The angle between L and the z-axis is found from

Transcription

1 Poblm 6 This is not a ifficult poblm but it is a al pain to tansf it fom pap into Mathca I won't giv it to you on th quiz, but know how to o it fo th xam Poblm 6 S Figu 6 Th magnitu of L is L an th z-componnt is L z () l ( l ) hba m l hba Th angl btwn L an th z-axis is foun fom L z cos( θ ) L Lt's o this fist fo l= Notic th is an hba in both numato an nominato of th quation fo cos(θ), so th hba's cancl Fo convninc, I will st hba= hba l m l, L z m l m Ll () () l ( l ) hba l hba θ m l, l 6 acos π L z m l Ll () θ m l, l = Ths a th angls in gs masu clockwis fom th positiv z-axis to th vcto L Now lt's o it fo l= l m l, θ m l, l =

2 Poblm 6 Th magntic quantum numb, ml, is limit to th valus, ñ, ñ, ñ l Th a a total of l + valus Fo an obital quantum numb of l =, th a nin possibl valus fo ml: m l,,,,,,,, Poblm 6 Th obital quantum numb l may tak th valus l =,,, n-, wh n is th pincipal quantum numb Each obital quantum numb has an associat st of magntic quantum numbs, ml Th magntic quantum numb ml is limit to th valus, ñ, ñ, ñ l Th a a total of l + valus fo ach valu of th obital quantum numb l Fo a pincipal quantum numb n =, l taks th valus l =,,, l = ml = l = ml = -, -, l = ml = -, -,,, l = ml = -, -, -,,,, Th total numb of obitals with n = is thfo N = = 6 Poblm 6 s p f l= L l ( l ) hba L z m l hba bcaus m l angs fom -l to l, th maximum valu of m l is l an th maximum valu of L z is l*hba L zmax l hba hba cancls out in th quation fo pcntag iffnc, so fo convninc, I will just st it qual to hba pcnt() l () l ( l ) hba l hba () l ( l ) hba

3 pcnt( ) = 989 pcnt( ) = 85 pcnt( ) = 97 Bis's finition of pcntag iffnc is not ncssaily what I woul call pcnt iffnc Bis ally ask "what pcnt of L is th maximum valu of Lz" H is th al finition of pcntag iffnc btwn two quantitis A an B: P A B ( A B ) Howv, on an xam o quiz, o it lik Bis i it Poblm 6 A sphically symmtic pobability-nsity istibution cospons to zo obital angula momntum This occus only fo l=; i, fo s wavfunctions In tabl 6, whn l=, thn ml=, an th wav functions φ an θ a inpnnt of angl, which is why th pobability-nsity istibution is inpnnt of angl Poblm 65 Th aial wav function fo th s lcton in hyogn is R s () Fom quation 6, P() ( R ) To fin th most pobabl, w maximiz P(); i tak its ivativ an st it qual to zo P()

4 a a a 8 Th only way fo th ight han si abov to qual zo is if Th abov quation has two oots: = an = Th answ h is =, which cospons to a maximum in P Th oot at = cospons to a minimum in P You can vify this by chcking th valus of th scon ivativ at = an = With Mathca, it is asi to just plot th function:, 5 5 In this poblm, I scal to b Thn = mans, fo xampl, that is qual to R s () P s () R s () P s ( ) R s ( ) 5 6 In this plot I show both th s aial wav function, R(), an th cosponing pobability nsity, P() Notic that P() is maximum at =, an minimum at = Rmmb, = ally mans = Mathca also has a built-in oot function an a symbolic solv which shoul b abl to fin th oot fo you Not much hlp on th xam, though, so b su you can o it lik I i abov

5 Poblm 66 Th p aial wav function cospons to n=, l=, m l = o m l =+- (th choics), but all th choics fo m l giv th sam R Fom tabl 6, R p 6 a To minimiz/maximiz, tak ivativ an st it qual to zo R 5 a 5 a 5 Th fist tm on th ight, with th xponntial in th numato, is nv zo, so which givs oots; = (th tims) an = As in poblm 65, you can vify using calculus that th maximum is at =, o you can chck it with a gaph 5

6 , 5 5 In this poblm, I scal to b Thn = mans, fo xampl, that is qual to R p () 6 a P p () R p () P p ( ) R p ( ) 8 6 Poblm 67 Th wavfunction is R 8 a a C Wh C is a constant To fin th most pobabl, tak th ivativ of P()= R an st it qual to zo P() R C a P() C a 6

7 C 6 5 a a 6 a C 6 5 a 6 C 5 6 a Th abov quation has 6 oots; = (5 tims) an 6 a Th oots at = all giv minima, so th maximum is at th solution to th abov quation, which is 9 a Again, lt's plot it I lik to SEE what is going on R () 8 a P () R (), 5 5 Notic how fo high-n wav functions, I hav to go futh an futh out in to s most of th wav function That's bcaus high-n wav functions a mo xtn in spac P ( ) R ( )

8 Poblm 68 Fin th two valus of at which P() fo a s lcton has a maximum R s a Lt A Thn w maximiz th pobability by solving this quation: R A a A Why i I tak th insi th ()? Answ: to simplify th application of th chain ful fo th ivativ In this fom, I hav two tms to which I apply th chain ul In th pvious quation, I hav th tms to which to apply th chain ul Th algba fo th th-tm chain ul ivativ is too much fo an oinay human lik m to o without making at last on mistak Evn with my simplification, this is on of th longst homwok poblms of Chapt 6 Now this is th quation w want to solv (th oots a th valus of fo which th function has an xtmum): ( ) I ivi out th A bcaus it is nv zo a 8

9 Bcaus th xponntial is nonzo, w onc again ivi both sis by th laing tm, laving 6 Th oots of th abov quation a foun fom: an 6 a a Th quation on th lft givs So that = an =a a two oots W will s in a minut a that ths oots cospon to pobability minima Th quation on th ight is a simpl quaatic quation, which has oots 6a 6 6 a = 7 Th two maxima thus occu at 6 76 an 56 9

10 Lt's plot again R s () a P a s () R s (), 5 5 P s ( ) R s ( ) You can claly s th two maxima in this plot Notic also how P() is minimum at = an at = 6 8 Lt's tak th ivativ an s btt wh P is zo D s () P s () D s ( ) D s ( ) 5 6 Ths two ivativ plots show th fist maximum, th minimum at =, an th scon maximum

11 Poblm 69 W simply want to calculat th atios P P an P P fo a s lcton Rmmb, P() R() an fo th s lcton, R() A wh A Thus, P P A A wh th tm in th numato is valuat at = an th tm in th nominato is valuat at = / Th atio is a P P a A A a a P P Th atio is - =/ Th scon pat is simila: P P A wh th tm in th numato is valuat at = an th tm in th nominato is valuat at = A

12 P P a A a A P P P P Poblm 6 I'll lt Mathca o it fist Aft that, I'll us Mathca to simulat th pap an pncil solution I alay fin lik this bfo, but I'll o it again h fo you to s R s () xpct R s () R s () xpct = 5 In ali vsions of Mathca, this was zo I won why? 5 xpct R s () R s () xpct = 5 I knw th answ, of cous Th intgan paks up shaply na zo, an gos xponntially to zo fo lativly small (a fw Boh aii) Whn Mathca intgats fom to, it os th intgal numically, ss an intgan of zo most vywh, an gts an answ of zo Whn I cut th intgal off at 5 Boh aii, Mathca happily tuns th coct answ of 5

13 Th "pap an pncil" solution: < s > sta R s R s R s Fom tabl of intgals: x n x x Γ ( n ) Wh Γ(n)=(n-)! fo n an intg (! is th factoial function) Lt x=/ Thn an With ths substitutions, 8 x x < s > x 8 x x a a a 6 x x x

14 Γ( ) a 5 ( ) Poblm 6 Pat a Calculat th pobabilituy of fining a s lcton at a istanc gat than fom th nuclus P > R Fom tabl of intgals: x β x x β x x β x β β

15 a a 8 a a a 8 Evaluat th abov xpssion at an an tak th iffnc to fin th valu of th intgal Not that lim( -/a )= so that th sult is zo at th upp limit, laving -> P > a 8 P > 5 P > 5 P > 68 That is, th is a 68% chanc of fining th lcton outsi Pat b Fin th pobabilituy that > This is intical to pat a xcpt that w now hav a low limit of on th intgal I will jump ictly to th stp wh w plugg into th xpssion in th intgal, xcpt now I will plug in P > a a a P > P > P > Th is a % chanc of fining th s lcton futh away than 5

16 Poblm 6 W n to calculat P s < an P p < P s < R s P p < R p Fo th s stat, th pobability is (plugging in th s aial wav function): P s < a 8a That /8 tm coms in bcaus I almost fogot to inclu th nomalization constant Lt x=/ so that x=/ With ths substitutions, th abov intgal bcoms P s < 8 x ( x) x x W us ths two intgals to o th calculation: x x x x x x x x x x x x 6x 6 Skipping a bit though th algba P s <

17 P s < A littl ov a % chanc Fo th p stat, th pobability is (plugging in th p aial wav function): P p < Lt x=/ as abov Thn P p < x x x W us th two intgals fom abov plus th on blow to o th calculation: x Th sult is x x x x x x x P p < 65 P p < 66 About a facto of lss than fo th s Poblm 6 A hyogn atom is in th p stat To what stat o stats can it go by aiating a photon in an allow tansition? Solution will b post bfo you a tst on this on 7

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