41 Quantum Mechanics CHAPTER OUTLINE 41.1 The Wave Function 41.2 Analysis Model: Quantum Particle Under Boundary Conditions 41.3 The Schrödinger Equation 41.4 A Particle in a Well of Finite Height 41.5 Tunneling Through a Potential Energy Barrier 41.6 Applications of Tunneling 41.7 The Simple Harmonic Oscillator * An asterisk indicates a question or problem new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ41.1 Answer (b) Fewer particles are reflected as the height of the potential barrier decreases and approaches the energy of the particles By Equations 41.22 and 41.23, the transmission coefficient T ≈ e −2CL , where C = 2m (U − E ) , increases as U − E decreases, so the reflection coefficient R = − T ≈ − e −2CL decreases as U − E decreases OQ41.2 The ranking is answer (b) > (a) > (c) > (e) > (d) From Equation 41.14, consider the quantity ⎛ h2 ⎞ E=⎜ n : ⎝ 8mL2 ⎟⎠ (a) ⎡ ⎤ ⎛ h2 ⎞ h2 nm −1 ⎟ ⎢ ⎥ ( 1) = ⎜ ⎝ 8m1 ⎠ ⎣ 8m1 ( nm ) ⎦ 948 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 OQ41.3 (b) ⎡ ⎤ ⎛ h2 ⎞ h2 nm −1 ⎟ ⎢ ⎥(2) = ⎜ ⎝ 8m1 ⎠ ⎣ 8m1 ( nm ) ⎦ (c) ⎡ ⎤ ⎛ h2 ⎞ h2 nm −1 ⎟ ⎢ ⎥ ( 1) = ⎜ 18 ⎝ 8m1 ⎠ ⎢⎣ ( 2m1 ) ( nm ) ⎥⎦ (d) ⎡ ⎤ ( )2 ⎢ ⎥ ( 1) = ⎣ 8m1 ( nm ) ⎦ (e) ⎡ ⎤ ⎛ h2 ⎞ h2 nm −1 ⎟ ⎢ ⎥ ( 1) = ⎜ 36 ⎝ 8m1 ⎠ ⎣ 8m1 ( nm ) ⎦ (a) True Examples: An electron has mass and charge, but it can also display interference effects (b) False An electron has rest energy ER = mec2 (c) True A moving electron possesses kinetic energy 949 (d) True p = meu OQ41.4 (e) True (a) True Examples: A photon behaves as a particle in the photoelectric effect and as a wave in double-slit interference (b) True A photon cannot have rest energy (mass) because it is never at rest: it travels at the speed of light (c) True E = hf (d) True p = E/c (e) True OQ41.5 Answer (d) The probability of finding the particle is at the antinodes (places of greatest amplitude) of the standing wave OQ41.6 Compare the ground state wave functions in Figures 41.4 and 41.7 in the text In the square well with infinitely high walls, the particle’s simplest wave function has strict nodes separated by the length L of the well The particle’s wavelength is 2L, its momentum h/2L, and its energy p2/2m = h2/8mL2 In the well with walls of only finite height, the wave function has nonzero amplitude at the walls, and it extends outside the walls (i) Answer (a) The ground state wave function extends somewhat outside the walls of the finite well, so the particle’s wavelength is longer © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 950 Quantum Mechanics (ii) Answer (b) The particle’s momentum in its ground state is smaller because p = h/λ and the wave function has a larger wavelength (iii) Answer (b) The particle has less energy because is has smaller momentum OQ41.7 Answer (e) From the relation between the square of the wave function and the probability P of finding the particle in the interval Δx = (7 nm − nm) = nm, we have ψ Δx = P OQ41.8 → ψ= P 0.48 = = 0.40 nm −1 Δx nm Answer (a) Because of the exponential tailing of the wave function within the barrier, the tunneling current is more sensitive to the width of the barrier than to its height Notice that the exponent term CL in the transmission coefficient T ≈ e −2CL , where C = 2m (U − E ) , decreases more if L decreases than if U decreases by the same percentage OQ41.9 Answer (c) Other points see a wider potential-energy barrier and carry much less tunneling current OQ41.10 Answer (d) The probability of finding the particle is greatest at the place of greatest amplitude of the wave function The next most likely place is point b, after that, points a and e appear to be equally probable The particle would never be found at point c ANSWERS TO CONCEPTUAL QUESTIONS CQ41.1 Consider the Heisenberg uncertainty principle It implies that electrons initially moving at the same speed and accelerated by an electric field through the same distance need not all have the same measured speed after being accelerated Perhaps the philosopher could have said “it is necessary for the very existence of science that the same conditions always produce the same results within the uncertainty of the measurements.” CQ41.2 Consider a particle bound to a restricted region of space If its minimum energy were zero, then the particle could have zero momentum and zero uncertainty in its momentum At the same time, the uncertainty in its position would not be infinite, but equal to the width of the region In such a case, the uncertainty product ΔxΔpx would be zero, violating the uncertainty principle This contradiction proves that the minimum energy of the particle is not zero © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 951 CQ41.3 The motion of the quantum particle does not consist of moving through successive points The particle has no definite position It can sometimes be found on one side of a node and sometimes on the other side, but never at the node itself There is no contradiction here, for the quantum particle is moving as a wave It is not a classical particle In particular, the particle does not speed up to infinite speed to cross the node CQ41.4 (a) ψ (x) becomes infinite as x → ∞ (b) ψ (x) is discontinuous and becomes infinite at x = π/2, 3π/2,… CQ41.5 A particle’s wave function represents its state, containing all the information there is about its location and motion The squared absolute value of its wave function tells where we would classically think of the particle as spending most its time Ψ is the probability distribution function for the position of the particle CQ41.6 In quantum mechanics, particles are treated as wave functions, not classical particles In classical mechanics, the kinetic energy is never negative That implies that E ≥ U Treating the particle as a wave, the Schrödinger equation predicts that there is a nonzero probability that a particle can tunnel through a barrier—a region in which E < U CQ41.7 Both (d) and (e) are not physically significant Wave function (d) is not acceptable because ψ is not single-valued Wave function (e) is not acceptable because ψ is discontinuous (as is its slope) CQ41.8 Newton’s 1st and 2nd laws are used to determine the motion of a particle of large mass The Schrödinger equation is not used to determine the motion of a particle of small mass; rather, it is used to determine the state of the wave function of a particle of small mass In particular, the states of atomic electrons are confined-wave states whose wave functions are solutions to the Schrödinger equation Anything that we can know about a particle comes from its wave function © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 952 Quantum Mechanics SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 41.1 P41.1 (a) The Wave Function The wave function, ψ ( x ) = Ae ( i 5×1010 x ) = A cos × 1010 x + iA sin × 1010 x ( ) ( ) will go through one full cycle between x1 = and (5.00 × 1010)x2 = π The wavelength is then λ = x − x1 = 2π = 1.26 × 10 –10 m 10 −1 5.00 × 10 m To say the same thing, we can inspect Ae wave number is k = 5.00 × 1010 m–1 = 2π/λ (b) ) to see that the Since λ = h/p, the momentum is p= (c) ( i × 1010 x h 6.626 × 10 –34 J ⋅ s = = 5.27 × 10 –24 kg ⋅ m/s λ 1.26 × 10 –10 m The electron’s kinetic energy is K= p2 mu2 = 2m ( 5.27 × 10 kg ⋅ m/s ) = ( 9.11 × 10 kg ) –24 –31 ⎛ ⎞ 1 eV ⎜⎝ 1.602 × 10 –19 J ⎟⎠ = 95.3 eV [We use u to represent the speed of a particle with mass in chapters 39, 40, and 41.] P41.2 (a) See ANS FIG P41.2 for a graph of −3 < x < a f (x) = e − x /a for the range A ANS FIG P41.2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 (b) 953 Normalization requires ∫ ψ dx = 1: all space ∞ ∞ −∞ −2 x /a dx = ∫ A e −2 x /a dx = ∫Ae − aA e −2 x /a ∞ = aA = → A = a (c) a −2 x /a a e −2 x /a e P= ∫ dx = ∫ dx = −e −2 x/a = −e −2 + = 0.865 a a −a (a) Normalization requires a P41.3 ∫ ψ dx = 1: all space 1.00 2 ∫ A x dx = A2 x 3 0.400 (b) 1.00 = A 23 =1 0.400 → A= P = ∫ 3x dx = x 0.300 = ( 0.400 ) − ( 0.300 ) = 0.037 2 0.300 (c) The expectation value is 1.00 x = ∫ ψ * xψ dx = ∫0 all space P41.4 3x 3x dx = 1.00 = 0.750 The probability is given by P= a ∫ ψ ( x) −a P= a a ⎛ a ⎞ ⎛ 1⎞ ⎛ x⎞ = ∫ dx = ⎜ ⎟ ⎜ ⎟ tan −1 ⎜ ⎟ 2 ⎝ ⎠ ⎝ ⎠ ⎝ a⎠ π a −a π (x + a ) a −a 1 ⎡π ⎛ π ⎞ ⎤ ⎡⎣ tan −1 − tan −1 ( −1) ⎤⎦ = ⎢ − ⎜ − ⎟ ⎥ = π π ⎣ ⎝ 4⎠⎦ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 954 Quantum Mechanics Section 41.2 P41.5 (a) Analysis Model: Quantum Particle Under Boundary Conditions The energy of a quantum particle confined to a line segment is En = h n2 8mL2 Here we have for the ground state (6.626 × 10 J · s ) (1) = ( 1.67 × 10 kg ) ( 2.00 × 10 –34 E1 –27 –14 m) = 8.22 × 10 –14 J = 0.513 MeV and for the first and second excited states, which are states and 3, E2 = 4E1 = 2.05 MeV (b) and E3 = 9E1 = 4.62 MeV They do; the MeV is the natural unit for energy radiated by an atomic nucleus Stated differently: Scattering experiments show that an atomic nucleus is a three-dimensional object always less than 15 fm in diameter This one-dimensional box 20 fm long is a good model in energy terms P41.6 From Equation 41.14, the allowed energy levels of a particle in a box is ⎛ h2 ⎞ En = ⎜ n , ⎝ 8mL2 ⎟⎠ (a) n = 1, 2, 3, For L = 1.00 nm, ⎛ h2 ⎞ En = ⎜ n ⎝ 8mL2 ⎟⎠ ⎤ 6.626 × 10−34 J ⋅ s ) ( ⎛ eV ⎞⎡ ⎢ ⎥ =⎜ n −19 ⎟ −31 −9 ⎝ 1.60 × 10 J ⎠ ⎢ ( 9.11 × 10 kg ) ( 1.00 × 10 m ) ⎥ ⎣ ⎦ = 0.377n2 = eV n≈4 P41.7 (b) For n = 4, En = 0.377 ( ) = 6.03 eV (a) From Equation 41.14, the allowed energy levels of an electron in a box is ⎛ h2 ⎞ En = ⎜ n ⎝ 8me L2 ⎟⎠ n = 1, 2, 3, © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 955 Substituting numerical values, ⎡ ⎤ 6.626 × 10−34 J ⋅ s ) ( ⎥ En = ⎢ n −31 −9 ⎢ ( 9.11 × 10 kg ) ( 0.100 × 10 m ) ⎥ ⎣ ⎦ = ( 6.02 × 10−18 J ) n2 = ( 37.7 eV ) n2 ANS FIG P41.7 (b) When the electron falls from higher level ni to lower level nf , it emits energy ⎛ h2 ⎞ ΔEn = ⎜ ni − n2f = ( 37.7 eV ) ni2 − n2f ⎝ 8me L2 ⎟⎠ ( ) ( ) by emitting a photon of wavelength hc 8me cL2 λ= = ΔEn h ni2 − n2f ( = ) ( 9.109 × 10−31 kg ) ( 2.998 × 108 m/s ) ( 0.100 × 10−9 m ) (6.626 × 10 −34 ( J ⋅ s ) ni2 − n2f ) ⎛ nm ⎞ × ⎜ −9 ⎟ ⎝ 10 m ⎠ = 33.0 nm (n i − n2f ) For example, for the transition → 3, the wavelength is λ= 33.0 nm = 4.71 nm ( )2 − ( )2 The wavelengths produced by all possible transitions are: Transition → → → → → → λ (nm) 4.71 2.75 2.20 6.59 4.12 11.0 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 956 P41.8 Quantum Mechanics The energy of the photon is E= hc 1 240 eV ⋅ nm ⎛ 1 mm ⎞ −4 = ⎜⎝ ⎟⎠ = 2.05 × 10 eV λ 6.06 mm 10 nm The allowed energies of the proton in the box are ⎛ h2 ⎞ En = ⎜ n ⎝ 8mL2 ⎟⎠ ⎡ ⎤⎛ 6.626 × 10−34 J ⋅ s ) ( 1 eV ⎞ ⎥ =⎢ ⎜ −19 ⎟n −27 −9 ⎢ ( 1.673 × 10 kg ) ( 1.00 × 10 m ) ⎥ ⎝ 1.602 × 10 J ⎠ ⎣ ⎦ = ( 2.05 × 10−4 eV ) n2 The smallest possible energy for a transition between states is from n = to n = 2, which has energy ΔEn = ( 2.05 × 10−4 eV ) ( 2 − 12 ) = 6.14 × 10−4 eV The photon does not have enough energy to cause this transition The photon energy would be sufficient to cause a transition from n = to n = 1, but the n = state does not exist for the particle in a box P41.9 From Equation 41.14, ΔE = hc ⎛ h2 ⎞ 3h2 ⎡ ⎤ =⎜ − = ⎦ 8m L2 λ ⎝ 8me L2 ⎟⎠ ⎣ e Solving for the length of the box then gives L= = 3hλ 8me c ( 6.626 × 10−34 J ⋅ s ) ( 694.3 × 10−9 m ) ( 9.11 × 10−31 kg ) ( 3.00 × 108 m/s ) = 7.95 × 10−10 m = 0.795 nm P41.10 From Equation 41.14, ΔE = hc ⎛ h2 ⎞ 3h2 ⎡ ⎤ =⎜ − = ⎦ 8m L2 λ ⎝ 8me L2 ⎟⎠ ⎣ e Solving for the length of the box then gives L= 3hλ 8me c © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 P41.11 957 From Equation 41.14, the allowed energy levels of a particle in a box is ⎛ h2 ⎞ En = ⎜ n = n2E1 ⎝ 8mL2 ⎟⎠ n = 1, 2, 3, … For a proton (m = 1.673 × 10–27 kg) in a 10.0-fm wide box: (6.626 × 10 J ⋅ s ) = ( 1.673 × 10 kg ) ( 10.0 × 10 −34 E1 −27 −15 m) ⎛ eV ⎞ = 3.28 × 10−13 J ⎜ = 2.05 × 106 eV = 2.05 MeV −19 ⎟ ⎝ 1.602 × 10 J ⎠ (a) The energy of the emitted photon is E = ΔEn = E2 − E1 = ( ) E1 − E1 = 3E1 = 6.14 MeV (b) The wavelength of the photon is hc 1 240 eV ⋅ nm = E 6.14 × 106 eV = 2.02 × 10−4 nm = 2.02 × 10−13 m = 202 × 10−15 m = 202 fm λ= (c) P41.12 This is a gamma ray, according to the electromagnetic spectrum chart in Chapter 34 The ground state energy of a particle (mass m) in a 1-dimensional box h2 of width L is E1 = 8mL2 (a) For a proton (m = 1.67 × 10–27 kg) in a 0.200-nm wide box: (6.626 × 10 J ⋅ s ) = ( 1.67 × 10 kg ) ( 2.00 × 10 −34 E1 −27 −10 m) = 8.22 × 10−22 J = 5.13 × 10−3 eV (b) For an electron (m = 9.11 × 10–31 kg) in the same size box: (6.626 × 10 J ⋅ s ) = ( 9.11 × 10 kg ) ( 2.00 × 10 −34 E1 −31 −10 m) = 1.51 × 10−18 J = 9.41 eV (c) The electron has a much higher energy because it is much less massive © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 981 Comparing this to K1, we see that this value is too large by 28.6% P41.54 Looking at Figure 41.7, we see that wavelengths for a particle in a finite well are longer than those for a particle in an infinite well Therefore, the energies of the allowed states should be lower for a finite well than for an infinite well As a result, the photons from the source have too much energy to be absorbed or, equivalently, the photons have a frequency that is too high In order to lower their apparent frequency using the Doppler shift, the source would have to move away from the particle in the finite square well, not toward it P41.55 (a) For a particle with wave function ⎧ −x a ⎪ e ψ ( x) = ⎨ a ⎪0 ⎩ for x > for x < The probability densities are ψ ( x) = and ψ ( x ) = for x < −2 x a e a for x > ANS FIG P41.55 shows a sketch of the probability density for this particle ANS FIG P41.55 (b) The probability is obtained from 0 Prob ( x < ) = ∫ ψ ( x ) dx = ∫ ( ) dx = −∞ (c) −∞ For the wave function to be normalized, we require ∞ ∞ ∫ ψ ( x ) dx = ∫ ψ dx + ∫ ψ dx = −∞ −∞ 2 Performing the integration gives ∞ ⎛ ⎞ −2 x a −2 x a ∫−∞ 0dx + ∫0 ⎜⎝ a ⎟⎠ e dx = − e ∞ = − ( e −∞ − 1) = © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 982 Quantum Mechanics (d) The probability is obtained from a a a ⎛ 2⎞ Prob( < x < a ) = ∫ ψ dx = ∫ ⎜ ⎟ e −2 x a dx = −e −2 x a ⎝ a⎠ 0 = − e −2 = 0.865 P41.56 (a) Taking Lx = Ly = L, we see that the expression for E becomes E= h2 nx2 + ny2 8me L ( ) The general form of the wave function is ⎛ nyπ y ⎞ ⎛ n π x⎞ ψ ∼ sin ⎜ x ⎟ sin ⎜ ⎝ L ⎠ ⎝ L ⎟⎠ For a normalizable wave function, neither nx nor ny can be zero, otherwise ψ = (b) The ground state corresponds to nx = ny = (c) The energy of the ground state is E1, = h2 h2 2 + = ( ) 4m L2 8me L2 e (d) For the first excited state, nx = and ny = 2, or nx = and ny = (e) For the second excited state, nx = and ny = (f) The second excited state, corresponding to nx = 2, ny = 2, has an energy of E2, = (g) h2 h2 2 + = ( ) 8me L2 me L2 The energy difference between the ground state and the second excited state is ΔE = E2, − E1, = (h) ΔE = 3h2 hc = 4me L λ h2 h2 3h2 − = me L2 4me L2 4me L2 → λ= 4me cL2 3h © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 P41.57 (a) 983 The expectation value is ∞ x ⎛ a⎞ = ∫ x⎜ ⎟ ⎝π⎠ −∞ 12 e − ax dx = since the integrand is an odd function of x (b) The expectation value is ∞ ⎛ 4a ⎞ x = ∫ x⎜ ⎝ π ⎟⎠ −∞ 12 x e − ax dx = since the integrand is an odd function of x (c) The expectation value is ∞ x 01 = ∫ x −∞ ∞ 1 ψ + ψ dx = x + x + ( 1) ∫ xψ ( x )ψ ( x ) dx 2 −∞ The first two terms are zero, from (a) and (b) Thus, ∞ ⎛ a⎞ x 01 = ∫ x ⎜ ⎟ ⎝π⎠ −∞ 14 ⎛ 2a ⎞ = 2⎜ ⎝ π ⎟⎠ 12 e − ax 2 ⎛ 4a ⎞ ⎜⎝ π ⎟⎠ 1⎛ π ⎞ ⎜ ⎟ ⎝ a3 ⎠ 14 xe − ax 2 ⎛ 2a ⎞ dx = ⎜ ⎝ π ⎟⎠ 12 ∞ 2 − ax ∫ x e dx 12 2a = Where we have used Table B.6 in the Appendix to evaluate the integral P41.58 With one slit open, P1 = ψ or P2 = ψ 2 With both slits open, P = ψ1 +ψ2 At a maximum, the wave functions are in phase Pmax = ( ψ + ψ ) At a minimum, the wave functions are out of phase, Pmin = ( ψ − ψ ) © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 984 Quantum Mechanics Now, P1 ψ = P2 ψ 2 = 25.0 , so ψ1 = 5.00 ψ2 and Pmax ( ψ + ψ = Pmin ( ψ − ψ ) = ( 5.00 ψ ) ( 5.00 ψ 2 2 + ψ2 − ψ2 ) ) 2 6.00 ) ( = ( 4.00)2 = 36.0 = 2.25 16.0 Challenge Problems P41.59 (a) The claim is that Schrödinger’s equation ∂ 2ψ 2m = − ( E − U )ψ ∂x has the solutions ψ = Ae ik1 x + Be − ik1 x [region I] ψ = Ce ik2 x [region II] ANS FIG P41.59(a) Check that the solution for region I satisfies Schrödinger’s equation: ∂ 2ψ 2m = − Eψ ∂x 2 ∂ ∂ 2m Ae ik1 x ) + ( Be − ik1 x ) = − E ( Ae ik1 x + Be − ik1 x ) ( ∂x ∂x 2m −k 12 ( Ae ik1 x ) − k12 ( Be − ik1 x ) = − E ( Ae ik1 x + Be − ik1 x ) 2m −k 12 ( Ae ik1 x + Be − ik1 x ) = − E ( Ae ik1 x + Be − ik1 x ) The last line is true if k 12 = p ( k1 ) E= = 2m 2m 2m E, which it is because 2 → k1 = 2mE Therefore, the equation is satisfied in region I © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 985 Check that the solution for region II satisfies Schrödinger’s equation: ∂ 2ψ 2m = − ( E − U )ψ 2 ∂x ∂ 2m Ce ik2 x ) = − ( E − U ) ( Ce ik2 x ) ( ∂x 2m −k 22 ( Ce ik2 x ) = − ( E − U ) ( Ce ik2 x ) 2m The last line is true if k 22 = ( E − U ) , which it is because ( k2 ) p2 E= +U = 2m 2m 2m ( E − U ) → k2 = Therefore, the equation is satisfied in region II We apply boundary conditions Matching functions and derivatives at x = 0, we find that (ψ )0 = (ψ )0 gives A + B = C, ⎛ dψ ⎞ ⎛ dψ ⎞ and ⎜ ⎟ = ⎜ ⎟ ⎝ dx ⎠ ⎝ dx ⎠ gives k1 ( A − B) = k2C Then B= − k k1 A + k k1 and C= A + k k1 Incident wave Aeikx reflects Be–ikx, with probability ( k1 − k ) B ( − k k1 ) R= = = A ( + k k ) ( k + k )2 (b) With E = 7.00 eV and U = 5.00 eV: k2 E −U = = k1 E 2.00 eV = 0.535 7.00 eV The reflection probability is P41.60 2 − 0.535 ) ( R= (1 + 0.535)2 = 0.092 0 T = − R = 0.908 (c) The probability of transmission is (a) The potential energy of the system is given by U= e ⎡⎛ 1⎞ ⎛ 1⎞ ⎤ ( − 3) e −1 + − + −1 + + −1 = ( ) ⎜ ⎟ ⎜ ⎟ ⎥⎦ 4π ∈ d 4π ∈0 d ⎢⎣⎝ 3⎠ ⎝ 2⎠ = − 7ke e 3d © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 986 Quantum Mechanics (b) There are two electrons, each with minimum energy E1 From Equation 41.14, the total energy is 2h2 h2 K = 2E1 = = 36me d 8me ( 3d ) (c) The total energy of the system is h2 7ke e E = K +U = − 36me d 3d For a minimum, we require dE = Differentiating, d (d) dE =0 d (d) d ⎛ h2 7ke e ⎞ − =0 d ( d ) ⎜⎝ 36me d 3d ⎟⎠ h2 7ke e − ( −1) =0 ( −2 ) 36me d 3d h2 7ke e = 18me d 3d 3h2 h2 d= = ( 18me ) ke e 42me ke e Substituting numerical values, (6.626 × 10 J ⋅ s ) kg ) ( 8.99 × 10 N ⋅ m /C ) ( 1.60 × 10 −34 d= ( 42 ) ( 9.11 × 10−31 2 −19 C) = 4.99 × 10−11 m = 49.9 pm (d) The lithium spacing is d and the number of atoms N in volume V Nm is related by Nd3 = V, and the density is , where m is the mass V of one atom We have: density = Nm Nm m = = V Nd d © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 987 From which we obtain mol ⎡ 6.94 g ⎛ ⎞⎤ 13 ⎜ ⎟ 23 ⎢ ⎛ m ⎞ mol ⎝ 6.022 × 10 atoms ⎠ ⎥ d=⎜ = ⎢ ⎥ g ⎝ density ⎟⎠ ⎢ ⎥ 0.530 cm ⎣⎢ ⎦⎥ 13 = 2.79 × 10−8 cm = 2.79 × 10−10 mm = 279 pm The lithium interatomic spacing of 280 pm is 5.59 times larger Therefore, it is of the same order of magnitude as the interatomic spacing 2d here P41.61 The wave functions and probability densities are the same as those shown in Active Figure 41.4 of the textbook From Equation 41.13, the wave functions are ψn = (a) ⎛ nπ x ⎞ where n = 1, 2, 3… sin ⎜ ⎝ L ⎟⎠ L For n = 1, P1 = 0.350 nm ∫ 0.150 nm ψ1 0.350 ⎛ πx ⎞ ⎛ ⎞ dx = ⎜ sin ⎜ dx ⎟ ∫ ⎝ 1.00 nm ⎠ 0.150 ⎝ 1.00 nm ⎟⎠ 0.350 nm ⎡ x 1.00 nm ⎛ 2π x ⎞ ⎤ = ( 2.00 nm ) ⎢ − sin ⎜ ⎝ 1.00 nm ⎟⎠ ⎥⎦ 0.150 nm 4π ⎣2 In the above result we used x ⎛ ⎞ ⎛ ⎞ ∫ sin ( ax ) dx = ⎜⎝ ⎟⎠ − ⎜⎝ 4a ⎟⎠ sin ( 2ax ) Therefore, 0.350 nm 1.00 nm ⎡ ⎛ 2π x ⎞ ⎤ P1 = ( 1.00 nm ) ⎢ x − sin ⎜ ⎝ 1.00 nm ⎟⎠ ⎥⎦ 0.150 nm 2π ⎣ ⎧ P1 = ( 1.00 nm ) ⎨0.350 nm − 0.150 nm ⎩ − ⎫ 1.00 nm sin ( 0.700π ) − sin ( 0.300π )]⎬ [ 2π ⎭ = 0.200 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 988 Quantum Mechanics 0.350 0.350 (b) P2 = ⎡ x 1.00 ⎛ 2π x ⎞ ⎛ 4π x ⎞ ⎤ sin ⎜ dx = 2.00 ⎢ − sin ⎜ ⎟ ∫ ⎝ 1.00 ⎠ ⎝ 1.00 ⎟⎠ ⎥⎦ 0.150 1.00 0.150 ⎣ 8π 0.350 1.00 ⎡ ⎛ 4π x ⎞ ⎤ P2 = 1.00 ⎢ x − sin ⎜ ⎝ 1.00 ⎟⎠ ⎥⎦ 0.150 4π ⎣ { = 1.00 ( 0.350 − 0.150 ) − } 1.00 [ sin (1.40π ) − sin ( 0.600π )] 4π = 0.351 n2 h Using En = , we find that 8mL2 P41.62 (c) E1 = 0.377 eV (d) E2 = 1.51 eV and (a) and (b) The Wave functions are shown in ANS FIG P41.62(a) and ANS FIG P41.62(b) ANS FIG P41.62(a) (c) ANS FIG P41.62(b) ψ is continuous and ψ → as x → ±∞ The function can be normalized It describes a particle bound near x = (d) Since ψ is symmetric, ∞ ∞ 2 ∫ ψ dx = ∫ ψ dx = −∞ or ∞ 2A ∫ e −2 α x ⎛ 2A ⎞ −∞ dx = ⎜ e − e ) = ( ⎟ ⎝ −2α ⎠ This gives A = α (e) The probability of finding the particle between –1/2 α and +1/2 α is P( −1 2α )→(1 2α ) = ( ) a = (1 − e −1 2α 2α ⎛ ⎞ − 2α 2α −2 α x − 1) ∫ e dx = ⎜⎝ −2α ⎟⎠ ( e x=0 )= 0.632 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 P41.63 (a) 989 Recall from Section 41.7 that the potential energy of a harmonic 1 oscillator is kx = mω x We can find the energy of the 2 oscillator E by substituting the wave function into the Schrödinger equation − d 2ψ + Uψ = Eψ 2m dx − d 2ψ + mω x 2ψ = Eψ 2m dx → − mω 2 ) x From ψ = Bxe ( , we have dψ ⎛ mω ⎞ − mω 2 )x − mω 2 )x = Be ( + Bx ⎜ − 2xe ( ⎟ ⎝ 2 ⎠ dx ⎛ mω ⎞ −( mω 2 )x2 − mω 2 )x = Be ( − B⎜ xe ⎝ ⎟⎠ d 2ψ ⎛ mω ⎞ −( mω 2 )x2 ⎛ mω ⎞ − mω 2 )x = Bx ⎜ − xe − B⎜ 2xe ( ⎟ ⎟ ⎝ ⎠ ⎝ ⎠ dx ⎛ mω ⎞ ⎛ mω ⎞ −( mω 2 )x2 − B⎜ x − xe ⎝ ⎟⎠ ⎜⎝ ⎟⎠ d 2ψ ⎛ mω ⎞ −( mω 2 )x2 ⎛ mω ⎞ −( mω 2 )x2 = −3B ⎜ xe + B⎜ xe ⎟ ⎝ ⎠ ⎝ ⎟⎠ dx Substituting the above into the Schrödinger equation, we have − d 2ψ + mω x 2ψ = Eψ 2m dx 2 − ⎡ ⎛ mω ⎞ −( mω 2 )x2 ⎛ mω ⎞ −( mω 2 )x2 ⎤ + B⎜ xe ⎟ xe ⎢ −3B ⎜⎝ ⎥ ⎝ ⎟⎠ 2m ⎣ ⎠ ⎦ − mω 2 )x ⎤ + mω x ⎡ Bxe ( ⎣ ⎦ − mω 2 )x ⎤ = E ⎡ Bxe ( ⎣ ⎦ ⎛ 3ω ⎞ ⎡ −( mω 2 )x ⎤+ ⎜⎝ ⎟⎠ ⎣ Bxe ⎦ ⎛ −( mω 2 )x 2 2⎞ ⎤ ⎜⎝ − mω x ⎟⎠ ⎡⎣ Bxe ⎦ ⎛1 ⎞ − mω 2 )x ⎤ + ⎜ mω x ⎟ ⎡ Bxe ( ⎦ ⎝2 ⎠⎣ − mω 2 )x ⎤ = E ⎡ Bxe ( ⎣ ⎦ ( ) ( ⎛ 3ω ⎞ −( mω 2 )x − mω 2 )x = E Bxe ( ⎜⎝ ⎟⎠ Bxe ) © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 990 Quantum Mechanics The last line is true if E = 3ω (b) We never find the particle at x = because ψ = there (c) ψ is maximized if dψ ⎛ mω ⎞ − ( mω 2 ) x2 = Be − ( mω 2 ) x − B ⎜ xe =0 ⎝ ⎟⎠ dx ⎛ mω ⎞ 1− ⎜ x =0 ⎝ ⎟⎠ mω which is true at x = ± ∞ ∫ ψ dx = : (d) We require −∞ ∞ = ∫ B2 x e − ( mω ) x dx = 2B2 ∫ x e − ( mω ) x dx 2 −∞ π = 2B ( mω )3 π1 ⎛ ⎞ =B ⎜ ⎟ ⎝ mω ⎠ 32 Then, 2 ⎛ mω ⎞ B= 14 ⎜ ⎟ π ⎝ ⎠ (e) 34 ⎛ 4m3ω ⎞ = ⎜ ⎝ π ⎟⎠ 14 At x = ( mω ) , the potential energy is 12 1 ⎛ 4 ⎞ mω x = mω ⎜ = 2ω ⎝ mω ⎟⎠ 2 3ω , so there is zero classical probability of finding the particle here This is larger than the total energy (f) The actual probability is given by ( P = ψ dx = Bxe − ( mω 2 ) x 2 −( mω )x P = δB x e 2 )δ ⎛ 4m3ω ⎞ =δ⎜ ⎝ π ⎟⎠ 12 ⎛ 4 ⎞ −( mω )x2 ⎜⎝ ⎟e mω ⎠ 12 ⎛ m3 2ω ⎞ ⎛ 4 ⎞ −( mω )4( mω ) ⎛ mω ⎞ =δ 12 ⎜ = 8δ ⎜ e −4 ⎜ ⎟e ⎝ π ⎟⎠ π ⎝ ⎟⎠ ⎝ mω ⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 L P41.64 (a) 991 To find the normalization constant, we note that ∫ ψ dx = , or L A ⎡ ∫ ⎢⎣ sin ⎛ π x⎞ ⎛ π x⎞ ⎛ 2π x ⎞ ⎤ ⎛ 2π x ⎞ ⎜⎝ ⎟⎠ + 16 sin ⎜⎝ ⎟⎠ + sin ⎜⎝ ⎟⎠ sin ⎜⎝ ⎟ dx = L L L L ⎠ ⎥⎦ Noting that L − cos ⎡⎣ (π x L ) ⎤⎦ ⎛ π x⎞ sin dx = dx ∫0 ⎜⎝ L ⎟⎠ ∫0 L ⎡ x L sin ( 2π x L ) ⎤ L =⎢ − ⎥ = ⎣2 π ⎦0 L the integral becomes L ⎡ ⎛ L⎞ ⎛ π x⎞ ⎛ 2π x ⎞ ⎤ ⎛ L⎞ ψ dx = A + 16 + sin ⎜ sin ⎜ dx ⎜⎝ ⎟⎠ ⎟ ⎢⎜⎝ ⎟⎠ ∫ ∫ ⎝ L ⎠ ⎝ L ⎟⎠ ⎥⎦ 0 ⎣ L ⎧⎛ L ⎞ ⎛ L⎞ = A ⎨⎜ ⎟ + 16 ⎜ ⎟ ⎝ 2⎠ ⎩⎝ ⎠ ⎛ π x⎞ ⎡ ⎛ π x⎞ ⎛ π x ⎞ ⎤ ⎫⎪ +8 ∫ sin ⎜ sin cos ⎟ ⎜ ⎟ ⎜⎝ ⎟ dx ⎬ ⎝ L ⎠ ⎢⎣ ⎝ L ⎠ L ⎠ ⎥⎦ ⎪⎭ L L ⎡ 17L ⎛ π x⎞ ⎛ π x⎞ ⎤ = A2 ⎢ + 16 ∫ sin ⎜ cos ⎜ ⎟ ⎟⎠ dx ⎥ ⎝ ⎠ ⎝ L L ⎣ ⎦ ⎡ 17L 16L ⎛ π x⎞ 1= A ⎢ + sin ⎜ ⎝ L ⎟⎠ 3π ⎢⎣ 2 → A= (b) ⎤ ⎛ 17L ⎞ ⎥ = A ⎜⎝ ⎟ ⎠ x=0 ⎥ ⎦ x=L 17L To determine the relationship between A and B, we note that a ∫ ψ dx = Therefore, −a a ⎡ ∫ ⎢⎣ A −a 2 ⎛ π x⎞ ⎛ π x⎞ cos ⎜ + B sin ⎜ ⎝ 2a ⎟⎠ ⎝ a ⎟⎠ ⎛ π x⎞ ⎛ π x⎞ ⎤ + A B cos ⎜ sin ⎜ dx = ⎝ 2a ⎟⎠ ⎝ a ⎟⎠ ⎥⎦ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 992 Quantum Mechanics Noting that a − cos ⎡⎣ (π x 2a ) ⎤⎦ ⎛ π x⎞ sin dx = dx ∫− a ⎜⎝ 2a ⎟⎠ −∫a a ⎡ x 2L sin (π x a ) ⎤ =⎢ + ⎥ =a ⎣2 π ⎦ −a a and a + cos ⎡⎣ (π x 2a ) ⎤⎦ ⎛ π x⎞ cos dx = dx ∫− a ⎜⎝ 2a ⎟⎠ −∫a a ⎡ x 2L sin (π x a ) ⎤ =⎢ + ⎥ =a ⎣2 π ⎦ −a a the integral becomes a A a+ B a+ ⎡ ∫ ⎢⎣ A −a ⎛ π x⎞ ⎛ π x⎞ ⎤ B cos ⎜ sin ⎜ dx = ⎟ ⎝ 2a ⎠ ⎝ a ⎟⎠ ⎥⎦ The third term is: a 2A B ⎛ π x⎞ ⎡ ⎛ π x⎞ ⎛ π x⎞ ⎤ ∫ cos ⎜⎝ 2a ⎟⎠ ⎢⎣ sin ⎜⎝ 2a ⎟⎠ cos ⎜⎝ 2a ⎟⎠ ⎥⎦ dx −a a =4A B ∫ cos −a ⎛ π x⎞ ⎛ π x⎞ ⎜⎝ ⎟⎠ sin ⎜⎝ ⎟ dx 2a 2a ⎠ a = 8a A B ⎛ π x⎞ cos ⎜ =0 ⎝ 2a ⎟⎠ − a 3π so the whole integral is ( a A + B ) = , giving 2 A + B = a © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 993 ANSWERS TO EVEN-NUMBERED PROBLEMS ; (c) 0.865 a P41.2 (a) See ANS FIG P41.2; (b) P41.4 P41.6 (a) n ≈ 4; (b) 6.03 eV P41.8 The photon does not have the smallest possible energy to cause the transition between states n = to n = P41.10 3hλ 8me c P41.12 (a) 5.13 × 10−3 eV; (b) 9.41 eV; (c) The electron has a much higher energy because it is much less massive P41.14 (a) 2.00 × 10−9 J; (b) 1.66 × 10−28 m; (c) No The length of the box would have to be much smaller than the size of a nucleus (~10−14 m) to confine the particle P41.16 (a) P41.18 See P41.18 for full explanation P41.20 (a) x = L/4, L/2, and L/4; (b) We look for sin (3π x/L) taking on its extreme values and –1 so that the squared wave function is as large as it can be The result can also be found by studying Figure 41.4b P41.22 (a) P41.24 See P41.24 for complete solution ; (b) /8mL2 ; (c) This estimate is too low by 4π ≈ 40 times, but 2L it correctly displays the pattern of dependence of the energy on the mass and on the length of the well ⎛ 2π ⎞ − sin ⎜ ; (b) See ANS FIG P41.22(b); (c) The wave function ⎝ L ⎟⎠ L 2π is zero for x < and for x > L The probability at = must be zero because the particle is never found at x < or exactly at x = The probability at = L must be for normalization: the particle is always found somewhere in the range < x < L; (d) 0.585L © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 994 P41.26 Quantum Mechanics (a) n = 1: ψ ( x ) = n = 2: ψ ( x ) = 2 ⎛ π x⎞ ⎛ π x⎞ cos ⎜ ; P1 ( x ) = ψ ( x ) = cos ⎜ , ⎟ ⎝ L ⎠ ⎝ L ⎟⎠ L L 2 ⎛ 2π x ⎞ ⎛ 2π x ⎞ sin ⎜ ; P2 ( x ) = ψ ( x ) = sin ⎜ , ⎟ ⎝ L ⎟⎠ ⎝ L ⎠ L L 2 ⎛ 3π x ⎞ ⎛ 3π x ⎞ cos ⎜ ; P3 ( x ) = ψ ( x ) = cos ⎜ ; ⎟ ⎝ L ⎟⎠ ⎝ L ⎠ L L (b) See ANS FIG P41.26(b) n = 3: ψ ( x ) = 2 ; (b) L2 m 15 ; (c) 0.580 16L P41.28 (a) P41.30 (a) See ANS FIG P41.30(a); (b) 2L P41.32 (a) 1.03 × 10−3 ; (b) 1.91 nm P41.34 1.35 P41.36 (a) See P41.36(a) for full explanation; (b) b = mω and ω ; 2 (c) first excited state m k P41.38 2π c P41.40 (a) See P41.40(a) for full explanation; (b) See P41.40(b) for full k explanation; (c) f = 2π µ P41.42 See P41.42 for full explanation P41.44 (a–b) See P41.44(a) and (b) for full explanations P41.46 See P41.46 for full explanation P41.48 (a) See P41.48(a) for full proof; (b) For to 1, λ = 1.38 µm, infrared; For to 2, λ = 827 nm, infrared; For to 1, λ = 275 nm, ultraviolet; For to 2, λ = 344 nm, near ultraviolet; For to 3, λ = 590 nm, yellow-orange visible P41.50 ∼10−10 P41.52 See P41.52 for full explanation 30 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 41 995 P41.54 Looking at Figure 41.7, we see that wavelengths for a particle in a finite well are longer than those for a particle in an infinite well Therefore, the energies of the allowed states should be lower for a finite well than for an infinite well As a result, the photons from the source have too much energy to be absorbed or, equivalently, the photons have a frequency that is too high In order to lower their apparent frequency using the Doppler shift, the source would have to move away from the particle in the finite square well, not toward it P41.56 (a) E = h2 h2 2 n + n ; (b) n = n = 1; (c) ; (d) nx = and ny = 2, or x y x y 8me L2 4me L2 ( ) nx = and ny = 1; (e) nx = and ny = 2; (f) P41.58 P41.60 P41.62 h2 3h2 4me cL2 ; (g) ; (h) me L2 4me L2 3h 2.25 h2 7ke e (a) − ; (b) ; (c) 49.9 pm; (d) The lithium interatomic 36me d 3d spacing of 280 pm is 5.59 times larger Therefore, it is of the same order of magnitude as the interatomic spacing 2d here (a) See ANS FIG P41.62(a); (b) See ANS FIG P41.62(b); (c) ψ is continuous and ψ → as x → ±∞ The function can be normalized It describes a particle bound near x = 0; (d) A = α ; (e) 0.632 P41.64 (a) A = 2 ; (b) A + B = a 17L © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part