SOLUTIONS MANUAL for Dopants and Defects in Semiconductors 2nd Edition by Matthew McCluskey, Eugene

Dopants and Defects in Semiconductors Second Edition

Matthew D. McCluskey and Eugene E. Haller

Solutions Manual

*Corrections or comments may be sent to mattmcc@alum.mit.edu 1

Chapter 1 1.1 Silicon density = 2.33 g/cm3.

1 mole = 28 g (2.33 g/cm3)(1 mole / 28 g)(6.02×1023 / mole) = 5×1022 cm-3 8/a3 = 5×1022 cm-3 a = 5.5×10-8 cm = 5.5 Å 2

d  a 2 42  a 2 4  a 2 42 

3 a = 2.39 Å 4

1.2 Germanium a = 5.66 Å = 5.66×10-8 cm.

Density = 8/a3 = 4.4×1022 cm-3

1.3 Strontium titanate (SrTiO3) has the perovskite structure. (a) Unit cube with the atoms:

(b) Each Ti atom is bonded to 6 O atoms. (c) Molecular mass = 183.49u a = 3.9 Å Number density = 1/a3 = 1.69×1022 cm-3 (1.69×1022 / cm3)(183.49 g / mole)(1 mole / 6.02×1023) = 5.14 g/cm3 1.4 Plot E versus kx along the [100] direction, up to a maximum energy of 52π2/2ma2.

2

2 2 2  2  2   2   2   E nx    k y  ny    kz  nz    k x  2m  a a a      

1. nx = ny = nz = 0, E 

2 2 kx 2m

2. nx = 0, ny or nz = ±1, E 

3. nx = ±1, ny = nz = 0,

E

2 2  22 kx  4 2m 2ma 2 2  2  kx   2m  a 

2

1.5 An intrinsic semiconductor has no defects, or few enough to not affect its properties significantly. An extrinsic semiconductor’s properties are determined by defects.

1 1  4.0 1.6 f E   E  E  / kT = 0.018 F e 1 e 1

1.7 The effective mass is the largest for the heavy-hole band.

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1.8 Conduction band:

E  ak 2  b

dE ak  dk ak 2  b d 2E

a

dk

ak  b

 2

2



ak

ak  b

At k = 0, this simplifies to a

3/ 2

2

2ak 

b.

m*   2 b a If b = 0, m* = 0.

1.9 Diamond has a large band gap, which transmits visible light. Few electrons can be thermally excited to the conduction band. InSb has a band gap in the mid-IR, so it absorbs visible light.

Electrons and holes are present at room temperature.

1.10 Si is an indirect-gap semiconductor. Radiative recombination of an electron and hole requires a phonon, which decreases the probability. The efficiency is therefore low.

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Chapter 2

2.1 GaAs:CAs

Single acceptor

Si:Se

Double donor

Ge:Cu

Triple acceptor

GaAs:NAs

Isoelectronic

GaAs:CuGa

Double acceptor

Si:Zn

Double acceptor

ZnO:AlZn

Single donor

ZnS:CuZn

Single acceptor

GaAs:SiGa

Single donor

2.2 Because Zn is a double acceptor, the net acceptor concentration is 1×1017 – 7×1016 cm-3 = 3×1016 cm-3. It is p-type.

2.3 The fraction of occupied donors (neutral) is

1 1 = 0.731 f E   E  E  / kT  1.0 F 1 e 1 e The fraction of unoccupied donors (positive) is 0.269.

2.4 Nd = 3×1016 cm-3 Na = 8×1016 cm-3 (a) [Nd+] = [Na-] = 3×1016 cm-3 (b) p-type 2.5 Nd = 5×1016 cm-3 Na = 1×1016 cm-3 The donor binding energy is 40 meV. (a) [Nd+] = 1×1016 cm-3 (b) [Na-] = 1×1016 cm-3 (c) Fermi level is at the donor level, 40 meV below the conduction-band minimum.

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2.6 Compensation: Donors give electrons to acceptors. Example: SiGa donors compensate CAs acceptors in GaAs. Passivation: An impurity such as hydrogen forms a neutral complex with a donor/acceptor.

Example: Hydrogen passivates the boron acceptor in silicon.

2.7 (a) Carbon acceptor in GaAs, (b) zinc-hydrogen complex in GaAs:

2.8 Threefold rotations “1” axis: (1,2,3,4) → (1,4,2,3) → (1,3,4,2) “2” axis:

→ (3,2,4,1) → (4,2,1,3)

“3” axis:

→ (4,1,3,2) → (2,4,3,1)

“4” axis:

→ (2,3,1,4) → (3,1,2,4)

Twofold rotations Axis bisecting 1,4: → (4,3,2,1) Axis bisecting 1,3: → (3,4,1,2) Axis bisecting 1,2: → (2,1,4,3) Reflections 6

1-4 plane:

→ (1,3,2,4)

1-3 plane:

→ (1,4,3,2)

1-2 plane:

→ (1,2,4,3)

2-3 plane:

→ (4,2,3,1)

3-4 plane:

→ (2,1,3,4)

2-4 plane:

→ (3,2,1,4)

Improper rotations Axis bisecting 1,4: → (3,1,4,2) → (2,4,1,3) Axis bisecting 1,3: → (2,3,4,1) → (4,1,2,3) Axis bisecting 1,2: → (4,3,1,2) → (3,4,2,1)

2.9 Burgers circuit and Burgers vector:

2.10 Ratio of the lattice constants = 5.65/5.45 = 1.0367. (a) (The lattice constant difference has been exaggerated for clarity)

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(b) Over n lattice constants, the two materials are in registry.

5.65(n – 1) = 5.45n 0.2n = 5.65 n = 28 5.45n = 153 Å. There is 1 dislocation in this interval.

1 dislocation / (1.53×10-8 cm)2 = 4×1011 cm-2

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Chapter 3

3.1 Work function = 4.6 eV, electron affinity = 4.2 eV, band gap = 1.0 eV:

3.2 For p-type silicon, Vbi ≈ Eg/3 = 0.37 V; N = 1015 cm-3 = 1021 m-3; ε = 11.7.

From Equation (3.7), W 

2 0 Vbi = 7×10-7 m = 0.7 μm eN

3.3 Minority carriers are more important in p-n junctions.

3.4 Vbi ≈ 0.7 V; ε = 16, N = 7.7×1012 cm-3.

From Equation (3.7), W 

2 0 Vbi = 1.3×10-5 m = 13 μm eN

3.5 (a) I = (6.25×1012 s-1)(1.6×10-19 C) = 10-6 A V = IR = (10-6 A)(100 Ohm) = 10-4 V (b) Each photon has E = hc/λ = 4.9×10-19 J (4.9×10-19 J)(6.25×1012 s-1) = 3.1×10-6 W

3.6 (a) p-type MOSFET, (b) band diagram given the inversion condition:

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3.7 CCD array for an n-type semiconductor:

3.8 A laser diode has a Fabry-Perot cavity and waveguide (low-index) layers.

3.9 (a) no valence-band offset, (b) conduction-band and valence-band offsets are both 0.5 eV:

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3.10 P-type AlAs (Eg = 2.2 eV), n-type GaAs (Eg = 1.42 eV):

11

Chapter 4 4.1 Impurity concentration: c s x   k 0 c0 1  x k0 1 B:

c0 = 1016 cm-3, k0 = 17

cs(0.1) = 3.15×1016 cm-3 cs(0.8) = 1.11×106 cm-3

P:

c0 = 1016 cm-3, k0 = 0.08

cs(0.1) = 8.8×1014 cm-3 cs(0.8) = 3.5×1015 cm-3

4.2 Impurity concentration:

c s x   k 0 c0 1  x k0 1 0.1 = 0.08(1 – x)-0.92 (0.1/0.08)-1/0.92 = 1 – x 0.78 = 1 – x x = 0.22 m

4.3 Impurity concentration:

cs(x) = c0[1 – (1 – k0)exp(–k0x/l)] = 1014 cm-3 [1 + 4.5exp(-1.1x)]

cs(0) = 5.5×1014 cm-3 cs(1) = 1.3×1014 cm-3

4.4 Czochralski apparatus for growing InP crystals:

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4.5 FZ growth has no crucible, so there is minimal carbon and oxygen contamination. The high surface tension and low density of silicon makes FZ feasible.

4.6 Workhorse for III-V devices such as LEDs:

MOCVD

Economical way to grow pure Si on a doped Si substrate: CVD Gives the ultimate atomic-layer control:

MBE

A very cheap way to grow pure GaAs thin films:

LPE

4.7 The peaks in the RHEED oscillations occur during full coverage, while the valleys occur during ~50% coverage.

4.8 Answers vary, see text.

4.9 Ion implantation: Well-defined dose; non-equilibrium process can yield high concentrations.

Diffusion: Relatively simple and inexpensive.

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4.10 (1015 neutrons/cm2)(3.25×10-24 cm2) = 3.25×10-9 [70Ge] = 8/(5.7Å)3 = 4.32×1022 cm-3 (3.25×10-9) [70Ge] = 1.4×1014 cm-3. This is the concentration of Ga acceptors.

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Chapter 5 5.1 Ed = (13.6 eV)(0.2)/92 = 33.6 meV r0 = (0.53 Å)(92/0.2) = 24 Å

5.2 S state wavefunctions are nonzero at the central cell (r = 0), so they are perturbed by it. P state wavefunctions are zero at r = 0, so they experience no first-order perturbation.

5.3 1s → 2p transition: (a) ¾ (28 meV) = 21 meV (b) The 1s state is shifted downward by 7 meV. 21 meV + 7 meV = 28 meV 5.4 ϕi = Aexp[-(r – ri)2/b2] The distance from the vacancy to an atom is d  a 2 4  a 2 4  a 2 4  3a / 4 A1: ψ = 4A exp(-3a2/16b2)

|ψ|2 = 16|A|2 exp(-3a2/8b2)

T2: ψ = |ψ|2 = 0 5.5 A1: ψ ~ ½ (eikx + e-ikx + eiky + e-iky + eikz + e-ikz) = cos(kx) + cos(ky) + cos (kz) E:

ψ ~ ½ (eikx + e-ikx - eiky - e-iky) = cos(kx) – cos(ky) ψ ~ eikz + e-ikz – ½ (eikx + e-ikx + eiky + e-iky) = 2cos(kz) – cos(kx) – cos(ky)

T2: ψ ~ ½ (eikx – e-ikx) = sin(kx) ψ ~ ½ (eiky – e-iky) = sin(ky) ψ ~ ½ (eikz – e-ikz) = sin(kz) 5.6 Nd – Na = 0.99×1016 cm-3 ≈ 1016 cm-3 Na = 1014 cm-3

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5.7 Na – Nd ≈ 2×1016 cm-3 Nd = 1013 cm-3

Eg = 1.4 eV Ea = 0.031 eV

log(e-E/2kT) = -(E/2kT)log(e) = -0.434 (E/2kT) 1 eV/k = 1160 K, so if E is expressed in eV, log(e-E/2kT) = -0.5 (E/2)(1000/T) Slope of intrinsic region = -0.5(0.7) = -0.35 Slope of half-slope region = -0.5(0.0155) = -0.008 Slope of full slope region = -0.016

5.8 Donor binding energy = 50 meV (a) The Fermi level is at the donor level (50 meV below the CBM, or 1.95 eV above the VBM). (b) The carrier mobility decreases when the sample is warmed due to phonon scattering.

5.9 Phonon scattering:

μ = AT-3/2 16

100 = A(324-3/2) A = 583,200 Ionized impurity scattering: μ = B′T3/2 270 = B′93/2 B′ = 10 At 100 K, 1/μ = 1003/2/A + 100-3/2/B′ μ = 551 cm2/Vs

5.10 B, because its Bohr radius is larger than Cu.

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Chapter 6 6.1 The interaction between two atoms is modeled by U(R) = –2a2R2 + b2R4. (a) Sketch of U(R):

(b) dU/dR = -4a2R + 4b2R3 = 0 a2 = b2R2 R = a/b (c) C = d2U/dR2 = -4a2 + 12b2R2 = -4a2 + 12b2(a2/b2) = 8a2 6.2 Since K = 0, we can omit the eiKsa terms and s subscripts. Similar to Equation (C.1), we have Mu  3C v  u   5C v  u  Mv  5C u  v   3C u  v 

Replacing the second derivatives with –ω2 yields

8C  M u  8Cv  0  8Cu  8C  M v  0 2

2

Setting the determinant (Equation B.6) equal to zero,

8C  M   64C  0 2 2

2

This is solved by 8C – Mω2 = ±8C ω2 = 0 or 16C/M

  0 or 4 C / M 18

6.3 TO  520  4C M For Si, M = 28 4C  2752

LVM  2C 10  4C 20 = 615 cm-1

6.4 Bi should not have a local vibrational mode because it is heavier than As. Its vibrational mode lies in the phonon band. (GaAs has no forbidden gap).

6.5 14N-H fundamental = 3151 cm-1 14

N-H overtone = 6133 cm-1

(a) For 15N-H,   3151

1 / 1  1 / 15 = 3144 cm-1 1 / 1  1 / 14

(b) ω1 – ω0 = ωe – 2ωexe = 3151 or 2ωe – 4ωexe = 6302 ω2 – ω0 = 2ωe – 6ωexe = 6133 Subtracting the last two equations yields 2ωexe = 169 Plugging this into the first equation, ωe = 3151 + 169 = 3320 cm-1 xe = 169/2ωe = 0.025

6.6 The peak shifts and broadens with increasing temperature, due to interactions with host atoms (phonons).

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6.7 F = qE = ma a = qE/m v = qEt/m (a) E = ½ mv2 = q2E2t2/2m (b) To find the amplitude A, set KE = PE. ½q2E2t2/m = ½CA2

A  qEt

mC

6.8 Equation (6.43): x = x0e–iωte–γt/2 Assuming x0 is real, taking the real part gives: x = x0cos(ωt)e–γt/2 v = dx/dt = x0[ωsin(ωt) + cos(ωt)(-γ/2)]e–γt/2 Since γ << ω0, we can neglect the second term and set ω = ω0: v = x0ω0sin(ω0t)e–γt/2 PE = ½Cx2 = ½Cx02cos2(ω0t)e–γt KE = ½mv2 = ½ mx02ω02sin2(ω0t)e–γt Substituting ω02 = C/m yields KE = ½Cx02sin2(ω0t)e–γt Total energy = PE + KE = ½Cx02e–γt

6.9 Td symmetry, nx = 1, ny = 0, nz = 1 excited vibrational state. (a) E = 2ω (b) ψ ~ xz exp(-r2/2) 20

The symmetry of this wavefunction is the same as xz. From Table 5.1, this is T2 symmetry. (c) ψ ~ xy exp(-r2/2) ψ ~ yz exp(-r2/2)

6.10 Not including the exponential functions, nx = 1, ny = 1

ψ11 ~ xy

nx = 2, ny = 0

ψ20 ~ 2x2 – 1

nx = 0, ny = 2

ψ02 ~ 2y2 – 1

Symmetry: E

ψ20 + ψ02 ~ (x2 + y2) – 1

Symmetry: A

ψ20 – ψ02 ~ x2 – y2

Symmetry: E

The E modes are degenerate.

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Chapter 7 7.1 Let the damping force be –γmv. The force due to the electric field is –|e|E. (a) F = –|e|E – γmv ma = –|e|E – γmv = 0 at steady state v = –|e|E/γm P = Force exerted by the electric field × Electron velocity = (–|e|E)(–|e|E/γm) = e2E2/γm (b) From Equation 7.23, the free-carrier absorption at high frequencies is ~ γ.

Mobility (μ) is inversely proportional to γ. (Recall v = μE) Low-frequency absorption is larger for high μ High-frequency absorption is larger for low μ 7.2 N = 3×1019 cm-3 = 3×1025 m-3, ɛ∞ = 5, m* = 0.2m. (a)  p 

Ne 2 0 m

*

= 3.1×1014 rad/s

(b) 1/λ = ω/(2πc) = 1637 cm-1 (c) Reflectivity versus frequency (cm-1):

7.3 From Equation (5.41), letting E at the conduction-band minimum equal zero, V  2m  DE   2  2e  2   

3/ 2

E 1/ 2

22

EF

V  2me   2  2 0 2   

(a) N el  

V  2m   2  2e     

N el 2   2  (b)   2m  V  e

3/ 2

E 1/ 2

3/ 2

EF 3/ 2

3/ 2

 EF 3/ 2

 2    EF n 2 / 34 / 3   2m   e  2   E = Eg + EF = E g  n 2 / 3  4 / 3   2m  e   (c) The gap is lowered by renormalization.

7.4 Band gap absorption spectrum for a direct-gap semiconductor:

7.5 Band gap 3.10 eV, LO phonon energy 35 meV, free exciton binding energy 50 meV, donor bound exciton has a binding energy 10 meV. (a) Band-to-band: 3.10 eV (b) Free exciton recombination: 3.10 – 0.05 = 3.05 eV (c) Donor bound exciton recombination: 3.05 – 0.01 = 3.04 eV (d) 3.040 – 0.035 = 3.005 eV 7.6 3.0 eV = 3.4 eV – 0.05 eV – Ea Ea = 0.35 eV 23

7.7 a = 5.45 Å, band gap 2.34 eV, ε = 10.7 SP donor binding energy = 80 meV, SiP acceptor binding energy = 60 meV (a) DAP for distant pairs = 2.34 – 0.08 – 0.06 = 2.20 eV (b) R = 10a = 5.45×10-9 m e2/(4πεε0R) = 2.47×10-2 eV 2.20 + 0.0247 = 2.22 eV 7.8 ψD = B exp[-(x2 + y2 + z2)/2a2} ψA = C δ(x-x0,y,z)

f x i    A* x D dxdydz







= BC  x exp  x 2  y 2  z 2 2a 2 x  x0 , y, z dxdydz = BC x0 exp(-x02/2a2) 7.9 [B] = 5 × 1016 cm-3, [Au] = 1 × 1014 cm-3. (a) The Au impurities compensate the B acceptors, so Au is positive (+). (b) When Au+ captures an electron, it becomes Au0. Therefore, the donor (0/+) level is the relevant one. vth = (3kBT/m)1/2 = 1.17×105 m/s = 1.17×107 cm/s 1/τ = σnNTvth = (10-14 cm2)(1014 cm-3)(1.17×107 cm/s) = 1.17×107 s-1

7.10 The optical absorption threshold = 1.1 eV. Donor level = 0.4 eV below conduction-band minimum. (a) Franck-Condon shift = 1.1 – 0.4 = 0.7 eV (b) S = 0.7 eV / ω = 0.7 / 0.050 = 14

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Chapter 8

8.1 NV = Nexp(-H/kBT) H = 2 eV T = 1209 K N = 8/a3 = 8/(5.66×10-8 cm)3 = 4.4×1022 cm-3 NV = (4.4×1022 cm-3)exp(-2·11600/1209) = 2×1014 cm-3

8.2 Band gap = 1.4 eV. At the valence-band maximum, the +1 charge state has an enthalpy 1.0 eV lower than the neutral charge state. (a) Formation enthalpy versus Fermi energy:

(b) Eg – 1.0 eV = 0.4 eV

8.3 Silicon has a band gap of 1.1 eV. The substitutional Au impurity can exist in three charge states: +, 0, and -. When the Fermi level is at the valence-band maximum, the enthalpy of the + state is 0.35 eV lower than the 0 state, while the enthalpy of the – state is 0.63 eV higher than the 0 state. (a) Formation enthalpy versus Fermi energy:

25

(b) (0/+) level = 0.35 eV (0/-) level = 0.63 eV

8.4 (0/-) level: 0.043 eV above valence band maximum.

(-/2-) level: 0.33 eV (2-/3-) level: 0.7 - 0.26 = 0.44 eV

8.5 For the neutral oxygen vacancy, Hf = Htot (ZnO:VO) – Htot (ZnO) + μO Zn-rich conditions: Hf = 3.7 eV (given) μO = μO(gas) – 3.6 eV (see text) O-rich conditions: 26

μO = μO(gas) The formation enthalpy is 3.6 eV higher than for Zn-rich conditions.

Hf = 3.7 + 3.6 = 7.3 eV 8.6 C = C0(1 + sinkx·e-γt) ∂C/∂t = C0sinkx(-γ)e-γt ∂2C/∂x2 = C0(-k2)sinkx·e-γt

C  2C D 2 t x C0sinkx(-γ)e-γt = DC0(-k2)sinkx·e-γt γ = Dk2

8.7 Frank-Turnbull mechanism:

8.8 D  0.01 e 0.48 eV kBT  cm2 s -1 T = 1373 K D = 0.01 exp(-0.48·11600/1373) = 1.73×10-4 cm2/s (Dt)1/2 = 0.1 cm t = 0.01 cm2 / 1.73×10-4 cm2/s = 58 s

8.9 Phosphorus in germanium diffuses by the vacancy mechanism. At 1210 K, the diffusion coefficient is D = 1 × 10-10 cm2/s. At 860 K, D = 1 × 10-14 cm2/s. (a) Vacancy diffusion mechanism:

27

(b) Expressing the migration enthalpy HM in eV, D = D0exp(-HM·11600/T) 10-10 = D0exp(-HM·11600/1210) 10-14 = D0exp(-HM·11600/860) Dividing the top equation by the bottom, 104 = exp[HM(-9.587 + 13.488)] ln(104) = HM(3.901) HM = 2.36 eV

8.10 From Equation (8.60), n-/n0 = ¼ exp[(EF – Ea)/kBT] = ¼ exp(0.1·11600/600) = 1.73 n+/n0 = ½ exp[(Ed – EF)/kBT] = ½ exp(0.3·11600/600) = 165 n0 + n- + n+ = 1010 n0 (1 + 1.73 + 165) = 1010 n0 = 6×107 cm-3 n- = 1.0×108 cm-3 n+ = 9.84×109 cm-3

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Chapter 9 9.1 n = 1016 cm-3, μ = 200 cm2/Vs, E = 20 V/cm. (a) vd = μE = 4000 cm/s (b) j = nvde = 6.4 A/cm2 (c) σ = j/E = 0.32 Ω-1cm-1

(σ = neμ gives the same answer)

(d) ρ = 1/σ = 3.125 Ω cm

9.2 From Equation (9.19), 



V23  I ln 2

 10 4 = 0.136 Ω cm 5  10 3 ln 2 1.5

9.3 Hall-effect experiment (Hall bar) for a p-type sample:

9.4 Sample dimensions: 0.5 mm × 0.5 mm × 5 mm I = 1.0 mA, p = 6.25×1015 cm-3, μ = 100 cm2/Vs (a) j 

10 3 A

0.05 cm

2

= 0.4 A/cm2

(b) σ = neμ = (6.25×1015 cm-3)(1.6×10-19 C)(100 cm2/Vs) = 0.1 Ω-1cm-1 (c) ρ = 1/σ = 10 Ω cm (d) E = j/σ = 0.4/0.1 = 4 V/cm V = Ed = (4)(0.5) = 2 V (e) vd = μE = 400 cm/s

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9.5 δ = 0.05 cm I12 = 5 mA, V34 = 0.40 V → R12,34 = 0.40 V / 5 mA = 80 Ω I23 = 5 mA, V14 = 0.05 V → R23,14 = 0.05 V / 5 mA = 10 Ω R12,34 / R23,14 = 8 From Figure 9.10, f(8) ≈ 0.8 From Equation (9.31),



0.05 450.8 = 8 Ω cm ln 2

9.6 ε = 11.7, A = 1 mm2 = 10-6 m2 1/C2 = (2.4×1020)V + 1.7×1020 = 0 when V = -1.7/2.4 = -0.71 V. Schottky barrier height = 0.71 V From Equation (9.42),

N

2 0 A e slope 2

= 5×1020 m3 = 5×1014 cm3

9.7 NT deep donors, Na compensating acceptors. (a) From Equation (9.44),

dn dt  en NT0  cn nNT For steady state,

0  en NT0  cn nNT en NT0  cn nNT en(NT – Na – n) = cnn(n + Na) (b) From Equation (5.56),

N a  n n

N T  N a  n 

 N C e  ET / kBT

en  N C e  ET / k BT cn

30

en  cn N C e  ET / kBT en   n vth N C e  ET / kBT

9.8 Consider Au in its neutral charge state (0) at room temperature. vth = (3kBT/m)1/2 = 1.17×105 m/s = 1.17×107 cm/s m k T  N C  2 e B2   2 

3/ 2

= 2.94×1019 cm-3

(we also use this for NV)

(a) Electron capture (0 to -) σn = 10-14 cm2 cn = vthσn = 1.17×10-7 cm3/s Hole capture (0 to +) σp = 4×10-15 cm2 cp = vthσp = 4.68×10-8 cm3/s (b) Electron emission (- to 0):

en  cn N C e  ET / kBT ET = 1.12 - 0.57 = 0.55 eV en = 2×103 s-1 Hole emission (+ to 0): ET = 0.35 eV ep = 2×106 s-1 9.9 cp = 10-7 cm3/s ep = 1/τe = 105 s-1  mk T  NV  2 B 2   2 

3/ 2

 1.0×1019 cm-3

e p  c p NV e  ET / kBT

10 5  10 71019 e  ET / kBT

 

ln 10 7   ET / k BT 31

ET = 16kBT = 0.2 eV

9.10 The positive lead of a voltmeter is attached to contact #1 of a semiconductor sample, while the ground lead is attached to contact #2. Contact #1 is heated to 40°C while contact #2 is 20°C. The voltmeter reads -0.5 mV. (a) S = ΔV/ΔT = -0.5 mV/-20 K = 2.5×10-5 V/K (b) Positive S → holes

32

Chapter 10

10.1 n = 2.5 1/(8×10-7 m) = 1.25×106 m-1 1/(5.33×10-7 m) = 1.88×106 m-1 1/(4×10-7 m) = 2.5×106 m-1 Δ(1/λ) = 0.625×106 m-1 Δ(1/λ) = 1/(2nx) 0.625×106 m-1 = 1/(5x) x = 3.2×10-7 m = 320 nm 10.2 λ = 500 nm a = 1 mm / 600 = 1667 nm Assuming first-order reflection (m = 1), a sinθ = λ (a) (1667 nm) sinθ = 500 nm sinθ = 0.3 θ = 17.5° (b) The dispersed light travels 0.25 m and exits a 100-μm wide slit. The angular spread is dθ = 10-4 m / 0.25 m = 4×10-4 rad λ = a sinθ dλ = a cosθ dθ = (1667 nm)(0.95)(4×10-4) = 0.6 nm

10.3 PL band centered at 2.1 eV, ZPL = 2.4 eV PLE onset = 2.8 eV Band gap = 3.5 eV (a) The ZPL (2.4 eV) gives the energy level relative to the conduction band minimum.

This is 3.5 – 2.4 = 1.1 eV above the valence-band maximum. (b) Stokes shift = 2.8 – 2.1 = 0.7 eV (c) Franck-Condon shift for the excited state = 2.8 – 2.4 = 0.4 eV (d) Franck-Condon shift for the ground state = 2.4 – 2.1 = 0.3 eV

33

10.4 In a Raman scattering experiment, light of wavelength 532 nm is incident on a GaN crystal. The Raman-scattered light has a wavelength of 512 nm. (a) The photon gained energy, so a phonon was destroyed. (b) 1/(5.32×10-5 cm) = 18,797 cm-1 1/(5.12×10-5 cm) = 19,531 cm-1 19,531 - 18,797 = 734 cm-1 (c) Raman spectrum:

10.5 From Equation (10.10) and (10.11), 001  xˆ   0  eˆ L  001  yˆ    2 6  001  zˆ    1 / 3 

010   xˆ   1 2    eˆ S  010   yˆ    1 6  010   zˆ    1 / 3   

From Equation (10.12),



IS ~ 0  2



 0 2

a   1 2     3  a   1 6   b   1 3 



6 1

6 1

 a 2    3  a 6   0  2a 6  b 3 = (b – a)/3  b 3   



10.6 kmax = 2π(1000 cm-1) = 6283 rad/s kmax

I ()   I 0 cosk dk I 0  sin k max   0

34

~ sin(6283·δ) / δ The zeros of this function occur when 6283·δ = mπ, or δ = 5 μm, 10 μm, …

10.7 I(δ) = Acos(k0δ)e–bδ (a) Interferogram:



(b) I 0 k   A  cosk 0   cosk e b d 0 











A   e ik0  e ik0 e ik  e ik e b d 40



A   e[i ( k0  k )b]  e[i ( k0 k )b]  e[i (  k0  k )b]  e[i (  k0 k )b] d 40

Evaluating this and discarding terms with (k + k0) in the denominator gives A  e[i ( k0 k )b] e[i (  k0  k )b]     4  i(k 0  k )  b i(k 0  k )  b 

=



0

 A 1 1    4  i (k 0  k )  b i (k 0  k )  b 

35

=

=

A   i (k 0  k )  b i (k 0  k )  b     4  (k 0  k ) 2  b 2 (k 0  k ) 2  b 

 A 2b   4  (k  k 0 ) 2  b 2 

10.8 g = 2.015, E = 9 GHz, A = 1.4 GHz E = gμBB ± A/2 9 GHz = (2.015)(14 GHz/T)B ± 0.7 GHz B = 0.344, 0.294 T

10.9 E = 24 GHz, B = 0.85 T E = gμBB 24 GHz = g (14 GHz/T) (0.85 T) g = 2.017 10.10 ΔB = 0.83347 – 0.8161 = 0.01737 T From the figure, ΔE ≈ 0.22 MHz ΔE = gNμN ΔB gNμN = 0.22 MHz / 0.01737 T = 13 MHz/T

36

Chapter 11

11.1 Proton: m = 1, silicon: M = 28. θ = 60°, sinθ = 0.866, cosθ = 0.5 2

E m  28 2  1(0.866 2 )  0.5    = 0.965 E 29  

E = 200 keV Em = (0.965)(200) = 193 keV

11.2 m = 4, M = 28 2

(a)

E m  28  4    = 0.5625 E  28  4 

 

 1  2 2 m Z ze (b) E    Nx 2 me E   64 0

E = 2 MeV = 1.6×10-13 J z = 4, Δx = 10-7 m, Z = 14, N = 5 × 1028 m-3 m/me = 7289 Plugging these in, ΔE = 0.0265 MeV (c) The backscattered kinetic energy is Em = (0.5625)(200 MeV) = 1.125 MeV. A He nucleus that travels 100 nm and then gets backscattered has 1.125 - 2·0.0265 = 1.07 MeV

37

2

 65.4  4  E  2 MeV  = 1.57 MeV  65.4  4 

11.3 Zn:

Relative yield ~ Z2 = 302 = 900 2

 32  4  E  2 MeV  = 1.21 MeV  32  4 

S:

Relative yield ~ Z2 = 162 = 256 ΔE = (230 eV/nm)(2·200 nm) = 0.09 MeV

11.4 Local symmetry of In impurities at low concentrations

PACS

Vacancy concentration

positron annihilation

Spatially resolved chemical composition of a thin film

EDS

Hydrogen donor level

μSR

Thin-film thickness

RBS

Near-surface (nm) chemical composition

XPS

Distance between atom and nearest neighbors

XAFS

11.5 (a) Si:

R = 0.527 μm Rp = 0.73R = 0.385 μm

(b) Ge: R = 0.501 μm Rp = 0.50R = 0.251 μm

38

11.6 1.3 (2/10) = 0.26 μm

11.7 R = 2.7 Å, Z = 30 (a) Eb ≈ 13.6(282) = 10.7 keV (b) From Equation (11.21), hν – Eb = 2k2/2m From Equation (11.24), we find the maxima of cos(2kR), which occur at k = 0, π/R, 2π/R hν – Eb = 0, 2π 2/2mR2, 22π 2/mR2 hν – Eb = 0, 5.2 eV, 21 eV (c) X-ray absorption spectrum:

11.8 1,487 eV - 554 eV = 993 eV

11.9 From Equation (11.26),

R

1 KE0 n 

where ρ = 5.3 g/cm3 K = 0.0428 n = 1.75 E0 = 60 keV Plugging in these values, R = 10.4 μm 39

11.10 One gamma ray is 515 keV. (a) E + 515 = 1022 keV E = 507 keV (b) p = (515 keV)/c – (507 keV)/c = 8 keV/c = 4.3×10-24 kg m/s

40

Chapter 12 12.1 NA = 0.7, λ = 460 nm r = (0.61)(460 nm)/0.7 = 401 nm ≈ 0.4 μm 12.2 The image in Figure 12.4 is roughly 100 μm × 100 μm = 10-4 cm2 There are ~10 dislocations in the image.

10/10-4 cm2 = 105 cm-2 12.3 E = 10 keV = 1.6×10-15 J (a) E = p2/2m

p  2mE = 5.4×10-23 kg m/s λ = h/p = 1.2×10-11 m (b) (0.61) λ / NA = (0.61)(1.2×10-11 m)/10-3 = 7.3 nm (c) Electrons scatter inside the material, spreading out over ~1 μm. This limits the resolution to about 1 μm.

12.4 Vb = 40 kV, Ib = 0.3 μA, Eg ~ 1.5 eV (a) Ei = 2.8Eg + 1 eV = 5.2 eV From Equation (12.7), VbIb(1 – γ)/Ei = (4×104)(3×10-7)(0.8)/[(5.2)(1.6×10-19)] = 1.15×1016 s-1 (b) r = 0.5 μm = 5×10-5 cm V = 4/3 πr3 = 5.2×10-13 cm3 G = 1.15×1016 s-1 / 5.2×10-13 cm3 = 2.2×1028 s-1cm-3 12.5 Generation rate G, minority-carrier lifetime τ, density n (a) dn/dt = G – n/τ (b) Let dn/dt = 0 0 = G – n/τ n = Gτ 41

12.6 v  10 6 m/s zˆ

B  (0.01 T)xˆ  (0.01 T)zˆ t = 10-9 s

 

(a) F  ev  B  e 10 6 0.01yˆ

 1.6  10 15 ŷ ay = -1.6×10-15/9.11×10-31 = -1.76×1015 m/s2 vy = ayt = -1.76×106 m/s







(b) F  ev  B   1.6  10 19  1.76  10 6 yˆ  0.01xˆ  0.01zˆ 





= 2.8  10 15 N  zˆ  xˆ  (Notice how this bends the electron in the x direction) 12.7 λ = 1.54 Å, (111) peak at 2θ = 29°, FWHM = 0.3°

d a

3

(a) 2d sinθ = λ d = 1.54 Å / [2 sin(14.5°)] = 3.08 Å a  3d = 5.33 Å

(b) From Equation (12.17), L = λ / (β cosθ) β = FWHM = 0.0052 rad L = 303 Å or 30 nm 12.8 2d sinθ = λ d = 5.43 Å / 4 = 1.36 Å

p  2mE = 7.6×10-23 kg m/s λ = h/p = 8.6×10-12 m sinθ = 0.03 θ = 1.8°

42

12.9 Bright (white) colors indicate high current. Ga atom orbitals are mostly empty (they comprise the conduction band). Figure (a) is positive sample bias, so electrons go into the Ga orbitals. Figure (b) is for negative sample bias, where electrons flow from the filled As orbitals.

12.10 Cantilever length = 100 μm = 105 nm Cantilever deflection = 10 nm This is an angle of 10 nm / 105 nm = 10-4 rad The beam is deflected by 2 × 10-4 rad (2×10-4 rad)(10 cm) = 2×10-3 cm or 20 μm

43