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-- n~n2 ) ; 21(b) (T) == T 1i2/mr~; (e) expand the function ~'1s (r) in terms of eigenfunctions of the operator k: ~)ts (r) = (2:rr) -3/2 i Cke i k r dk, where Ck = (2:rr)-3/2 ) '¢1s (r) e- i k r dV. The latter integral is to be computed in spherical coordinates with the polar axis being directed along the vector k: 1 eJ\ sin (nr/ro> ikr cos t} 2 d . ~<\, d~<\' dCk == , / \ e- r r sm tr ·v <p(2n)2 J" ro r v; sin kro k (n 2 - k2r8) • Therefore the probability ~density of the given wave vector is: ICkl2 = k 2 [;2s~;:'i)2 . To find the probability of (the modulus of the wave vector being between the values k and k +dk, integrate the latter expression with respect to all possible- directions of the vector k, i ,e, mi1tiply it by 4nk2 dk: Finally we get w (k) dk = 4nro sin2 kro dk. (n 2- k2r8)2 4.68. (a) The solutions of the Schr6dinger equation for the- function X (r) are r<ro, Xt=Asin(kr+a), k=l/2inE/1i; r ;» ro, 'X2 = Bexr +Ce:>: x = V2m (Vo- Evlh, From the finiteness of the function '¢ (r) throughout the space, it follows that a = 0 and B = O. Thus, Ilh -A sinkr. Ilh -C e- x r ~1- -r-' ~2- -r-· From the condition of continuity of 'tjJ and 'P' at the point r = ro,· we obtain tan kro = -k/x, or sin kro = +V 1i2/2mr~Uokro. This equation, as it is shown in the solution of Problem 3.47, defines the discrete-spectrum of energy eigenvalues. (b) n2Ji2/8m < r~Uo< 9n2Ji2/8m. (c) In this case there is a single 3 V3 2 211 2 1i2level: sinkro=-4-kro; kr o= -3 n: E===-o-·--2. From theJt ~J mro condition a(r2'¢2)18r == 0, we find r p r = 3ro/4; 34%. .4.69. iJiJ2~ +~ aaR ·t(E+~- l(lt 1)) R=O, p=rlrt, E= p P P P P ~E/Ei· 4.70. (a) Neglecting the small values, reduce the Schrodinger equation to the form X" - X2X = 0, X = V 2m IE I/h. Its solution is X (r) = Aexr + Berw. From the finiteness of R (r), it follows that A = 0 and R (r) 0: e-xr/r. (b) Transform the Schrodinger equation to the form X" - l (l t 1) X = O. I ts solution is to be found in the r form X = Ara . After the substitution into the equation, we find -two values of a: 1 + land -l. The function R (r) is finite only if o: = 1 + l. Hence, R (r) 0: r l • 4.71. (a) Substituting this function into the Schr6dinger equa- tion, we obtain B (a, ex, E) + rC (a, ex, E) + ~D (a, ex) = 0, where r B, C, and D are certain polynomials. This equality holds for any values of r only when B = C = D = 0, whence a = a = -1/2r1 = -me2/21i2 ; E = -me4/81i2• (b) A = -} (2nr~) -1/2; rt is the first Bohr radius. 4.72. (a) rp r = r1 , where r1 is the first Bohr radius; 32.3%;(b) 23.8%. 4.73. (a) (r) == 3rl/2; (r2 ) == 3r~; «L\r)2) == (r2) - (r)2 == 3r~/4; r. is the first Bohr radius; (b) (F) ~ 2e2/r~; (U) ~ -e2/rt; (c) (T) = = J1~f'¢ dT = me"12/i2; V(v2) = e2//i = 2.2.106 m/s. 4.74. (a) 4rt and 9r1; (b) 5r~ and 15.75r;; r1 is the first Bohr .radius. 4.75. GOo "CO' ) P~r) 4nr2dr = :1' where p (r) = e'¢rs (r) is the -vclume density of charge; rt is the first Bohr radius. 4.76. Write Poisson's equation in spherical coordinates: i.. a0 22 (rcpe) = 4~ecpr8 (r), e ;» O. r r Integrating this equation twice, we get q>e (r) = ( :1 + ; ) e- 2r/T1 + .4+ ~ , where r 1 is the first Bohr radius. A and B are the integration con- .stants. Choose these constants so that cP e (00) = 0 and cP e (0) be finite. Hence, A = 0, B = -e. Adding the potential induced by the nuc- ·176 j 5/2 3/2 1/2 3/2 Fig. 66 J I 2 J 4 5 ~ =2 n= Ieus to the expression obtained, we get cp (r) ~ ( ..:.. +~ ) e - 2 r [r 1. , rl r 4.77. See the solution of Problem 4.67, (c), w(k)dk=== 32rtk2 dk . . = 1t (1+k2r;)-l , where r 1 IS the first Bohr radi us. 5.1. 5.14 and 2.1 V. 5.2. 0.41, 0.04, and 0.00. 5.3. Having calculated the quantum defect of S terms, we find E b = 5.4 eVe 5.4. (a) 6; (h) 12. 5.5. O.2i and 0.05; 0.178 urn. 5.6. a == 1.74; n = 2. 5.7. 7.2.10-3 eV; 1.62 -v. 5.8. 555 cm -1. 5.tO. (a) ~T === a 2RZ4 (n - 1)/n4 === 5.85. 2.31 and 1.10 cm r-; (b) 1.73 and 0.58 em -I (three sublevels). 5.11. ~A. == a 2/9R = 5.4.10-3 A (equal for Hand He+). 5.12. Z === 3, i.e. Li"". 5.13. (a) See Fig,: 66; dVS1 = vs-v t === =7.58 em-I, L\A.s1 = 0.204 A; (b) ~v === = 2.46 em-I, dA = 0.54 A.~ 5.14. A/fJA. > v/(v3 - v2) === 4.2.105(see Fig. 66). 5.15. In units of n. V3tj/2, V15/2, n and 1/3/2 (4.P); 2 V5, 2 V3, VEi, V2, and 0 (5D). 5.16. (a) 1Pt and 3Po, t, 2; (b) ipf, 1D2 , 1F3, 3Po. 1,2' 3D1, 2. 3' 3F2 , 3, 4.; (c) 2P1/ 2 , 3/ 2 , 2D3 / 2 , 5 / 2 , 2F5/ 2 , 7/ 2 , 4.P1/ 2 , 3/2. 5/2, 4.D1/2 , 3/2,5/2,7/2, 4F3/ 2 , 5/2,7/2, f)/2· .5.17. 20 (5 singlet and 15 triplet types). 5.18. ISo, IP1 , ID 2 , 3S1 , 3PO, l , 2 ' 3D1,2t3. 5.19. (a) 2,4, 6,8; (b) respectively, 2; 1, 3; 2, 4; 1, 3, 5. 5.20. V301i. 5.21. Respectively, Ps>V2nand Ps:::= V2/i. 5.22. (a) 35.2°; (b) 34.4°. 5.23. 10 (the number of states with different values of mJ). 5.24. Jtr 30 n ; 5113 , 5.25. 125°15'. 5.26. (a) ~ (2/ + 1) = (28 + 1)· (2£ + 1); (b) 2 (2l 1 + 1) X J X 2 (2l 2 + 1) = 60; (c) the number of states with identical quan- 12-0339 177 tum numbers nand l is N = 2 (2l + 1). While distributing k electrons over these states, the Pauli exclusion principle should he taken into account. Consequently, the problem reduces to finding the number of combinations of N elements taken k at a time: Ck _ N(N-1)(N-2) ... (N-k+1) -120N- kf -. 5.27. (a) 15; (b) 46. 5.28. (a) 2 (2l + 1); (b) 2n2• 5.29. (a) C: 1s22s22p2(3Po); N: 1s22s22p3(4S3/2); (b) S: 1s22s22p63s23p4(3P2); CI: 1S22s22pG3s23p5(2P3/2). 5.30. (a) 3F2; (b) 4F3/2. 5.31. 685/ 2 • 5.32. The basic term 5D". The degeneracy 2J + 1 = 9. 5.33. Let us compile the table of possible distributions of elec- trons over quantum states (numbers) with the Pauli exclusion principle taken into account (Tables 1 and 2). While doing this, we can leave out those distributions that provide the negative values of the sum of projections M Land M s; such distributions do not submit anything new, which can be proved directly. To illustrate, let us denote the spin projection m ; of each electron by the arrow pointing either up (if m , = +1/2) or down (if m , = -1/2). Table 1 +1 t t t t t.J. - 0 t - .J. - - tt -1 - t - t - - Ms 1 1 0 0 0 0 M L 1 0 1 0 2 0 Table 2 +1 t t t t t+ t+ t 0 t t t t t - t+ -1 t .J. t t - t - Ms 3/2 1/2 1/2 1/2 1/2 1/2 1/2 M L 0 0 0 0 2 1 t (a) See Table 1. The presence of the state with M L === 2 and Al s == 0 indicates that there is a term ID; consequently, there must also be two other states: M L == 1 and M L == 0 (for both ill s == 0). From other distributions the state with M L = 1 and M s = 1 points to the existence of the 3p term; therefore, there must be still another state with M L = 0 and M s = 1. The last remaining state with M L = 0 and M s = 0 belongs to the IS term. Consequently, three types of terms correspond to the given configuration: 18, ID, and 3P. (h) See Table 2. Via similar reasoning, we get 2D, 2p, and 48. (e) 18, ID, IG, 3p, and 3F. 178 5.34. Both configurations have the following identical type! of terms: (a) 2p; (b) IS, ID, and 'JP; (c) 2D. This follows from the fact that the absence of an electron in a snbshell can be treated as "a hole" whose state is determined by the same quantum numbers as those of the absent electron. 5.35. Let us compile the table of possible distributions of elec- trons over quantum states, taking into account that the Pauli exclusion principle imposes limitations only on equivalent elec- trons. (a) See Table 3 in which the thin arrows indicate spin projections of a p-electron and heavy arrows those of an s-electron. Table 3 +1 t t t t ... t t+ -I 0 tt tt t -!- it t t t-!-t -1 - - t J.. - -!- - - ills 3/2 If' 1/2 3/2 1 .) 1 ") 1/2 1/2 1/2, .... / .... ~,,, L 1 1 0 () 1 0 2 0 t' The possible types of terms: 2D, 2p, 28, and 4P. (b) 28,