(b)For E =0.499,0.500,0.501,the values of y/co at x=4 are 0.684869, 0.000335463.and-0.68198. 4.24 (a)With the harmonic-oscillator approximation for the molecular vibration,Eq.(4.61) gives the molecular vibration frequency as=8.65x103 s From(4.59).k=4u and u=mm/(m+mz).From Table A.3 in the Appendix, -0mD=1a7io“e 1 S0k=4π2v24=4π2(8.65×1015s21.627×1027kg)=481N/m (b)号hm-0.5(6.626×1034Js8.65×1013s=2.87×10-20J. (c)From the last equation in(4.59),the force constant k of a molecule is found from the U(R)function.The electronic energy function Uis found by repeatedly solving the electronic Schrodinger equation at fixed nuclear locations.The nuclear masses do no occur in the electronic Schrodinger equation,so the function U is independent of the nuclear masses and is the same for is the same for these two molecules.From the first equation in (4.59),=(),where 2 and 1 refer to2HCl and HCl,respectively.From Table A.3, -2014(34.97g 2.014+34.976022×103-3162x1024g S032=(41/42)24=(1.627/3.162)2(8.65×103s=6.20×103s 4.25 (a)Putting v2=1 and v2=2 in the result of Prob.4.27b,we have 2885.98cm=。-2ex。and5667.98cm=2m。-6vx。.Subtracting twice the first equation from the second,we get-103.98 cm-=-2x and=51.99 cm-1.The first equation then givesv=2885.98 cm+2(51.99 cm)=2989.96 cm-Also, 。=.c=(2989.96cm2.99792×100cm/s)=8.96366×103 s and vx=exc= (51.99cm-1)2.99792×100cm/s)=1.559×102s (b)With v2=3,the result of Prob.4.27b becomes=3-1 32989.96cm-12(51.99cm=8346.00cm. 4.26 (a)Using the harmonic-oscillator approximation,the energy difference between these two vibrational levels is =hc=(6.)(1359 cm(2./s)= 2.70 x 10 J.The Boltzmann distribution law (4.63)for these nondegenerate levels gives N,/N。=exp[-2.70×1020J)/(1.381×1023J/K)298K】=0.0014at25℃and N,/N0=exp[(-2.70×10-20J)1.381×10-23J/K)473K)=0.016at200℃. 4-8 Copyright2014 Pearson Education.Inc
4-8 Copyright © 2014 Pearson Education, Inc. (b) For 0.499, 0.500, 0.501, Er = the values of 0 ψ /c at 4 r x = are 0.684869, 0.000335463, and –0.68198. 4.24 (a) With the harmonic-oscillator approximation for the molecular vibration, Eq. (4.61) gives the molecular vibration frequency as 13 1 ν 8.65 10 s− = × . From (4.59), 2 2 k = 4π ν μ and 12 1 2 μ = + mm m m /( ). From Table A.3 in the Appendix, 24 23 1.008(34.97) g 1 1.627 10 g (1.008 34.97) 6.022 10 μ − = =× + × So 2 2 k = 4π ν μ = 4π 2 (8.65 ×1013 s–1) 2 (1.627 × 10–27 kg) = 481 N/m. (b) 1 2 hν = 0.5(6.626 × 10–34 J s)(8.65 × 1013 s–1) = 2.87 × 10–20 J. (c) From the last equation in (4.59), the force constant k of a molecule is found from the U(R) function. The electronic energy function U is found by repeatedly solving the electronic Schrödinger equation at fixed nuclear locations. The nuclear masses do not occur in the electronic Schrödinger equation, so the function U is independent of the nuclear masses and is the same for 2 H35Cl as for 1 H35Cl. Hence k is the same for these two molecules. From the first equation in (4.59), 1/2 21 1 2 ν / (/ ) ν μμ = , where 2 and 1 refer to 2 H35Cl and 1 H35Cl, respectively. From Table A.3, 24 2 23 2.014(34.97) g 1 3.162 10 g (2.014 34.97) 6.022 10 μ − = =× + × So 1/2 2 12 1 ν = = (/) μμ ν (1.627/3.162)1/2(8.65 ×1013 s–1) = 6.20 ×1013 s–1. 4.25 (a) Putting 2 v = 1 and 2 v = 2 in the result of Prob. 4.27b, we have 1 2885.98 cm 2 e ee ν ν x − = − and 1 5667.98 cm 2 6 e ee ν ν x − = − . Subtracting twice the first equation from the second, we get 1 103.98 cm 2 e e ν x − − =− and 1 ν e e x 51.99 cm− = . The first equation then gives 11 1 2885.98 cm 2(51.99 cm ) 2989.96 cm . ν e − − − =+ = Also, e e ν = = ν c ( 1 2989.96 cm− )(2.99792 ×1010 cm/s) = 8.96366 × 1013 s–1 and ee ee ν x = = ν x c ( 1 51.99 cm− )(2.99792 ×1010 cm/s) = 1.559 × 1012 s–1. (b) With 2 v = 3, the result of Prob. 4.27b becomes light 3 12 e ee ν = ν ν − = x 3(2989.96 cm–1) – 12(51.99 cm–1) = 8346.00 cm–1. 4.26 (a) Using the harmonic-oscillator approximation, the energy difference between these two vibrational levels is h hc ν = = ν (6.626 × 10–34 J s)(1359 cm–1)(2.998 × 1010 cm/s) = 2.70 × 10–20 J. The Boltzmann distribution law (4.63) for these nondegenerate levels gives 20 23 1 0 N N/ exp[( 2.70 10 J)/(1.381 10 J/K)(298 K)] − − = −× × = 0.0014 at 25°C and 20 23 1 0 N N/ exp[( 2.70 10 J)/(1.381 10 J/K)(473 K)] − − = −× × = 0.016 at 200°C
(b)hm=hmc=(6.626×1034Js)381cm2.998×1010cm/s)=7.57×1021J N/N0=expl(-757×10-21JM1381x10-23jK298K】=0.16at25Cand N/N0=exp-7.57×10-21Jy1,381x10-23JK473K】=0.31at200℃ 4.27(a eu=(E2-E)h=h{2+号加。-(w2+2hm。-[+h。-(W+}x]}= (v2-vi )v+vx[(V-v)+(V -v2)](Eq.1).Use of the selection rule v2-v=1 gives Vlight Ve+Vex[vi -(V +1)-I]=V.-2vx(Vi+1). (b)Putting v=0 in Eq.I of part (a),we get vlight =v2V-v+V2) 4.28 The Taylor series(4.85)of Prob.4.1 with x=R,f(x)=U(R).and a=R.gives U(R)=U(R)/0I+U"(R(R-R)/1I+U"(R)(R-R)2/21+U"(R)R-R)3/31+ Since R.occurs at the minimum in the U(R)curve,we have U'(R.)=0.From(4.59). U"(R)=k.The zero of potential energy can be chosen wherever we please,so we can take U(R.)=0,as in Fig.4.6.Neglecting the (R-R.)3 term and higher terms,we thus have U(R)k(R-R)2=kx2,where x=R-R. 4.29 (a)Putting R=o and then R=R.in the Morse function,we get U()=D.and U(R.)=0.So U(co)-U(R.)=D. (b)From(4.59),k=U"(R.).For the Morse function, U'-2D.[-e-a(R-Rae-R-R)-2aD.le-(R-R)-e-2a(R-R)]and U"=2aD.[-ae-4(R-R)+2ae-2(R-RJ.Then k=U(R,)=2aD.(-a+2a)=2a'D.so a-(k/2D)2 4.30 We begin by finding combinations of m.l,and h that have dimensions of energy and of length.The reduced energy and x coordinate are E,E/A and x,=x/B. Let A=m1h.Using (4.71)and (4.70).we have [A]=ML2T2=[m"Pm ]ML (ML2T-)MateLb+2eT-,so a+c=1,b+2c=2,-c=-2.Hence c=2,a=-1,b=-2 and EE/(h2 /ml2). Let B=mh/.We have [B]=L=MdL(ML2T-)=Md+/Le+2/T-,so d+f=0,e+2f=1,-f=0.Hence,f=0,d=0,e=1,and x,=x/l,as is obvious without doing the detailed analysis.From (4.78)and (479).and "="B52=1P.The Schrodinger equation-(h2/2m"=Ev becomes 4-9 Copyright2014 Pearson Education.Inc
4-9 Copyright © 2014 Pearson Education, Inc. (b) h hc ν = = ν (6.626 × 10–34 J s)(381 cm–1)(2.998 × 1010 cm/s) = 7.57 × 10–21 J. 21 23 1 0 N N/ exp[( 7.57 10 J)/(1.381 10 J/K)(298 K)] − − = −× × = 0.16 at 25°C and 21 23 1 0 N N/ exp[( 7.57 10 J)/(1.381 10 J/K)(473 K)] − − = −× × = 0.31 at 200°C. 4.27 (a) { } 12 2 1 1 11 light 2 1 2 2 1 1 2 2 22 ( )/ ( ) ( ) ( ) ( ) ν E E h h h hx h hx ν ν νν e ee e ee − = − = + − + − + −+ ⎡ ⎤ ⎣ ⎦ vv vv = 2 2 21 1 2 12 ( ) [( ) ( )] e ee vv v v vv − + − +− ν ν x (Eq. 1). Use of the selection rule 2 1 v v − = 1 gives 2 2 light 1 1 1 [ ( 1) 1] 2 ( 1). e ee e ee ν = + − + −= − + νν ν ν x x vv v (b) Putting 1 v = 0 in Eq. 1 of part (a), we get 2 light 2 2 2 ( ) e ee ν =− + v vv ν ν x . 4.28 The Taylor series (4.85) of Prob. 4.1 with x = R , f ( ) ( ), x UR = and e a R = gives 2 3 ( ) ( ) / 0! ( )( ) / 1! ( )( ) /2! ( )( ) /3! UR UR U R R R U R R R U R R R e ee ee ee = + −+ − + − + ′ ′′ ′′′ ". Since Re occurs at the minimum in the ( ) U R curve, we have ( ) 0. U Re ′ = From (4.59), () . UR k e ′′ = The zero of potential energy can be chosen wherever we please, so we can take ( ) 0 U Re = , as in Fig. 4.6. Neglecting the 3 ( ) R − Re term and higher terms, we thus have 1 1 2 2 2 2 () ( ) , U R k R R kx ≈ −= e where e x ≡ R R − . 4.29 (a) Putting R = ∞ and then R = Re in the Morse function, we get ( ) U D ∞ = e and ( ) 0. U Re = So ( ) ( ) . U UR D ∞− = e e (b) From (4.59), ( ). e e k UR = ′′ For the Morse function, ( ) ( ) ( ) 2( ) 2 [1 ] 2 [ ] ee e e aR R aR R aR R aR R U D e ae aD e e e e −− −− −− − − ′ =− = − and ( ) 2( ) 2 [ 2 ]. e e aR R aR R U aD ae ae e −− − − ′′ =− + Then 2 ( ) 2 ( 2) 2 eee e k U R aD a a a D = = −+ = ′′ , so 1/2 ( /2 ) . e e akD = 4.30 We begin by finding combinations of m, l, and = that have dimensions of energy and of length. The reduced energy and x coordinate are / E EA r ≡ and / . r x ≡ x B Let . ab c A ml = = Using (4.71) and (4.70), we have [A] = ML2 T–2 = [ ab c m l = ] = 21 2 M L (ML T ) M L T , a b c ac b c c − ++ − = so ac b c c + = + = − =− 1, 2 2, 2. Hence ca b = 2, 1, 2 =− =− and 2 2 / ( / ). E E ml r = = Let de f B = m l = . We have 21 2 [ ] L M L (ML T ) M L de f df e f f B T − ++ − == = , so df e f f += + = −= 0, 2 1, 0. Hence, 0, 0, 1, f de = = = and / , r x = x l as is obvious without doing the detailed analysis. From (4.78) and (4.79), 1/2 1/2 ψ r = = ψ ψ B l and 5/2 5/2 ψ ψ ψ r r B l − − ′′ ′′ ′′ = = . The Schrödinger equation 2 − = ( /2 ) = m E ψ′′ ψ becomes
-(h2/2m)=(h2/mP)EI or w=-2E.To put this equation in the form of Eq.(4.66)and the first equation in(4.82),we define G,=-2E,to give=G The formula in cell B7 and the cells below it in Fig.4.9 becomes =-2SBS3.There is no penetration into the classically forbidden region,so we omit steps(c)and (d)at the endof Sec.4.4.The variable x,=x/l runs from 0 to 1.We take the interval s,as 0.01.We enter .0001 in C8.The formulas in column C are the same as in Fig 4.9.The Solve is set to make C107 equal to zero by varying B3.The lowest three E,=42E/(h2/m) eigenvalues are found to be 4.9348021805,19.7392075201,and 44.41320519866.(For maximum accuracy,use the Options button in the Solver to reduce the Precision to 10-)These E,values correspond to E values of h2/m/2 times 0.12499999949. 0.4999999675,and 1.124999630,as compared with the true values of h2/mltimes n2/8=0.125,0.500,and1.125. 4.31 (a)As in Prob.4.30,we take combinations of m,and h that have dimensions of energy and of length;the reduced energy and x coordinate are E=E/4=E/(h2/m) and x,=x/B =x/I.The Schrodinger equation is-(/2m)"+K(h /=Ev 202 becomes -(h212m)sPw+K(h2 /ml2)w (h2/ml2)EI-Pw,or =(2K-2EThe bound-state reduced energies are less than 20,so the maximum reduced energy we are interested in is 20.For reduced energies less than 20,the classically forbidden regions are regions I and III in Fig.2.5.Reasonable starting and ending points are 1.5 units into each of the classically forbidden regions,so we shall take x,to run from-1.5 to 2.5.A reasonable interval iss=0.02 or 0.01.For greater accuracy,we shall use 0.01.The K value for regions I and III is entered into cell B2 of Fig.4.9.In column B.x values in regions I(from-1.5 to 0)and III (from I to 2.5) contain the formula 2*$BS2-2*$B$3 and x,values in region II (from 0 to 1)contain the formula-2*$SBS3.The formulas in column C are the same as in Fig.4.9.The Solver is set to make C407 equal to zero by varying B3.The Options button in the Solver is used to set the Precision at 108.The bound-state E.=4z2E/(h2/m/2)eigenvalues are found to be 2.772515720011 and 10.6051190761.(A value of 20.213299 is also obtained,but the graph shows that the solution for this energy does not go to zero asymptotically in the forbidden region.) (b)The spreadsheet of part(a)is modified by cell B2 from 20 to 50.The Solver gives the E,values3.3568218287,13.256836483275,29.003101429782,and 47.66519784181 4-10 Copyright2014 Pearson Education.Inc
4-10 Copyright © 2014 Pearson Education, Inc. 2 5/2 2 2 1/2 ( /2 ) ( / ) m l ml E l ψ r rr ψ − − − = = = ′′ or 2 . ψ r rr ′′ = − E ψ To put this equation in the form of Eq. (4.66) and the first equation in (4.82), we define 2 G E r r ≡ − to give . ψ r rr ′′ = G ψ The formula in cell B7 and the cells below it in Fig. 4.9 becomes =-2*$B$3. There is no penetration into the classically forbidden region, so we omit steps (c) and (d) at the end of Sec. 4.4. The variable / r x = x l runs from 0 to 1. We take the interval rs as 0.01. We enter 0.0001 in C8. The ψ r formulas in column C are the same as in Fig. 4.9. The Solver is set to make C107 equal to zero by varying B3. The lowest three 2 22 4 /( / ) E E h ml r = π eigenvalues are found to be 4.9348021805, 19.7392075201, and 44.41320519866. (For maximum accuracy, use the Options button in the Solver to reduce the Precision to 14 10 . − ) These Er values correspond to E values of 2 2 h ml / times 0.12499999949, 0.4999999675, and 1.124999630, as compared with the true values of 2 2 h ml / times 2 n /8 = 0.125, 0.500, and 1.125. 4.31 (a) As in Prob. 4.30, we take combinations of m, l, and = that have dimensions of energy and of length; the reduced energy and x coordinate are / E EA r ≡ = 2 2 E/( / ) = ml and / r x ≡ x B = x/l . The Schrödinger equation is 2 22 −+ = ( /2 ) ( / ) = = m K ml E ψ′′ ψ ψ , where K = 20 in regions I and III of Fig. 2.5, and K = 0 in region II. From (4.78) and (4.79), 1/2 1/2 ψ r = = ψ ψ B l and 5/2 5/2 ψ ψ ψ r r B l − − ′′ ′′ ′′ = = . The Schrödinger equation becomes 2 5/2 2 2 1/2 2 2 1/2 ( /2 ) ( / ) ( / ) m l K ml l ml E l ψ r r rr ψ ψ −− − −+ = = == ′′ or (2 2 ) . ψ r rr ′′ = − K E ψ The bound-state reduced energies are less than 20, so the maximum reduced energy we are interested in is 20. For reduced energies less than 20, the classically forbidden regions are regions I and III in Fig. 2.5. Reasonable starting and ending points are 1.5 units into each of the classically forbidden regions, so we shall take r x to run from –1.5 to 2.5. A reasonable interval is 0.02 rs = or 0.01. For greater accuracy, we shall use 0.01. The K value for regions I and III is entered into cell B2 of Fig. 4.9. In column B, r x values in regions I (from –1.5 to 0) and III (from 1 to 2.5) contain the formula 2*$B$2-2*$B$3 and r x values in region II (from 0 to 1) contain the formula -2*$B$3. The ψ r formulas in column C are the same as in Fig. 4.9. The Solver is set to make C407 equal to zero by varying B3. The Options button in the Solver is used to set the Precision at 8 10− . The bound-state 2 22 4 /( / ) E E h ml r = π eigenvalues are found to be 2.772515720011 and 10.6051190761. (A value of 20.213299 is also obtained, but the graph shows that the solution for this energy does not go to zero asymptotically in the forbidden region.) (b) The spreadsheet of part (a) is modified by changing cell B2 from 20 to 50. The Solver gives the Er values 3.3568218287, 13.256836483275, 29.003101429782, and 47.66519784181
(e)Substitution of Vo=20h2/ml2 in(2.34)for b gives b=6.3245553203 and b/=2.0132,so there are three bound states.The Solver shows the roots of Eq.(2.35)to be=0.1407215,0.5375806,0.9995981.From(2.34),En=eo=20e= 2.814429,10.75161,19.99196.The eigenvalues found in(a)are rather inaccurate indicating that we need to go further into the classically forbidden regions and decrease the interval.For Vo =50h2/ml2,one finds b=10; 6=0.06827142,0.26951445,0.58904615,0.9628693; E=50=3.413571,13.47572,29.452308,48.143464 The eigenvalues in(b)are rather inaccurate 4.32 We begin by finding combinations of m,c,and h that have dimensions of energy and of length.c has dimensions of energy divided by length',so [c]=MLT2=MT2L2. The reduced energy and x coordinate are E =E/A and x.=x/B. Let =mc Using (4.71)and (4.70),we have [A]=ML2T2=[m ]M(ML2T-1)(MT-21-2)d=Ma+b+dL26-24T-b-2d.so a+b+d=1,2b-2d=2.-b-2d=-2.Adding twice the third equation to the second we get -6d =-2 and d=.Then b=and a=-2.So E.E/A=E/m 2/3h413cl3. Let B=mh/.We have [B]=L=M(ML2T-)(MT-2L2)=M+/+L2/-28T-/-28.so e+f+g=0,2f-2g=1,-f-2g=0.Subtracting the third equation from the second,we getf.Theng=andeSo=B=x/m-6.The Schrodinger equation is(mEw.From (4.78)and (4.79). and =B-2B2=B-2m2.The Schrodinger equation becomes (2m)B-V2m+cxmcB =mE,B-,and =(2x-E,=G,W,where G=2x-E,.Let us find eigenvalues with Es10.Setting this maximum E,equal to V.we have 10=and the classically allowed region is bounded by x=+1.78.We shall start well into the classically forbidden region at x=-3.5 and go to=+3.5 in steps of 0.05.Cell B7 of Fig.4. contains the formula 2*A704-2*SB$3 and this is copied to other column B cells.With 0.001 in cell C8,a suitable Precision(set by clicking the Solver Options button)is 0.1. The Solver gives the lowest three eigenvalues as E,=E1m23n4c5=0.667986133,2.39364258,4.69678795 4.33 Proceeding similarly as in Prob.4.32,we have [a]=ML2T-2/13=MT-2L6 E,=E/A and x =x/B.Let A=mnad.Then [A]=ML2T2=[mh ad ] 4-11 Education.Inc
4-11 Copyright © 2014 Pearson Education, Inc. (c) Substitution of 2 2 0 V ml = 20 / = in (2.34) for b gives b = 6.3245553203 and b/ 2.0132, π = so there are three bound states. The Solver shows the roots of Eq. (2.35) to be 0.1407215, 0.5375806, 0.9995981. ε = From (2.34), 0, 20 E V r r = ε = = ε 2.814429, 10.75161, 19.99196. The eigenvalues found in (a) are rather inaccurate, indicating that we need to go further into the classically forbidden regions and decrease the interval. For 2 2 0 V ml = 50 / = , one finds b = 10; ε = 0.06827142, 0.26951445, 0.58904615, 0.9628693; 50 3.413571, 13.47572 29.452308, 48.143464. , Er = = ε The eigenvalues in (b) are rather inaccurate. 4.32 We begin by finding combinations of m, c, and = that have dimensions of energy and of length. c has dimensions of energy divided by length4 , so 2 24 22 [ ] ML T /L MT L c − −− = = . The reduced energy and x coordinate are / E EA r ≡ and / . r x ≡ x B Let abd Am c = = . Using (4.71) and (4.70), we have [A] = ML2 T–2 = [ abd m c = ] = 2 1 2 2 22 2 M (ML T ) (MT L ) M L T , a b d abd b d b d − − − ++ − −− = so 1, 2 2 2, 2 2 abd b d b d + + = − = − − =− . Adding twice the third equation to the second, we get − =− 6 2 d and 1 3 d = . Then 4 3 b = and 2 3 a = − . So 2/3 4/3 1/3 // . E EA Em c r − = = = Let efg B = m c = . We have 2 1 22 22 2 [ ] L M (ML T ) (MT L ) M L T e f g efg f g f g B − − − ++ − −− == = , so efg f g f g ++= − = −− = 0, 2 2 1, 2 0 . Subtracting the third equation from the second, we get 1 3 f = . Then 1 6 g = − and 1 6 e = − . So 1/6 1/3 1/6 / / r x xB xm c − − = = = . The Schrödinger equation is 2 4 − += ( /2 ) = m cx E ψ′′ ψ ψ . From (4.78) and (4.79), 1/2 ψ r =ψ B and 1/2 2 1/2 1/3 2/3 1/3 ψ ψ ψ r r BB Bm c −− − − ′′ ′′ ′′ = = = . The Schrödinger equation becomes 2 1/2 1/3 2/3 1/3 4 2/3 4/3 2/3 1/2 2/3 4/3 1/3 1/2 ( /2 ) m B m c cx m c B m c E B ψ ψψ rr r r r − − − −− − − − += == = = ′′ and 4 (2 2 ) r r r r rr ψ′′ =− = xE G ψ ψ , where 4 2 2 GxE rr r ≡ − . Let us find eigenvalues with 10. Er ≤ Setting this maximum Er equal to Vr , we have 4 10 r = x and the classically allowed region is bounded by 1.78 r x = ± . We shall start well into the classically forbidden region at 3.5 r x = − and go to 3.5 r x = + in steps of 0.05. Cell B7 of Fig. 4.9 contains the formula 2*A7^4-2*$B$3 and this is copied to other column B cells. With 0.001 in cell C8, a suitable Precision (set by clicking the Solver Options button) is 0.1. The Solver gives the lowest three eigenvalues as 2/3 4/3 1/3 / 0.667986133, 2.39364258, 4.69678795. E Em c r − ≡ = = 4.33 Proceeding similarly as in Prob. 4.32, we have 2 2 8 26 [ ] ML T / L MT L a − − − = = . / E EA r ≡ and / . r x ≡ x B Let bcd Am a = = . Then [A] = ML2 T–2 = [ bcd m a = ] =
M(ML2T-1)(MT-2L6)d=MbetdL26-6dT-24,so b+c+d=l,2c-6d=2-c-2d=-2.d=3,c=g,b=- E,=EIA=E/m55a5.Let B=mhl as.We have [B]=L=M(ML2T-1)/(MT-2L-6)F Me+/+EL2/-68T-/-28,so e+f+g=0,2f-6g=l-f-2g=0.g=-,f=,e=-0 x,=x/B=x/m-oa-n.The Schrodinger equation is-(h2/2m)y"+ax=Ev From (4.78)and (4.79).W=wBR and w"=w"B-2B2=B-2m5h25a.The Schrodinger equation becomes -(h2/2m)B-2m5h-25a/5w,+am45h5a45B-1/y,=m4/5h5a5E,B-1/2w,and =(2x=G,where G-2E,.Let us find eigenvalues with E,10.Setting this maximum E,equal to V,we have 10=and the classically allowed region is bounded byx=1.33.We shall start well into the classically forbidden region at x =-3 and go to x=+3 in steps of 0.02.Cell B7 of Fig.4.9 contains the formula 2A78-2"SBS3 and this is copied to other column Bcells.With 0.001 in cell C8,a suitable Precision(set by clicking the Solver Options button)is 0.1 The Solver gives the lowest three eigenvalues as E=E/A=E/= 0.70404876,2.731532,5.884176 4.34 Proceeding similarly as in Prob.4.32,we have [b]=MLT2/L=MT-2L E.=E/A and x =x/B.Let 4=m"hbd.Then [A]=ML2T2=Im"hb]= M(MLT(MT-L)=ML-24,so a+c+d=1,2c+d=2 -c-2d=-2.d=c=a=-3.So E=ElA=Elm-1Bh2136213.Let B=mh!bs We have [B]=L=M(ML2T-y(MT-2L)5=M+/+L2/+T-2,so e+f+g=0,2f+g=l,-f-2g=0andg=-,f=,e=- x=x/B=x/m-32-3.The Schrodinger equation is -(/2m)"+bxw =Ew. From (4.78)and (4.79).B andwB=B-2mThe Schrodinger equation becomes -(2/2m)B-12m23h43b2Bwr+bx,m-1323b-13B-12y,=m-13n23b23EB-2g,and V=(2x,-2E,=G,W,where G,=2x,-2E,.Let us find eigenvalues with E.s8.Setting this maximum E.equal to V,we have 8=x,and the classically allowed region is 0sx s8.We shall go from x =0 to 10 in steps of 0.05.Cell B7 of Fig.4.9 contains the formula 2*A7-2*$BS3 and this is copied to other column B cells The Solver gives the lowest four eigenvalues as 1.85575706,3.24460719,4.38167006 5.38661153. 4-12 Copyright2014 Pearson Education.Inc
4-12 Copyright © 2014 Pearson Education, Inc. 2 1 2 6 26 2 M (ML T ) (MT L ) M L T , b c d bcd c d c d − − − ++ − −− = so bcd c d c d + + = − = − − =− 1, 2 6 2, 2 2 . 1 4 8 55 5 dcb = , , = =− . 4/5 8/5 1/5 // . E EA Em a r − = = = Let efg B = m a = . We have 2 1 26 26 2 [ ] L M (ML T ) (MT L ) M L T e f g efg f g f g B − − − ++ − −− == = , so efg f g f g ++= − = −− = 0, 2 6 1, 2 0 . 11 1 10 5 10 g fe = − = =− , . 1/10 1/5 1/10 / / r x xB xm a − − = = = . The Schrödinger equation is 2 8 − += ( /2 ) = m ax E ψ′′ ψ ψ . From (4.78) and (4.79), 1/2 ψ r =ψ B and 1/2 2 1/2 1/5 2/5 1/5 ψ ψ ψ r r BB Bm a −− − − ′′ ′′ ′′ = = = . The Schrödinger equation becomes 2 1/2 1/5 2/5 1/5 8 4/5 8/5 4/5 1/2 4/5 8/5 1/5 1/2 ( /2 ) m B m a ax m a B m a E B ψ rr r r r ψ ψ − − − −− − − − += == = = ′′ and 8 (2 2 ) r r r r rr ψ′′ =− = xE G ψ ψ , where 8 2 2 GxE rrr ≡ − . Let us find eigenvalues with 10. Er ≤ Setting this maximum Er equal to Vr , we have 8 10 r = x and the classically allowed region is bounded by 1.33 r x = ± . We shall start well into the classically forbidden region at 3 r x = − and go to 3 r x = + in steps of 0.02. Cell B7 of Fig. 4.9 contains the formula 2*A7^8-2*$B$3 and this is copied to other column B cells. With 0.001 in cell C8, a suitable Precision (set by clicking the Solver Options button) is 0.1. The Solver gives the lowest three eigenvalues as 4/5 8/5 1/5 / / E EA Em a r − = = = = 0.70404876, 2.731532, 5.884176. 4.34 Proceeding similarly as in Prob. 4.32, we have 22 2 [ ] ML T / L MT L b − − = = . / E EA r ≡ and / . r x ≡ x B Let acd A = m b = . Then [A] = ML2 T–2 = [ acd m b = ] = 21 2 2 2 M (ML T ) (MT L) M L T , a c d acd cd c d − − ++ + −− = so acd + + = 1, 2 2, c d + = − − =− c d2 2. 22 1 33 3 d ca = = =− , . So 1/3 2/3 2/3 // . Er EA Em b − = = = Let efg B = m b = . We have 21 2 2 2 [ ] L M (ML T ) (MT L) M L T e f g efg fg f g B − − ++ + −− == = , so efg fg ++= += 0, 2 1, −− = f g2 0 and 12 1 33 3 g fe = − = =− , . 1/3 2/3 1/3 / / r x xB xm b − − = = = . The Schrödinger equation is 2 − += ( /2 ) = m bx E ψ′′ ψ ψ . From (4.78) and (4.79), 1/2 ψ r =ψ B and 1/2 2 1/2 2/3 4/3 2/3 ψ ψ ψ r r BB Bm b −− − − ′′ ′′ ′′ = = = . The Schrödinger equation becomes 2 1/2 2/3 4/3 2/3 1/3 2/3 1/3 1/2 1/3 2/3 2/3 1/2 ( /2 ) m B m b bx m b B m b E B ψ rr r r r ψ ψ − − − −− − − − += == = = ′′ and (2 2 ) r r r r rr ψ′′ =− = xE G ψ ψ , where 2 2 GxE rrr ≡ − . Let us find eigenvalues with 8. Er ≤ Setting this maximum Er equal to Vr , we have 8 r = x and the classically allowed region is 0 8 r ≤ ≤ x . We shall go from 0 r x = to 10 in steps of 0.05. Cell B7 of Fig. 4.9 contains the formula 2*A7-2*$B$3 and this is copied to other column B cells. The Solver gives the lowest four eigenvalues as 1.85575706, 3.24460719, 4.38167006, 5.38661153