LN–3 IDLE MIND SOLUTIONS 1. Let us first look in most general ...

LN–3LN–3 IDLE MIND SOLUTIONS1. Let us first look in most general terms at the optical properties of solids with bandgaps (E g ) of less than 4 eV, semiconductors by definition.The band gap energy (E g ) can also be depicted as an “optical band edge”, as abarrier beyond which photons will excite electrons from the valence to theconduction band.absorp.E g = hν = hcνabsorp.transmission(transparent)absorption(opaque)absorption(opaque)transmission(transparent)photoenergyλ* λ(hc/λ* = E hν = E g )In the present case:E g + 3eVx 1.6 x 10*19 J + 4.8 x 10*19J + hceV l *l * + hc + 4.14 x 10 *7 m4.8 x 10 *19The optical band edge is at 4140Å, which means that from the daylight spectrum(sunlight) all λ’s less than 4140Å will be absorbed, all λ’s larger than 4140Å will betransmitted. The visible spectrum is normally given as ranging from 4000 to about8000Å (in some texts from 4200Å to 7200Å). According to either definition, a λ of4140Å is on the far end of the violet, bordering on the UV. This suggests that thesolid is essentially transparent to the visible radiation and in transmission mayhave a slight orange tinge, provided there are no impurities or other defects in thematerial which provide for electronic transitions other than the v.b. to c.b.transitions.2. The answer to this question is subject to the same restrictions as above.“Assuming” a defect–free and pure solid, then the color will be controlled by c.b. tov.b. transitions. A solid that appears green in transmitted light will absorb in the red(leaving yellow ––green –– blue → green). The amount of blue tinge in the greenwill be determined by the red cut–off location. Taking the optical band edge in themiddle of the red, λ* ± 6500Å, we obtain for a band gap (E g ):

LN–32. Continued.E g + E hn+ hc +l *6.63 x 10 *34 x3x10 86.5 x 10 *7+ 3.06 x 10 *19 Jx 1eV + 1.9 eV1.6 x 10 *19 J3. Both compounds are of the III–V family, which hybridize and form “adamantine”(diamond–like) structures which places them into the category of semiconductor.AlNGaSbΔEN = 1.43. The covalent radii of the constituents are small and,combined with the large ΔEN, the bonds (polar covalencies) are verystrong – the semiconductor is expected to exhibit a large band gap(likely transparent).ΔEN = 0.24. The covalent radii of both constituents are significantly larger(than those of AlN), the ionic contribution to bonding is small – thesemiconductor is expected to exhibit a much smaller band gap thanAlN.(AlN, E g = 3.8 eV,GaSb, E g = 0.8 eV)4. (a) Na 3s valence band is half–filled. EElectrons may acquire additionalincremental energy and moveabout (conduct).filledΔE(b) Mg 3s valence band is completely filled;however, the 3p band, which isempty, partly overlaps the filledvalence band and provides forelectrical conductivity.filledemptyΔE3s}}3p

LN–35. Considering electronic transitions only, the answer in principle is the same for allthree materials: there is no limit. But rephrased (λ = hc/E) – which is the lowestenergy radiation still capable of being transmitted through a material? – thequestion makes sense; we are concerned about a cut–off wavelength beyondwhich the radiation is being absorbed. Absorption means the energy is beingtransferred to an electron. Such a transfer can only take place if E (photon) ≥ ΔE.The critical energy (E = hν = hc/λ) is presently given by the value of the energygap (E g ).(a) Si: E g = 1.1eV = 1.1 x 1.6 x 10 –19 J = 1.76 x 10 –19 J = hc/λl crit+hc1.76 x 10 *19 + 1.13 x 10 *6 m + 1.13 mm(b) Ge: E g = 0.7eV = 1.12 x 10 –19 Jl crit+ 1.77 x 10 *6 m + 1.8 mm(c) GaAs: E g = 1.43eV = 2.29 x 10 –19 Jl crit+ 0.87 x 10 *6 m + 0.87 mm6. An extrinsic semiconductor is a semiconductor which contains foreign elementscapable of contributing mobile charge carriers, electrons, to the conduction band(n–type) or holes to the valence band (p–type). An intrinsic semiconductorcontains no foreign elements.7. The question is answered by determining (a) the volume of 50g Si and (b) thenumber of P atoms in 3 mg = 3x10 –3 g P.r Si + 2.33 gcm 3 50 g;+ 21.46 cm 32.33 gcm 330.97 g P + 6.02 x 10 23 atoms atomic weight of P = 30.79 g3mg + 3x10 *3 g + 6.02 x 1023 x3x10 *3 + 5.83 x 10 19 atoms30.97Number P atomscm 3 silicon + 5.83 x 1019 + 2.7 x 10 18 cm 321.46

LN–38.E••••E diagram for K}3d band empty4s band half filleddist.E•••E diagram for Be}dist.2s filled}2p overlappingIn K, each atom contributes one electron and one orbital to the conduction band(4s band). According to the Pauli exclusion principle, each “molecular orbital”formed in the band (energy state) can accommodate two electrons. As aconsequence, the conduction band is only half–filled with electrons – whichprovides for electrical conduction.In Be, each atom contributes two electrons and one orbital to the conduction band(2s band). With two electrons per orbital (from each atom) the 2s conduction bandis filled. The observed electrical conductivity is due to the overlapping 2p band,which is empty and thus provides empty energy states required for electronicconduction.9. (a) E g = 3.2eV = 3.2 x 1.6 x 10 –19 J = 5.12 x 10 –19 JThe full band gap (energy) corresponds to 5.12 x 10 –19 J. If we look for thecorresponding wavelength of radiation, we have:E + hn + hc and l + hc + 6.6 x 10 *34 x3x10 8l E 5.12 x 10 *19l + 3.879 x 10 *7 m +3880ÅThe visible spectrum ranges from 4000–7500Å. Since only radiation withE ≥ E g will be absorbed (λ < 3880Å), the material is transparent to visiblelight.

LN–39. (b) Most oxide semiconductors are either doped to create extrinsic defects orare annealed under conditions in which they become nonstoichiometric.When ZnO becomes nonstoichiometric, it is commonly because of an excessmetal content, for example, Zn 1+x O. This Zn 1+x O is a negative carrier(n–type) oxide conductor and is sometimes called an “excess”semiconductor.10. (a) [T in thermally activated processes is the absolute temperature –T o K = (273.16 + t o C); Boltzmann’s constant = k = 1.38 x 10 –23 J/ o K]T = 293.16K:n i+ 9.7 x 10 15 x 293.16 32 xe * 0.72x1.6x10 *192x1.38x10 *23 x 293.16+ 9.7 x 10 15 x 5019 x 6.6 x 10 *7n i+ 3.21 x 10 13 cm 3(b) 200 o C = 473.16Kn i+ 9.7 x 10 15 x 473.16 2xe0.72x1.6x10**193 2x1.38x10 *23 x 473.16n i(200K) + 9.7 x 10 15 x 1.03 x 10 4 x 1.47 x 10 *4n i+ 1.47 x 10 16 cm 3The number of conducting electrons (in the conduction band) at 200 o C is byabout five orders of magnitude less than that of a good conductor. Thematerial will not be a good conductor. (There are additional factors whichcontribute to the “relatively poor conductivity” of Ge at this temperature.)11. E th+ 3kT E g + 0.72 x 1.6 x 10 *19 J2T + 0.72 x 1.6 x 10*19 x2 + 5565K +3 x 1.38 x 10 *235.3 x 10 3o CThe temperature would have to be 5.3 x 10 3 o C (about 4400 o C above the meltingpoint).12. (a) From our limited knowledge of the conduction behavior, we must assumethat in semiconductors the conductivity will increase with T since the numberof electrons in the conduction band increases.

LN–312. (b) As a first approximation, we can say that the number of free electrons isindependent of T; that, however, the atoms in the lattice will, with increasingtemperature, be subject to increased oscillations about their (relatively) fixedposition. These oscillations provide greater opportunity for “scattering” ofconducting electrons and thus will “reduce” the mobility of the electrons. Wecan expect the electrical conductivity of metals to decrease with increasingtemperature.13. (a) The optical properties of ZnSe can be explained when comparing the“energy band” of the visible spectrum with the energy band diagram of ZnSe.E(hν)E–c.b.E g = 2.3eVabsorption takes place(photo–excitation) forall radiation withE (hν) ≥ 2.3 eV.+v.b.λ(Å) →40005397↓7000blue–greenyellow–red3.1↑2.32.26 1.77← E(eV)

LN–313. (a) Continued.absorp.1.0blue–greenabsorbedyellow–redtransmitted0.05397λ(Å) wavelengthFrom the energy distribution of the visible spectrum we recognize that theblue–green portion has photon energies in excess of the band gap(E g = 2.3eV) and thus will be absorbed. The yellow–red potion, on the otherhand, has photon energies less than the band gap – it will be transmitted.ZnSe, therefore, is expected to exhibit a yellowish–red color.(b)In principle there are two ways to increase the electrical conductivity ofZnSe:(1) A temperature rise. Any rise in temperature will increase the numberof “thermally activated” charge carriers in the conduction band(electrons) and in the valence band (holes) and, thus, the electronicconductivity.[The electrical conductivity of solids, demonstrated by the flow ofelectronic charge carriers under an applied electric field (E), can beformulated through Ohm’s law, J = σE, which states that the currentdensity (J = number of charges transported through a unit area in aunit time) is proportional (σ = conductivity) to the applied electric field.Accordingly:J = N e v dwhere N = number of charge carriers/unit volume, e = electroniccharge and v d = average drift velocity of charge carriers in an appliedelectric field. We thus obtain:σ = (N e v d )/E

LN–313. (b) Continued.(1) continued –and if we define (v d /E) = µ, the charge carrier mobility, we have:σ = N e µIn intrinsic semiconductors we have both electrons and holescontributing to conduction:σ = N e eµ e + N n eµ h = Ne(µ e + µ h )since N e = N h . Taking the number of thermally generated chargecarriers, the relationshipN + AT 32 exp *E g 2kTwe obtain the temperature dependence of the conductivity as:s + AT 32 exp *E g2kTxe(m e ) m h)To assess the temperature dependence of electrical conductivity wemust take into consideration that, because of increased vibration ofthe atoms about their lattice positions, the charge carrier mobility willdecrease (increased scattering of charge carriers) with increasingtemperature. This effect explains why the electronic conductivity inmetals, where N is constant, will decrease with increasingtemperature. In semiconductors, where N increases with temperature,the accompanying mobility effect is not apparent at low temperatures(conductivity increases), but becomes pronounced at hightemperatures (conductivity decreases).](2) Introduction of shallow impurity (or defect) states close to theconduction or valence band. This is accomplished by theincorporation of appropriate dopant elements into the crystal matrix. Ifthese impurities are shallow ( ~ 0.01eV from the conduction or valenceband), they will be totally ionized at room temperature and each willcontribute an electron (donor dopant: K, Na) or holes (acceptordopant: G, Br), thus increasing the electrical conductivity without thenecessity of a temperature rise. (Be aware that certain defects in thecrystal lattice may also increase the electronic conductivity.)

LN–314. (a) This crystal is extrinsic of p–type conductivity – i.e. holes in the valence band(majority charge carriers) are primarily responsible for electronic conduction.p–type conductivity results from the fact that the substitutional dopingelement, boron, has only three valence electrons. Upon incorporation, boronhas a tendency to attract an additional electron from the matrix (valenceband) to satisfy the bonding requirements (formation of four covalent bonds).In doing so, each boron atom assumes a negative charge (which is notmobile) and simultaneously creates a mobile hole.(b)Approach: determine the number of atoms in 28 mg boron and the volume of100 g silicon.ρ Si= 2.33 g/cm 3At.Wt. B= 10.81 g/moleVolume of 100 g Si = (100/2.33) = 42.9 cm 3# atoms per 28mg B+6.02 x 10 23 x28x10 *3 +10.811.56 x 10 21When incorporated uniformly into a volume of 42.9 cm 3 of silicon, thenumber of boron atoms per cm 3 (the number of extrinsic charge carriers –i.e. the number of holes generated) is:1.56 x 10 21 + 3.6 x 10 1942.9[It should be specified that this number applies to silicon at roomtemperature. Assume the temperature were approaching 0 K; then thenumber of extrinsic charge carriers would approach “zero” since the impuritylevels are, by a finite amount (~0.01 eV), above the top of the valence bandand the acquired charges will drop back to the lowest possible energy state.]EB E– – – –+ + + +T→ o KvalencebandBvalenceband

LN–315. “White light” contains radiation in wavelength ranging from about 4000Å (violet) to7000Å (deep red). A material appearing red in transmission has the followingabsorption characteristics:absorp.blue–yellowabsorbed4000 6500redtransmittedλ (Å)(a)(b)If the material is pure (no impurity states present), then it must be classifiedas a semiconductor since it exhibits a finite “band gap” – i.e. to activatecharge carriers, photons with energies in excess of “red” radiation arerequired.Taking λ = 6500Å as the optical absorption edge for this material, we have:E hn+ hcl+ 3.05 x 10 *29 Jx 1eV1.6 x 10 *19 J+ 1.9 eVAccordingly, the band gap for the material is E g = 1.9 eV.16. (a) The antimony–doped Ge is n–type because Sb (with electronic configuration[Kr]4d 10 5s 2 5p 3 ) has five valence electrons, and, therefore, donates one tothe Ge ([Ar]3d 10 4s 2 4p 2 ).(b)Assume uniform doping of Ge with Sb. Each Sb atom contributes oneelectron to germanium’s conduction band.27 mg +2.7(10 *2 )g121.75 gmole + 2.2(10*4 )moleSbSo the antimony addition contributes2.2(10 –4 ) x 6.02(10 23 ) = 1.32(10 20 ) electrons

LN–316. (b) Continued.The volume of germanium is:110 g+ 20.7 cm 35.32 gcm 3n e + 1.32(1020 ) + 6.4(10 18 *3)cm20.7 cm 317. Need to find E g :ν = 5.631 x 10 5 m –1λ = 1/ν = 1.776 x 10 –6 mE g = (hc)/λ = (hc)ν = 0.698 eV[Looks like Ge.]T = 250 o C = 523Kn i+ AT 32 e *E g 2kTcm *30.698+ (5 x 10 15 ) (523) 32 exp* 2 x (8.63 x 10 *5 ) (523)= [5 x 10 15 ] [1.196 x 10 4 ] exp [–7.688]= 2.74 x 10 16 cm –318. (a) From the lecture notes you know that a group III element, substitutionallyincorporated into silicon, acts as an “acceptor” dopant, and by acquiringelectrons from the lattice (valence band) will generate “positive holes” – thecrystal will have an excess of (positive) holes – and it will be p–type.

LN–318. (b) First we determine the volume of 100g Si, then the number of Al atoms per33 mg Al and divide this number by the volume of silicon. This gives us thenumber of Al atoms per cm 3 Si – which is the same as the number ofextrinsic charge carriers, since each Al generates one carrier.Vol. Si +100 g2.33 gcm 3+ 42.9 cm 3 100 g33 mg Al contain:6.02 x 10 23 atoms x33x10 *3 g + 7.36 x 10 20 atoms26.98 g# Al atoms per cm 3 = # charge carriers per cm 3+ 7.36 x 1020 + 1.72 x 10 19 Alcm 3 Si42.9= 1.72 x 10 19 extrinsic holes per cm 3 Si19. Since E = hν, we find the material to be transparent to photons with energies lessthan:(6.63 x 10 *34 Js) x (1.3 x 10 14 s *1 )x 1eV + 0.54 eV1.6 x 10 *19 Jwhich constitutes the absorption edge, ie the energy band gap.Econduction bandE g = 0.54 eVvalence band

LN–320. (An inadvertent error occurred in the relationship for thermally activated chargecarriers. It should be:n e + AT 32 e *E g 2kTcm 3Also, the dimension of A should be [cm –3 x ( o K) –3/2 ].)This trivia question therefore has the answer:n e + 5x10 15 x (500 ) 273.15)2 xe3*1.1x1.6x10 *192x1.38x10 *23 x 773.15n e = 5 x 10 15 x 773.15 3/2 x e –8.25 cm –3n e = 2.8 x 10 16 cm –321. (a)E5 x 10 18 extrinsic carriers (c.b.)– – – – – – –↑ ↑ ↑↑↑↑↑+ + + + + + + ←donor impurity levelE g = 0.7 eV(v.b.)(b)The band gap of E g = 0.7 eV x 1.6 x 10 –19 J = 1.1 x 10 –19 J corresponds to acut–off frequency (E g = hν):n +1.1 x 10*19+6.63 x 10 *341.66 x 10 14 s*1The indicated frequency (1 x 10 15 s –1 ) is higher – the photon energy is largerthan E g . The radiation will be absorbed by germanium and the crystal isopaque to this radiation.

LN–322.23. 6.63x10 *34 Js x 3x10 8 msE + hc ++ 6.63 x 10300m*38 photonJphotonlThe energy of the radio transmitter = Power x time= 50 x 10 3 J/s x 60s = 3 x 10 6 JNumber of photons =3x10 6 J= 4.52 x 10 33 photons6.63x10 *28 Jphoton24. (a) l + c + 3x108 ms + 6x10 *7 m + 600nmn 5x10 14 s(b)The color will be orange or “reddish”.

LN–324. (c) The band gap energy E g = E photon = hν= 6.63x10 –34 Js x 5x10 14 /s= 3.315x10 –19 J x 1eV/1.6x10 –18 J= 2.07 eV25. (Determine the volume of 30g Ge – from that determine the mg of As/cm 3 andconvert that quantity into # of As atoms (carriers)/cm 3 .)30gGe x 1cm3 + 5.64cm 3 ; 5.4x10 *2 gAs+ 9.57x10*3 gAs5.32g5.64cm 3 cm 39.57x10 *3 gAs x 6.02x1023 atoms = 7.69 x 10 19 carriers/cm 374.92g26.27. (a) 1 1 visiblel + + mrad.= 0.62µm100n 1.613x10 6 0 0.4 0.6 0.7 λ [µm]Abs [%]bluered(b)The color will be reddish–orange.