Chapter 1 The Schrodinger Equation 1.1 (a)F:(b)T:(c)T. 1.2 (a)Ephoton=hm=hcl元=(6.626×104Js2.998×108m/s)/1064×109m) 1.867×1019J (b)E=(5×10Js2×108s)=0.1J=m1.867×1019)andn=5×1017 1.3 Use of Ephoton hc/a gives E-602x102X6626×104Js298×10m9-399J 300×109m 1.4(a)）Tma=hw-Φ= (6.626×104Js2.998×108ms)200×10m)-(2.75eV1.602×1019J/eV)= 5.53×1019J=3.45eV and1=4.51×10m=451m. (c)Since the impure metal has a smaller work function,there will be more energy left over after the electron escapes and the maximum T is larger for impure Na. 1.5 (a)At high frequencies,we have e>>1and the-1 in the denominator of Planck's formula can be neglected to give Wien's formula. (b)The Taylor series for the exponential function is ex=+x+x/1+.Forx< we can neglect x and higher powers to give.Takingxh/,we have for Planck's formula at low frequencies ✉-m 2πhm3 1.61=h/mw=137h/mc=137(6.626×104Js)9.109×1031kg)2.998×103m/s)= 3.32×100m=0.332nm. 1.1 Copyright2014 Pearson Education.Ine
1-1 Copyright © 2014 Pearson Education, Inc. Chapter 1 The Schrödinger Equation 1.1 (a) F; (b) T; (c) T. 1.2 (a) photon E h hc == = ν /λ (6.626 × 10–34 J s)(2.998 × 108 m/s)/(1064 × 10–9 m) = 1.867 × 10–19 J. (b) E = (5 × 106 J/s)(2 × 10–8 s) = 0.1 J = n(1.867 × 10–19 J) and n = 5 × 1017. 1.3 Use of photon E hc = /λ gives 23 34 8 9 (6.022 10 )(6.626 10 J s)(2.998 10 m/s) 399 kJ 300 10 m E − − ×× × = = × 1.4 (a) T h max = −Φ= ν (6.626 × 10–34 J s)(2.998 × 108 m/s)/(200 × 10–9 m) – (2.75 eV)(1.602 × 10–19 J/eV) = 5.53 × 10–19 J = 3.45 eV. (b) The minimum photon energy needed to produce the photoelectric effect is (2.75 eV)(1.602 × 10–19 J/eV) = hν =hc/λ = (6.626 × 10–34 J s)(2.998 × 108 m/s)/λ and λ = 4.51 × 10–7 m = 451 nm. (c) Since the impure metal has a smaller work function, there will be more energy left over after the electron escapes and the maximum T is larger for impure Na. 1.5 (a) At high frequencies, we have / 1 b T e ν >> and the −1 in the denominator of Planck’s formula can be neglected to give Wien’s formula. (b) The Taylor series for the exponential function is 2 1 /2! . x e xx =+ + +" For x << 1, we can neglect 2 x and higher powers to give 1 . x e x − ≈ Taking x ≡ h kT ν / , we have for Planck’s formula at low frequencies 3 3 32 / 2/ 2 2 2 22 1 ( 1) ( / ) b T h kT a h h kT e c e c h kT c ν ν ν π ν π ν πν ν = ≈= − − 1.6 λ = h m/ v = = 137 / h mc 137(6.626 × 10–34 J s)/(9.109 × 10–31 kg)(2.998 × 108 m/s) = 3.32 × 10–10 m = 0.332 nm
1.7 Integration gives x=g+(go+vo)+c2.If we know that the particle had position xo at time to:then xo=-gto+(glo+Vo)to+c2 and c2=xo-gto-Volo.Substitution of the expression for c2 into the equation for x givesx=g()2+vo(r-t). 1.8 -h/0ag1a）=-22mXa2Ψar2)+y.ForΨ=ae,we find aΨ/al=-ibΨ，aΨ/ax=-2bmhxΨ，anda2Ψ/ax2=-2bmh-lΨ-2bmh-x(a平/ax) =-2 bmhΨ-2bmh'x(-2bmh'xΨ)=-2bmhlΨ+4b2m2h2x2Ψ.Substituting into the time-dependent Schrodinger equation and then dividing by we get -(h/i)(-ib)=-(h2/2m)(-2bmh+4b2m2h2x2)+and V=2b2mx? 1.9 (a)F;(b)F.(These statements are valid only for stationary states.) 1.10 satisfies the time-independent Schrodinger (1.19)./x=be2bcxe; wx=-2bcxe-4bcxe+b=-6bcxe+4bxe Equation (1.19)becomes (-h2/2mX-6bcxe+4bc2xe)+(2c2h2x2/m)bxe-=Ebxe Thex'terms cancel and E=3h2c/m= 36.626×10-34Js)22.00(109m)214π2(1.00×10-30kg)=6.67×10-20J 1.11 Only the time-dependent equation. 1.12(a)|ΨPdk=(2/b3)x2e24bdk= 2(3.0×109m)3(0.90×109m)2e20.90m3.0nm(0.0001×109m)=3.29×106 (b)For x20,we havex=x and the probability is given by (1.23)and (A.7)as r本=2/6)jre2bx=2/be2(-bx2n-b22-61④呢m= -e2(x21b2+xb+1/26m=-e8(4/9+23+/2)+/2=0.0753. (c)is zero atx=0,and this is the minimum possible probability density. (@∫IΨPk=(2Ib)f°x2e2wb+(2b)小0re2 ds.Let w=-in the first integral on the right This integral becomes which equals the second integral on the right [see Eq.(4.10)].Hence ∫IΨ&=(4/b)j0xe2bk=(4/b'2/(b/2月=1，where(A8)in the Appendix was used. 1-2 Copyright2014 Pearson Education.Inc
1-2 Copyright © 2014 Pearson Education, Inc. 1.7 Integration gives 1 2 2 00 2 x =− + + + gt gt t c ( ). v If we know that the particle had position 0x at time 0t , then 1 2 0 0 0 00 2 2 x =− + + + gt gt t c ( ) v and 1 2 2 0 0 00 2 c x gt t =− − v . Substitution of the expression for 2c into the equation for x gives 1 2 0 0 00 2 x =− − + − x gt t t t ( ) ( ). v 1.8 2 22 − ∂Ψ ∂ =− ∂ Ψ ∂ + Ψ ( / )( / ) ( /2 )( / ) = = i t m xV . For 2 / , ibt bmx ae e − − Ψ = = we find ∂Ψ ∂ = − Ψ / t ib , 1 /2 , x bm x − ∂Ψ ∂ = − Ψ = and 22 1 1 / 2 2 ( /) x bm bm x x − − ∂ Ψ ∂ = − Ψ − ∂Ψ ∂ = = = 1 1 1 1 2 2 22 2 2 (2 ) 2 4 bm bm x bm x bm b m x − −− − − − Ψ− − Ψ =− Ψ+ Ψ = == = = . Substituting into the time-dependent Schrödinger equation and then dividing by Ψ, we get 2 1 2 2 22 ( / )( ) ( /2 )( 2 4 ) i ib m bm b m x V − − − − Ψ =− − + Ψ+ Ψ = = == and 2 2 V b mx = 2 . 1.9 (a) F; (b) F. (These statements are valid only for stationary states.) 1.10 ψ satisfies the time-independent Schrödinger (1.19). 2 / cx ψ x be− ∂ ∂= 2 2 2 cx bcx e− − ; 2 2 2 / 2 cx ψ x bcxe− ∂ ∂ =− − 2 2 2 3 4 4 cx cx bcxe bc x e − − + = 2 2 2 3 6 4 cx cx bcxe bc x e − − − + . Equation (1.19) becomes 2 2 2 23 ( /2 )( 6 4 ) cx cx m bcxe bc x e − − −− + = + 2 2 222 (2 / ) cx cx c x m bxe Ebxe − − = = . The x 3 terms cancel and 2 E cm = = 3 / = 3(6.626 × 10–34 J s)2 2.00(10–9 m)–2/4π 2 (1.00 × 10–30 kg) = 6.67 × 10–20 J. 1.11 Only the time-dependent equation. 1.12 (a) 2 3 2 2| |/ | | (2/ ) x b dx b x e dx − Ψ= = 9 3 9 2 2(0.90 nm)/(3.0 nm) 9 2(3.0 10 m) (0.90 10 m) (0.0001 10 m) e −− − − − ×× × = 3.29 × 10–6. (b) For 0, x ≥ we have | | x = x and the probability is given by (1.23) and (A.7) as 2 nm 2 nm 2 3 2 2 / 3 2 / 2 2 3 2 nm 0 0 0 | | (2 / ) (2 / ) ( /2 /2 /4) | x b xb dx b x e dx b e bx xb b − − Ψ= = − − − ∫ ∫ = 2 / 2 2 2 nm 0 ( / / 1/2) | x b e x b xb − − ++ = 4/3 e (4/9 2/3 1/2) 1/2 − − ++ + = 0.0753. (c) Ψ is zero at x =0, and this is the minimum possible probability density. (d) 0 2 3 22/ 3 2 2/ 0 | | (2/ ) (2/ ) . xb xb dx b x e dx b x e dx ∞ ∞ − −∞ −∞ Ψ= + ∫∫ ∫ Let w = –x in the first integral on the right. This integral becomes 0 2 2/ 2 2/ 0 () , wb wb w e dw w e dw ∞ − − ∞ − = ∫ ∫ which equals the second integral on the right [see Eq. (4.10)]. Hence 2 3 2 2/ 3 3 0 | | (4 / ) (4 / )[2!/ ( / 2) ] x b dx b x e dx b b ∞ ∞ − −∞ Ψ= = ∫ ∫ = 1, where (A.8) in the Appendix was used
1.13 The interval is small enough to be considered infinitesimal(since changes negligibly whthis interval.)At1=0,we have1YP在=g2/元cyr2e-2rie在- [32/x2.00A)1(2.00A)2e2(0.001A)=0.000216. 4r=mae2aok=eaa2=(em+em2= 1.5001nm 4.978×106 1.15 (a)This function is not real and cannot be a probability density. (b)This function is negative whenxand cannot be a probability density. (c)This function is not normalized(unless b=)and can't be a probability density. 1.16 (a)There are four equally probable cases for two children:BB,BG,GB,GG,where the first letter gives the gender of the older child.The BB possibility is eliminated by the given information.Of the remaining three possibilities BG.GB.GG.only one has two girls,so the probability that they have two girls is 1/3. (b)The fact that the older child is a girl eliminates the BB and BG cases,leaving GB and GG.so the probability is 1/2 that the younger child is a girl. 1.17 The 138 peak arises from the case CCF,whose probability is(0.9889)2=0.9779 The 139 peak arises from the cases CCFsandCCF whose probability is (0.9889x0.011)+(0.011IX0.9889=0.02195.The140 peak arises from5ccF6 whose probability is(0.0111)2=0.000123.(As a check,these add to 1.)The 139 peak height is(0.02195/0.9779)100=2.24.The140 peak height is(0.000123/0.9779)100= 0.0126. 1.18 There are 26 cards,2 spades and 24 nonspades,to be distributed between B and D. Imagine that 13 cards,picked at random from the 26,are dealt to B.The probability that every card dealt to B is a nonspade is装器号}告是是==÷.Likewise,the probability that Dgets 13 nonspades isIf B does not get all nonspades and D does not get all nonspades,then each must get one of the two spades and the probability that each gets one spade is 1-=13/25.(A commonly given answer is:There are four possible outcomes,namely,both spades to B,both spades to D,spade I to B and spade 2 to D.spade 2 to B and spade 1 to D.so the probability that each gets one spade is 2/4= 1/2.This answer is wrong,because the four outcomes are not all equally likely.) 1-3 Copyright2014 Pearson Education.Inc
1-3 Copyright © 2014 Pearson Education, Inc. 1.13 The interval is small enough to be considered infinitesimal (since Ψ changes negligibly within this interval). At t = 0, we have 2 2 2 6 1/2 2 2 / | | (32 / ) x c dx c x e dx π − Ψ = = [32/π(2.00 Å)6 ] 1/2(2.00 Å)2 e –2 (0.001 Å) = 0.000216. 1.14 2 1 2/ 2/ 1.5001 nm 1.5001 nm 1.5000 nm 1.5000 nm | | /2| b xa xa a dx a e dx e −− − Ψ = =− = ∫ ∫ (–e –3.0002 + e –3.0000)/2 = 4.978 × 10–6. 1.15 (a) This function is not real and cannot be a probability density. (b) This function is negative when x < 0 and cannot be a probability density. (c) This function is not normalized (unless ) b = π and can’t be a probability density. 1.16 (a) There are four equally probable cases for two children: BB, BG, GB, GG, where the first letter gives the gender of the older child. The BB possibility is eliminated by the given information. Of the remaining three possibilities BG, GB, GG, only one has two girls, so the probability that they have two girls is 1/3. (b) The fact that the older child is a girl eliminates the BB and BG cases, leaving GB and GG, so the probability is 1/2 that the younger child is a girl. 1.17 The 138 peak arises from the case 12C12CF6, whose probability is (0.9889)2 = 0.9779. The 139 peak arises from the cases 12C13CF6 and 13C12CF6, whose probability is (0.9889)(0.0111) + (0.0111)(0.9889) = 0.02195. The 140 peak arises from 13C13CF6, whose probability is (0.0111)2 = 0.000123. (As a check, these add to 1.) The 139 peak height is (0.02195/0.9779)100 = 2.24. The 140 peak height is (0.000123/0.9779)100 = 0.0126. 1.18 There are 26 cards, 2 spades and 24 nonspades, to be distributed between B and D. Imagine that 13 cards, picked at random from the 26, are dealt to B. The probability that every card dealt to B is a nonspade is 24 22 21 14 12 23 13 6 13(12) 26 25 24 23 16 15 14 26(25) 25 " = = . Likewise, the probability that D gets 13 nonspades is 6 25 . If B does not get all nonspades and D does not get all nonspades, then each must get one of the two spades and the probability that each gets one spade is 6 6 25 25 1 13 /25 −−= . (A commonly given answer is: There are four possible outcomes, namely, both spades to B, both spades to D, spade 1 to B and spade 2 to D, spade 2 to B and spade 1 to D, so the probability that each gets one spade is 2/4 = 1/2. This answer is wrong, because the four outcomes are not all equally likely.)
1.19 (a)The Maxwell distribution of molecular speeds:(b)the normal (or Gaussian) distribution. 1.20 (a)Real;(b)imaginary;(c)real;(d)imaginary;(e)imaginary:(f)real; (g)real;(h)real;(1)real. 1.21 (a)A point on thex axis three units to the right of the origin. (b)A point on theyaxis one unit below the origin. (c)A point in the second quadrant withx coordinate-2 andy coordinate +3. 1.23(a）2=-1.(b)=i2=i(-1)=-i.(c)=(2)2=(-1)2=1. (d)i=(-)i=1 (e)(1+5i)2-30)=2+101-31-152=17+7i. 0-6201-0 1.24(a4(b)2,(d6-3元，(d2e5 1.25(a)1,90°：(b)2,π/3 (e)-2e=2(-)e3.Since-1 has absolute value I and phase we have =2e=2e()=re,so the absolute value is2 and the phase is 4/3 radians (d1zl=(x2+y2)2=2+(-2}2]2=52;tan6=y/x=-2/1=-2and 0=-63.4°=296.6°=5.176 radians. 1.26 On a circle of radius 5.On a line starting from the origin and making an angle of 45 with the positivex axis. 1.27(a)i=ler12,(b)-1=le; (c)Using the answers to Prob.1.25(d),we have 5276 (dr=[(-1)2+(-1)2]2=22;0=180°+45°=225°=3.927rad,22e3.927 1.28 (a)Using Eq.(1.36)with n=3,we have e=1, e2m/)=cos(2π/3)+isin(2r/3)=-0.5+i5/2,ande4m/3)-0.5-i5/2. 1-4 Education.Inc
1-4 Copyright © 2014 Pearson Education, Inc. 1.19 (a) The Maxwell distribution of molecular speeds; (b) the normal (or Gaussian) distribution. 1.20 (a) Real; (b) imaginary; (c) real; (d) imaginary; (e) imaginary; (f) real; (g) real; (h) real; (i) real. 1.21 (a) A point on the x axis three units to the right of the origin. (b) A point on the y axis one unit below the origin. (c) A point in the second quadrant with x coordinate –2 and y coordinate +3. 1.22 2 1 1 1 ii i i i ii i = = = =− − 1.23 (a) 2 i = −1. (b) 3 2 i ii i i = = − =− ( 1) . (c) 4 22 2 i i = ( ) ( 1) 1. =− = (d) * ( ) 1. ii ii =− = (e) 2 (1 5 )(2 3 ) 2 10 3 15 17 7 . + − =+ − − = + i i ii i i (f) 1 3 1 3 4 2 4 14 6 2 14 0.1 0.7 . 4 2 4 2 4 2 16 8 8 4 20 i ii i i i i i i ii − − − − − −− = = = =− − + + − +−+ 1.24 (a) –4 (b) 2i; (c) 6 – 3i; (d) /5 2 . i e π 1.25 (a) 1, 90°; (b) 2, π/3; (c) /3 /3 2 2( 1) . i i ze e π π =− = − Since –1 has absolute value 1 and phase π, we have /3 (4 /3) 22 , ii i i z e e e re π π πθ = == so the absolute value is 2 and the phase is 4π/3 radians. (d) 2 2 1/2 2 2 1/2 1/2 | | ( ) [1 ( 2) ] 5 ; z xy = + = +− = tan / 2 / 1 2 θ = y x =− =− and θ = –63.4° = 296.6° = 5.176 radians. 1.26 On a circle of radius 5. On a line starting from the origin and making an angle of 45° with the positive x axis. 1.27 (a) /2 1 ; i i e π = (b) 11 ; i e π − = (c) Using the answers to Prob. 1.25(d), we have 1/2 5.176 5 ;i e (d) 2 2 1/2 1/2 r = − + − = = °+ °= °= [( 1) ( 1) ] 2 ; 180 45 225 3.927 θ rad; 1/2 3.927 2 .i e 1.28 (a) Using Eq. (1.36) with n = 3, we have 0 1, i e ⋅ = (2 /3) cos(2 /3) sin(2 /3) 0.5 3 /2, i e ii π = + =− + π π and (4 /3) 0.5 3 /2. i e i π =− −
(b)We see that in(1.36)satisfies *=e=1,so the nth roots of I all have absolute value 1.When k in(1.36)increases by 1,the phase increases by 2/n. 1.29e-e0 cisinc()isin()cos0+isin-(cos0-isin)sin0 2i 2i 2i where(2.14)was used. cos+isin+cos(-0)+isin(-0)cos0+isin+cos-isincos0 2 2 1.30 (a)From f=ma,1N=1 kg m/s2. (b)1J=1kg m2/s2 151F-=-cgw09-0sN 21.602×10-19C)791.602×10-19C) where 2 and 79 are the atomic numbers of He and Au. 1.32(a)4xsin(3x4)+2x212xr3)cos(3x4)=4xsin(3x4)+24x3cos(3x4). (b)(x3+x)=(8+2)-(1+1)-8. 133(a)T,(b)F;(c)F;(dT(e)F,(国T. 1-5 Copyright2014 Pearson Education.Inc
1-5 Copyright © 2014 Pearson Education, Inc. (b) We see that ω in (1.36) satisfies 0 ωω* 1, = e = so the nth roots of 1 all have absolute value 1. When k in (1.36) increases by 1, the phase increases by 2π/n. 1.29 cos sin [cos( ) sin( )] cos sin (cos sin ) 22 2 i i ee i i i i ii i θ θ θ θ θ θ θθθθ − − + − −+ − + − − = = = sin θ, where (2.14) was used. cos sin [cos( ) sin( )] cos sin cos sin 22 2 i i ee i i i i θ θ θ θ θ θ θ θθ θ − + + + −+ − + + − = = = cos θ. 1.30 (a) From , f = ma 1 N = 1 kg m/s2 . (b) 1 J = 1 kg m2 /s2 . 1.31 F = 19 19 1 2 2 12 2 2 13 2 0 2(1.602 10 C)79(1.602 10 C) 4 4 8.854 10 C /N-m )(3.00 10 m) Q Q πε π r − − − − × × = (× × = 0.405 N, where 2 and 79 are the atomic numbers of He and Au. 1.32 (a) 4 23 4 4 5 4 4 sin(3 ) 2 (12 )cos(3 ) 4 sin(3 ) 24 cos(3 ). x x xx x x x x x + =+ (b) 3 2 1 ( ) | (8 2) (1 1) 8. x x + = + −+= 1.33 (a) T; (b) F; (c) F; (d) T; (e) F; (f) T