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solutions HW5

# solutions HW5 - P213 In this problem you will use the...

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Unformatted text preview: P213) In this problem, you will use the variational method to ﬁnd the optimal ls wave function for the hydrogen atom starting from the trial function (I) (r) = 9“" with or as the variational parameter. You will minimize )_I— <13 Ha) d1: I— (I) (I) 0’1: with respect to on. a) Show that A 2 ad) 2 2 2 HCD=— h izi r2 (r) — e (DIr)= ah2IZr—ar2)e_m— e 3“" 2m? r a r a F 47:80? 2:11; 47580.?” . 39"” 8026“” 2 . We start w1th a = wag—“r and a = 2r?"rr —ar 9“”. Usmg these results, r r 2 2 2 2 I-l'CD :— h 2“"(21‘61 —arze_m)— e 9—“? = cm 2 (2r—(Jc’r1)a_aKr — e 9“" 2m r2 47:60!“ 2mg?” ﬁlmsﬂr on 2 2 b) Obtain the result I CD’H'ZD dz" = 47rI rztiFHCD dr- — Eh 3 using the standard 2m 0: 4.900:2 integrals in the Math Supplement. 05112 E I (I) HCD dr=2 mIz: I as I s1n6d€ I r e—M’dr— a I da I sm€d6 I 9—Way] —4 eIdéIsinGdQIre‘zmdr mu 0 2 co = 47m: 2I r e'gmdr — aIr2 e'zwdr — e—I re'gwdr 2mg 0 o _2:rm:a2 21!_a 2! _§ 1! _ 3:112 _ :22 mg 22052 230:3 so 220:2 2mg 480052 C) Show that IthCI) dz" = 47rIr2(I)*(I) dr = 13 using the standard integrals in the Math 0: Supplement. I o‘er) dz- : Tats)? sin ads? rig-2‘" dr 1] 0 0 °° 2! at = 47! race—2‘” dr = 47! = — l 23053 a3 722052 820! . . . . . . d) You now have the result E (a) = — . Minimize thls function With respect to 2m 47:80 9 a: and obtain the optimal value of a. d [#12072 920: J 92 F120: do: 2mg 47:80 47:80 m If? 92 a . = 9 l' "pm 47:80h2 e) Is E (an F mm, ) equal to or greater than the true energy? Why? _ 132052 320: _ 112 meg 2 32 met-:2 _ mge‘1 2m 4m, 2mg 47:3th 42:30 47:30? 32752.93?" 9 El“) E (ammo!) is equal to the true energy because our trial wave function has the same form as the true wave function. P21.11) In this problem you will show that the charge density of the ﬁlled n = 2, I = l subshell is spherically symmetrical and that therefore L = 0. The angular distribution of the electron charge is simply the sum of the squares of the magnitude of the angular part ofthe wave ﬁmctions for l = 1 and m; = —1, 0, and 1. a) Given that the angular part of these wave functions is 3 1’1" (ﬂat) = [a] cos 9 3 1:2 F11 (9,95) = [a] Siﬂl‘El 8'-95 3 U2 171—1 (9&5) = [a] Sim? 3—” 2 write an expression for l}? (t9,¢u)|2 +‘ﬁ1(9,¢)‘2 +‘Yl‘1(t9,¢l)| 2 =icos26l+isin2 €+isi1126 WW +lY1‘ (WV +lYf‘ (ml 471' 87: 87: b) Show that ‘11" (9,¢)‘2 + ‘Yll (6,53%)‘2 + ‘I’l'l “9’le does not depend on 6* and a). 1cos2 6' +isin2 6 +isin2 t9 = 1(cos2 6 + sin2 19) = i. This is not a function of 47: 87: 87: 47: 47: t9 and (33'. c) Why does this result show that the charge density for the ﬁlled n = 2, Z = 1 subshell is spherically symmetrical? If a function is independent of 6 and a, then it has the same value for all 9 and 45. This is what we mean by being spherically symmetrical. P21.15) Calculate the terms that can arise from the conﬁguration uplnivlﬂ i 11'. Compare your results with those derived in the text for npg. Which conﬁguration has more terms and why? Because the principle quantum number is different, any combination of m; and ms is allowed. Therefore, S = 0, 1 and I. = 0, 1, and 2. Any combination of the two quantum numbers is allowed. This leads to 1S, 1P, and 1D terms as well as 33, 3P, and 3D terms. The npln’pl , :1 ¢ n’ conﬁguration has more terms because some of the possible terms listed above are not allowed if n = n'because of the Pauli principle. P2117) How many ways are there to place three electrons into an f subshell? What is the ground-state term for the f 3 conﬁguration, and how many states are associated with this term? See Problem P21.16. The ﬁrst electrons can have any combination of T m; and 2 m; values so that n = 14 and i m = 3. The number of states is L = 364. 31(14 — 3) 1 Using the method discussed in Example Problem 21.7, MLM = 6 and MM“ = 312. Therefore the ground state term is 4I, which has (21. + 1)(2S + 1) = (2X 6 + 1)(4) = 52 states. P2126) Derive the ground—state term symbols for the following atoms or ions: a) H, b) F, c) F", (1) Na, e) Na+, t) P, g) Sc We use Hund’s rule that the term with the highest multiplicity is the lowest in energy to get the left superscript. For the right subscript, if there are several choices for J, the lowest J value gives the lowest energy if the subshell is less than half full, and the highest J value gives the lowest energy if the subshell is exactly or more than half full. Applying these rules gives rise to the following term symbols. Atom Conﬁguration Ground state term symbol H 131 F 2322;? 1? 2322p6 Na 331 Mar 2.5722106 P 3;? So 3d‘ P233) Using I; as a variational parameter in the normalized function 3 g:- E _ Wm; = % [51\$] 9 ”0 allows one to vary the size of the orbital. Show this by calculating 0 the probability of ﬁnding the electron inside a sphere of radius 2 on for different values of g using the standard integral 2 2 [rile—Md): = —e'"" [—3 + 2i2 + x—J a a a a) Obtain an expression for the probability as a function of If 2 Using the standard integral Ixze'ﬂdx = —e"” [33 + 2 i2 + x—J a a a 3 24? 25’1“ 2 * 2&0 Iwﬂlstlsdr=4—E[£]I r26 a" =—e a" [1+2n;+2[r§]} 7: an 0 an .510 Evaluating this expression between the limits r = 0 and r = 2 on gives 290 x WHISy/thr:—e4§ (1+4; +s;2)+1 =1—e49' (1+4;+8;2) 0 b) Evaluate the probability for Q' = 1, 2, and 3. 1—e4§(1+4§+s;2)= 0.762 for g“ :1; 0.986 for C = 2; 0.999 for Q’ = 3 ...
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