) N2?

(a) HF - ionic bonding. (b) KBr - ionic bonding. (c) N2 - covalent bonding.

5* ∙ What kind of bonding mechanism would you expect for (a) the N2 molecule, (b) the KF molecule, (c) Ag atoms in a

solid?

(a) N2 - covalent bonding. (b) KF - ionic bonding. (c) Ag (solid) - metallic bonding.

6 ∙ Calculate the separation of Na+ and Cl- ions for which the potential energy is -1.52 eV.

Ue = -ke2/r; solve for r (see Problem 37-70) r = (1.44/1.52) nm = 0.947 nm

7 ∙ The dissociation energy of Cl2 is 2.48 eV. Consider the formation of an NaCl molecule according to the reaction

Na + 1/2Cl2 → NaCl

Does this reaction absorb energy or release energy? How much energy per molecule is absorbed or released?

Find the net energy change, ∆E. Note that the

dissociation energy per Cl atom is 1.24 eV.

∆E < 0, therefore energy is released

∆E = 1.24 eV - (Binding energy of NaCl - see p. 1205)

= (1.24 - 4.27) eV = -3.03 eV

Reaction is exothermic

8 ∙ The dissociation energy is sometimes expressed in kilocalories per mole (kcal/mol). (a) Find the relation between

the units eV/molecule and kcal/mol. (b) Find the dissociation energy of molecular NaCl in kcal/mol.

(a) 1 eV/molecule = (1 eV/molecule)(1 kcal/4184 J)(6.02×1023 molecules/mole)(1.6×10-19 J/eV) = 23.0 kcal/mol

(b) Dissociation energy of NaCl = (4.27×23) kcal/mol = 98.3 kcal/mol

9* ∙ The equilibrium separation of the HF molecule is 0.0917 nm and its measured electric dipole moment is 6.40×10-30

Chapter 38 Molecules and Solids

C⋅⋅⋅⋅m. What percentage of the bonding is ionic ?

1. Find the dipole moment for 100% ionic bonding

2. Percent ionic bonding = 100(pmeas/p100)p100 = er = 1.6×10-19 ×9.17×10-11 C.m = 1.47×10-29 C.m

Ionic bonding = 43.6%

10 ∙ Do Problem 9 for CsCl, for which the equilibrium separation is 0.291 nm and the measured electric dipole moment

is 3.48×10-29 C⋅⋅⋅⋅m.

1. Find the dipole moment for 100% ionic bonding

2. Percent ionic bonding = 100(pmeas/p100)p100 = 1.6×10-19 ×2.91×10-10 C.m = 4.656×10-29 C.m

Ionic bonding = 74.7%

11 ∙∙ The dissociation energy of LiCl is 4.86 nm and the equilibrium separation is 0.202 nm. The electron affinity of

chlorine is 3.62 eV, and the ionization energy of lithium is 5.39 eV. Determine the core-repulsion energy of LiCl.

1. Determine ∆E (see Example 38-1)

2. Determine Ue = -ke2/r (see Problem 37-70)

3. Find Urep = -(∆E + Ue + Ed)

∆E = (5.39 - 3.62) eV = 1.77 eV

Ue = -(1.44/0.202) eV = -7.13 eV

Urep = -(1.77 - 7.13 + 4.86) eV = 0.50 eV

12 ∙∙ The equilibrium separation of the K+ and Cl- ions in KCl is about 0.267 nm. (a) Calculate the potential energy of

attraction of the ions assuming them to be point charges at this separation. (b) The ionization energy of potassium is 4.34

eV and the electron affinity of Cl is 3.62 eV. Find the dissociation energy neglecting any energy of repulsion. (See

Figure 38-1.) The measured dissociation energy is 4.49 eV. What is the energy due to repulsion of the ions at the

equilibrium separation?

(a) Ue = -ke2/r; ke2 = 1.44 eV.nm

(b) 1. Ed,calc = -(Ue + ∆E)

2. Urep = Ed,calc - Ed,meas

Ue = -(1.44/0.267) eV = -5.39 eV

∆E = (4.34 - 3.62) eV = 0.72 eV; Ed,calc = 4.67 eV

Urep = (4.67 - 4.49) eV = 0.18 eV

13* ∙∙ Indicate the mean value of r for two vibration levels in the potential-energy curve for a diatomic molecule and

show that because of the asymmetry in the curve, rav increases with increasing vibration energy, and therefore

solids expand when heated.

We shall assume that U(r) is of the form given in Problem

16 with n = 6. The potential energy curve is shown in the

figure. The turning points for vibrations of energy E1 and

E2 are at the values of r where the energies equal U(r). It

is apparent that the average value of r depends on the

energy, and that r2,av is greater than r1,av.

14 ∙∙ Calculate the potential energy of attraction between the Na+ and Cl- ions at the equilibrium separation r0 = 0.236 nm

and compare this result with the dissociation energy given in Figure 38-1. What is the energy due to repulsion of the ions


at the equilibrium separation?

1. Ue = -ke2/r

2. Dissociation energy = 4.27 eV

3. ∆E = 1.52 eV (see Fig. 38.1); find Urep

Ue = -(1.44/0.236) eV = -6.10 eV

Urep + ∆E = (6.10 - 4.27) eV = 1.83 eV

Urep = 0.31 eV

15 ∙∙ The equilibrium separation of the K+ and F- ions in KF is about 0.217 nm. (a) Calculate the potential energy of

attraction of the ions, assuming them to be point charges at this separation. (b) The ionization energy of potassium is

4.34 eV and the electron affinity of F is 3.40 eV. Find the dissociation energy neglecting any energy of repulsion. (c)

The measured dissociation energy is 5.07 eV. Calculate the energy due to repulsion of the ions at the equilibrium

separation.

(a) Ue = -ke2/r

(b) Ed,calc = -(Ue + ∆E)

(c) Urep = Ed,calc - Ed,meas

Ue = -(1.44/0.217) eV = -6.64 eV

∆E = (4.34 - 3.40) eV = 0.94 eV; Ed,calc = 5.70 eV

Urep = (5.70 - 5.07) eV = 0.63 eV

16 ∙∙∙ Assume that the core repulsion can be represented by a potential energy of the form Urep = C/rn. So the total potential

energy is

Use the fact that dU/dr = 0 at r = r0, and the results for Urep at r = r0 from Problem 11, to calculate C and n.

With Urep = C/rn, the repulsive force is Frep = -nC/rn+1. The electrostatic force of attraction is Fe = ke2/r2. At r = r0, the net

force is zero. Consequently, nC/r0n = ke2/r0 = Ue(r0) = nUrep(r0), and n = Ue(r0)/Urep(r0) . Then the constant C =

Urep(r0) r0n.

From Problem 11, Ue(r0) = -7.13 eV and Urep(r0) = 0.5 eV; so n = 14 .26 and C = 6.21×10-11 eV.nm-14.26.

17*∙∙∙ (a) Find Urep at r = r0 for NaCl. (b) Assume Urep = C/rn and find C and n for NaCl. (See Problem 16.)

(a) 1. Find Ue = -ke2/r0

2. Find Urep = -(Ue + Ed + ∆E)

(b) Using the procedure indicated in Problem 16,

n = Ue/Urep and C = Urep r0n

Ue = -(1.44/0.236) eV = -6.10 eV

Urep = -(-6.10 + 4.27 + 1.52) eV = 0.31 eV

n = 19.7; C = 1.37×10-13 eV.nm-19.7

18 ∙ Find other elements with the same subshell electron configuration in the two outermost orbitals as carbon. Would

you expect the same type of hybridization for these elements as for carbon?

Elements similar to carbon in outer shell configurations are silicon, germanium, tin, and lead. We would expect the

same hybridization for these as for carbon, and this is indeed the case for silicon and germanium whose crystal

E+r

C+

rke_ = U

n

2

∆


structure is the diamond structure. Tin and lead, however, are metallic and here the metallic bond is dominant.

19 ∙ How does the effective force constant calculated for the CO molecule in Example 38-4 compare with the force

constant of an ordinary spring?

A stiff spring will have a force constant of about 2×103 N/m, about the same as the force constant obtained in

Example 38-4.

20 ∙ Explain why the moment of inertia of a diatomic molecule increases slightly with increasing angular momentum.

As the angular momentum increases, the separation between the nuclei also increases (the effective force between

the nuclei is similar to that of a stiff spring). Consequently, the moment of inertia also increases.

21* ∙ The characteristic rotational energy E0r for the rotation of the N2 molecule is 2.48×10-4 eV. From this find the

separation distance of the N atoms in N2.

1. From Equ. 38-13, I = h_ 2/2E0r

2. I = 2MN(r/2)2 = MN r2/2; MN = 14mp

I = (1.05×10-34 J.s)2/7.94×10-23 J = 1.39×10-46 J.s2

r = [(2×1.39×10-46)/(14×1.67×10-27)]1/2 m = 0.109 nm

22 ∙ The separation of the O atoms in O2 is actually slightly greater than the 0.1 nm used in Example 38-3, and the

characteristic energy of rotation E0r is 1.78×10-4 eV rather than the result obtained in that example. Use this value to

calculate the separation distance of the O atoms in O2.

I = MO r2/2 = h

_ 2/2E0r; r = h_/ E r0OM

r = 1.05×10-34/(16×1.67×10-27×2.85×10-23)1/2 m = 0.12 nm

23 ∙∙ Show that the reduced mass is smaller than either mass in a diatomic molecule, and calculate it for (a) H2, (b) N2, (c)

CO, and (d) HCl. Express your answers in unified mass units.

The reduced mass is given by µ = m1m2/(m1 + m2) = m1/(1 + m1/m2) = m2/(1 + m2/m1). In each of the latter

expressions the denominator is greater than 1. Consequently, µ < m1 and µ < m2; Q.E.D.

(a) Here, m1 = m2 = 1 u

(b) As in part (a), mr = m1/2

(c) Let m1 = 12 u, m2 = 16 u

(d) Let m1 = 1 u, m2 = 35.5 u

µ = 0.5 u

µ = 7 u

µ = 6.86 u

µ = 0.973 u

24 ∙∙ The equilibrium separation between the nuclei of the LiH molecule is 0.16 nm. Determine the energy separation

between the l = 3 and l = 2 rotational levels of this diatomic molecule.

1. Find µ and E0r = h_ 2/2I = h

_ 2/2µr2

Note: I = µr2 (see Equ. 38-14)

2. Use Equ. 38-12;∆E = 6E0r

µ = (6.94/7.94) u = 0.874 u = 1.45×10-27 kg;

E0r = (1.05×10-34)2/[2.9×10-27(1.6×10-10)2] J = 0.928 meV

∆E = 5.57 meV

25*∙∙ Repeat Problem 24 for LiD, where D is the symbol for deuterium. Note that replacing the proton by the

deuteron does not change the equilibrium separation between the nuclei of the molecule.

1. Find µ and E0r = h_ 2/2I = h

_ 2/2µr2

Note: I = µr2 (see Equ. 38-14)

µ = (2×6.94/8.94) u = 1.55 u = 2.58×10-27 kg;

E0r = (1.05×10-34)2/[5.16×10-27(1.6×10-10)2] J


2. Use Equ. 38-12;∆E = 6E0r

= 0.522 meV

∆E = 3.13 meV

26 ∙∙ Derive Equations 38-14 and 38-15 for the moment of inertia in terms of the reduced mass of a diatomic molecule.

Let the origin of coordinates be at the point mass m1 and point mass m2 be at a distance r0 from the origin. Then the

center of mass is at rCM = m2r0/(m1 + m2). The system rotates about the CM and the distances of m1 and m2 from the

CM are r1 = rCM and r2 = r0 - rCM = m1r0(m1 + m2). The moment of inertia about the CM is I = m1r12 + m2r2

2 =

m1m2r02/(m1 + m2) = µr0

2; Q.E.D.

27 ∙∙ Use the separation of the K+ and Cl- ions given in Problem 12 and the reduced mass of KCl to calculate the

characteristic rotational energy E0r .

Find µ and E0r = h_ 2/2I = h

_ 2/2µr2

Note: I = µr2 ( see Equ. 38-14)

µ = (39.1×35.5/74.6) u = 18.6 u = 3.09×10-26 kg;

E0r = (1.05×10-34)2/[6.18×10-26(2.67×10-10)2] J

= 2.5×10-24 J = 0.0156 meV

28 ∙∙ The central frequency for the absorption band of HCl shown in Figure 38-17 is at f = 8.66×1013 Hz, and the absorption

peaks are separated by about ∆f = 6×1011 Hz. Use this information, to find (a) the lowest (zero-point) vibrational energy

for HCl, (b) the moment of inertia of HCl, and (c) the equilibrium separation of the atoms.

(a) From Equ. 38-18, E0 = hf/2

(b) For ∆ ! = ±1, ∆El = ! h_ 2/I = ! h∆f; I = h/4π2∆f

(c) I = µr2; µ = mH mCl /(mH + mCl ); solve for r

E0 = (6.63×10-34 ×8.66×1013/2×1.6×10-19) eV = 0.179 eV

I = [6.63×10-34/(4π2×6×1011)] kg.m2 = 2.80×10-47 kg.m2

µ = 0.973 u = 1.62×10-27 kg; r = 0.132 nm

29*∙∙ Calculate the effective force constant for HCl from its reduced mass and the fundamental vibrational frequency

obtained from Figure 38-17.

From Equ. 38-19, K = 4π 2µf 2; µ = 0.973 u K = [4π 2×0.973×1.66×10-27 (8.66×1013)2] N/m = 480 N/m

30 ∙∙ Two objects of mass m1 and m2 are attached to a spring of force constant K and equilibrium length r0 . (a) Show that

when m1 is moved a distance ∆r1 from the center of mass, the force exerted by the

spring is

r)m

m +mK( = F 12

21 ∆−

(b) Show that the frequency of oscillation is f = (1/2π) µK/ 0 where µ is the reduced mass.

(a) For a two-mass and spring system on which no external forces are acting, the center of mass must remain fixed.

Consequently, if m1 moves a distance ∆r1 from (or toward) the CM, then m2 must move a distance ∆r2 = ∆r1(m1/m2)

from (or toward) the CM. Then the force exerted by the spring is F = -K∆r = -K(∆r1 + ∆r2) = -K∆r1(m1 + m2)/m2.

(b) From (a) we see that a displacement ∆r1 of m1 results in restoring force F = -K∆r1(m1 + m2)/m2 = -Keff ∆r1,

where Keff = K(m1 + m2)/m2. Writing ω = mK 1eff / 0 and recalling that µ = m1m2/(m1 + m2 ) we obtain ω = µK/ 0.


Note: In the first printing of the textbook the problem statement for part (b) reads, “Show that the angular

frequency of oscillation…” The adjective “angular” should have been omitted.

31 ∙∙∙ Calculate the reduced mass for the H35Cl and H37Cl molecules and the fractional difference ∆µ/µ. Show that the

mixture of isotopes in HCl leads to a fractional difference in the frequency of a transition from one rotational state to

another given by ∆f/f = ∆µ/µ. Compute ∆f/f and compare your result with Figure 38-17.

We shall express the results in unified mass units. For H35Cl, µ = 35/36 u = 0.9722 U; for H37Cl, µ = 37/38 u = 0.9737 u.

The quantity ∆µ/µ = (36×37 - 38×35)/(38×36) = 0.00146. The rotational frequency is proportional to 1/I, where I is the

moment of inertia of the system. Since I is proportional to µ, f = C/µ and df/f = -dµ/µ, or ∆f/f = -∆µ/µ. For the HCl system,

∆f/f = -0.00146. From Figure 38-17, ∆f is about 0.01×1013 = 1011 Hz, where f = 8.4×1013 Hz. Thus ∆f/f ≅ 0.0012, in fair

agreement with the calculated result. (Note that ∆f is difficult to determine precisely from Figure 38-17.)

32 ∙∙∙ In calculating the rotational energy levels of a diatomic molecule, we did not consider rotation of the molecule about

the line joining the atoms. (a) Estimate the moment of inertia of the H2 molecule about this line. (b) Use your results of

(a) to estimate the typical rotational energy E0r for rotation about the line joining the atoms. (c) Compare your answer in

(b) with the typical thermal energy kT at T = 300 K.

(a) Assume protons are point particles and the electrons are at a distance of a0 from the axis.

I = 2me a02

(b) Use Equ. 38-13

(c) at T = 300 K, kT 0.025 eV; E0r >> kT

I = 5.1×10-51 kg.m2

E0r = 1.08×10-18 J = 6.8 eV

Rotation about the axis cannot be excited thermally at

temperatures below the dissociation temperature.

33* ∙ Suppose that hard spheres of radius R are located at the corners of a unit cell with a simple cubic structure. (a)

If the hard spheres touch so as to take up the minimum volume possible, what is the size of the unit cell? (b) What

fraction of the volume of the cubic structure is occupied by the hard spheres?

(a) The cube has a length a = 2R

(b) “Packing fraction” = Vsphere /Vunit cell

The unit cell is the cube; its volume is 8R 3

Fraction = [(4πR 3/3)/(8R 3)] = 0.5236

34 ∙ Calculate the distance r0 between the K + and the Cl - ions in KCl, assuming that each ion occupies a cubic volume

of side r0. The molar mass of KCl is 74.55 g/mol and its density is 1.984 g/cm3.

1. Find the volume/mole of KCl

2. Vmol = 2NA r03 since cube length/ion = r0

Vmol = MMMM /ρ = (74.55/1.984) cm3 = 37.58 cm3

r0 = (37.58/12.04×1023)1/3 cm = 0.315 nm

35 ∙ The distance between the Li + and Cl - ions in LiCl is 0.257 nm. Use this and the molecular mass of LiCl (42.4

g/mol) to compute the density of LiCl.

1. Find the volume of the unit cell

2. Unit cell has 4 molecules; find mass of unit cell

3. Find the density of solid; ρ = Munit cell /Vunit cell

Vunit cell = 8(0.257×10-7 cm)3 = 1.358×10-22 cm3

Munit cell = 4(42.4/6.02×1023) = 2.82×10-22 g

ρ = 2.075 g/cm3

≅


36 ∙ Find the value of n in Equation 38-32 that gives the measured dissociation energy of 741 kJ/mol for LiCl, which has

the same structure as NaCl and for which r0 = 0.257 nm.

1. Use Equs. 38-32 and 38-33 to write U(r0)2. Solve for n

7.68 = 1.7476(1.44/0.257)(1 - 1/n)

n = 4.64

37*∙∙ Suppose identical bowling balls of radius R are packed into a hexagonal close-packed structure. What fraction

of the available volume of the unit cell is filled by the bowling balls?

In the adjacent drawing we show the base layer of six spheres (solid circles) and

the next layer of spheres (dashed circles) placed in the intersteces. From the

drawing (see also Fig. 38-20) we see that the three spheres of the lower layer and

one of the spheres in the next layer form a regular tetrahedron. The next layer

above the spheres (shown dashed) overlaps the hexagonal layer (shown solid). It

follows that the distance between the layer of solid spheres and that of the dashed

spheres is a 2/3 , where a = 2R. The unit cell is a cylinder whose base is the

hexagon formed by the centers of the solid spheres and whose height is the

distance between two overlapping layers. The base of the unit cell has an area of

6(a2 3 /4) = 3a2 3 /2 and the height of the unit cell is 2a 2/3 . Thus the unit

cell volume is 3a3 2 . Each of the six solid spheres on the hexagon is shared by 3

adjacent unit cells so that these six spheres contribute 2 spheres to the unit cell. In

addition there is the sphere at the center of the base and the three spheres in the

“dashed” layer. Thus the unit cell contains a total of six spheres whose total

volume is 6[(4π/3)(a3/8)] = πa3. The volume occupied by the spheres is therefore

π/(3 2 ) of the unit cell volume or 0.74 of the unit cell volume.

38 ∙ Which of the following elements are most likely to serve as acceptor impurities in germanium? (a) Bromine, (b)

Gallium, (c) Silicon, (d) Phosphorus, (e) Magnesium.

(b)

39 ∙ Which of the following elements are most likely to serve as donor impurities in germanium? (a) Bromine, (b)

Gallium, (c) Silicon, (d) Phosphorus, (e) Magnesium.

(d)

40 ∙ What type of semiconductor is obtained if silicon is doped with (a) aluminum and (b) phosphorus? (See Table 37-1

for the electron configurations of these elements.)

(a) p- type (b) n-type

41* ∙ What type of semiconductor is obtained if silicon is doped with (a) indium and (b) antimony? (See Table 37-1

for the electron configurations of these elements.)

(a) p-type (b) n-type

42 ∙ The donor energy levels in an n-type semiconductor are 0.01 eV below the conduction band. Find the temperature


for which kT = 0.01 eV.

k = 8.62×10-5 eV/K; T = ∆E/k T = (10-2/8.62×10-5) K = 116 K

43 ∙∙ The relative binding of the extra electron in the arsenic atom that replaces an atom in silicon or germanium can be

understood from a calculation of the first Bohr orbit of this electron in these materials. Four of arsenic's outer electrons

form covalent bonds, so the fifth electron sees a singly charged center of attraction. This model is a modified hydrogen

atom. In the Bohr model of the hydrogen atom, the electron moves in free space at a radius a0 given by

When an electron moves in a crystal, we can

approximate the effect of the other atoms by

replacing ε0 with κε0 and me with an effective

mass for the electron. For silicon κ is 12 and the

effective mass is about 0.2me. For germanium κ is 16 and the effective mass is about 0.1me. Estimate the Bohr radii for

the outer electron as it orbits the impurity arsenic atom in silicon and germanium.

Here aB = (κme/meff)a0 Si: aB = 31.7 nm; Ge: aB = 84.6 nm

44 ∙∙ The ground-state energy of the hydrogen atom is given by

h2emk = E

42

1 220

e4

2 8me = ε

−−"

Modify this equation in the spirit of Problem 43 by replacing ε0 by κε0 and me by an effective mass for the electron to

estimate the binding energy of the extra electron of an impurity arsenic atom in (a) silicon and (b) germanium.

Here E1 = (meff /meκ2)E0 (a) E1 = (0.2/144)×13.6 eV = 19 meV

(b) E1 = (0.1/256)×13.6 eV = 5.3 meV

45*∙∙ A doped n-type silicon sample with 1016 electrons per cubic centimeter in the conduction band has a resistivity of

5×10-3 Ω⋅⋅⋅⋅m at 300 K. Find the mean free path of the electrons. Use the effective mass of 0.2me for the mass of the

electrons. (See Problem 43.) Compare this mean free path with that of conduction electrons in copper at 300 K.

1. Use Equ. 27-8 to find vav

2. Use Equ. 27-7; λ = en

m2

e

aveff v

ρ

3. uF = 1.57×106 m/s (Example 27-4);

ρ = 1.7×10-8 Ω.m; ne = 8.47×1028 m-3

vav = 102.62 = m/s 5××

×−

−

)100.2(9.11

)(300)103(1.3831

23

m/s

λ = m ) 10)(1.610)(10(5

)10)(2.62110.2(9.219223

531

−−

−

×××

= 37.3 nm

λCu = m ) 10)(1.610)(8.4710(1.7

)10)(1.5710(9.11219288

631

−−

−

×××××

0 = 38.8 nm

em

h = a2

00 2

eπε


The mean free paths are about the same

46 ∙∙ The measured Hall coefficient of a doped silicon sample is 0.04 V⋅⋅⋅⋅m/A⋅⋅⋅⋅T at room temperature. If all the doping

impurities have contributed to the total charge carriers of the sample find (a) the type of impurity (donor or acceptor)

used to dope the sample and (b) the concentration of these impurities.

(a), (b) R = 1/nq (see Problem 28-70) R > 0, so q > 0, i.e., hole conduction; acceptor impurities

n = 1/Re = 15.6×1020 m-3

47 ∙ When a pnp junction transistor is used as an amplifier, a small signal in the __________ current results in a large

signal in the __________ current.

(a) collector; base (b) base; collector (c) emitter; base (d) emitter; collector (e) collector; base

(b)

48 ∙ When light strikes the p-type semiconductor in a pn junction solar cell,

(a) only free electrons are created.

(b) only positive holes are created.

(c) both electrons and holes are created.

(d) positive protons are created.

(e) none of these is correct.

(c)

49*∙∙ Simple theory for the current versus the bias voltage across a pn junction yields the equation

1)e(I = I /kTeV0

b −

Sketch I versus Vb for both positive and negative values of Vb using this equation.

The sketch is shown.

Here we plot I/I0 versus eV/kT where we have used V to

represent the bias voltage Vb. Except for the region of

reverse breakdown, this curve is identical to Figure 38-

27.

50 ∙∙ For a temperature of 300 K, use the equation in Problem 49 to find the bias voltage Vb for which the exponential

term has the value (a) 10 and (b) 0.1.


(a), (b) Vb = (kT/e) ln(value); k/e = 8.62×10-5 V/K (a) Vb = 0.0595 V (b) Vb = -0.0595 V

51 ∙∙ In Figure 38-34 for the pnp-transistor amplifier, suppose Rb = 2 kΩ and RL = 10 kΩ. Suppose further that a 10-

µA ac base current generates a 0.5-mA ac collector current. What is the voltage gain of the amplifier?

Voltage gain = icRL /ibRb Voltage gain = 5/0.02 = 250

52 ∙∙ Germanium can be used to measure the energy of incident particles. Consider a 660-keV gamma ray emitted

from 137Cs. (a) Given that the band gap in germanium is 0.72 eV, how many electron–hole pairs can be generated as

this gamma ray travels through germanium? (b) The number of pairs N in part (a) will have statistical fluctuations

given by ± N . What then is the energy resolution of this detector in this photon energy region?

(a) Number of pairs = E/Eg = N

(b) Resolution = ∆E/E = ∆N/N

N = 9.17×105 electron-hole pairs

Resolution = 1/(9.17×105)1/2 = 1.04×10-3 ≅ 0.1%

53*∙∙ Make a sketch showing the valence and conduction band edges and Fermi energy of a pn-junction diode when

biased (a) in the forward direction and (b) in the reverse direction.

The sketch of the pn junction biased in the (a) forward

and (b) reverse directions is shown below.

The nearly full valence band is shown shaded.

The Fermi level is shown by the dashed line.

54 ∙∙ A "good" silicon diode has the current–voltage characteristic given in Problem 49. Let kT = 0.025 eV (room

temperature) and the saturation current I0 = 1 nA. (a) Show that for small reverse-bias voltages, the resistance is 25

MΩ. Hint: Do a Taylor expansion of the exponential function, or use your calculator and enter small values for Vb.

(b) Find the dc resistance for a reverse bias of 0.5 V. (c) Find the dc resistance for a 0.5-V forward bias. What is the

current in this case? (d) Calculate the ac resistance dV/dI for a 0.5-V forward bias.

(a) ex - 1 = x for x << 1; I ≅ I0(eVb /kT)

(b) Use Problem 49 expression

(c) Repeat (b) with Vb = +0.5 V

(d) dV/dI = (dI/dV)-1 = [(eI0/kT)eeV/kT]-1

R = Vb /I = kT/eI0 = (0.025/10-9 ) Ω = 25 MΩI = 10-9(e-20 - 1) A = -10-9 A; R = 0.5×109 Ω = 500 MΩI = 10-9(e20 - 1) A = 0.485 A; R = (0.5/0.485) Ω = 1.03 ΩRac = (25 MΩ)e-20 = 0.0515 Ω

55 ∙∙ A slab of silicon of thickness t = 1.0 mm and width w = 1.0 cm is placed in a magnetic field B = 0.4 T. The slab is in

the xy plane, the magnetic field points in the positive z direction. When a current of 0.2 A flows through the sample in

the positive x direction, a voltage difference of 5 mV develops across the width of the sample with the electric field in


the sample pointing in the positive y direction. Determine the semiconductor type (n or p) and the concentration of

charge carriers.

1. Use Equ. 28-19: n = IB/teVH

2. See Fig. 28-28; now B points out of the paper

n = [(0.2×0.4)/(10-3×1.6×10-19×5×10-3 )] m-3 = 1023 m-3

E is in the y direction, therefore carriers are holes; p-type

56 ∙ Why would you expect the separation distance between the two protons to be larger in the H2+ ion than in the H2

molecule?

For H2, the concentration of negative charge between the two protons holds the protons together. In the H2+ ion,

there is only one electron that is shared by the two positive charges such that most of the electronic charge is again

between the two protons. However, the negative charge in the H2+ ion is not as effective as the larger charge in the

H2 molecule, and the protons should be farther apart. The experimental values support our argument. For H2, r0 =

0.074 nm, while for H2+, r0 = 0.106 nm.

Note: In the first printing of the textbook the problem statement reads, “Why … to be smaller in …?” There is no

correct answer to this question since the exact opposite is true.

57* ∙ What kind of bonding mechanism would you expect for (a) the HCl molecule (b) the O2 molecule, (c) Cu atoms

in a solid?

(a) Ionic bonding (b) Covalent bonding (c) Metallic bonding

58 ∙ Why does an atom usually absorb radiation only from the ground state whereas a diatomic molecule can absorb

radiation from many different rotational states?

The energy of the first excited state of an atom is orders of magnitude greater than kT at ordinary temperatures.

Consequently, practically all atoms are in the ground state. By contrast, the energy separation between the ground

rotational state and nearby higher rotational states is less than or roughly equal to kT at ordinary temperatures, and

so these higher states are thermally excited and occupied.

59 ∙ The equilibrium separation of CsF is 0.2345 nm. If its bonding is 70% ionic, what is its electric dipole moment.

1. p100 = er0; pexp = 0.70p100 pexp = (0.70×1.6×2.345×10-29) C.m = 2.63×10-29 C.m

60 ∙ Show that when one atom in a diatomic molecule is much more massive than the other the reduced mass is

approximately equal to the mass of the lighter atom.

From Problem 23, µ = m1m2/(m1 + m2) = m1/(1 + m1/m2). If m2 >> m1, then m1/m2 << 1 and so µ ≅ m1.

61* ∙∙ The equilibrium separation between the nuclei of the CO molecule is 0.113 nm. Determine the energy

difference between the l = 2 and l = 1 rotational energy levels of this molecule.

1. Find µ and I = µr02

2. ∆E1,2 = 2h_ 2/I

µ = (16×12/28) u = 6.86 u; I = 1.45×10-46 kg.m2

∆E1,2 = 1.52×10-22 J = 0.948 meV

62 ∙∙ When a thin slab of semiconducting material is illuminated with monochromatic light most of the light is

transmitted through the slab if the wavelength is greater than 1.85 µm. For wavelengths less than 1.85 µm, most of

the incident light is absorbed. Determine the energy gap of this semiconductor.


Eg = hc/λ Eg = (1240/1850) eV = 0.67 eV

63 ∙∙ Show that when an intrinsic semiconductor carries a current in a transverse magnetic field no Hall voltage is

developed across the sample.

In an intrinsic semiconductor, the electron and hole concentrations are equal. Each group of charge carriers

contributes the same Hall voltage, but they are in opposite directions. Consequently, the net voltage is zero.

64 ∙∙ The semiconducting compound CdSe is widely used for light emitting diodes (LEDs). The energy gap in CdSe

is 1.8 eV. What is the frequency of the light emitted by a CdSe LED?

λ = hc/Eg λ = (1240/1.8) nm = 689 nm (red light)

65* ∙∙ The resistivity of a sample of pure silicon diminishes drastically when it is irradiated with infrared light of

wavelength less than 1.13 µm. What is the energy gap of silicon?

Eg = hc/λ Eg = (1240/1130) eV = 1.10 eV

66 ∙∙ The effective force constant for the HF molecule is 970 N/m. Find the frequency of vibration for this molecule.

Note: We shall assume that the force constant is 970 N/m. (In Problem 30 Keff was defined in terms of the

masses.)

From Problem 30, f = (1/2π) µK/ 0 µ = 0.95 u = 1.58×10-27 kg; f = 1.25×1014 Hz

67 ∙∙ The frequency of vibration of the NO molecule is 5.63 × 1013 Hz. Find the effective force constant for NO.

From Problem 30, K = 4π 2f 2µ µ = 7.47 u = 1.24×10-26 kg; K = 1550 N/m

68 ∙∙ The force constant of the hydrogen bond in the H2 molecule is 580 N/m. Obtain the energies of the four lowest

vibrational levels of H2, HD, and D2 molecules, and the wavelengths of photons resulting from transitions between

adjacent vibrational levels of these molecules.

1. Find µ for H2, HD, and D2; µ = m1m2/(m1 + m2)

2. Use Equs. 38-18 and 38-19; Eν = (ν + 1/2)h_ µK/

µ(H2) = 0.837×10-27 kg; µ(HD) = 1.12×10-27 kg;

µ(D2) = 1.674×10-27 kg

For the energies, in eV, see the tabulation below.

ν H2 HD D2

0 0.272 eV 0.235 eV 0.192 eV

1 0.814 eV 0.706 eV 0.576 eV

2 1.358 eV 1.176 eV 0.960 eV

3 1.902 eV 1.646 eV 1.344 eV

3. λ = hc/∆E = (1240 eV.nm)/(∆E eV) λ(H2) = 2.28 µm; λ(HD) = 2.64 µm; λ(D2) = 3.23 µm

69* ∙∙ The potential energy between two atoms in a molecule can often be described rather well by the Lenard-Jones

potential, which can be written


−

r

a2

r

a U = U

612

0

where U0 and a are constants. Find the interatomic separation r0 in terms of a for which the potential energy is a

minimum. Find the corresponding value of Umin. Use Figure 38-4 to obtain numerical values of r0 and U0 for the H2

molecule and express your answers in nanometers and electron volts.

1. Set dU/dr = 0 and solve for r = r0

2. From Figure 38-4, r0 = 0.074 nm, Umin = -U0

a = r 0; = r

a12

r

a12

rU =

dr

dU0

6120

−

− 0

a = 0.074 nm; Umin = -4.52 eV; U0 = 4.52 eV

70 ∙∙ In this problem you are to find how the van der Waals force between a polar and a nonpolar molecule depends on

the distance between the molecules. Let the dipole moment of the polar molecule be in the x direction and the nonpolar

molecule be a distance x away.(a) How does the electric field due to an electric dipole depend on distance x ? (b) Use the

fact that the potential energy of an electric dipole of moment p in an electric field E is U = -p ⋅⋅⋅⋅ E, and that the induced

dipole moment of the nonpolar molecule is proportional to E to find how the potential energy of interaction of the two

molecules depends on separation distance. (c) Using Fx = - dU/dx find the x dependence of the force between the two

molecules.

(a) From Equ. 22-10, E is proportional to 1/x3.

(b) The induced dipole moment is proportional to the field that induces the moment. Hence, p ∝ 1/x3 and the

potential energy of interaction U = -p .E ∝ 1/x6.

(c) Fx = -dU/dx ∝ 1/x7.

71 ∙∙ Find the dependence of the force on separation distance between two polar molecules. (See Problem 70.)

In the case of two polar molecules, p does not depend on the field E. Consequently, U ∝ 1/x3 and Fx ∝ 1/x4.

72 ∙∙ Use the infrared absorption spectrum of HCl in Figure 38-17 to obtain (a) the characteristic rotational energy E0r (in

eV) (b) the vibrational frequency f and the vibrational energy hf (in eV).

(a) See Problem 28b and Equs. 38-12 and 38-13

(b) See Problem 28a; hf = 2E0

E0r = h∆f/2 = (4.136×10-15 eV.s)(3×1011 Hz) = 1.24 meV

hf = 2×0.179 eV = 0.358 eV

73* ∙∙ For a molecule such as CO, which has a permanent electric dipole moment, radiative transitions obeying the

selection rule ∆ ! = ±1 between two rotational energy levels of the same vibrational level are allowed. (That is, the

selection rule ∆ν = ±1 does not hold.) (a) Find the moment of inertia of CO and calculate the characteristic

rotational energy E0r (in eV). (b) Make an energy level diagram for the rotational levels for ! = 0 to ! = 5 for some

vibrational level. Label the energies in electron volts starting with E = 0 for ! = 0. (c) Indicate on your diagram

transitions that obey ∆ ! = -1 and calculate the energy of the photon emitted (d) Find the wavelength of the

photons emitted for each transition in (c). In what region of the electromagnetic spectrum are these photons?


ω"

(a) 1. Find µ and I; µ = m1m2/(m1 + m2); I = µr02

Note: r0 = 0.113 nm; see Problem 61

2. E0r = h_ 2/2I

µ = 6.86 u; I = (6.86×1.66×10-27)(0.113×10-9)2 kg.m2

= 1.45×10-46 kg.m2

E0r = [(1.05×10-34)2/2.9×10-46] J = 0.238 meV

(b), (c) The energy level diagram is shown in the

adjacent figure. Note that ∆El,l -1, the energy

difference between adjacent levels for ∆ ! = -1,

is ∆E 1, −!! = 2 ! E0r.

(d) λl,l -1 = 620/l E0r; λ in nm and E0r in eV. We find:

λ1,0 = 2605 µm; λ2,1 = 1303 µm; λ3,2 = 868 µm;

λ4,3 = 651 µm; λ5,4 = 521 µm

These wavelengths fall in the microwave region

of the spectrum.

74 ∙∙∙ Use the results of Problem 16 to calculate the vibrational frequency of the LiCl molecule. To do this, expand the

potential about r = r0 where r0 is the equilibrium separation, in a Taylor series. Retain only the term proportional to

(r- r0)2. Recall that the potential energy of a simple harmonic oscillator is given by V = 1/2mω2x2. What is the

wavelength resulting from transitions between adjacent harmonic oscillator levels of this molecule?

From Problem 16 we have U(r) = -(ke2/r) + C/rn, neglecting the constant term ∆E. Also, nUrep(r0) = ke2/r0 = nC/r0n. The

constant C = Urep(r0)r0n. So, C/rn = ke2r0

n-1/nrn. We now let r = r0 + δ, where δ << r0. Using the binomial expansion one

can write rn = r0n + nδr0

n-1 + n(n - 1)δ 2r0n-2/2 + . . .. Then C/rn = ke2/[nr0(1 + nδ/r0 + n(n - 1)δ2/2r0

2)] for r = r0 + δ. Next,

using (1 + ε)-1 = 1 - ε + ε2 + . . . and keeping only terms of order (δ/r0)2, we can write -(ke2/r) = -(ke2/r0)[1 - δ/r0 + (δ/r0)

2]

and C/rn = (ke2/nr0)[1 - nδ/r0 + n(n + 1)δ2/2r02]. To lowest order in δ/r0 the potential energy is U = U0 + (ke2/2r0

3)[(n +

1)δ2]. We now set (ke2/2r03)[(n + 1)δ2] = 1/2mω 2δ 2 and solve for ω. We obtain ω = r /me k 1) + (n 3

02 .

For the specific case of LiCl, we must use the reduced mass µ = 6.94×35.5/42.44 u = 5.81 u for m, r0 = 0.202 nm,

and set n = 14.26 The result is f = 2.11×1014 s-1 and ∆E = = 0.139 eV. Thus λ = 1240/0.139 nm = 8.93 µm.

75 ∙∙∙ Repeat Problem 74 for the NaCl molecule.


ω"

We use the result of the preceding problem; ω = r /me k 1) + (n 30

2 .

For NaCl we must use the reduced mass µ = 23×35.5/58.5 u = 13.96 u for m, r0 = 0.236 nm, and n = 19.7. We then

find that f = 1.25×1014 s-1 and ∆E = = 0.0826 eV. Thus λ = 1240/0.0821 nm = 15.1 µm

76 ∙∙∙ A one-dimensional model of an ionic crystal consists of a line of alternating positive and negative ions with distance

r0 between each ion. (a) Show that the potential energy of attraction of one ion in the line is

−−−− ... + +

r

ke2 = V

0

2

5

1

4

1

3

1

2

11

(b) Using the result that

... + 4x

3x +

2x x = x) + (

432

−−1ln

show that the Madelung constant for this one-dimensional model is α = 2 ln 2 = 1.386.

(a) The figure shows the one dimensional model.

Here the open and shaded circles represent ions of

charges +e and -e, respectively. The separation

between adjacent ions is r0. We can now write the

potential energy of the ion at position 0.

−−−−−− ## + +

r

ke = + r

ke r

ke + r

keV = 5

1

4

1

3

1

2

11

2

3

2

2

22

0

2

0

2

0

2

0

2

0.

(b) Note that when x = 1, the expansion for ln(1 + x) is just equal to the series in the parenthesis above. Therefore,

we can write V = -(ke2/r0)(2 ln 2). From the definition of the Madelung constant, Equ. 38-27, α = 2 ln 2 = 1.386.

Ch38F

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