Physics 312 First Pledged Problem Set

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Clyde O’Connor’
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1 Physics 31 First Pldgd Problm St 1. Th ground stat of hydrogn is dscribd by th wavfunction whr a is th Bohr radius. (a) Comput th charg dnsity Ã (r) = 1 p ¼ µ 1 a 3 r=a ; ½ (r) = jã (r)j : and plot 4¼r ½ as a function of th dimnsionlss variabl r=a. Dos th plot pak at r = a? Thus W hav µ ½ (r) = p r=a ¼ a = 1 ¼ A plot of this function is blow. W s that th plot dos pak at r = a. a 3 r=a : 4¼r ½(r) = 4 a µ r a r a : (b) Comput th avrag valu of r. Is it qual to a? W hav hri = = Z Z Ã y (r)rã (r) d 3 r r jã (r)j d 3 r Z 1 = r 1 µ 3 1 r=a 4¼r dr ¼ a Z 1 = 4a 3 d ;

2 x 1.5 Figur 1: whr = r=a. Th rmaining intgral computd by hand, lookd up, or don by computr, and is found to b qual to 3!= 4 = 3=8. Thus W s that hri is not qual to a. hri = 3 a : (c) Dos th lctron hav a dfinit valu of r? Dos it hav a dfinit spd? Dos it hav a dfinit angular momntum? Dos it hav a dfinit nrgy? If th answr to any of ths qustions is ys, giv th valu of th quantity. Th Hamiltonian that dscribs th lctron is H = p m + V (r) = ~ r 1 m 4¼² r : W know from basic quantum mchanics that an ignfunction of a systm dscribd by a Hamiltonian H will also b an ignfunction of an oprator O if and only if O commuts with H. On can chck that r and v = p=m do not commut with H, whil L = L x + L y +L z and H itslf do commut with H. Thrfor th lctron has a dfinit nrgy and angular momntum, but not a dfinit radius or spd. Th valu of th angular momntum is L = p l (l + 1)~ = and th valu of th nrgy is E = 13:6 V. (d) Comput th lctric fild du to this charg dnsity vrywhr in spac. (Hint: us Gauss s thorm.) Add it to th fild of th proton. Plot th total (radial) fild as a function of r=a for r > a = to chck that it dcrass xponntially. W hav from Gauss s law that Z E (r) da = 1 Z S ½ (r) d 3 r; for any closd surfac S. Bcaus th charg dnsity is sphrically symmtric, it follows that th lctric fild will b radial. Thus if w choos for S a sphr of radius r cntrd at th origin, w find that E (r) 4¼r = 1 Z r ½(r ) 4¼r dr ; ² so Z r

3 = ¼² Z r Z R µ r r =a r dµ = R R dr ; ¼² whr R = r=a and R = r =a. Th rmaining intgral can b don by Mapl or th lik, but is also asily don by hand. First w prform a chang of variabl: Z R R R dr = d ; ¼² 8¼² whr is simply qual to R. Now w prform two intgrations by parts: Z R Thrfor w hav d = R + a Z R Z R d = (R) R R + a Z R = 4R R 4R R + R = 1 + R + R R : E (r) = R + R R r 8¼² µ 1 a = 4¼² a r 1 + a r + a r=a : r Th fild du to th proton is just th Coulomb fild E (r) = 1 4¼² r ; so adding this to our prvious rsult w find that th total fild is E tot (r) = 1 µ 4¼² a 1 + a r + a r r=a : A plot of this fild is blow. () Comput th potntial by intgrating th fild from r to infinity. Using Mapl or th lik to do th intgration, w find that V (r) = Z 1 r = 1 4¼² a E tot (r ) dr Ã! 1 + r a r xp( r=a ): a 3

4 r Figur :. Rad th short sction on impurity lvls on pag 368 of Kittl and Kromr and look up th dilctric constants of Si and G, as wll as th ffctiv masss for th valnc and conduction bands. Thn: (a) Comput th ffctiv Bohr radius for donor and accptor ground stats. Th asy way to do this is to scal th usual Bohr radius by th appropriat factors. From Tabl 13.1 on pag 357 of Kittl and Kromr w obtain th following valus: Elmnt Effctiv Mass of Hols Effctiv Mass of Elctrons Dilctric Constant Si.58 m 1.6 m 11.7 G.35 m.56 m 15.8 Th Bohr radius for hydrogn is givn (in MKS units) by a ;hydrogn = 4¼² ~ m ; which is qual to.59 nm. As a rough approximation, w can trat th accptor and donor atoms as hydrogn-lik atoms by rplacing m by th ffctiv lctron mass (m ff ) and ² by ², whr is th dilctric constant of th matrial. Thus w hav a = 4¼ ² ~ m ff = (m ff =m ) a ;hydrogn: Substituting, w find th following valus for th ffctiv Bohr radii: Elmnt a for Hols a for Elctrons Si. a 11. a G 45.1 a 8. a (b) Comput th ionization nrgis and compar with Tabl Th ionization nrgy of th hydrogn atom is givn by m 4 E ion;hydrogn = (4¼² ) ~ ; which is qual to 13.6 V. Prforming th sam approximations as bfor w find that E ion = m ff 4 (4¼ ² ) ~ = (m ff=m ) E ion;hydrogn :

5 mor accurat than that. 3. Comput th bias voltag of a grmanium p-n junction at 77 K, at room tmpratur, and at 5 K. Commnt on th ffct of incrasd tmpratur on th prformanc of such a junction. W hav from pag 17 of Mlissinos that µ ND N ln A V bi = k BT and w hav from pag 8 that n i = N s Eg k BT : Combining ths two rlations w hav that V bi = k BT = E g + k BT ln n i ; µ ND N A ln Ns Eg=kBT µ ND N A Th xact valus of N A and N D will dpnd on th doping, but for rprsntativ valus w can us th ons givn in Figurs 1.8(b) and 1.8(c) on pag 15 of Mlissinos. Ths ar N A = 1 16 cm 3 and N D = 1 15 cm 3. On pag 9 of Mlissinos, w larn that for grmanium n i ¼ 1 13 cm 3 and E g = :7 V at room tmpratur (3 K). From thr data w dtrmin that N s ¼ cm 3. Plugging ths valus into th xprssion for th bias voltag w find that V bi = :596 V, :95 V, and :5 V at 77 K, room tmpratur (3 K), and 5 K, rspctivly. Sinc th bias voltag dcrass with tmpratur, th prformanc of th junction bcoms wors as th tmpratur dcrass. N s : 4. Short qustions. (a) Using just th lattic paramtr, comput th Frmi nrgy for aluminum and compar with th valu givn, for instanc, in Tiplr. (Hint: how many valanc lctrons pr atom ar thr in Al?) If thr is a discrpancy, it indicats that th ffctiv mass of lctrons in Al is diffrnt fromth fr lctron mass. Aluminum has a fac-cntrd cubic lattic structur, and so it has four atoms pr unit cll. Th dnsity and atomic wight (as found in th CRC Handbook of Chmistry and Physics ) ar.7 g/cm 3 and 6.98 g/mol, rspctivly. W larn from a priodic tabl that aluminum has thr valnc lctron pr atom. Thrfor th volum occupid by on atom is (6.98 g/mol)(1 mol/ atoms)/(.7 g/cm 3 ) = cm 3 /atom. Sinc thr ar four atoms pr unit cll, th volum of th unit cll is four tim this, or

6 cm 3, and th lattic paramtr is givn by cm. Sinc thr ar thr valnc lctrons pr atom, th numbr dnsity of valnc lctrons is simply n = 3/( cm 3 ) = lctrons/cm 3. From quation 1.4 on pag 6 of Mlissinos w s that E F = ~ 3¼ =3 n : m Plugging in th appropriat quantitis, w find that E F = J =.46 V. In Tabl 39-1 on pag 195 of Tiplr (volum ), th Frmi nrgy of aluminum is givn as 11.7 V. Thrfor th ffctiv mass of an lctron in aluminummust b considrably diffrnt from that of a fr lctron. (b) Using th masurd rsistivitis, givn by Tiplr on pag 7, compar th rsistanc (or, if you wish, th conductanc) of an Al wir to that of a Cu wir having th sam lngth and cross sction. From pag 7 of Tiplr w hav ½ Al = :8 1 8 =m and ½ Cu = 1:7 1 8 =m. On pag 71 of Tiplr, w larn that R = ½ L A ; so if both wirs hav th sam lngth and cross sction thn R Al = ½ Al R Cu ½ Cu = 1:65 (c) Compar th rsistanc (or, if you wish, th conductanc) of an Al wir to that of a Cu wir having th sam lngth and wight. Sinc th wirs hav th sam wight w hav d Al A Al L = d Cu A Cu L, or A Al =A Cu = d Cu =d Al, whr d rprsnts th dnsity of th matrial. Thrfor, R Al = ½ Al A Cu R Cu ½ Cu A Al = ½ Al ½ Cu d Al d Cu : From Tabl 11-1 on pag 333 of Tiplr w s that d Al =d Cu = :3, so w hav finally that R Al =R Cu = :5. (d) You should hav found from (b) and (c) that, at last by som critria, Al is a bttr conductor than Cu. It is also much chapr. Why thn is Cu usd for most lctrical wiring? 6

7 On rason coppr is usd rathr than aluminum is that aluminum oxid dos not conduct lctricity, whil coppr oxid dos. Lik most mtals, aluminum and coppr quickly dvlop a thin layr of mtal oxid whn xposd to air. This mans that if you strip two nds of an aluminum wir and twist thm togthr, currnt will not want to flow across th splic. If th sam procdur is rpatd with coppr, howvr, on finds that currnt radily flows across th splic. Thus, dspit th fact that aluminum is chapr, in most applications it is mor convnint to us coppr instad. Anothr rason is that aluminum tnds to xpand mor whn hatd, which significantly rducs th liftim of splics. () Why is th filamnt of an incandscnt light bulb mad of tungstn, rathr than stl or coppr? From pags B-14, B-4, and E-88 of th CRC Handbook of Chmistry and Physics, 66th Edition, w larn that th mlting points of coppr, tungstn, and stl ar 183 ± C, 341 ± C, and 15 ± C, rspctivly. W also larn from pag 78 of Bloomfild that tungstn sublims vry slowly at tmpraturs blow its mlting point. Thus tungstn filamnts can b run much hottr than filamnts mad of coppr and stl bfor th filamnt mlts or th rat of sublimation bcoms intolrabl. Bcaus it can b run hottr, th tungstn filamnt also producs a richr, whitr light that is mor similar to th light of th sun. For this rason, tungstn filamnts ar usd in incandscnt light bulbs instad of coppr or stl. (f) Why has silicon bcom th standard smiconductor matrial for most purposs? What ar possibl altrnativ matrials? On rason for th prvalnc of silicon in smiconductor dvics is that silicon is a radily availabl matrial. Sand (i.. quartz) is simply silicon dioxid. Anothr rason is that whn popl first startd to work with smiconductors thy chos to work with silicon, and ovrth yars manufacturrs hav dvlopd tricks to mak smiconductor componnts smallr, fastr, tc. that ar spcific to silicon. It would cost a grat dal of mony to modify ths tchniqus for us with othr lmnts, so manufacturrs continu to work with th traditional silicon. Altrnativ matrials that ar usd includ grmanium and gallium arsnid. (g) Why is silicon not suitabl for a light mitting diod? What matrials ar usd for this purpos? A matrial such as gallium arsnid is much mor suitabl than silicon for LEDs bcaus GaAs is dirct gap smiconductor, whil Si is an indirct gap smiconductor. In both matrials th stats nar th top of th valnc band hav zro momntum, but in GaAs th stats at th bottom of th conduction band also hav zro momntum, whil in Si thy hav nonzro momntum. This is bcaus th wavfunction changs sign in going from on Si atom to th nxt on th dg of th cubic cll. (S th pictur of th nrgy bands of Si on 7

8 pag 38 of th Physicist s Dsk Rfrnc.) This mans that th momntum of an lctron in GaAs changs vry littl whn it mits or absorbs a photon, and thus light mission can occur dirctly in this matrial. In Si, howvr, photon mission is an indirct procss that cannot occur unlss th conduction band lctron has a scondary procss by which it can dispos of its momntum. (Th trms dirct gap and indirct gap rfr to whthr or not th bottom of th conduction band lins up dirctly abov th top of th valnc band in a plot vrsus momntum.) Gallium arsnid has a bandgap of 1:4 V, which corrsponds to a rd photon. Thus GaAs is usd to mak rd LEDs. A widr bandgap, rsulting in th mission of yllow or grn light, is obtaind by rplacing som of th gallium atoms with aluminum, or by rplacing som of th arsnic atoms with phosphorus. (h) In a p-n junction at zro xtrnal bias, what is th sign of th nt lctrical charg on th p sid? On th n sid? Th nt lctric charg is ngativ on th p sid and positiv on th n sid. This is bcaus whnth two pics of smiconductor firstmak contact, thr is a highr concntration of positiv carrirs on on sid of th junction and a highr concntration of ngativ carrirs on th othr. Thus thr is a diffusiv currnt and positiv carrirs flow from th p sid to th n sid and ngativ carrirs flow from th n sid to th p sid. Onc th lctrons cross into th p sid of th junction thy annihilat with a hol and similarly for th hols that cross into th n sid. Th nw charg distribution producs an lctric via Gauss s law, and th dirction of th lctric fild is such that it opposs th diffusion currnt. Th currnt continus to flow until th systm rachs quilibrium, whr th currnt du to diffusion is xactly th opposit of that du to th lctric fild. 8

The pn junction: 2 Current vs Voltage (IV) characteristics

The pn junction: 2 Current vs Voltage (IV) characteristics Th pn junction: Currnt vs Voltag (V) charactristics Considr a pn junction in quilibrium with no applid xtrnal voltag: o th V E F E F V p-typ Dpltion rgion n-typ Elctron movmnt across th junction: 1. n