12 Quantum theory: techniques and applications Solutions to exercises Discussion questions E12.1(b) The correspondence principle states that in the limit of very large quantum numbers quantum mechanics merges with classical mechanics An example is a molecule of a gas in a box At room temperature, the particle-in-a-box quantum numbers corresponding to the average energy of the gas molecules ( 21 kT per degree of freedom) are extremely large; consequently the separation between the levels is relatively so small (n is always small compared to n2 , compare eqn 12.10 to eqn 12.4) that the energy of the particle is effectively continuous, just as in classical mechanics We may also look at these equations from the point of view of the mass of the particle As the mass of the particle increases to macroscopic values, the separation between the energy levels approaches zero The quantization disappears as we know it must Tennis balls not show quantum mechanical effects (Except those served by Pete Sampras.) We can also see the correspondence principle operating when we examine the wavefunctions for large values of the quantum numbers The probability density becomes uniform over the path of motion, which is again the classical result This aspect is discussed in more detail in Section 12.1(c) The harmonic oscillator provides another example of the correspondence principle The same effects mentioned above are observed We see from Fig 12.22 of the text that probability distribution for large values on n approaches the classical picture of the motion (Look at the graph for n = 20.) E12.2(b) The physical origin of tunnelling is related to the probability density of the particle which according to the Born interpretation is the square of the wavefunction that represents the particle This interpretation requires that the wavefunction of the system be everywhere continuous, event at barriers Therefore, if the wavefunction is non-zero on one side of a barrier it must be non-zero on the other side of the barrier and this implies that the particle has tunnelled through the barrier The transmission probability depends upon the mass of the particle (specifically m1/2 , through eqns 12.24 and 12.28): the greater the mass the smaller the probability of tunnelling Electrons and protons have small masses, molecular groups large masses; therefore, tunnelling effects are more observable in process involving electrons and protons E12.3(b) The essential features of the derivation are: (1) The separation of the hamiltonian into large (unperturbed) and small (perturbed) parts which are independent of each other (2) The expansion of the wavefunctions and energies as a power series in an unspecified parameters, λ, which in the end effectively cancels or is set equal to (3) The calculation of the first-order correction to the energies by an integration of the perturbation over the zero-order wavefunctions (4) The expansion of the first-order correction to the wavefunction in terms of the complete set of functions which are a solution of the unperturbed Schrodinger equation (5) The calculation of the second-order correction to the energies with use of the corrected first order wavefunctions See Justification 12.7 and Further reading for a more complete discussion of the method INSTRUCTOR’S MANUAL 186 Numerical exercises E= E12.4(b) n2 h 8me L2 h2 (6.626 × 10−34 J s)2 = = 2.678 × 10−20 J 8me L2 8(9.109 × 10−31 kg) × (1.50 × 10−9 m)2 Conversion factors E kJ mol−1 = NA E/J 103 eV = 1.602 × 10−19 J cm−1 = 1.986 × 10−23 J E3 − E1 = (9 − 1) (a) h2 = 8(2.678 × 10−20 J) 8me L2 = 2.14 × 10−19 J = 129 kJ mol−1 = 1.34 eV = 1.08 × 104 cm−1 h2 8me L2 = 13(2.678 × 10−20 J) E7 − E6 = (49 − 36) (b) = 3.48 × 10−19 J = 210 kJ mol−1 = 2.17 eV = 1.75 × 104 cm−1 E12.5(b) The probability is ψ ∗ ψ dx = P = where L sin2 x nπ x nπ x dx = sin2 L L L x = 0.02L and the function is evaluated at x = 0.66L (a) For n = P = 2(0.02L) sin (0.66π) = 0.031 L (b) For n = P = E12.6(b) 2(0.02L) sin [2(0.66π )] = 0.029 L The expectation value is pˆ = ˆ dx ψ ∗ pψ but first we need pψ ˆ pψ ˆ = −i d 1/2 nπ x sin dx L L = −i 1/2 nπ nπ x cos L L L QUANTUM THEORY: TECHNIQUES AND APPLICATIONS so pˆ = 187 nπ x −2i nπ L nπ x dx = for all n cos sin L L L h2 n2 pˆ = 2m Hˆ = 2mEn = 4L2 for all n So for n = pˆ = E12.7(b) h2 L2 n=5 1/2 5πx sin L L 5πx P (x) ∝ ψ52 ∝ sin2 L ψ5 = Maxima and minima in P (x) correspond to d dψ 5π x P (x) ∝ ∝ sin dx dx L cos dP (x) =0 dx 5π x L ∝ sin 10π x L (2 sin α cos α = sin 2α) sin θ = when θ = 0, π, 2π, = n π (n = 0, 1, 2, ) 10πx = n π n ≤ 10 L nL x= 10 Minima at x = 0, x = L Maxima and minima alternate: maxima correspond to n = 1, 3, 5, 7, E12.8(b) x= L 3L L 7L 9L , , , , 10 10 10 10 The energy levels are En1 ,n2 ,n3 = (n21 + n22 + n23 )h2 8mL2 = E1 (n21 + n22 + n23 ) where E1 combines all constants besides quantum numbers The minimum value for all the quantum numbers is 1, so the lowest energy is E1,1,1 = 3E1 The question asks about an energy 14/3 times this amount, namely 14E1 This energy level can be obtained by any combination of allowed quantum numbers such that n21 + n22 + n23 = 14 = 32 + 22 + 12 The degeneracy, then, is , corresponding to (n1 , n2 , n3 ) = (1, 2, 3), (2, 1, 3), (1, 3, 2), (2, 3, 1), (3, 1, 2), or (3, 2, 1) INSTRUCTOR’S MANUAL 188 E12.9(b) E = 23 kT is the average translational energy of a gaseous molecule (see Chapter 20) E = 23 kT = (n21 + n22 + n23 )h2 8mL2 = n2 h2 8mL2 E = 23 × (1.381 × 10−23 J K−1 ) × (300 K) = 6.214 × 10−21 J n2 = 8mL2 E h2 If L3 = 1.00 m3 , L2 = 1.00 m2 (6.626 × 10−34 J s)2 h2 = = 1.180 × 10−42 J 0.02802 kg mol−1 8mL (8) × 6.022×1023 mol−1 × (1.00 m ) 6.214 × 10−21 J n = 7.26 × 1010 = 5.265 × 1021 ; 1.180 × 10−42 J E = En+1 − En = E7.26×1010 +1 − E7.26×1010 n2 = E = (2n + 1) × h2 8mL2 = [(2) × (7.26 × 1010 + 1)] × h2 8mL2 = 14.52 × 1010 h2 8mL2 = (14.52 × 1010 ) × (1.180 × 10−42 J) = 1.71 × 10−31 J The de Broglie wavelength is obtained from λ= h h = [Section 11.2] p mv The velocity is obtained from EK = 21 mv = 23 kT = 6.214 × 10−21 J 6.214 × 10−21 J v2 = λ= −1 0.02802 kg mol × 6.022×10 23 mol−1 = 2.671 × 105 ; v = 517 m s−1 6.626 × 10−34 J s = 2.75 × 10−11 m = 27.5 pm (4.65 × 10−26 kg) × (517 m s−1 ) The conclusion to be drawn from all of these calculations is that the translational motion of the nitrogen molecule can be described classically The energy of the molecule is essentially continuous, E ≪ E E12.10(b) The zero-point energy is E0 = ω= k 1/2 = 21 (1.0546 × 10−34 J s) × m = 3.92 × 10−21 J 285 N m−1 5.16 × 10−26 kg 1/2 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 189 E12.11(b) The difference in adjacent energy levels is k 1/2 m E= ω= so k= m( E)2 = (2.88 × 10−25 kg) × (3.17 × 10−21 J)2 (1.0546 × 10−34 J s)2 k = 260 N m−1 E12.12(b) The difference in adjacent energy levels, which is equal to the energy of the photon, is E = ω = hν and λ = hc k 1/2 hc = λ m so m 1/2 k 1/2 = 2πc m k −1 = 2π(2.998 × 10 m s )× (15.9949 u) × (1.66 × 10−27 kg u−1 ) 544 N m−1 1/2 λ = 1.32 × 10−5 m = 13.2 µm E12.13(b) The difference in adjacent energy levels, which is equal to the energy of the photon, is E = ω = hν and λ = hc so hc k 1/2 = m λ m 1/2 k 1/2 = 2πc m k Doubling the mass, then, increases the wavelength by 21/2 So taking the result from Ex 12.12(b), the new wavelength is λ = 21/2 (13.2 µm) = 18.7 µm E12.14(b) ω= g 1/2 [elementary physics] l E = ω = hν (a) (b) E = hν = (6.626 × 10−34 J Hz−1 ) × (33 × 103 Hz) = 2.2 × 10−29 J k 1/2 1 = + with m1 = m2 meff meff m1 m2 For a two-particle oscillator meff , replaces m in the expression for ω (See Chapter 16 for a more complete discussion of the vibration of a diatomic molecule.) E= ω= E= 2k 1/2 = (1.055 × 10−34 J s) × m = 3.14 × 10−20 J (2) × (1177 N m−1 ) (16.00) × (1.6605 × 10−27 kg) 1/2 INSTRUCTOR’S MANUAL 190 E12.15(b) The first excited-state wavefunction has the form ψ = 2N1 y exp − 21 y mω 1/2 To see if it satisfies Schrăodingers equation, where N1 is a collection of constants and y ≡ x we see what happens when we apply the energy operator to this function Hˆ ψ = − d2 ψ + 21 mω2 x ψ 2m dx 2 We need derivatives of ψ dψ dy mω 1/2 dψ (2N1 ) × (1 − y ) × exp − 21 y = = dx dy dx and dy mω mω × (2N1 ) × (−3y + y ) × exp − 21 y = × (y − 3)ψ = dx d2 ψ d2 ψ = dx dy So Hˆ ψ = − 2m × mω × (y − 3)ψ + 21 mω2 x ψ = − 21 ω × (y − 3) × ψ + 21 ωy ψ = 23 ωψ Thus, ψ is an eigenfunction of H (i.e it obeys the Schrăodinger equation) with eigenvalue E = 23 hω E12.16(b) The zero-point energy is k 1/2 meff E0 = 21 ω = 21 For a homonuclear diatomic molecule, the effective mass is half the mass of an atom, so E0 = 21 (1.0546 × 10−34 J s) × 2293.8 N m−1 −27 kg u−1 ) (14.0031 u) × (1.66054 × 10 E0 = 2.3421 × 10−20 J E12.17(b) Orthogonality requires that ∗ ψn dτ = ψm if m = n Performing the integration ∗ ψn dτ = ψm 2π N e−imφ Neinφ dφ = N 2π ei(n−m)φ dφ 1/2 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 191 If m = n, then ∗ ψm ψn dτ = N2 ei(n−m)φ i(n − m) 2π = N2 (1 − 1) = i(n − m) Therefore, they are orthogonal E12.18(b) The magnitude of angular momentum is Lˆ 1/2 = (l(l + 1))1/2 = (2(3))1/2 (1.0546 × 10−34 J s) = 2.58 × 10−34 J s Possible projections on to an arbitrary axis are Lˆ z = ml where ml = or ±1 or ±2 So possible projections include 0, ±1.0546 × 10−34 J s and 2.1109 × 10−34 J s E12.19(b) The cones are constructed as described in Section 12.7(c) and Fig 12.36 of the text; their edges are of length {6(6 + 1)}1/2 = 6.48 and their projections are mj = +6, +5, , −6 See Fig 12.1(a) The vectors follow, in units of From the highest-pointing to the lowest-pointing vectors (Fig 12.1(b)), the values of ml are 6, 5, 4, 3, 2, 1, 0, −1, −2, −3, −4, −5, and −6 Figure 12.1(a) Figure 12.1(b) INSTRUCTOR’S MANUAL 192 Solutions to problems Solutions to numerical problems P12.4 E= l(l + 1) l(l + 1) [12.65] = 2I 2meff R = [I = meff R , meff in place of m] l(l + 1) × (1.055 × 10−34 J s)2 (2) × (1.6605 × 10−27 kg) × (160 × 10−12 m)2 × 1 + 1.008 126.90 1 = + m2 meff m1 The energies may be expressed in terms of equivalent frequencies with ν = E = 1.509 × 1033 E h Therefore, E = l(l + 1) × (1.31 × 10−22 J) = l(l + 1) × (198 GHz) Hence, the energies and equivalent frequencies are l P12.6 22 10 E/J 2.62 7.86 15.72 ν/GHz 396 1188 2376 Treat the gravitational potential energy as a perturbation in the energy operator: H (1) = mgx The first-order correction to the ground-state energy, E1 , is: L E1 (1) L (0)∗ (1) H = E1 (1) = E1 (1) 2mg L (0) dx = L x sin2 2mg = L 1/2 πx mgx sin L L 1/2 πx dx, sin L L πx dx, L πx x2 xL πx πx L2 cos2 − cos sin − 2π L L L 4π L , E1 (1) = 21 mgL Not surprisingly, this amounts to the energy perturbation evaluated at the midpoint of the box For m = me , E1 (1) /L = 4.47 × 10−30 J m−1 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 193 Solutions to theoretical problems P12.8 − 2m × ∂2 ∂2 ∂2 + + ∂x ∂y ∂z2 ψ = Eψ [V = 0] We try the solution ψ = X(x)Y (y)Z(z) − − 2m (X Y Z + XY Z + XY Z ) = EXY Z X Y Z + + Z X Y 2m =E X depends only on x; therefore, when x changes only this term changes, but the sum of the three X X terms is constant Therefore, must also be constant We write X − X = EX , 2m X with analogous terms for y, z Hence we solve   X = E X X   2m    X Y Z Y − Y = E Y E = E + E + E , 2m     Z  − Z = E Z 2m − ψ = XY Z The three-dimensional equation has therefore separated into three one-dimensional equations, and we can write E= ψ= h2 8m n23 n22 + + L21 L22 L23 n21 1/2 n1 π x sin L1 L2 L3 L1 n1 , n2 , n3 = 1, 2, 3, sin n2 πy L2 sin n3 π z L3 For a cubic box E = (n21 + n22 + n23 ) P12.10 h2 8mL2 The wavefunctions in each region (see Fig 12.2(a)) are (eqns 12.22–12.25): ψ1 (x) = eik1 x + B1 e−ik2 x ψ2 (x) = A2 ek2 x + B2 e−k2 x ψ3 (x) = A3 eik3 x with the above choice of A1 = the transmission probability is simply T = |A3 |2 The wavefunction coefficients are determined by the criteria that both the wavefunctions and their first derivatives w/r/t INSTRUCTOR’S MANUAL 194 x be continuous at potential boundaries ψ1 (0) = ψ2 (0); ψ2 (L) = ψ3 (L) dψ1 (0) dψ2 (0) dψ2 (L) dψ3 (L) = ; = dx dx dx dx These criteria establish the algebraic relationships: + B − A2 − B2 = (−ik1 − k2 )A2 + (−ik1 + k2 )B2 + 2ik1 = A2 ek2 L + B2 e−k2 L − A3 eik3 L = A2 k2 ek2 L − B2 k2 e−k2 L − iA3 k3 eik3 L = V2 V V3 V1 x L Figure 12.2(a) Solving the simultaneous equations for A3 gives A3 = 4k1 k2 eik3 L (ia + b) ek2 L − (ia − b) e−k2 L where a = k22 − k1 k3 and b = k1 k2 + k2 k3 since sinh(z) = (ez − e−z )/2 or ez = sinh(z) + e−z , substitute ek2 L = sinh(k2 L) + e−k2 L giving: A3 = 2k1 k2 eik3 L (ia + b) sinh(k2 L) + b e−k2 L T = |A3 |2 = A3 A¯ = 4k12 k22 (a + b2 ) sinh2 (k2 L) + b2 where a + b2 = (k12 + k22 )(k22 + k32 ) and b2 = k22 (k1 + k3 )2 (b) In the special case for which V1 = V3 = 0, eqns 12.22 and 12.25 require that k1 = k3 Additionally, k1 ε E = where ε = E/V2 = k2 V2 − E 1−ε 2 a +b = (k12 + k22 )2 = k24 1+ k1 k2 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 195 b2 = 4k12 k22 a2 + b2 b2 T = T = = 2 k22 + kk21 4k12 = b2 b2 + (a 1+ + b2 ) sinh2 (k sinh2 (k2 L) 4ε(1 − ε) L) 4ε(1 − ε) = −1 = 1+ 1+ a +b2 b2 sinh2 (k2 L) (ek2 L − e−k2 L )2 16ε(1 − ε) −1 This proves eqn 12.28a where V1 = V3 = In the high wide barrier limit k2 L This implies both that e−k2 L is negligibly small k2 L and that is negligibly small compared to e2 k2 L /{16ε(1 − ε)} The previous compared to e equation simplifies to T = 16 ε(1 − ε)e−2 k2 L (c) [eqn12.28b] E = 10 kJ/mol, V1 = V3 = 0, L = 50 pm 0.25 0.2 0.15 T 0.1 0.05 1.2 1.4 1.6 1/ (i.e., V2/E) P12.12 d2 ψ + 21 kx ψ = Eψ 2m dx 2 dψ and we write ψ = e−gx , so = 2gxegx dx The Schrăodinger equation is 2 d2 ψ = −2ge−gx + 4g x e−gx = −2gψ + 4g x ψ dx 1.8 Figure 12.2(b) INSTRUCTOR’S MANUAL 196 2g 2 g2 m ψ− m 2g x ψ + 21 kx ψ = Eψ 2k −E ψ + m − 2 g2 m x2ψ = This equation is satisfied if 2g E= and m 2 g = 21 mk, or mk 1/2 g = 21 Therefore, k 1/2 = 21 ω m E = 21 P12.14 x n = αn y n = αn x3 ∝ x +∞ −∞ = α5 +∞ −∞ if ω = k 1/2 m ψy n ψ dx = α n+1 +∞ −∞ ψ y n dy [x = αy] ψ y dy = by symmetry [y is an odd function of y] +∞ −∞ ψy ψ dy y ψ = y N Hv e−y /2 y Hv = y 21 Hv+1 + vHv−1 = y 21 21 Hv+2 + (v + 1)Hv + v 21 Hv + (v − 1)Hv−2 = y 41 Hv+2 + v + 21 Hv + v(v − 1)Hv−2 = y 41 21 Hv+3 + (v + 2)Hv+1 + v + 21 × 21 Hv+1 + vHv−1 +v(v − 1) × 21 Hv−1 + (v − 2)Hv−3 = y 18 Hv+3 + 43 (v + 1)Hv+1 + 23 v Hv−1 + v(v − 1) × (v − 2)Hv−3 Only yHv+1 and yHv−1 lead to Hv and contribute to the expectation value (since Hv is orthogonal to all except Hv ) [Table 12.1]; hence y Hv = 43 y{(v + 1)Hv+1 + 2v Hv−1 } + · · · = 43 (v + 1) 21 Hv+2 + (v + 1)Hv + 2v 21 Hv + (v − 1)Hv−2 + ··· = 43 {(v + 1)2 Hv + v Hv } + · · · = 43 (2v + 2v + 1)Hv + · · · Therefore +∞ −∞ ψy ψ dy = 43 (2v + 2v + 1)N +∞ −∞ Hv2 e−y dy = (2v + 2v + 1) 4α and so x = (α ) × 4α × (2v + 2v + 1) = 43 (2v + 2v + 1)α QUANTUM THEORY: TECHNIQUES AND APPLICATIONS P12.17 197 e2 · [13.5 with Z = 1] = αx b 4πε0 r Since T = b V [12.45, T ≡ EK ] V =− with b = −1 [x → r] T =− V Therefore, T = − 21 V P12.18 In each case, if the function is an eigenfunction of the operator, the eigenvalue is also the expectation value; if it is not an eigenfunction we form ψ ∗ ˆ ψ dτ [11.39] = (a) (b) (c) (d) d iφ e = eiφ ; hence Jz = + i dφ d −2iφ e lˆz e−2iφ = = −2 e−2iφ ; hence Jz = −2 i dφ lˆz eiφ = 2π 2π d cos φ dφ ∝ − cos φ sin φ dφ = i dφ i 0 2π d × (cos χ eiφ + sin χ e−iφ ) dφ lz = N (cos χ eiφ + sin χ e−iφ )∗ i dφ lz ∝ = i cos φ N2 = N2 2π 2π (cos χe−iφ + sin χ eiφ ) × (i cos χ eiφ − i sin χ e−iφ ) dφ (cos2 χ − sin2 χ + cos χ sin χ [e2iφ − e−2iφ ]) dφ = N (cos2 χ − sin2 χ ) × (2π ) = 2π N cos 2χ N2 2π (cos χ eiφ + sin χ e−iφ )∗ (cos χ eiφ + sin χ e−iφ ) dφ = N2 2π (cos2 χ + sin2 χ + cos χ sin χ [e2iφ + e−2iφ ]) dφ = 2πN (cos2 χ + sin2 χ ) = 2π N = if N = 2π Therefore lz = cos 2χ [χ is a parameter] d2 Jˆ2 For the kinetic energy we use Tˆ ≡ Eˆ K = z [12.47] = − [12.52] 2I dφ 2I (a) Tˆ eiφ = − (b) Tˆ e−2iφ = − (c) Tˆ cos φ = − 2I (i2 eiφ ) = 2I 2I 2I eiφ ; (2i)2 e−2iφ = (−cos φ) = hence −2iφ e ; 2I 2I cos φ; T = hence hence 2I 2 I T = T = 2I INSTRUCTOR’S MANUAL 198 Tˆ (cos χ eiφ + sin χ e−iφ ) = − (d) 2I (−cos χ eiφ − sin χ e−iφ ) = 2I (cos χ eiφ + sin χ e−iφ ) and hence T = 2I Comment All of these functions are eigenfunctions of the kinetic energy operator, which is also the total energy or Hamiltonian operator, since the potential energy is zero for this system π 2π P12.20 0 ∗ Y3,3 Y3,3 sin θ dθ dφ = π 64 × 35 π = 64 × 2π 35 dφ [Table 12.3] sin6 θ sin θ dθ π × (2π ) −1 (1 − cos2 θ)3 d cos θ [sin θ dθ = d cos θ, sin2 θ = − cos2 θ] 35 = 32 −1 (1 − 3x + 3x − x ) dx [x = cos θ] 35 32 35 = 32 x − x + 35 x − 17 x × = = −1 32 35 P12.22 ∇2 = ∂2 ∂2 ∂2 + + ∂x ∂y ∂z2 ∂2 f = −a f ∂x ∂2 f = −b2 f ∂y ∂2 f = −c2 f ∂y and f is an eigenfunction with eigenvalue −(a + b2 + c2 ) P12.25 (a) Suppose that a particle moves classically at the constant speed v It starts at x = at t = and L at t = τ is at position x = L v = and x = vt τ τ x dt = τ t=0 v τ = t dt = τ t=0 x = = τ vt dt τ t=0 v 2τ t 2τ t=0 L vτ vτ = = x = 2 2τ x2 = τ v2 τ x dt = t dt τ t=0 τ t=0 v2 = t 3τ τ = t=0 L x 1/2 = 1/2 (vτ )2 L2 = 3 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS nπ x 1/2 sin L L ψn = (b) L 199 for ≤ x ≤ L [12.7] nπ x L dx x sin2 L L x=0 x=L  x x x sin 2nπ cos 2nπ L L  x2  = − − L 4(nπ/L) 8(nπ/L)2 x n= ψn∗ xψn dx = x=0 L L2 = = = x n L This agrees with the classical result L x2 n = x=0 ψn∗ x ψn dx =  = x3 2 − L x2 4(nπ/L) L = L2 − 4(nπ/L)2 = − 8(nπ/L)3 sin dx 2nπ x L − x x cos 2nπ L 8(nπ/L)2 x=L  x=0 L L3 − 8(nπ/L)2 = 1/2 x2 n L 2 nπ x x sin L x=0 L L2 − 4(nπ/L)2 1/2 L = 1/2 This agrees with the classical result 1/2 lim x n n→∞ P12.27 (a) The energy levels are given by: h2 n2 , 8mL2 and we are looking for the energy difference between n = and n = 7: En = h2 (72 − 62 ) 8mL2 Since there are 12 atoms on the conjugated backbone, the length of the box is 11 times the bond length: E= L = 11(140 × 10−12 m) = 1.54 × 10−9 m, (6.626 × 10−34 J s)2 (49 − 36) = 3.30 × 10−19 J 8(9.11 × 10−31 kg)(1.54 × 10−9 m)2 (b) The relationship between energy and frequency is: so E= E = hν so ν= 3.30 × 10−19 J E = 4.95 × 10−14 s−1 = h 6.626 × 10−34 J s INSTRUCTOR’S MANUAL 200 (c) The frequency computed in this problem is about twice that computed in problem 12.26b, suggesting that the absorption spectrum of a linear polyene shifts to lower frequency as the number of conjugated atoms increases The reason for this is apparent if we look at the terms in the energy expression (which is proportional to the frequency) that change with the number of conjugated atoms, N The energy and frequency are inversely proportional to L2 and directly proportional to (n + 1)2 − n2 = 2n + 1, where n is the quantum number of the highest occupied state Since n is proportional to N (equal to N/2) and L is approximately proportional to N (strictly to N − 1), the energy and frequency are approximately proportional to N −1 P12.29 In effect, we are looking for the vibrational frequency of an O atom bound, with a force constant equal to that of free CO, to an infinitely massive and immobile protein complex The angular frequency is ω= k 1/2 , m where m is the mass of the O atom m = (16.0 u)(1.66 × 10−27 kg u−1 ) = 2.66 × 10−26 kg, and k is the same force constant as in problem 12.2, namely 1902 N m−1 : ω= 1902 N m−1 2.66 × 10−26 kg 1/2 = 2.68 × 1014 s−1 ... 3, 2, 1, 0, −1, −2, −3, −4, −5, and −6 Figure 12.1(a) Figure 12.1(b) INSTRUCTOR S MANUAL 192 Solutions to problems Solutions to numerical problems P12.4 E= l(l + 1) l(l + 1) [12.65] = 2I 2meff... m−1 ) (16.00) × (1.6605 × 10−27 kg) 1/2 INSTRUCTOR S MANUAL 190 E12.15(b) The first excited-state wavefunction has the form ψ = 2N1 y exp − 21 y mω 1/2 To see if it satisfies Schrăodingers equation,.. .INSTRUCTOR S MANUAL 186 Numerical exercises E= E12.4(b) n2 h 8me L2 h2 (6.626 × 10−34 J s)2 = = 2.678 × 10−20 J 8me L2 8(9.109 × 10−31 kg) × (1.50 × 10−9 m)2 Conversion factors E kJ