c hc h c h. Chapter Since E n L 2 in Eq. 39-4, we see that if L is doubled, then E 1 becomes (2.6 ev)(2) 2 = 0.65 ev.

Transcription

1 Capter 39 Since n L in q 39-4, we see tat if L is doubled, ten becoes (6 ev)() = 065 ev We first note tat since = J s and c = /s, 34 8 c J sc / s c 40eV n J / ev 0 / n c c Using te c value for an electron fro Table 37-3 (5 0 3 ev), q 39-4 can be rewritten as n Te energ to be absorbed is terefore c e e bg n n c L 8 8 c L c eV n 4 903eV 3 8 c L ev 050n 3 We can use te c value for an electron fro Table 37-3 (5 0 3 ev) and c = 40 ev n b writing q 39-4 as n bg n n c L 8 8 c L c For n = 3, we set tis expression equal to 47 ev and solve for L: L bg n c 8cc 3 n 4 Wit = p = kg, we obtain 3b40eV ng c evb evg n Js n 390 J 0006 ev 7 8(670 kg)

2 54 CHAPTR 39 Alternativel, we can use te c value for a proton fro Table 37-3 ( ev) and c = 40 ev n b writing q 39-4 as n bg n n c 8 c L dp i Tis alternative approac is peraps easier to plug into, but it is recoended tat bot approaces be tried to find wic is ost convenient 5 To estiate te energ, we use q 39-4, wit n =, L equal to te atoic diaeter, and equal to te ass of an electron: J s n 3070 J=90MeV 9 GeV kg (a) Te ground-state energ is 8 8(90 kg) eV J s 8 n L 3 e (b) Wit p = kg, we obtain 5 0 J J s n L 7 p 8 8(670 kg) ev 85 0 J 7 According to q 39-4 n L As a consequence, te new energ level ' n satisfies n n F G I H J L LK F H G I K J L L, wic gives L L Tus, te ratio is L / L 4 8 Let te quantu nubers of te pair in question be n and n +, respectivel Ten Letting n+ n = (n + ) n = (n + )

3 55 n b g b4 3g c, n n we get n + =, or n = 0 Tus, (a) te iger quantu nuber is n + = 0 + =, and (b) te lower quantu nuber is n = 0 (c) Now letting n b g b4 3g c, n n we get n + = 4, wic does not ave an integer-valued solution So it is ipossible to find te pair of energ levels tat fits te requireent 9 Let te quantu nubers of te pair in question be n and n +, respectivel We note tat bn g n b n g n n Terefore, n+ n = (n + ) Now n 5 5 n n 5 wic leads to n + = 5, or n = Tus, (a) Te iger quantu nuber is n+ = + = 3 (b) Te lower quantu nuber is n = b g, (c) Now let b g, n n n wic gives n + = 36, or n = 75 Tis is not an integer, so it is ipossible to find te pair tat fits te requireent 0 Te energ levels are given b n = n /, were is te Planck constant, is te ass of an electron, and L is te widt of te well Te frequenc of te ligt tat will excite te electron fro te state wit quantu nuber n i to te state wit quantu nuber n f is f nf ni and te wavelengt of te ligt is

4 56 CHAPTR 39 c f c n n df i i We evaluate tis expression for n i = and n f =, 3, 4, and 5, in turn We use = J s, = kg, and L = 50 0, and obtain te following results: (a) for n f =, (te longest wavelengt) (b) for n f = 3, (te second longest wavelengt) (c) for n f = 4, (te tird longest wavelengt) We can use te c value for an electron fro Table 37-3 (5 0 3 ev) and c = 40 ev n b rewriting q 39-4 as n bg n n c L 8 8 c L c (a) Te first excited state is caracterized b n =, and te tird b n' = 4 Tus, c 40eV n n n 3 8 c L ev 050n 7eV 4 60eV 6 4 Now tat te electron is in te n' = 4 level, it can drop to a lower level (n'') in a variet of was ac of tese drops is presued to cause a poton to be eitted of wavelengt n c n c c 8 c L c n n For exaple, for te transition n' = 4 to n'' = 3, te poton eitted would ave wavelengt 3 c b g 8 50 ev 0 50n 40eV n 4 3 b gc ḣ 9 4 n, and once it is ten in level n'' = 3 it igt fall to level n''' = eitting anoter poton Calculating in tis wa all te possible potons eitted during te de-excitation of tis sste, we obtain te following results: (b) Te sortest wavelengt tat can be eitted is l 4 37n

5 57 (c) Te second sortest wavelengt tat can be eitted is l 4 7n (d) Te longest wavelengt tat can be eitted is l 687 n (e) Te second longest wavelengt tat can be eitted is l 3 4n (f) Te possible transitions are sown next Te energ levels are not drawn to scale (g) A wavelengt of 94 n corresponds to 4 3 transition Tus, it could ake eiter te 3 transition or te pair of transitions: 3 and Te longest wavelengt tat can be eitted is l 687 n () Te sortest wavelengt tat can next be eitted is l 3 58n Te frequenc of te ligt tat will excite te electron fro te state wit quantu nuber n i to te state wit quantu nuber n f is f n n f i and te wavelengt of te ligt is c f c n n df i i Te widt of te well is L c n ( f ni ) 8c Te longest wavelengt sown in Figure 39-8 is 8078 n wic corresponds to a jup fro ni to n 3 Tus, te widt of te well is f

6 58 CHAPTR 39 c( n f ni ) (8078 n)(40ev n)(3 ) L 0350n 350 p 3 8c 8(50 ev) z 3 Te probabilit tat te electron is found in an interval is given b P dx, were te integral is over te interval If te interval widt x is sall, te probabilit can be approxiated b P = x, were te wave function is evaluated for te center of te interval, sa For an electron trapped in an infinite well of widt L, te ground state probabilit densit is F x G I sin H J, L L K so P F GH IJ K FGI HJ K x x sin L L (a) We take L = 00 p, x = 5 p, and x = 50 p Ten, L P M N O P Q L M N O b50 pg 5 P b pg Q 00p sin p (b) We take L = 00 p, x = 50 p, and x = 50 p Ten, L P M N O P Q L M N O b50 pg 50 P b pg Q 00p sin 00 00p (c) We take L = 00 p, x = 90 p, and x = 50 p Ten, L P M N O P Q L M N O b5 0 pg 90 P b pg Q 00p sin p 4 We follow Saple Proble 39-3 in te presentation of tis solution Te integration result quoted below is discussed in a little ore detail in tat Saple Proble We note tat te arguents of te sine functions used below are in radians (a) Te probabilit of detecting te particle in te region 0 x L/ 4 is /4 L /4 sin d L sin 009 (b) As expected fro setr,

7 59 L sin L sin d 009 /4 4 /4 (c) For te region L / 4 x 3 L / 4, we obtain 3 / 4 L 3 / 4 sin d L /4 4 /4 sin 08 wic we could also ave gotten b subtracting te results of part (a) and (b) fro ; tat is, (009) = 08 5 Te position of axiu probabilit densit corresponds to te center of te well: x L/ (00 p) / 00 p (a) Te probabilit of detection at x is given b q 39-: n n p( x) n ( x) dx sin x dx sin xdx L L L L For n 3, L 00 p and dx 00 p detection at xl/ 00 p is (widt of te probe), te probabilit of 3 L 3 p x L dx dx dx L L L L 00 p ( / ) sin sin 00 p 000 (b) Wit N 000 independent insertions, te nuber of ties we expect te electron to be detected is n Np (000)(000) 0 6 Fro q 39-, te condition of zero probabilit densit is given b n n sin x 0 x L L were is an integer Te fact tat x 0300L and x 0400L ave zero probabilit densit iplies sin 0300n sin 0400n 0 wic can be satisfied for n 0, were,, However, since te probabilit densit is non-zero between x 0300L and x 0400L, we conclude tat te electron is in te n 0 state Te cange of energ after aking a transition to n 9 is ten equal to

8 530 CHAPTR J s 3 0 n n J kg According to Fig 39-9, te electron s initial energ is 06 ev After te additional energ is absorbed, te total energ of te electron is 06 ev ev = 506 ev Since it is in te region x > L, its potential energ is 450 ev (see Section 39-5), so its kinetic energ ust be 506 ev 450 ev = 56 ev 8 Fro Fig 39-9, we see tat te su of te kinetic and potential energies in tat particular finite well is 33 ev Te potential energ is zero in te region 0 < x < L If te kinetic energ of te electron is detected wile it is in tat region (wic is te onl region were tis is likel to appen), we sould find K = 33 ev 9 Scrödinger s equation for te region x > L is d 8 U 0 0 dx If = De kx, ten d /dx = 4k De kx = 4k and d 8 8 U k U dx Tis is zero provided b k U0 ġ Te proposed function satisfies Scrödinger s equation provided k as tis value Since U 0 is greater tan in te region x > L, te quantit under te radical is positive Tis eans k is real If k is positive, owever, te proposed function is psicall unrealistic It increases exponentiall wit x and becoes large witout bound Te integral of te probabilit densit over te entire x-axis ust be unit Tis is ipossible if is te proposed function 0 Te sallest energ a poton can ave corresponds to a transition fro te nonquantized region to Since te energ difference between 3 3 and 4 is ev 40 ev 50 ev, te energ of te poton is poton K 00 ev 500 ev 700 ev Using c / (40eV n)/, te energies associated wit a, b and c are

9 53 Te ground-state energ is a b c c 40eV n 8500 ev 4588 n a c 40eV n 560 ev n b c 40eV n 460 ev 908 n c 4 c 4500 ev 460 ev 40 ev Since a, te energ of te first excited state is a 40 ev 850 ev 09 ev We can use te c value for an electron fro Table 37-3 (5 0 3 ev) and c = 40 ev n b writing q 39-0 as For n x = n =, we obtain F G I bgf H K J c G H I K J n n c n n x x nx, n 8 Lx L 8 c Lx L 40eV n 0734 ev ev 0800n 600n, 3 We can use te c value for an electron fro Table 37-3 (5 0 3 ev) and c = 40 ev n b writing q 39- as F n n n I c F n n x z x nz nx, n, nz G H L L L J bg I 8 K 8 c G H L L L J K For n x = n = n z =, we obtain x z c 40eV n 3 ev ev 0800n 600n 0390n, 4 We are looking for te values of te ratio x z

10 53 nx, n L F F HG n n x nx n HG I Lx LK J 4 I K J CHAPTR 39 and te corresponding differences (a) For n x = n =, te ratio becoes 5 4 bg (b) For n x = and n =, te ratio becoes One can ceck (b coputing oter (n x, n ) values) tat tis is te next to lowest energ in te sste (c) Te lowest set of states tat are degenerate are (n x, n ) = (, 4) and (, ) Bot of tese states ave tat ratio equal to 4bg bg (d) For n x = and n = 3, te ratio becoes One can ceck (b coputing oter (n x, n ) values) tat tis is te lowest energ greater tan tat coputed in part (b) Te next iger energ coes fro (n x, n ) = (, ) for wic te ratio is 4 4bg 4 5 Te difference between tese two values is = 00 5 Te energ levels are given b n x L M O P L N x Q M N n n L L L n n x, n x were te substitutions L x = L and L = L were ade In units of /, te energ levels are given b n n /4 Te lowest five levels are, = 5,, = 00,,3 = x 35,, = 45, and, =,4 = 500 It is clear tat tere are no oter possible values for te energ less tan 5 Te frequenc of te ligt eitted or absorbed wen te electron goes fro an initial state i to a final state f is f = ( f i )/, and in units of / is sipl te difference in te values of n n /4 for te two states Te possible frequencies are as follows: 075,,,00,3,,300,,, 375,,,5,3,,5,,,300,,,00,,3, 75,,3,075,,, all in units of / (a) Fro te above, we see tat tere are 8 different frequencies (b) Te lowest frequenc is, in units of /, 075 (,,) (c) Te second lowest frequenc is, in units of /, 00 (,,3) (d) Te tird lowest frequenc is, in units of /, 5 (, 3,) x O Q P P

11 533 (e) Te igest frequenc is, in units of /, 375 (,,) (f) Te second igest frequenc is, in units of /, 300 (,,) or (,,) (g) Te tird igest frequenc is, in units of /, 5 (,,) 6 We are looking for te values of te ratio F H G L n n nz L L L nx, n, nz x x and te corresponding differences z I J K d (a) For n x = n = n z =, te ratio becoes + + = 300 n n n x z (b) For n x = n = and n z =, te ratio becoes = 900 One can ceck (b coputing oter (n x, n, n z ) values) tat tis is te tird lowest energ in te sste One can also ceck tat tis sae ratio is obtained for (n x, n, n z ) = (,, ) and (,, ) (c) For n x = n = and n z = 3, te ratio becoes = 00 One can ceck (b coputing oter (n x, n, n z ) values) tat tis is tree steps up fro te lowest energ in te sste One can also ceck tat tis sae ratio is obtained for (n x, n, n z ) = (, 3, ) and (3,, ) If we take te difference between tis and te result of part (b), we obtain = 00 (d) For n x = n = and n z =, te ratio becoes = 600 One can ceck (b coputing oter (n x, n, n z ) values) tat tis is te next to te lowest energ in te sste One can also ceck tat tis sae ratio is obtained for (n x, n, n z ) = (,, ) and (,, ) Tus, tree states (tree arrangeents of (n x, n, n z ) values) ave tis energ (e) For n x =, n = and n z = 3, te ratio becoes = 40 One can ceck (b coputing oter (n x, n, n z ) values) tat tis is five steps up fro te lowest energ in te sste One can also ceck tat tis sae ratio is obtained for (n x, n, n z ) = (, 3, ), (, 3, ), (,, 3), (3,, ) and (3,, ) Tus, six states (six arrangeents of (n x, n, n z ) values) ave tis energ 7 Te ratios coputed in Proble 39-6 can be related to te frequencies eitted using f = /, were eac level is equal to one of tose ratios ultiplied b / Tis effectivel involves no ore tan a cancellation of one of te factors of Tus, for a transition fro te second excited state (see part (b) of Proble 39-6) to te ground state (treated in part (a) of tat proble), we find f b F gg I H J F Kb g G I H J K L L i

12 534 CHAPTR 39 In te following, we oit te / factors For a transition between te fourt excited state and te ground state, we ave f = = 900 For a transition between te tird excited state and te ground state, we ave f = = 800 For a transition between te tird excited state and te first excited state, we ave f = = 500 For a transition between te fourt excited state and te tird excited state, we ave f = = 00 For a transition between te tird excited state and te second excited state, we ave f = = 00 For a transition between te second excited state and te first excited state, we ave f = = 300, wic also results fro soe oter transitions (a) Fro te above, we see tat tere are 7 frequencies (b) Te lowest frequenc is, in units of /, 00 (c) Te second lowest frequenc is, in units of /, 00 (d) Te tird lowest frequenc is, in units of /, 300 (e) Te igest frequenc is, in units of /, 900 (f) Te second igest frequenc is, in units of /, 800 (g) Te tird igest frequenc is, in units of /, Te stateent tat tere are tree probabilit densit axia along x L x / iplies tat n 3 (see for exaple, Figure 39-6) Since te axia are separated b 00 n, te widt of L is L n (00 n) 600 n Siilarl, fro te inforation given along L /, we find nx 5 and Lx nx(300 n) 50 n Tus, using q 39-0, te energ of te electron is n 0 J 34 nx (6630 J s) nx, n Lx L 8(90 kg) (3000 ) (000 ) 0 9 Te discussion on te probabilit of detection for one-diensional case found in Section 39-4 can be readil extended to two diensions In analog to q 39-0, te noralized wave function in two diensions can be written as nx n n, (, ) ( ) ( ) sin sin x n x n x x n x Lx Lx L L 4 n n sin x sin x LxL L x L

13 535 Te probabilit of detection b a probe of diension x placed at ( x, ) is Wit 4( x ) n n p x x x x x (, ) n, (, ) sin sin x n LxL L x L L L L 50 p and x 500 p, te probabilit of detecting an x electron in ( n, n ) (,3) state b placing a probe at (000 L, 0800 L ) is x 4( x ) n n 4(500 p) 3 p sin xsin sin 000L sin 0800L L L L L L L x x x (50 p) 500 p 4 sin 000 sin p In analog to q 39-0, te noralized wave function in two diensions can be written as nx n n, (, ) ( ) ( ) sin sin x n x n x x n x Lx Lx L L 4 n n sin x sin x LxL L x L Te probabilit of detection b a probe of diension x placed at ( x, ) is 4( x ) n n p x x x x x (, ) n, (, ) sin sin x n LxL L x L A detection probabilit of of a ground-state electron ( n n ) b a probe of area x L L iplies x 400 p placed at (, ) ( /8, /8) x 4(400 p ) L L 0 p sin sin 4 sin L L 8 L 8 L 8 Solving for L, we get L 76 p 3 (a) We use q At r = a bg r 3 a a e 3 a a e 59 0 F H G I K J c n e 3 9n 3

14 536 CHAPTR 39 (b) We use q At r = a P r 4 a a e a a 4e 4e 3 a 59 0 bg 3 (a) We use q At r = 0, P(r) r = 0 0 n n (b) At r = a (c) At r = a P r P r 4 a a e a a 4e 4e 3 a 59 0 bg bg b g 4 a a e e a e 3 a a n n 554 n n 33 If kinetic energ is not conserved, soe of te neutron s initial kinetic energ is used to excite te drogen ato Te least energ tat te drogen ato can accept is te difference between te first excited state (n = ) and te ground state (n = ) Since te energ of a state wit principal quantu nuber n is (36 ev)/n, te sallest excitation energ is 36eV 36eV 0eV Te neutron does not ave sufficient kinetic energ to excite te drogen ato, so te drogen ato is left in its ground state and all te initial kinetic energ of te neutron ends up as te final kinetic energies of te neutron and ato Te collision ust be elastic 34 (a) Te energ level corresponding to te probabilit densit distribution sown in Fig 39- is te n = level Its energ is given b 36 ev 34 ev (b) As te electron is reoved fro te drogen ato te final energ of te protonelectron sste is zero Terefore, one needs to suppl at least 34 ev of energ to te sste in order to bring its energ up fro = 34 ev to zero (If ore energ is supplied, ten te electron will retain soe kinetic energ after it is reoved fro te ato) 35 (a) Since energ is conserved, te energ of te poton is given b = i f, were i is te initial energ of te drogen ato and f is te final energ Te electron energ is given b ( 36 ev)/n, were n is te principal quantu nuber Tus,

15 537 36eV 36eV ev 3 3 (b) Te poton oentu is given b 9 b evgc 60 0 J ev 7 p kg s 8 c s (c) Using c = 40 ev n, te wavelengt is 36 Fro q 39-6, c 40eV n 0n ev c c 5 4 f 44 0 evs 6 0 Hz 6eV 37 Te energ of te poton eitted wen a drogen ato jups fro a state wit principal quantu nuber n to a state wit principal quantu nuber n is given b A n n were A = 36 ev Te frequenc f of te electroagnetic wave is given b f = / and te wavelengt is given b = c/f Tus, f A c c c n n Te sortest wavelengt occurs at te series liit, for wic n = For te Baler series, n and te sortest wavelengt is B = 4c/A For te Lan series, n and te sortest wavelengt is L = c/a Te ratio is B / L = Te difference between te energ absorbed and te energ eitted is poton absorbed c c poton eitted absorbed eitted Tus, using c = 40 ev n, te net energ absorbed is FG I H J F K G H I K J c b40ev ng 7 ev 375n 580 n

16 538 CHAPTR (a) We take te electrostatic potential energ to be zero wen te electron and proton are far reoved fro eac oter Ten, te final energ of te ato is zero and te work done in pulling it apart is W = i, were i is te energ of te initial state Te energ of te initial state is given b i = ( 36 ev)/n, were n is te principal quantu nuber of te state For te ground state, n = and W = 36 ev (b) For te state wit n =, W = (36 ev)/() = 340 ev 40 (a) = (36 ev)(4 ) = 8 ev (b) Tere are 6 possible energies associated wit te transitions 4 3, 4, 4, 3, 3 and (c) Te greatest energ is 4 8eV 3 36eV 3 ev (d) Te second greatest energ is 36eV 0eV (e) Te tird greatest energ is eV ev (f) Te sallest energ is 3 36eV 3 89eV (g) Te second sallest energ is 4 36eV 4 55eV () Te tird sallest energ is 4 According to Saple Proble 39-8, te probabilit te electron in te ground state of a drogen ato can be found inside a spere of radius r is given b c x p( r) e x x were x = r/a and a is te Bor radius We want r = a, so x = and p( a) e ( ) 5e 0 33 Te probabilit tat te electron can be found outside tis spere is 033 = 0677 It can be found outside about 68% of te tie 4 Using q 39-6 and c = 40 ev n, we find c 40eV n poton 0 ev 6 n

17 539 Terefore, n low =, but wat precisel is n ig? 36eV 36eV ig low 0eV n wic ields n = (tis is confired b te calculation found fro Saple Proble 39-6) Tus, te transition is fro te n = to te n = state (a) Te iger quantu nuber is n = (b) Te lower quantu nuber is n = (c) Referring to Fig 39-8, we see tat tis ust be one of te Lan series transitions 43 Te proposed wave function is 3 a e r a were a is te Bor radius Substituting tis into te rigt side of Scrödinger s equation, our goal is to sow tat te result is zero Te derivative is so and d dr r d dr a r e r a 5 e r a 5 a FG I H J L K M O L N QP M N d 5 r dr r d dr a r a e r a a r a O QP 4 Te energ of te ground state is given b e 8 and te Bor radius is given b a 0 e, so e 8 a Te potential energ is given b U e 4 r, so 8 8 e e 8 e U 8 a 4 r 8 a r L M O N QP e a r a a r L N M 0 O P L Q M N L M O N QP Te two ters in Scrödinger s equation cancel, and te proposed function satisfies tat equation O QP

18 540 CHAPTR (a) Te calculation is sown in Saple Proble 39-6 Te difference in te values obtained in parts (a) and (b) of tat Saple Proble is n 94 n 3 n (b) We use q 39- For te Lan series, f s s 8 0 (c) Fig 39-9 sows tat te widt of te Baler series is 6563 n 3646 n 9 n 09 (d) Te series liit can be obtained fro te transition: 4 Hz s 9980 s f Hz 37 0 Hz 45 (a) and (b) Letting a = 59 0 be te Bor radius, te potential energ becoes c c 9 9 e N C 60 0 C U = 4 a 59 0 Te kinetic energ is K = U = ( 36 ev) ( 7 ev) = 36 ev J 7 ev 46 Conservation of linear oentu of te ato-poton sste requires tat f precoil ppoton pvrecoil c were we use q 39-7 for te poton and use te classical oentu forula for te ato (since we expect its speed to be uc less tan c) Tus, fro q 39-6 and Table 37-3, 36eV4 4 vrecoil 4 s 6 8 c c c 9380 ev 9980 s p p 47 Te radial probabilit function for te ground state of drogen is P(r) = (4r /a 3 )e r/a,

19 54 were a is te Bor radius (See q 39-44) We want to evaluate te integral z P( r) dr q 5 in te integral table of Appendix is an integral of tis for We set n = and replace a in te given forula wit /a and x wit r Ten z z r/ a P( r) dr r e dr 3 3 a a ( a) 48 (a) Since = 085 ev and = 36 ev + 0 ev = 34 ev, te poton energ is poton = = 085 ev ( 34 ev) = 6 ev 0 (b) Fro we obtain ( 36 ev ) 6 ev n n F H G I K J 6 ev 3 n n 36 ev 6 4 Tus, n = 4 and n = So te transition is fro te n = 4 state to te n = state One can easil verif tis b inspecting te energ level diagra of Fig 39-8 Tus, te iger quantu nuber is n = 4 (c) Te lower quantu nuber is n = 49 (a) 0 is real Squaring it, we obtain te probabilit densit: 4 r r / a r r / a P0( r) 0 (4 r ) e cos (4 r ) e cos a a (b) ac of te oter functions is ultiplied b its coplex conjugate, obtained b replacing i wit i in te function Since e i e i = e 0 =, te result is te square of te function witout te exponential factor: and / r sin 5 64 a e r a r r/ a e sin 5 64a Te last two functions lead to te sae probabilit densit:

20 54 CHAPTR 39 4 r r / a r r / a P ( r) (4 r ) e sin (4 r ) e sin 5 5 a 6a (c) Te total probabilit densit for te tree states is te su: P ( r) P ( r) P ( r) (4 r ) r r / a r r / a e cos sin sin e 5 5 a 8a Te trigonoetric identit cos + sin = is used We note tat te total probabilit densit does not depend on or ; it is spericall setric 50 Fro Saple Proble 39-8, we know tat te probabilit of finding te electron in te ground state of te drogen ato inside a spere of radius r is given b c x p( r) e x x were x = r/a Tus te probabilit of finding te electron between te two sells indicated in tis proble is given b x x x x p( a r a) p( a) p( a) e x x e x x Since r is sall, we a calculate te probabilit using p = P(r) r, were P(r) is te radial probabilit densit Te radial probabilit densit for te ground state of drogen is given b q 39-44: were a is te Bor radius P( r) (a) Here, r = 0500a and r = 000a Ten, 4 r a F H G I K J 3 e r / a 4r r r/ a 3 3 P e 4(0500) (000) e a (b) We set r = 00a and r = 000a Ten, 4r r r/ a 3 3 P e 4(00) (000) e a

21 543 5 Using q 39-6 and c = 40 ev n, we find c 40 ev n poton 09 ev 066 n Terefore, n low =, but wat precisel is n ig? 36eV 36eV ig low 09eV n wic ields n = 3 Tus, te transition is fro te n = 3 to te n = state (a) Te iger quantu nuber is n = 3 (b) Te lower quantu nuber is n = (c) Referring to Fig 39-8, we see tat tis ust be one of te Lan series transitions 53 According to Fig 39-5, te quantu nuber n in question satisfies r = n a Letting r = 0, we solve for n: n r a (a) Te plot sown below for 00 (r) is to be copared wit te dot plot of Fig 39- We note tat te orizontal axis of our grap is labeled r, but it is actuall r/a (tat is, it is in units of te paraeter a) Now, in te plot below tere is a ig central peak between r = 0 and r a, corresponding to te densel dotted region around te center of te dot plot of Fig 39- Outside tis peak is a region of near-zero values centered at r = a, were 00 = 0 Tis is represented in te dot plot b te ept ring surrounding te central peak Furter outside is a broader, flatter, low peak wic reaces its axiu value at r = 4a Tis corresponds to te outer ring wit near-unifor dot densit wic is lower tan tat of te central peak

22 544 CHAPTR 39 (b) Te extrea of (r) for 0 < r < a be found b squaring te given function, differentiating wit respect to r, and setting te result equal to zero: ( r a)( r 4a) / 6 3 a e r a 0 wic as roots at r = a and r = 4a We can verif directl fro te plot above tat r = 4a is indeed a local axiu of 00 ( r ) As discussed in part (a), te oter root (r = a) is a local iniu (c) Using q and q 39-4, te radial probabilit is (d) Let x = r/a Ten FG I H J K r r P r r r e r / a 00( ) 4 00( ) 3 8a a r r / r a x 4 3 x 0 00( ) ( ) ( 4 4 ) P r dr e dr x x e dx x x x e dx 8a a 8 [4! 4(3!) 4(!)] 8 were we ave used te integral forula z n x 0 x e dx n! 55 Te radial probabilit function for te ground state of drogen is P(r) = (4r /a 3 )e r/a, were a is te Bor radius (See q 39-44) Te integral table of Appendix a be used to evaluate te integral r avg z0 forula wit /a (and x wit r), we obtain rp( r) dr Setting n = 3 and replacing a in te given 4 3 r/ a 4 6 ravg rp( r) dr r e dr 5 a a a a 56 We can use te c value for an electron fro Table 37-3 (5 0 3 ev) and c = 40 ev n b writing q 39-4 as n bg n n c L 8 8 c L c

23 545 (a) Wit L = n, te energ difference is c c c (b) Since (n + ) n = n +, we ave b g bg ev c n n n n 8 c L Setting tis equal to 0 ev, we solve for n: c b ġ c L 4 50 ev 300 n 0eV 9 0 n c 40eV n (c) At tis value of n, te energ is Tus, (d) Since n / 40 n c c c n 60 ev c 50 ev c?, te energ is indeed in te relativistic range 8 ev 57 (a) and (b) Scrödinger s equation for te region x > L is d 8 U 0 0, dx were U 0 < 0 If (x) = Ce kx, ten (x) = C'e kx, were C' is anoter constant satisfing C' = C Tus, d kx 4k Ce 4k dx and d 8 8 U k U 0 0 dx Tis is zero provided tat k 8 U 0

24 546 CHAPTR 39 Te quantit on te rigt-and side is positive, so k is real and te proposed function satisfies Scrödinger s equation If k is negative, owever, te proposed function would be psicall unrealistic It would increase exponentiall wit x Since te integral of te probabilit densit over te entire x axis ust be finite, diverging as x would be unacceptable Terefore, we coose k U0 0 b g 58 (a) and (b) In te region 0 < x < L, U 0 = 0, so Scrödinger s equation for te region is d 8 0 dx were > 0 If (x) = B sin kx, ten (x) = B' sin kx, were B' is anoter constant satisfing B' = B Tus, d k Bsin kx k ( x) dx and d 8 8 k dx Tis is zero provided tat k 8 Te quantit on te rigt-and side is positive, so k is real and te proposed function satisfies Scrödinger s equation In tis case, tere exists no psical restriction as to te sign of k It can assue eiter positive or negative values Tus, k 59 (a) Te allowed values of l for a given n are 0,,,, n Tus tere are n different values of l (b) Te allowed values of l for a given l are l, l +,, l Tus tere are l + different values of l (c) According to part (a) above, for a given n tere are n different values of l Also, eac of tese l s can ave l + different values of l [see part (b) above] Tus, te total nuber of l s is n l 0 (l ) n

25 (a) Te allowed energ values are given b n = n / Te difference in energ between te state n and te state n + is and b g adj n n n n adj L N M O Q PF H G I K J b n g n n n b n g b g As n becoes large, n n and n n n n n (b) No As n, and do not approac 0, but adj / does adj (c) No See part (b) (d) Yes See part (b) (e) adj / is a better easure tan eiter adj or alone of te extent to wic te quantu result is approxiated b te classical result 6 Fro q 39-4, 6 For n = n n bn g L n 8 L F H G I K J F H G I K J F H G I K J bn ġ kg60 C e e 36eV F J s 60 0 J ev 63 (a) We recall tat a derivative wit respect to a diensional quantit carries te (reciprocal) units of tat quantit Tus, te first ter in q 39-8 as diensions of ultiplied b diensions of x Te second ter contains no derivatives, does contain, and involves several oter factors tat turn out to ave diensions of x : 8 kg U x Js J assuing SI units Recalling fro q 7-9 tat J = kg /s, ten we see te above is indeed in units of (wic eans diensions of x )

26 548 CHAPTR 39 (b) In one-diensional Quantu Psics, te wave function as units of / as Saple Proble 39- sows Tus, since eac ter in q 39-8 as units of ultiplied b units of x, ten tose units are / = 5 64 (a) Te oe-base energ level for te Baler series is n = Tus te transition wit te least energetic poton is te one fro te n = 3 level to te n = level Te energ difference for tis transition is b g FG H I K J 3 36 ev 889 ev 3 Using c = 40 ev n, te corresponding wavelengt is c 40eV n 658n 889 ev (b) For te series liit, te energ difference is b g FG H I K J 36 ev 340 ev Te corresponding wavelengt is ten c 40eV n 366n 340 ev