13 Atomic structure and atomic spectra Solutions to exercises Discussion questions E13.1(b) (1) The principal quantum number, n, determines the energy of a hydrogenic atomic orbital through eqn 13.13 (2) The azimuthal quantum number, l, determines the magnitude of the angular momentum of a hydrogenic atomic orbital through the relation {l(l + 1)}1/2 h ¯ (3) The magnetic quantum number, ml , determines the z-component of the angular momentum of a hydrogenic orbital through the relation ml h ¯ (4) The spin quantum number, s, determines the magnitude of the spin angular momentum through the relation {s(s + 1)}1/2 h ¯ For a hydrogenic atomic orbitals, s can only be 1/2 (5) The spin quantum number, ms , determines the z-component of the spin angular momentum ¯ For hydrogenic atomic orbitals, ms can only be ±1/2 through the relation ms h E13.2(b) (a) A boundary surface for a hydrogenic orbital is drawn so as to contain most (say 90%) of the probability density of an electron in that orbital Its shape varies from orbital to orbital because the electron density distribution is different for different orbitals (b) The radial distribution function gives the probability that the electron will be found anywhere within a shell of radius r around the nucleus It gives a better picture of where the electron is likely to be found with respect to the nucleus than the probability density which is the square of the wavefunction E13.3(b) The first ionization energies increase markedly from Li to Be, decrease slightly from Be to B, again increase markedly from B to N, again decrease slightly from N to O, and finally increase markedly from N to Ne The general trend is an overall increase of I1 with atomic number across the period That is to be expected since the principal quantum number (electron shell) of the outer electron remains the same, while its attraction to the nucleus increases The slight decrease from Be to B is a reflection of the outer electron being in a higher energy subshell (larger l value) in B than in Be The slight decrease from N to O is due to the half-filled subshell effect; half-filled sub-shells have increased stability O has one electron outside of the half-filled p subshell and that electron must pair with another resulting in strong electron–electron repulsions between them E13.4(b) An electron has a magnetic moment and magnetic field due to its orbital angular momentum It also has a magnetic moment and magnetic field due to its spin angular momentum There is an interaction energy between magnetic moments and magnetic fields That between the spin magnetic moment and the magnetic field generated by the orbital motion is called spin–orbit coupling The energy of interaction is proportional to the scalar product of the two vectors representing the spin and orbital angular momenta and hence depends upon the orientation of the two vectors See Fig 13.29 The total angular momentum of an electron in an atom is the vector sum of the orbital and spin angular momenta as illustrated in Fig 13.30 and expressed in eqn 13.46 The spin–orbit coupling results in a splitting of the energy levels associated with atomic terms as shown in Figs 13.31 and 13.32 This splitting shows up in atomic spectra as a fine structure as illustrated in Fig 13.32 INSTRUCTOR’S MANUAL 202 Numerical exercises E13.5(b) The energy of the photon that struck the Xe atom goes into liberating the bound electron and giving it any kinetic energy it now possesses Ephoton = I + Ekinetic I = ionization energy The energy of a photon is related to its frequency and wavelength Ephoton = hν = hc λ and the kinetic energy of an electron is related to its mass and speed Ekinetic = 21 me s So hc hc − me s = I + 21 me s ⇒ I = λ λ I = (6.626 × 10−34 J s) × (2.998 × 108 m s−1 ) 58.4 × 10−9 m − 21 (9.11 × 10−31 kg) × (1.79 × 106 m s−1 )2 = 1.94 × 10−18 J = 12.1 eV E13.6(b) The radial wavefunction is [Table 13.1] 2Zr R3,0 = A − 2ρ + 19 ρ e−ρ/6 where ρ ≡ , and A is a collection of constants Differentiating a0 with respect to ρ yields dR3,0 = = A(6 − 2ρ + 19 ρ ) × − 16 e−ρ/6 + −2 + 29 ρ Ae−ρ/6 dρ = Ae−ρ/6 − ρ54 + 59 ρ − This is a quadratic equation = aρ + bρ + c where a = − , b = , and c = −3 54 The solution is ρ= so r = √ −b ± (b2 − 4ac)1/2 = 15 ± 2a 15 3(71/2 ) ± 2 a0 Z Numerically, this works out to ρ = 7.65 and 2.35, so r = 11.5a0 /Z and 3.53a0 /Z Substituting Z = and a0 = 5.292 × 10−11 m, r = 607 pm and 187 pm The other maximum in the wavefunction is at r = It is a physical maximum, but not a calculus maximum: the first derivative of the wavefunction does not vanish there, so it cannot be found by differentiation ATOMIC STRUCTURE AND ATOMIC SPECTRA E13.7(b) 203 The radial wavefunction is [Table 13.1] R3,1 = A − 13 ρ ρe−ρ/6 where ρ= 2Zr a0 The radial nodes occur where the radial wavefunction vanishes This occurs at r=0 ρ = 0, and when − 13 ρ = 0, ρ = 4, or or ρ = 12 ρa0 12a0 ρa0 = = = 6a0 = 3.18 × 10−10 m 2Z 2 Normalization requires then r = E13.8(b) |ψ|2 dτ = = = N2 ∞ ∞ π 2π [N (2 − r/a0 )e−r/2a0 ]2 dφ sin θ dθ r dr e−r/a0 (2 − r/a0 )2 r dr π 2π sin θ dθ dφ Integrating over angles yields ∞ = 4πN ∞ = 4πN e−r/a0 (2 − r/a0 )2 r dr e−r/a0 (4 − 4r/a0 + r /a02 )r dr = 4π N (8a03 ) ∞ In the last step, we used So N = E13.9(b) e−r/k r dr = 2k , ∞ e−r/k r dr = 6k , and ∞ e−r/k r dr = 24k 2πa03 The average kinetic energy is Eˆ K = ψ ∗ Eˆ K ψ dτ where ψ = N (2 − ρ)e h ¯ ∇2 Eˆ K = − 2m −ρ/2 with N = Z3 2π a03 1/2 and ρ ≡ Zr a0 here a ρ sin θ dρ dθ dφ dτ = r sin θ dr dθ dφ = Z3 In spherical polar coordinates, three of the derivatives in ∇ are derivatives with respect to angles, so those parts of ∇ ψ vanish Thus ∇2ψ = ∂ 2ψ ∂ψ ∂ 2ψ + = r ∂r ∂r ∂ρ ∂ρ 2Z + ρa0 ∂r ∂ψ ∂ρ ∂ρ = ∂r Z × a0 ∂ 2ψ ∂ψ + ρ ∂ρ ∂ρ INSTRUCTOR’S MANUAL 204 ∂ρ = N (2 − ρ) × − 21 e−ρ/2 − N e−ρ/2 = N 21 ρ − e−ρ/2 ∂r ∂ 2ψ = N 21 ρ − × − 21 e−ρ/2 + 21 N e−ρ/2 = N 23 − 41 ρ e−ρ/2 ∂ρ Z N e−ρ/2 (−4/ρ + 5/2 − ρ/4) a0 ∇2ψ = and Eˆ K = π ∞ 2π 0 Z × a0 N (2 − ρ)e−ρ/2 −¯h2 2m a dφ sin θ dθ ρ dρ × N e−ρ/2 (−4/ρ + 5/2 − ρ/4) Z3 The integrals over angles give a factor of 4π , so ∞ a0 h ¯2 × − Eˆ K = 4πN Z 2m (2 − ρ) × −4 + 25 ρ − 41 ρ ρe−ρ dρ The integral in this last expression works out to −2, using Z3 Eˆ K = 4π × 32πa03 a0 × Z h ¯2 e−ρ ρ n dρ = n! for n = 1, 2, and So h ¯ Z2 = m ∞ 8ma02 The average potential energy is V = and V = ψ ∗ V ψ dτ ∞ π where V = − 2π 0 Z e2 Ze2 =− 4π ε0 r 4π ε0 a0 ρ N (2 − ρ)e−ρ/2 − a ρ sin θ dρ dθ dφ N (2 − ρ)e−ρ/2 Z3 Z e2 4π ε0 a0 ρ The integrals over angles give a factor of 4π , so V = 4πN − Z e2 4πε0 a0 × ∞ a03 Z3 (2 − ρ)2 ρe−ρ dρ ∞ The integral in this last expression works out to 2, using V = 4π Z3 32πa03 × − Z e2 4πε0 a0 × a03 Z3 e−ρ ρ n dρ = n! for n = 1, 2, 3, and So × (2) = − Z e2 16π ε0 a0 E13.10(b) The radial distribution function is defined as P = 4πr ψ so P3s = 4πr (Y0,0 R3,0 )2 , P3s = 4πr 4π × where ρ ≡ 2Zr 2Zr = na0 3a0 243 here × Z × (6 − 6ρ + ρ )2 e−ρ a0 ATOMIC STRUCTURE AND ATOMIC SPECTRA 205 But we want to find the most likely radius, so it would help to simplify the function by expressing it in terms either of r or ρ, but not both To find the most likely radius, we could set the derivative of P3s equal to zero; therefore, we can collect all multiplicative constants together (including the factors of a0 /Z needed to turn the initial r into ρ ) since they will eventually be divided into zero P3s = C ρ (6 − 6ρ + ρ )2 e−ρ Note that not all the extrema of P are maxima; some are minima But all the extrema of (P3s )1/2 correspond to maxima of P3s So let us find the extrema of (P3s )1/2 d(P3s )1/2 d =0= Cρ(6 − 6ρ + ρ )e−ρ/2 dρ dρ = C[ρ(6 − 6ρ + ρ ) × (− 21 ) + (6 − 12ρ + 3ρ )]e−ρ/2 = C(6 − 15ρ + 6ρ − 21 ρ )e−ρ/2 so 12 − 30ρ + 12ρ − ρ = Numerical solution of this cubic equation yields ρ = 0.49, 2.79, and 8.72 corresponding to r = 0.74a0 /Z, 4.19a0 /Z, and 13.08a0 /Z Comment If numerical methods are to be used to locate the roots of the equation which locates the extrema, then graphical/numerical methods might as well be used to locate the maxima directly That is, the student may simply have a spreadsheet compute P3s and examine or manipulate the spreadsheet to locate the maxima E13.11(b) Orbital angular momentum is ¯ (l(l + 1))1/2 Lˆ 1/2 = h There are l angular nodes and n − l − radial nodes (a) n = 4, l = 2, so Lˆ 1/2 = 61/2 h ¯ = 2.45 × 10−34 J s angular nodes radial node (b) n = 2, l = 1, so Lˆ 1/2 = 21/2 h ¯ = 1.49 × 10−34 J s angular node radial nodes (c) n = 3, l = 1, so Lˆ 1/2 = 21/2 h ¯ = 1.49 × 10−34 J s angular node radial node E13.12(b) For l > 0, j = l ± 1/2, so (a) l = 1, so j = 1/2 or 3/2 (b) l = 5, so j = 9/2 or 11/2 E13.13(b) Use the Clebsch–Gordan series in the form J = j1 + j2 , j1 + j2 − 1, , |j1 − j2 | Then, with j1 = and j2 = J = 8, 7, 6, 5, 4, 3, INSTRUCTOR’S MANUAL 206 E13.14(b) The degeneracy g of a hydrogenic atom with principal quantum number n is g = n2 The energy E of hydrogenic atoms is E=− hcZ RH hcZ RH =− g n so the degeneracy is g=− hcZ RH E (a) g=− (b) g=− (c) g=− hc(2)2 RH = −4hcRH hc(4)2 RH − 41 hcRH = 64 hc(5)2 RH = 25 −hcRH E13.15(b) The letter F indicates that the total orbital angular momentum quantum number L is 3; the superscript is the multiplicity of the term, 2S + 1, related to the spin quantum number S = 1; and the subscript indicates the total angular momentum quantum number J E13.16(b) The radial distribution function varies as P = 4πr ψ = r e−2r/a0 a0 The maximum value of P occurs at r = a0 since dP 2r ∝ 2r − a0 dr e−2r/a0 = at r = a0 and Pmax = −2 e a0 P falls to a fraction f of its maximum given by f = 4r −2r/a0 e a03 −2 a0 e r2 = e2 e−2r/a0 a0 and hence we must solve for r in f 1/2 r = e−r/a0 e a0 f = 0.50 r 0.260 = e−r/a0 solves to r = 2.08a0 = 110 pm and to r = 0.380a0 = 20.1 pm a0 (b) f = 0.75 r 0.319 = e−r/a0 solves to r = 1.63a0 = 86 pm and to r = 0.555a0 = 29.4 pm a0 (a) In each case the equation is solved numerically (or graphically) with readily available personal computer software The solutions above are easily checked by substitution into the equation for f The radial distribution function is readily plotted and is shown in Fig 13.1 ATOMIC STRUCTURE AND ATOMIC SPECTRA 207 0.15 0.10 0.05 0.00 0.0 0.5 1.0 1.5 2.0 2.5 Figure 13.1 E13.17(b) (a) 5d → 2s is not an allowed transition, for (b) 5p → 3s is allowed , since (c) 5p → 3f is not allowed, for l = −2 ( l must equal ±1) l = −1 l = +2 ( l must equal ±1) E13.18(b) For each l, there are 2l + values of ml and hence 2l + orbitals—each of which can be occupied by two electrons, so maximum occupancy is 2(2l + 1) (a) 2s: l = 0; maximum occupancy = (b) 4d: l = 2; maximum occupancy = 10 (c) 6f : l = 3; maximum occupancy = 14 (d) 6h: l = 5; maximum occupancy = 22 E13.19(b) V2+ : 1s 2s 2p 3s 3p 3d = [Ar]3d The only unpaired electrons are those in the 3d subshell There are three S = 23 and 23 − = 21 For S = 23 , MS = ± 21 and ± 23 for S = 21 , MS = ± 21 E13.20(b) (a) Possible values of S for four electrons in different orbitals are 2, 1, and ; the multiplicity is 2S + 1, so multiplicities are 5, 3, and respectively (b) Possible values of S for five electrons in different orbitals are 5/2, 3/2, and 1/2 ; the multiplicity is 2S + 1, so multiplicities are 6, 4, and respectively E13.21(b) The coupling of a p electron (l = 1) and a d electron (l = 2) gives rise to L = (F), (D), and (P) terms Possible values of S include and Possible values of J (using Russell–Saunders coupling) are 3, 2, and (S = 0) and 4, 3, 2, 1, and (S = 1) The term symbols are F3 ; F4 , F3 , F2 ; D2 ; D3 , D2 , D1 ; P1 ; P2 , P1 , P0 Hund’s rules state that the lowest energy level has maximum multiplicity Consideration of spin–orbit coupling says the lowest energy level has the lowest value of J (J + 1) − L(L + 1) − S(S + 1) So the lowest energy level is F2 INSTRUCTOR’S MANUAL 208 E13.22(b) (a) D has S = and L = 2, so J = 3, 2, and are present J = has states, with MJ = 0, ±1, ±2, or ±3; J = has states, with MJ = 0, ±1, or ±2; J = has states, with MJ = 0, or ±1 (b) D has S = 3/2 and L = 2, so J = 7/2, 5/2, 3/2, and 1/2 are present J = 7/2 has possible states, with MJ = ±7/2, ±5/2, ±3/2 or ±1/2; J = 5/2 has possible states, with MJ = ±5/2 ±3/2 or ±1/2; J = 3/2 has possible states, with MJ = ±3/2 or ±1/2; J = 1/2 has possible states, with MJ = ±1/2 (c) G has S = 1/2 and L = 4, so J = 9/2 and 7/2 are present J = 9/2 had 10 possible states, with MJ = ±9/2, ±7/2, ±5/2, ±3/2 or ±1/2; J = 7/2 has possible states, with MJ = ±7/2, ±5/2, ±3/2 or ±1/2 E13.23(b) Closed shells and subshells not contribute to either L or S and thus are ignored in what follows (a) Sc[Ar]3d 4s : S = 21 , L = 2; J = 25 , 23 , so the terms are D5/2 and D3/2 (b) Br[Ar]3d 10 4s 4p We treat the missing electron in the 4p subshell as equivalent to a single “electron” with l = 1, s = 21 Hence L = 1, S = 21 , and J = 23 , 21 , so the terms are P3/2 and P1/2 Solutions to problems Solutions to numerical problems P13.2 All lines in the hydrogen spectrum fit the Rydberg formula = RH λ n21 − n2 13.1, with ν˜ = λ RH = 109 677 cm−1 Find n1 from the value of λmax , which arises from the transition n1 + → n1 1 2n1 + = 2− = 2 λmax RH (n + 1) n1 n1 (n1 + 1)2 n2 (n1 + 1)2 = (656.46 × 10−9 m) × (109 677 × 102 m−1 ) = 7.20 λmax RH = 2n1 + and hence n1 = 2, as determined by trial and error substitution Therefore, the transitions are given by ν˜ = = (109 677 cm−1 ) × λ 1 − n2 , n2 = 3, 4, 5, The next line has n2 = 7, and occurs at ν˜ = = (109 677 cm−1 ) × λ 1 − 49 = 397.13 nm The energy required to ionize the atom is obtained by letting n2 → ∞ Then ν˜ ∞ = = (109 677 cm−1 ) × λ∞ − = 27 419 cm−1 , or 3.40 eV ATOMIC STRUCTURE AND ATOMIC SPECTRA 209 (The answer, 3.40 eV, is the ionization energy of an H atom that is already in an excited state, with n = 2.) Comment The series with n1 = is the Balmer series 610 nm I 460 nm 413 nm The lowest possible value of n in 1s nd is 3; thus the series of D terms correspond to 1s 3d, 1s 4d, etc Figure 13.2 is a description consistent with the data in the problem statement 670 nm P13.4 Figure 13.2 If we assume that the energies of the d orbitals are hydrogenic we may write E(1s nd , D) = − hcR n2 [n = 3, 4, 5, ] Then for the D → P transitions ν˜ = |E(1s 2p , P)| R − = hc λ n E = hν = hc E = hcν˜ , ν˜ = λ hc from which we can write  R   +   −7  610.36 × 10 cm   |E(1s 2p , P)| R R = + = +  hc λ n 16 460.29 × 10−7 cm     R   + 25 413.23 × 10−7 cm  (b) − (a) solves to R = 109 886 cm−1 Then (a) − (c) solves to R = 109 910 cm−1 Mean = 109 920 cm−1  (b) − (c) solves to R = 109 963 cm−1 The binding energies are therefore R = −12 213 cm−1 − 12 213 cm−1 = −28 597 cm−1 E(1s 2p, P) = − 610.36 × 10−7 cm − 28 597 cm−1 = −43 505 cm−1 E(1s 2s , S) = − 670.78 × 10−7 cm E(1s 3d , D) = (a) (b) (c) INSTRUCTOR’S MANUAL 210 Therefore, the ionization energy is I (1s 2s ,2 S) = 43 505 cm−1 , P13.5 or 5.39 eV The 7p configuration has just one electron outside a closed subshell That electron has l = 1, s = 1/2, and j = 1/2 or 3/2, so the atom has L = 1, S = 1/2, and J = 1/2 or 3/2 The term symbols are P1/2 and P3/2 , of which the former has the lower energy The 6d configuration also has just one electron outside a closed subshell; that electron has l = 2, s = 1/2, and j = 3/2 or 5/2, so the atom has L = 2, S = 1/2, and J = 3/2 or 5/2 The term symbols are D3/2 and D5/2 , of which the former has the lower energy According to the simple treatment of spin–orbit coupling, the energy is given by El,s,j = 21 hcA[j (j + 1) − l(l + 1) − s(s + 1)] where A is the spin–orbit coupling constant So E(2 P1/2 ) = 21 hcA[ 21 (1/2 + 1) − 1(1 + 1) − 21 (1/2 + 1)] = −hcA and E(2 D3/2 ) = 21 hcA[ 23 (3/2 + 1) − 2(2 + 1) − 21 (1/2 + 1)] = − 23 hcA This approach would predict the ground state to be D3/2 Comment The computational study cited above finds the P1/2 level to be lowest, but the authors caution that the error of similar calculations on Y and Lu is comparable to the computed difference between levels P13.7 RH = kµH , RD = kµD , R = kµ [18] where R corresponds to an infinitely heavy nucleus, with µ = me me mN [N = p or d] Since µ = me + m N RH = kµH = kme R me = me 1+ m + mp p R me where mp is the mass of the proton and md the mass of the deuteron The 1+ m d two lines in question lie at Likewise, RD = = RH − 41 = 43 RH λH = RD − 41 = 43 RD λD and hence λD ν˜ H RH = = RD λH ν˜ D Then, since 1+ RH = RD 1+ me md me mp , md = me 1+ me mp RH RD −1 ATOMIC STRUCTURE AND ATOMIC SPECTRA 211 and we can calculate md from me md = me 1+ m p λD λH −1 = me me 1+ m p ν˜ H ν˜ D −1 9.10939 × 10−31 kg = 1+ 9.10939×10−31 kg 1.67262×10−27 kg × 82 259.098 cm−1 82 281.476 cm−1 −1 = 3.3429 × 10−27 kg Since I = Rhc, ID RD ν˜ D 82 281.476 cm−1 = = = = 1.000 272 IH RH ν˜ H 82 259.098 cm−1 P13.10 If we assume that the innermost electron is a hydrogen-like 1s orbital we may write r∗ = 52.92 pm a0 [Example 13.3] = = 0.420 pm Z 126 Solutions to theoretical problems P13.12 Consider ψ2pz = ψ2,1,0 which extends along the z-axis The most probable point along the z-axis is where the radial function has its maximum value (for ψ is also a maximum at that point) From Table 13.1 we know that R21 ∝ ρe−ρ/4 dR = − 41 ρ e−ρ/4 = when ρ = dρ 2a0 2a0 Therefore, r ∗ = , and the point of maximum probability lies at z = ± = ±106 pm Z Z Comment Since the radial portion of a 2p function is the same, the same result would have been obtained for all of them The direction of the most probable point would, however, be different and so P13.13 In each case we need to show that all space ψ1∗ ψ2 dτ = ∞ π 2π (a) 0 ? ψ1s ψ2s r dr sin θ dθ dφ = 1/2 ψ1s = R1,0 Y0,0 Y0,0 = [Table 12.3] ψ2s = R2,0 Y0,0 4π Since Y0,0 is a constant, the integral over the radial functions determines the orthogonality of the functions ∞ R1,0 R2,0 r dr R1,0 ∝ e−ρ/2 = e−Zr/a0 ρ= R2,0 ∝ (2 − ρ/2)e−ρ/4 = − 2Zr a0 Zr a0 e−Zr/2a0 ρ= 2Zr a0 INSTRUCTOR’S MANUAL 212 ∞ R1,0 R2,0 r dr ∝ = = ∞ ∞ e−Zr/a0 − Zr a0 e−Zr/2a0 r dr 2e−(3/2)Zr/a0 r dr − × 2! Z a0 Z a0 − × ∞ 3! Z a0 Z −(3/2)Zr/a0 e r dr a0 = Hence, the functions are orthogonal (b) We use the px and py orbitals in the form given in Section 13.2(f ), eqn 25 px ∝ x, py ∝ y Thus all space px py dx dy dz ∝ +∞ −∞ +∞ +∞ −∞ −∞ xy dx dy dz This is an integral of an odd function of x and y over the entire range of variable from −∞ to +∞, therefore, the integral is zero More explicitly we may perform the integration using the orbitals in the form (Section 13.2(f ), eqn 13.25) px = f (r) sin θ cos φ all space py = f (r) sin θ sin φ ∞ px py r dr sin θ dθ dφ = f (r)2 r dr π sin2 θ dθ 2π cos φ sin φ dφ π The third factor is zero Therefore, the product of the integrals is zero and the functions are orthogonal The first factor is nonzero since the radial functions are normalized The second factor is P13.14 − h ¯2 2µ d d2 + r dr dr + Veff R = ER [13.11] (1) l(l + 1)¯h2 l(l + 1)¯h2 Ze2 Z¯h2 + + =− 4πε0 r µa0 r 2µr 2µr Using ρ = Zr/a0 , the derivative term of the Hamiltonian can be written in the form where Veff = − d2 d = + r dr dr d2 d Z × + a0 ρ dρ dρ ≡ Dop (2) To determine E2s and E2p , we will evaluate the left side of (1) and compare the result to the right side 2s orbital R2s = N2s (2 − ρ)e−ρ/2 where ρ ≡ Zr/a0 here dR2s = N2s −1 − 21 (2 − ρ) e−ρ/2 = N2s dρ ρ−4 e−ρ/2 = ρ−4 R2s 2(2 − ρ) d2 R2s ρ −ρ/2 6−ρ e R2s = N2s 21 − 41 (ρ − 4) e−ρ/2 = N2s 23 − = 4(2 − ρ) dρ ATOMIC STRUCTURE AND ATOMIC SPECTRA − h ¯2 h ¯2 Dop R2s = − 2µ 2µ = − Veff R2s = − − h ¯2 2µ 213 h ¯2 Dop + Veff R2s = 2µ =− 2p orbital ρ−8 4ρ × × Z¯h2 Z R2s = − × µa0 r a0 Therefore E2s = − 41 Z R2s a0 ρ−4 6−ρ + × 4(2 − ρ) ρ(2 − ρ) Z R2s a0 h ¯2 µ R2s Z h ì a0 2à Z2h × ρ−8 + R2s 4ρ ρ R2s 2µa02 Z2h ¯2 (3) 2µa02 R2p = N2p ρe−ρ/2 where ρ ≡ Zr/a0 here dR2p ρ −ρ/2 2−ρ = N2p − e R2p = dρ 2ρ d2 R2p ρ = N2p − 21 − 21 − dρ − h ¯2 h ¯2 Dop R2p = − 2µ 2µ =− Veff R2p = − = − h ¯2 2µ ρ−4 R2p 4ρ e−ρ/2 = ρ − 4 − 2ρ + 4ρ 2ρ ρ − 8ρ + 4ρ × × Z R2p a0 Z R2p a0 Z¯h2 Z h ì + R2p = àa0 r a0 àr Z h ì a0 2à h ¯2 Dop + Veff R2p = 2µ =− Therefore E2p = 41 ì 2àa02 + Z ì a0 h R2p 2( − 1) R2p ρ2 Z h ¯2 × − a0 2µ ¯2 Z2h h ¯2 µ ρ2 4ρ × ρ − 8ρ + 2(ρ − 1) R2p + 4ρ ρ2 R2p = − Z2h ¯2 2µa02 Comparison of eqns (3) and (4) reveals that E2s = E2p ¯2 Z2h 2µa02 R2p (4) INSTRUCTOR’S MANUAL 214 P13.15 |ψ3px |2 dτ = The integrations are most easily performed in spherical (a) We must show that coordinate (Fig 11) 2π π ∞ ψ3px dτ = ψ3px 2 r sin(θ ) dr dθ dφ 0 2π π ∞ = Y1−1 − Y11 √ R31 (ρ) 0 r sin(θ ) dr dθ dφ (Table 13.1, eqn 13.25) where ρ = 2r/a0 , r = ρa0 /2, dr = (a0 /2) dρ = 2π π ∞ a0 3/2 − ρ ρe−ρ/6 a0 27(6)1/2 0 1/2 sin(θ ) cos(φ) ρ sin(θ ) dρ dθ dφ 8π × = 46 656π 2π π ∞ − ρ ρe−ρ/6 sin(θ ) cos(φ) ρ sin(θ ) dρ dθ dφ 0 = 46 656π π 2π cos (φ) dφ ∞ sin (θ ) dθ 0 π =1 − ρ ρ e−ρ/3 dρ 4/3 34992 Thus, ψ3px is normalized to We must also show that ψ3px ψ3dxy dτ = Using Tables 12.3 and 13.1, we find that ψ3px = 54(2π)1/2 ψ3dxy = R32 = 3/2 − ρ ρe−ρ/6 sin(θ ) cos(φ) a0 Y22 − Y2−2 √ 2i 32(2π)1/2 3/2 −ρ/6 ρ e sin (θ ) sin(2φ) a0 where ρ = 2r/a0 , r = ρa0 /2, dr = (a0 /2)dρ ∞ ψ3px ψ3dxy dτ = constant × ρ e π 2π −ρ/3 dρ sin4 (θ )dθ cos(φ) sin(2φ)dφ 0 Since the integral equals zero, ψ3px and ψ3dxy are orthogonal ATOMIC STRUCTURE AND ATOMIC SPECTRA 215 (b) Radial nodes are determined by finding the ρ values (ρ = 2r/a0 ) for which the radial wavefunction equals zero These values are the roots of the polynomial portion of the wavefunction √ √ For the 3s orbital − 6ρ + ρ = when ρnode = + and ρnode = − The 3s orbital has these two spherically symmetrical modes There is no node at ρ = so we conclude that there is a finite probability of finding a 3s electron at the nucleus For the 3px orbital (4 − ρ)(ρ) = when ρnode = and ρnode = There is a zero probability of finding a 3px electron at the nucleus For the 3dxy orbital ρnode = is the only radial node (c) |R10 Y00 |2 r dτ = r 3s = ∞ 2π π R10 r dr = a0 888 |Y00 |2 sin(θ ) dθ dφ 0 = |R10 Y00 |2 r sin(θ ) dr dθ dφ ∞ − 2ρ + ρ /9 ρ e−ρ/3 dρ 52488 r 3s = 27a0 (d) Radial distribution functions of atomic hydrogen 0.12 3px 3dxy 0.1 3s r 2R2a0 0.08 0.06 0.04 0.02 0 10 15 r /a0 20 25 30 Figure 13.3(a) The plot shows that the 3s orbital has larger values of the radial distribution function for r < a0 This penetration of inner core electrons of multi-electron atoms means that a 3s electron INSTRUCTOR’S MANUAL 216 experiences a larger effective nuclear charge and, consequently, has a lower energy than either a 3px or 3dxy electron This reasoning also lead us to conclude that a 3px electron has less energy than a 3dxy electron E3s < E3px < E3dxy (e) Polar plots with θ = 90◦ The s Orbital 90 120 60 150 30 180 00.20.40.60.8 210 330 300 240 270 The p Orbital 90 120 60 30 150 180 0 0.2 0.4 0.6 0.8 210 330 300 240 270 The d Orbital 90 120 60 30 150 180 0 0.2 0.4 210 330 300 240 270 Figure 13.3(b) ATOMIC STRUCTURE AND ATOMIC SPECTRA 217 Boundary surface plots s - Orbital boundary surface p - Orbital boundary surface d - Orbital boundary surface f - Orbital boundary surface Figure 13.3(c) P13.20 Ze2 · 4π ε0 r (angular momentum)2 (n¯h)2 The repulsive centrifugal force = = [postulated] me r me r The two forces balance when The attractive Coulomb force = Ze2 ¯2 n2 h × = , 4πε0 r me r The total energy is E = EK + V = = n2 h ¯2 2me × implying that r= ¯ ε0 4π n2 h Ze2 me Ze2 n2 h ¯2 (angular momentum)2 Ze2 − [postulated] × = − 2I 4π ε0 r 4π ε0 r 2me r Ze2 me 4πn2 h ¯ 2ε − Ze2 4π ε0 × Ze2 me 4π n2 h ¯ 2ε = − Z e me ¯2 32π ε02 h × n INSTRUCTOR’S MANUAL 218 P13.21 (a) The trajectory is defined, which is not allowed according to quantum mechanics ¯ , not by n¯h In (b) The angular momentum of a three-dimensional system is given by {l(l + 1)}1/2 h the Bohr model, the ground state possesses orbital angular momentum (n¯h, with n = 1), but the actual ground state has no angular momentum (l = 0) Moreover, the distribution of the electron is quite different in the two cases The two models can be distinguished experimentally by (a) showing that there is zero orbital angular momentum in the ground state (by examining its magnetic properties) and (b) examining the electron distribution (such as by showing that the electron and the nucleus come into contact, Chapter 18) P13.25 Justification 13.5 noted that the transition dipole moment, µfi had to be non-zero for a transition to be allowed The Justification examined conditions that allowed the z component of this quantity to be non-zero; now examine the x and y components f ∗ x i dτ µx,fi = −e µy,fi = −e and f ∗ y i dτ As in the Justification, express the relevant Cartesian variables in terms of the spherical harmonics, Yl,m Start by expressing them in spherical polar coordinates: x = r sin θ cos φ and y = r sin θ sin φ Note that Y1,1 and Y1,−1 have factors of sin θ They also contain complex exponentials that can be related to the sine and cosine of φ through the identities (eqns FI1.20 and FI1.21) cos φ = 1/2(eiφ + e−iφ ) and sin φ =1/2i(eiφ −e−iφ ) These relations motivate us to try linear combinations Y1,1 +Y1,−1 and Y1,1 +Y1,−1 (form Table 12.3; note c here corresponds to the normalization constant in the table): Y1,1 + Y1,−1 = −c sin θ (eiφ + e−iφ ) = −2c sin θ cos φ = −2cx/r, so x = −(Y1,1 + Y1,−1 )r/2c; Y1,1 − Y1,−1 = c sin θ (eiφ − e−iφ ) = 2ic sin θ sin φ = 2icy/r, so y = (Y1,1 − Y1,−1 )r/2ic Now we can express the integrals in terms of radial wavefunctions Rn,l and spherical harmonics Yl,ml e µx,fi = 2c ∞ π 2π Y ∗ lf ,mlf (Y1,1 + Y1,−1 )Yli ,mli sin θ dθ dφ Rnf ,lf rRni ,li r dr 0 The angular integral can be broken into two, one of which contains Y1,1 and the other Y1,−1 According to the “triple integral” relation below Table 12.3, the integral π 2π Y ∗ lf ,mlf Y1,1 Yli ,mli sin θ dθ dφ 0 vanishes unless lf = li ± and mf = mi ± The integral that contains Y1,−1 introduces no further constraints; it vanishes unless lf = li ± and mlf = mli ± Similarly, the y component introduces no further constraints, for it involves the same spherical harmonics as the x component The whole set of selection rules, then, is that transitions are allowed only if l = ±1 and ml = or ± ATOMIC STRUCTURE AND ATOMIC SPECTRA P13.26 219 (a) The speed distribution in the molecular beam is related to the speed distribution within the chamber by a factor of v cos θ as shown in Fig 13.4 Since an integration over all possible θ must be performed, the cos θ factor may be absorbed into the constant of proportionality fbeam (v) = Cvfchamber (v) where C is to be determined Molecular beam Chamber Figure 13.4 By normalization over the possible beam speeds (0 < vbeam < ∞) fbeam = Cv v e−(mv /2kT ) = Cv e−(mv /2kT ) ∞ fbeam dv = = C v=0 ∞ v=0 v e−(mv /2kT ) dv = C 2(m/2kT )2 C = 2(m/2kT ) v2 = ∞ v=0 v fbeam (v) dv = C v e−(mv /2kT ) dv (m/2kT )2 = (m/2kT )3 (m/2kT )3 4kT = m m m 4kT v = = 2kT EK = 2 m =C x= (b) or 2µB L2 4EK dB dz dB 4EK x 4(2kT ) x = = dz 2µB L 2µB L2 = = 4kT x µ B L2 4(1.3807 × 10−23 J K−1 ) × (1000 K) × (1.00 × 10−3 m) (9.27402 × 10−24 J T−1 ) × (50 × 10−2 m)2 dB = 23.8 T m−1 dz .. .INSTRUCTOR S MANUAL 202 Numerical exercises E13.5(b) The energy of the photon that struck the Xe atom goes into liberating the bound electron and giving... a0 ATOMIC STRUCTURE AND ATOMIC SPECTRA 205 But we want to find the most likely radius, so it would help to simplify the function by expressing it in terms either of r or ρ, but not both To find... equal to zero; therefore, we can collect all multiplicative constants together (including the factors of a0 /Z needed to turn the initial r into ρ ) since they will eventually be divided into zero