Chapter 1 The Schrödnger Equaton 1.1 (a) F; () T; (c) T. 1. (a) Ephoton = hν = hc/ λ =(6.66 1 34 J s)(.998 1 8 m/s)/(164 1 9 m) = 1.867 1 19 J. () E = (5 1 6 J/s)( 1 8 s) =.1 J = n(1.867 1 19 J) and n = 5 1 17. 1.3 Use of Ephoton = hc/ λ gves 3 34 8 (6. 1 )(6.66 1 J s)(.998 1 m/s) E = = 399 kj 9 3 1 m 1.4 (a) Tmax = hν Φ = (6.66 1 34 J s)(.998 1 8 m/s)/( 1 9 m) (.75 ev)(1.6 1 19 J/eV) = 5.53 1 19 J = 3.45 ev. () The mnmum photon energy needed to produce the photoelectrc effect s (.75 ev)(1.6 1 19 J/eV) = hν =hc/λ = (6.66 1 34 J s)(.998 1 8 m/s)/λ and λ = 4.51 1 7 m = 451 nm. (c) Snce the mpure metal has a smaller work functon, there wll e more energy left over after the electron escapes and the maxmum T s larger for mpure Na. / T 1.5 (a) At hgh frequences, we have e ν >> 1 and the 1 n the denomnator of Planck s formula can e neglected to gve Wen s formula. () The Taylor seres for the exponental functon s x e = 1 + x+ x /! +. For x << 1, we can neglect x and hgher powers to gve e 1 x. Takng x hν / kt, we have for Planck s formula at low frequences 3 3 3 a ν h h kt = π ν π ν = πν ν/ T hν/ kt e 1 c ( e 1) c ( hν / kt) c x 1.6 λ = hm / v = 137 hmc / = 137(6.66 1 34 J s)/(9.19 1 31 kg)(.998 1 8 m/s) = 3.3 1 1 m =.33 nm. 1-1 Copyrght 14 Pearson Educaton, Inc.
1.7 Integraton gves x at tme t, then 1 x = gt + gt + t + c ( v ). If we know that the partcle had poston 1 x = gt + ( gt + v ) t + c and c 1 = x gt v t. Susttuton of the expresson for c nto the equaton for x gves 1 v x = x g( t t ) + ( t t ). 1.8 ( / )( Ψ/ t) = ( / m)( Ψ/ x ) + VΨ. For 1 Ψ= ae t e mx /, we fnd 1 1 Ψ / t = Ψ, Ψ / x = m xψ, and Ψ/ x = m Ψ m x( Ψ/ x) 1 1 1 1 = m Ψ m x( m xψ ) = m Ψ+ 4 m x Ψ. Susttutng nto the tme-dependent Schrödnger equaton and then dvdng y Ψ, we get 1 ( / )( Ψ ) = ( / m)( m + 4 m x ) Ψ+ VΨ and V = mx. 1.9 (a) F; () F. (These statements are vald only for statonary states.) 1.1 ψ satsfes the tme-ndependent Schrödnger (1.19). 3 ψ / x = cxe 4cxe + 4c x e = (1.19) ecomes cx 3 ψ / x = e cx e ; 3 6cxe + 4c x e. Equaton ( / m)( 6cxe + 4 c x e ) + ( c x / m) xe = Exe. The x 3 terms cancel and E = 3 c/ m = 3(6.66 1 34 J s).(1 9 m) /4π (1. 1 3 kg) = 6.67 1 J. 1.11 Only the tme-dependent equaton. 1.1 (a) 3 x / Ψ dx = (/ ) x e dx = 9 3 9 (.9 nm)/(3. nm) 9 (3. 1 m) (.9 1 m) e (.1 1 m) = 3.9 1 6. () For x, we have x = x and the proalty s gven y (1.3) and (A.7) as nm 3 nm / 3 / 3 nm (/ ) x x Ψ dx = x e dx = (/ ) e ( x / x / /4) x / nm = e ( x / + x/ + 1/) = e 4/3 (4/9 + /3 + 1/) + 1/ =.753. (c) Ψ s zero at x =, and ths s the mnmum possle proalty densty. (d) 3 x / 3 x/ dx x e dx x e dx w/ w/ = Ψ = (/ ) + (/ ). Let w = x n the frst ntegral on the rght. Ths ntegral ecomes we ( dw) we dw, whch equals the second ntegral on the rght [see Eq. (4.1)]. Hence 3 x / 3 3 Ψ dx = (4 / ) x e dx = (4 / )[!/ ( / ) ] = 1, where (A.8) n the Appendx was used. 1- Copyrght 14 Pearson Educaton, Inc.
1.13 The nterval s small enough to e consdered nfntesmal (snce Ψ changes neglgly wthn ths nterval). At t =, we have Ψ dx = (3/ π c ) x e dx = [3/π(. Å) 6 ] 1/ (. Å) e (.1 Å) =.16. 6 1/ x / c 1.14 1.51 nm 1 / / 1.51 nm dx a e x a dx e x a / a 1.5 nm 1.5 nm Ψ = = = ( e 3. + e 3. )/ = 4.978 1 6. 1.15 (a) Ths functon s not real and cannot e a proalty densty. () Ths functon s negatve when x < and cannot e a proalty densty. (c) Ths functon s not normalzed (unless = π ) and can t e a proalty densty. 1.16 (a) There are four equally proale cases for two chldren: BB, BG, GB, GG, where the frst letter gves the gender of the older chld. The BB posslty s elmnated y the gven nformaton. Of the remanng three possltes BG, GB, GG, only one has two grls, so the proalty that they have two grls s 1/3. () The fact that the older chld s a grl elmnates the BB and BG cases, leavng GB and GG, so the proalty s 1/ that the younger chld s a grl. 1.17 The 138 peak arses from the case 1 C 1 CF 6, whose proalty s (.9889) =.9779. The 139 peak arses from the cases 1 C 13 CF 6 and 13 C 1 CF 6, whose proalty s (.9889)(.111) + (.111)(.9889) =.195. The 14 peak arses from 13 C 13 CF 6, whose proalty s (.111) =.13. (As a check, these add to 1.) The 139 peak heght s (.195/.9779)1 =.4. The 14 peak heght s (.13/.9779)1 =.16. 1.18 There are 6 cards, spades and 4 nonspades, to e dstruted etween B and D. Imagne that 13 cards, pcked at random from the 6, are dealt to B. The proalty that 3 13 13(1) 6 every card dealt to B s a nonspade s 4 1 14 1 = = Lkewse, the 6 5 4 3 16 15 14 6(5) 5. proalty that D gets 13 nonspades s 6. If B does not get all nonspades and D does not 5 get all nonspades, then each must get one of the two spades and the proalty that each 6 6 gets one spade s 1 = 13/5. (A commonly gven answer s: There are four 5 5 possle outcomes, namely, oth spades to B, oth spades to D, spade 1 to B and spade to D, spade to B and spade 1 to D, so the proalty that each gets one spade s /4 = 1/. Ths answer s wrong, ecause the four outcomes are not all equally lkely.) 1-3 Copyrght 14 Pearson Educaton, Inc.
1.19 (a) The Maxwell dstruton of molecular speeds; () the normal (or Gaussan) dstruton. 1. (a) Real; () magnary; (c) real; (d) magnary; (e) magnary; (f) real; (g) real; (h) real; () real. 1.1 (a) A pont on the x axs three unts to the rght of the orgn. () A pont on the y axs one unt elow the orgn. (c) A pont n the second quadrant wth x coordnate and y coordnate +3. 1. 1 1 = = = = 1 1.3 (a) = 1. () (d) * = ( ) = 1. (e) 3 = = ( 1) =. (c) (1 + 5 )( 3 ) = + 1 3 15 = 17 + 7. 4 = ( ) = ( 1) = 1. (f) 1 3 1 3 4 4 14 6 14 = = = =.1.7. 4+ 4+ 4 16+ 8 8+ 4 1.4 (a) 4 () ; (c) 6 3; (d) /5. 1.5 (a) 1, 9 ; (), π/3; π/3 π/3 (c) z = e = ( 1) e. Snce 1 has asolute value 1 and phas, we have π π/3 (4 π/3) z = e e = e = re, so the asolute value s and the phase s 4π/3 radans. (d) 1/ 1/ 1/ z = ( x + y ) = [1 + ( ) ] = 5 ; tan θ = yx / = /1 = and θ = 63.4 = 96.6 = 5.176 radans. 1.6 On a crcle of radus 5. On a lne startng from the orgn and makng an angle of 45 wth the postve x axs. / 1.7 (a) = 1 ; () 1= 1 ; 1/ 5.176 (c) Usng the answers to Pro. 1.5(d), we have 5 e ; 1/ 1/ (d) r = [( 1) + ( 1) ] = ; θ = 18 + 45 = 5 = 3.97 rad; 1/ 3.97 e. 1.8 (a) Usng Eq. (1.36) wth n = 3, we have ( π /3) e = 1, e = cos( π/3) + sn( π/3) =.5 + 3 /, and e (4 π /3) 1-4 Copyrght 14 Pearson Educaton, Inc. =.5 3 /.
1.9 () We see that ω n (1.36) satsfes ωω * = e = 1, so the nth roots of 1 all have asolute value 1. When k n (1.36) ncreases y 1, the phase ncreases y π/n. e e cosθ + sn θ [cos( θ) + sn( θ)] cosθ + sn θ (cosθ sn θ) = = = sn θ, where (.14) was used. e + e cosθ + sn θ + [cos( θ) + sn( θ)] cosθ + snθ + cosθ snθ = = = cos θ. 1.3 (a) From f = ma, 1 N = 1 kg m/s. () 1 J = 1 kg m /s. 19 19 QQ 1 (1.6 1 C)79(1.6 1 C) 1.31 F = = 4πε r 4π ( 8.854 1 C /N-m )(3. 1 m) 1 13 where and 79 are the atomc numers of He and Au. =.45 N, 1.3 (a) () 4 3 4 4 5 4 4xsn(3 x ) + x (1 x )cos(3 x ) = 4xsn(3 x ) + 4x cos(3 x ). 3 x 1 ( x + ) = (8+ ) (1+ 1) = 8. 1.33 (a) T; () F; (c) F; (d) T; (e) F; (f) T. 1-5 Copyrght 14 Pearson Educaton, Inc.