M40 YOUN7066 13 ISM c40 tủ tài liệu training

40 QUANTUM MECHANICS 40.1 IDENTIFY: Using the momentum of the free electron, we can calculate k and ω and use these to express its wave function SET UP: Ψ( x, t ) = Aeikx e − iω t , k = p/ , and ω = k /2m EXECUTE: k = ω= p =− 4.50 × 10−24 kg ⋅ m/s = −4.27 × 1010 m −1 k (1.055 × 10−34 J ⋅ s)(4.27 × 1010 m −1 ) = = 1.05 × 1017 s −1 2m 2(9.108 × 10−31 kg) 10 40.2 1.055 × 10−34 J ⋅ s −1 17 −1 Ψ( x, t ) = Ae− i[4.27 ×10 m ) x e−i[1.05 ×10 s ]t EVALUATE: The wave function depends on position and time IDENTIFY: Using the known wave function for the particle, we want to find where its probability function is a maximum Ψ( x, t ) = A [eikxe− iωt − e2ikxe−4iωt ][e− ikx e+ iωt − e−2ikx e+4iω t ] SET UP: Ψ( x, t ) = A (2 − [e− i ( kx − 3ωt ) + e + i ( kx − 3ωt ) ]) = A (1 − cos(kx − 3ωt )) 2 2 EXECUTE: (a) For t = 0, Ψ( x, t ) = A (1 − cos(kx)) Ψ( x, t ) this happens when kx = (2n + 1)π , n = 0,1,… Ψ( x, t ) (b) t = 2π ω 2 is a maximum when cos( kx) = −1 and is a maximum for x = π 3π k , k , etc and 3ωt = 6π Ψ( x, t ) = A (1 − cos( kx − 6π )) Maximum for kx − 6π = π , 3π , , which gives maxima when x = 7π 9π , k k (c) From the results for parts (a) and (b), vav = ω − ω1 7π /k − π /k 3ω with ω = 4ω , ω1 = ω , = vav = k 2π /ω k2 − k1 3ω k EVALUATE: The expressions in part (c) agree IDENTIFY: Use the wave function from Example 40.1 k2 = 2k and k1 = k gives vav = 40.3 2 Ψ( x, t ) = A {1 + cos[( k2 − k1 ) x − (ω − ω1 )t ]} k2 = 3k1 = 3k ω = SET UP: k2 , so ω = 9ω1 = 9ω 2m Ψ( x, t ) = A {1 + cos(2kx − 8ωt )} 2 EXECUTE: (a) At t = 2π /ω , Ψ( x, t ) = A {1 + cos(2kx − 16π )} Ψ( x, t ) is maximum for cos(2kx − 16π ) = This happens for 2kx − 16π = 0, 2π , Smallest positive x where Ψ( x, t ) is a 8π maximum is x = k © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-1 40-2 Chapter 40 8π /k 4ω ω − ω1 8ω 4ω = vav = = = 2π /ω k k2 − k1 2k k EVALUATE: The two expressions agree IDENTIFY: We have a free particle, described in Example 40.1 ω − ω1 ( k22 − k12 ) (k2 + k1)(k2 − k1) p SET UP and EXECUTE: vav = = = = ( k2 + k1 ) = av k2 − k1 2m k2 − k1 2m k2 − k1 2m m EVALUATE: This is the same as the classical physics result, v = p/m = mv/m = v (b) From the result of part (a), vav = 40.4 40.5 IDENTIFY and SET UP: ψ ( x) = A sin kx The position probability density is given by ψ ( x) = A2 sin kx EXECUTE: (a) The probability is highest where sin kx = so kx = 2π x/λ = nπ /2, n = 1, 3, 5,… x = nλ /4, n = 1, 3, 5,… so x = λ /4, 3λ /4, 5λ /4,… (b) The probability of finding the particle is zero where ψ = 0, which occurs where sin kx = and kx = 2π x/λ = nπ , n = 0, 1, 2,… x = nλ /2, n = 0,1, 2,… so x = 0, λ /2, λ , 3λ /2,… EVALUATE: The situation is analogous to a standing wave, with the probability analogous to the square of the amplitude of the standing wave 40.6 IDENTIFY and SET UP: Ψ = Ψ ∗Ψ 2 EXECUTE: Ψ ∗ = ψ ∗ sin ωt , so Ψ = Ψ ∗Ψ = ψ ∗ψ sin ωt = ψ sin ωt Ψ is not time-independent, so Ψ is not the wavefunction for a stationary state EVALUATE: Ψ = ψ eiωφ = ψ (cos ωt + i sin ωt ) is a wavefunction for a stationary state, since for it 2 Ψ = ψ , which is time independent 40.7 IDENTIFY: Determine whether or not − SET UP: − d 2ψ 2m dx 2 d 2ψ 2m dx + U ψ = E1ψ and − + U ψ is equal to Eψ , for some value of E d 2ψ 2m dx + U ψ = E2ψ d 2ψ + U ψ = BE1ψ + CE2ψ If ψ were a solution with energy E, then 2m dx BE1ψ + CE2ψ = BEψ + CEψ or B ( E1 − E )ψ = C ( E − E2 )ψ This would mean that ψ is a constant EXECUTE: − multiple of ψ , and ψ and ψ would be wave functions with the same energy However, E1 ≠ E2 , so this is not possible, and ψ cannot be a solution to Eq (40.23) EVALUATE: ψ is a solution if E1 = E2 ; see Exercise 40.9 40.8 IDENTIFY: Apply the Heisenberg Uncertainty Principle in the form ΔxΔpx ≥ /2 SET UP: The uncertainty in the particle position is proportional to the width of ψ ( x ) EXECUTE: The width of ψ ( x) is inversely proportional to α This can be seen by either plotting the function for different values of α or by finding the full width at half-maximum The particle’s uncertainty in position decreases with increasing α (b) Since the uncertainty in position decreases, the uncertainty in momentum must increase EVALUATE: As α increases, the function A(k ) in Eq (40.19) must become broader 40.9 IDENTIFY: Determine whether or not − d 2ψ 2m dx + U ψ is equal to Eψ SET UP: ψ and ψ are solutions with energy E means that − − d 2ψ 2m dx 2 d 2ψ 2m dx + U ψ = Eψ and + U ψ = Eψ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics EXECUTE: Eq (40.23): 40-3 − d 2ψ + U ψ = Eψ Let ψ = Aψ + Bψ 2m dx − d2 ( Aψ + Bψ ) + U ( Aψ + Bψ ) = E ( Aψ + Bψ ) 2m dx 2 2 ⎛ ⎞ ⎛ ⎞ d ψ1 d ψ2 ⇒ A⎜ − + Uψ − Eψ ⎟ + B ⎜ − + Uψ − Eψ ⎟ = But each of ψ and ψ satisfy 2 ⎜ 2m dx ⎟ ⎜ 2m dx ⎟ ⎝ ⎠ ⎝ ⎠ Schrödinger’s equation separately so the equation still holds true, for any A or B EVALUATE: If ψ and ψ are solutions of the Schrodinger equation for different energies, then ⇒ ψ = Bψ + Cψ is not a solution (Exercise 40.7) 40.10 IDENTIFY: To describe a real situation, a wave function must be normalizable ψ dV is the probability that the particle is found in volume dV Since the particle must be SET UP: somewhere, ψ must have the property that ∫ψ dV = when the integral is taken over all space EXECUTE: (a) For normalization of the one-dimensional wave function, we have 1= ∫ ∞ −∞ ∞ −∞ −∞ ψ dx = ∫ ( Aebx )2dx + ∫ (Ae −bx )2 dx = ∫ ∞ A2e2bx dx + ∫ A2e−2bx dx ∞⎫ ⎧ 2bx e−2bx ⎪ A2 1= A ⎨ + , which gives A = b = 2.00 m −1 = 1.41 m –1/2 ⎬= ⎪⎩ 2b −∞ −2b ⎪⎭ b (b) The graph of the wavefunction versus x is given in Figure 40.10 ⎪e (c) (i) P = ∫ +5.00 m −0.500 m ψ dx = 2∫ +5.00 m A2e−2bx dx, where we have used the fact that the wave function is an even function of x Evaluating the integral gives − A2 −2b(0.500 m) − (2.00 m −1 ) −2.00 (e − 1) = (e − 1) = 0.865 b 2.00 m −1 There is a little more than an 86% probability that the particle will be found within 50 cm of the origin 0 A2 2.00 m −1 (ii) P = ∫ (Aebx )2 dx = ∫ A2e2bx dx = = = = 0.500 − −∞ −∞ 2b 2(2.00 m ) P= There is a 50-50 chance that the particle will be found to the left of the origin, which agrees with the fact that the wave function is symmetric about the y-axis (iii) P = ∫ 1.00 m 0.500 m A2e−2bx dx A2 −2(2.00 m−1 )(1.00 m) −2(2.00 m−1 )(0.500 m) −e (e ) = − (e−4 − e−2 ) = 0.0585 −2b EVALUATE: There is little chance of finding the particle in regions where the wave function is small = Figure 40.10 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-4 Chapter 40 40.11 IDENTIFY and SET UP: The energy levels for a particle in a box are given by En = EXECUTE: (a) The lowest level is for n = 1, and E1 = (b) E = (1)(6.626 × 10−34 J ⋅ s)2 8(0.20 kg)(1.3 m) n2h2 8mL2 = 1.6 × 10−67 J 2E 2(1.2 × 10−67 J) = = 1.3 × 10−33 m/s If the ball has this speed the time it mv so v = m 0.20 kg would take it to travel from one side of the table to the other is 1.3 m t= = 1.0 × 1033 s 1.3 × 10−33 m/s (c) E1 = 40.12 h2 , E2 = E1, so ΔE = E2 − E1 = 3E1 = 3(1.6 × 10−67 J) = 4.9 × 10−67 J 8mL (d) EVALUATE: No, quantum mechanical effects are not important for the game of billiards The discrete, quantized nature of the energy levels is completely unobservable IDENTIFY: Solve Eq (40.31) for L SET UP: The ground state has n = EXECUTE: L = 40.13 EVALUATE: The value of L we calculated is on the order of the diameter of a nucleus IDENTIFY: An electron in the lowest energy state in this box must have the same energy as it would in the ground state of hydrogen nh SET UP: The energy of the n th level of an electron in a box is En = 8mL2 EXECUTE: An electron in the ground state of hydrogen has an energy of −13.6 eV, so find the width corresponding to an energy of E1 = 13.6 eV Solving for L gives L= 40.14 h (6.626 × 10−34 J ⋅ s) = = 6.4 × 10−15 m −27 −19 8mE1 8(1.673 × 10 kg)(5.0 × 10 eV)(1.602 × 10 J/eV) h (6.626 × 10−34 J ⋅ s) = = 1.66 × 10−10 m 8mE1 8(9.11 × 10−31 kg)(13.6 eV)(1.602 × 10−19 J/eV) EVALUATE: This width is of the same order of magnitude as the diameter of a Bohr atom with the electron in the K shell c IDENTIFY and SET UP: The energy of a photon is E = hf = h The energy levels of a particle in a box λ are given by Eq (40.31) EXECUTE: (a) E = (6.63 × 10−34 J ⋅ s) L= (3.00 × 108 m/s) (122 × 10 −9 m) = 1.63 × 10−18 J Δ E = h2 8mL2 (n12 − n22 ) h (n12 − n22 ) (6.63 × 10−34 J ⋅ s) (22 − 12 ) = = 3.33 × 10−10 m 8mΔ E 8(9.11 × 10−31 kg)(1.63 × 10−18 J) (b) The ground state energy for an electron in a box of the calculated dimensions is h2 (6.63 × 10−34 J ⋅ s) E= = = 5.43 × 10−19 J = 3.40 eV (one-third of the original −31 −10 8mL 8(9.11 × 10 kg)(3.33 × 10 m) photon energy), which does not correspond to the −13.6 eV ground state energy of the hydrogen atom 40.15 EVALUATE: (c) Note that the energy levels for a particle in a box are proportional to n , whereas the energy levels for the hydrogen atom are proportional to − 12 A one-dimensional box is not a good model n for a hydrogen atom IDENTIFY and SET UP: Eq (40.31) gives the energy levels Use this to obtain an expression for E2 − E1 and use the value given for this energy difference to solve for L © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics EXECUTE: Ground state energy is E1 = h2 4h E = The energy ; first excited state energy is 8mL2 8mL2 separation between these two levels is Δ E = E2 − E1 = L = 6.626 × 10−34 J ⋅ s 40.16 3h 2 8mL This gives L = h 8(9.109 × 10 −31 40-5 kg)(3.0 eV)(1.602 × 10−19 J/1 eV) = 8mΔ E = 6.1 × 10−10 m = 0.61 nm EVALUATE: This energy difference is typical for an atom and L is comparable to the size of an atom IDENTIFY: The energy of the absorbed photon must be equal to the energy difference between the two states 9π 2 The ground state energy is SET UP and EXECUTE: The second excited state energy is E3 = 2mL2 E1 = π2 4π 2 hc E E = 1.00 eV, so E = 9.00 eV For the transition Δ = = Δ E λ mL2 2mL2 hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s) = = 1.55 × 10−7 m = 155 nm ΔE 8.00 eV EVALUATE: This wavelength is much shorter than those of visible light IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation ⎛ nπ x ⎞ −iEnt / sin ⎜ SET UP: We must substitute the equation Ψ ( x, t ) = into the one-dimensional ⎟e L ⎝ L ⎠ λ= 40.17 Schrödinger equation − d 2ψ ( x) dx 2m + U ( x)ψ ( x) = Eψ ( x) EXECUTE: Taking the second derivative of Ψ( x, t ) with respect to x gives Substituting this result into − 2m d 2ψ ( x) dx + U ( x)ψ ( x) = Eψ ( x), we get dx 2 ⎛ nπ ⎞ = −⎜ ⎟ Ψ ( x, t ) ⎝ L ⎠ ⎛ nπ ⎞ ⎜ ⎟ Ψ ( x, t ) = E Ψ ( x, t ) 2m ⎝ L ⎠ ⎛ nπ ⎞ ⎜ ⎟ , the energies of a particle in a box 2m ⎝ L ⎠ EVALUATE: Since this process gives us the energies of a particle in a box, the given wave function is a solution to the Schrödinger equation IDENTIFY: Find x where ψ is zero and where it is a maximum which gives En = 40.18 2 d Ψ ( x, t ) ⎛πx⎞ sin ⎜ ⎟ L ⎝ L ⎠ EXECUTE: (a) The wave function for n = vanishes only at x = and x = L in the range ≤ x ≤ L (b) In the range for x, the sine term is a maximum only at the middle of the box, x = L/2 EVALUATE: (c) The answers to parts (a) and (b) are consistent with the figure IDENTIFY and SET UP: For the n = first excited state the normalized wave function is given by SET UP: ψ = 40.19 2 ⎛ 2π x ⎞ ⎛ 2π x ⎞ sin ⎜ ⎟ ψ ( x ) dx = sin ⎜ ⎟ dx Examine ψ ( x ) dx and find where L L L L ⎝ ⎠ ⎝ ⎠ it is zero and where it is maximum ⎛ 2π x ⎞ EXECUTE: (a) ψ dx = implies sin ⎜ ⎟=0 ⎝ L ⎠ Eq (40.35) ψ ( x) = 2π x = mπ , m = 0, 1, 2, … ; x = m( L/2) L For m = 0, x = 0; for m = 1, x = L/2; for m = 2, x = L The probability of finding the particle is zero at x = 0, L/2, and L © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-6 Chapter 40 ⎛ 2π x ⎞ (b) ψ dx is maximum when sin ⎜ ⎟ = ±1 ⎝ L ⎠ 2π x = m(π /2), m = 1, 3, 5, … ; x = m( L/4) L For m = 1, x = L/4; for m = 3, x = 3L/4 The probability of finding the particle is largest at x = L/4 and 3L/4 (c) EVALUATE: The answers to part (a) correspond to the zeros of ψ shown in Figure 40.12 in the textbook and the answers to part (b) correspond to the two values of x where ψ 40.20 IDENTIFY: Evaluate SET UP: d ψ in the figure is maximum and see if Eq (40.25) is satisfied ψ ( x) must be zero at the walls, where U → ∞ dx d d sin kx = k cos kx cos kx = − k sin kx dx dx d 2ψ 2m = − k 2ψ , and for ψ to be a solution of Eq (40.25), k = E dx (b) The wave function must vanish at the rigid walls; the given function will vanish at x = for any k , but to vanish at x = L, kL = nπ for integer n EXECUTE: (a) 40.21 n 2π 2 nπ , so kn = and ψ = A sin kx is the same as ψ n in L 2mL2 Eq (40.32), except for a different symbol for the normalization constant (a) IDENTIFY and SET UP: ψ = A cos kx Calculate dψ /dx and substitute into Eq (40.25) to see if this EVALUATE: From Eq (40.31), En = equation is satisfied EXECUTE: Eq (40.25): − h d 2ψ 8π 2m dx = Eψ dψ = A(− k sin kx) = − Ak sin kx dx d 2ψ dx = − Ak ( k cos kx ) = − Ak cos kx Thus Eq (40.25) requires − This says h2k 2 8π m = E; k = h2 8π 2m (− Ak cos kx) = E ( A cos kx) 2mE 2mE = (h/2π ) ψ = A cos kx is a solution to Eq (40.25) if k = 2mE (b) EVALUATE: The wave function for a particle in a box with rigid walls at x = and x = L must satisfy the boundary conditions ψ = at x = and ψ = at x = L ψ (0) = A cos0 = A, since cos0 = Thus ψ is not at x = and this wave function isn’t acceptable because it doesn’t satisfy the required 40.22 boundary condition, even though it is a solution to the Schrödinger equation IDENTIFY: The energy levels are given by Eq (40.31) The wavelength λ of the photon absorbed in an hc atomic transition is related to the transition energy ΔE by λ = ΔE SET UP: For the ground state n = and for the third excited state n = EXECUTE: (a) The third excited state is n = 4, so ΔE = (42 − 1) h2 8mL2 = 15(6.626 × 10−34 J ⋅ s) 8(9.11 × 10−31 kg)(0.125 × 10−9 m) = 5.78 ì 1017 J = 361 eV â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics 40-7 hc (6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s) = = 3.44 nm ΔE 5.78 × 10−17 J EVALUATE: This photon is an x ray As the width of the box increases the transition energy for this transition decreases and the wavelength of the photon increases h h IDENTIFY and SET UP: λ = = The energy of the electron in level n is given by Eq (40.31) p 2mE (b) λ = 40.23 EXECUTE: (a) E1 = h2 8mL is twice the width of the box p1 = 4h (b) E2 = p2 = h λ2 (c) E3 = h ⇒ λ1 = 2mh /8mL2 h λ1 = = L = 2(3.0 × 10−10 m) = 6.0 × 10−10 m The wavelength (6.63 × 10−34 J ⋅ s) 6.0 × 10−10 m = 1.1 × 10−24 kg ⋅ m/s ⇒ λ = L = 3.0 × 10−10 m The wavelength is the same as the width of the box 8mL2 = p1 = 2.2 × 10−24 kg ⋅ m/s 9h 2 8mL L = 2.0 × 10−10 m The wavelength is two-thirds the width of the box ⇒ λ3 = p3 = p1 = 3.3 × 10−24 kg ⋅ m/s EVALUATE: In each case the wavelength is an integer multiple of λ /2 In the n th state, pn = np1 40.24 IDENTIFY: To describe a real situation, a wave function must be normalizable ψ SET UP: dV is the probability that the particle is found in volume dV Since the particle must be somewhere, ψ must have the property that ∫ψ dV = when the integral is taken over all space EXECUTE: (a) In one dimension, as we have here, the integral discussed above is of the form ∞ ∫−∞ ψ ( x) dx = (b) Using the result from part (a), we have ∞ ∫−∞ (eax ) dx = ∫ ∞ −∞ e2ax dx = e2 ax 2a ∞ = ∞ Hence this wave −∞ function cannot be normalized and therefore cannot be a valid wave function (c) We only need to integrate this wave function of to ∞ because it is zero for x < For normalization we have = ∫ ∞ −∞ 40.25 ∞ ∞ 0 ψ dx = ∫ (Ae-bx )2 dx = ∫ A2e−2bx dx = A2e−2bx −2b ∞ = A2 A2 , which gives = 1, so A = 2b 2b 2b EVALUATE: If b were negative, the given wave function could not be normalized, so it would not be allowable 2 d ψ IDENTIFY: Compare − + U ψ to Eψ and see if there is a value of k for which they are equal 2m dx SET UP: EXECUTE: d2 dx sin kx = − k sin kx (a) Eq (40.23): Left-hand side: − d 2ψ + U ψ = Eψ 2m dx 2 ⎛ 2k ⎞ − d2 k + = + = + U ⎟ψ But ( A sin kx ) U A sin kx A sin kx U A sin kx ⎜⎜ 0 ⎟ 2m dx 2m m ⎝ ⎠ 2 2 k k + U > U > E if k is real But + U should equal E This is not the case, and there is no k 2m 2m for which this ψ is a solution © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-8 Chapter 40 40.26 2 k + U = E is consistent and so ψ = A sin kx is a solution of Eq (40.23) for this case 2m EVALUATE: For a square-well potential and E < U , Eq (40.23) with U = U applies outside the well and the wave function has the form of Eq (40.40) h p2 IDENTIFY: λ = p is related to E by E = +U p 2m (b) If E > U , then SET UP: For x > L, U = U For < x < L, U = EXECUTE: For < x < L, p = 2mE = 2m(3U ) and λin = h h = Thus, the ratio of the m( E − U ) 2m(2U ) p = 2m( E − U ) = 2m(2U ) and λout = wavelengths is 40.27 h For x > L, 2m(3U ) 2m(3U ) λout = = λin 2m(2U ) EVALUATE: For x > L some of the energy is potential and the kinetic energy is less than it is for < x < L, where U = Therefore, outside the box p is less and λ is greater than inside the box IDENTIFY: Figure 40.15b in the textbook gives values for the bound state energy of a square well for which U = E1-1DW SET UP: E1-1DW = π2 2mL2 EXECUTE: E1 = 0.625E1-1DW = 0.625 π2 2mL2 ; E1 = 2.00 eV = 3.20 × 10−19 J 1/2 40.28 ⎛ ⎞ 0.625 −10 L = π ⎜⎜ m ⎟⎟ = 3.43 × 10 −31 −19 2(9.109 10 kg)(3.20 10 J) × × ⎝ ⎠ EVALUATE: As L increases the ground state energy decreases IDENTIFY: The energy of the photon is the energy given to the electron SET UP: Since U = E1-1DW we can use the result E1 = 0.625 E1-1DW from Section 40.4 When the electron is outside the well it has potential energy U , so the minimum energy that must be given to the electron is U − E1 = 5.375 E1-1DW EXECUTE: The maximum wavelength of the photon would be hc hc 8mL2c 8(9.11 × 10−31 kg)(1.50 × 10−9 m) (3.00 × 108 m/s) = = = λ= U − E1 (5.375)(h /8mL2 ) (5.375)h (5.375)(6.63 × 10−34 J ⋅ s) 40.29 = 1.38 × 10−6 m EVALUATE: This photon is in the infrared The wavelength of the photon decreases when the width of the well decreases d 2ψ 2mE IDENTIFY: Calculate and compare to − ψ dx d d SET UP: sin kx = k cos kx cos kx = − k sin kx dx dx EXECUTE: Eq (40.37): ψ = Asin 2mE x + B cos 2mE x d 2ψ 2mE 2mE −2mE ⎛ 2mE ⎞ ⎛ 2mE ⎞ = − A ⎜ ⎟ sin x − B ⎜ ⎟ cos x= (ψ ) This is Eq (40.38), so this ψ is a dx ⎝ ⎠ ⎝ ⎠ solution EVALUATE: ψ in Eq (40.38) is a solution to Eq (40.37) for any values of the constants A and B © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics 40.30 40-9 IDENTIFY: The longest wavelength corresponds to the smallest energy change h2 SET UP: The ground level energy level of the infinite well is E1-1DW = , and the energy of the 8mL2 photon must be equal to the energy difference between the two shells EXECUTE: The 400.0 nm photon must correspond to the n = to n = transition Since U = E1-1DW , we have E2 = 2.43E1-1DW and E1 = 0.625 E1-1DW The energy of the photon is equal to the energy difference between the two levels, and E1-1DW = Eγ = E2 − E1 ⇒ L= 40.31 hc λ = (2.43 − 0.625) E1-1DW = h2 8mL2 1.805 h 8mL2 , which gives Solving for L gives (1.805) hλ (1.805)(6.626 × 10−34 J ⋅ s)(4.00 × 10−7 m) = = 4.68 × 10−10 m = 0.468 nm 8mc 8(9.11 × 10−31 kg)(3.00 × 108 m/s) EVALUATE: This width is approximately half that of a Bohr hydrogen atom IDENTIFY: Find the transition energy ΔE and set it equal to the energy of the absorbed photon Use E = hc/λ , to find the wavelength of the photon SET UP: U = E1-1DW , as in Figure 40.15 in the textbook, so E1 = 0.625 E1-1DW and E3 = 5.09 E1-1DW with E1-1DW = π2 2mL2 In this problem the particle bound in the well is a proton, so m = 1.673 × 10−27 kg EXECUTE: E1-1DW = π2 2 2mL = π (1.055 × 10−34 J ⋅ s) 2(1.673 × 10−27 kg)(4.0 × 10−15 m) = 2.052 × 10−12 J The transition energy is Δ E = E3 − E1 = (5.09 − 0.625) E1-1DW = 4.465 E1-1DW Δ E = 4.465(2.052 × 10−12 J) = 9.162 × 10−12 J The wavelength of the photon that is absorbed is related to the transition energy by ΔE = hc/λ , so hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) = = 2.2 × 10−14 m = 22 fm ΔE 9.162 × 10−12 J EVALUATE: The wavelength of the photon is comparable to the size of the box λ= 40.32 IDENTIFY: The tunneling probability is T = Ge −2κ L , with G = 16 E ⎛ E ⎞ T = 16 ⎜1 − ⎟e U0 ⎝ U0 ⎠ −2 m (U − E ) L 2m(U − E ) E ⎛ E ⎞ so ⎜1 − ⎟ and κ = U0 ⎝ U0 ⎠ SET UP: U = 30.0 × 106 eV, L = 2.0 × 10−15 m, m = 6.64 × 10 −27 kg EXECUTE: (a) U − E = 1.0 × 106 eV (E = 29.0 × 106 eV), T = 0.090 (b) If U − E = 10.0 × 106 eV (E = 20.0 × 106 eV), T = 0.014 EVALUATE: T is less when U − E s 10.0 MeV than when U − E is 1.0 MeV 40.33 IDENTIFY: The tunneling probability is T = 16 SET UP: E ⎛ E ⎞ −2 L ⎜1 − ⎟e U0 ⎝ U0 ⎠ m (U − E ) / E 6.0 eV = and E − U = eV = 8.0 × 10−19 J U 11.0 eV EXECUTE: (a) L = 0.80 × 10−9 m: ⎛ 6.0 eV ⎞⎛ 6.0 ev ⎞ −2(0.80 × 10−9 m) T = 16 ⎜ ⎟⎜1 − ⎟e ⎝ 11.0 eV ⎠⎝ 11.0 eV ⎠ 2(9.11 × 10−31 kg)(8.0 × 10−19 J) /1.055 × 10−34 J ⋅ s = 4.4 × 10−8 (b) L = 0.40 × 10−9 m: T = 4.2 × 10−4 EVALUATE: The tunneling probability is less when the barrier is wider © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-10 40.34 Chapter 40 IDENTIFY: The transmission coefficient is T = 16 E ⎛ E ⎞ −2 ⎜1 − ⎟e U0 ⎝ U0 ⎠ m (U − E ) L / SET UP: E = 5.0 eV, L = 0.60 × 10−9 m, and m = 9.11 × 10−31 kg EXECUTE: (a) U = 7.0 eV ⇒ T = 5.5 × 10−4 (b) U = 9.0 eV ⇒ T = 1.8 × 10−5 (c) U = 13.0 eV ⇒ T = 1.1 × 10−7 40.35 EVALUATE: T decreases when the height of the barrier increases IDENTIFY and SET UP: Use Eq (39.1), where K = p /2m and E = K + U EXECUTE: λ = h/p = h/ 2mK , so λ K is constant λ1 K1 = λ K ; λ1 and K1 are for x > L where K1 = 2U and λ and K are for < x < L where K = E − U = U K2 U0 λ1 = = = K1 λ2 2U 40.36 EVALUATE: When the particle is passing over the barrier its kinetic energy is less and its wavelength is larger IDENTIFY: The probability of tunneling depends on the energy of the particle and the width of the barrier E ⎛ E ⎞ SET UP: The probability of tunneling is approximately T = Ge−2κ L , where G = 16 ⎜ − ⎟ and U0 ⎝ U0 ⎠ κ= 2m(U − E ) EXECUTE: G = 16 κ= 2m(U − E ) E ⎛ E ⎞ 50.0 eV ⎛ 50.0 eV ⎞ ⎜1 − ⎟ = 16 ⎜1 − ⎟ = 3.27 U0 ⎝ U0 ⎠ 70.0 eV ⎝ 70.0 eV ⎠ = 2(1.67 × 10−27 kg)(70.0 eV − 50.0 eV)(1.60 × 10−19 J/eV) (6.63 × 10−34 J ⋅ s)/2π = 9.8 × 1011 m −1 Solving T = Ge−2κ L for L gives 1 ⎛ 3.27 ⎞ −12 L= ln(G /T ) = ln ⎜ m = 3.6 pm ⎟ = 3.6 × 10 11 − 2κ 2(9.8 × 10 m ) ⎝ 0.0030 ⎠ If the proton were replaced with an electron, the electron’s mass is much smaller so L would be larger EVALUATE: An electron can tunnel through a much wider barrier than a proton of the same energy 40.37 IDENTIFY and SET UP: The probability is T = Ae−2κ L , with A = 16 E ⎛ E ⎞ ⎜1 − ⎟ and κ = U0 ⎝ U0 ⎠ 2m(U − E ) E = 32 eV, U = 41 eV, L = 0.25 × 10−9 m Calculate T EXECUTE: (a) A = 16 κ= κ= E ⎛ E ⎞ 32 ⎛ 32 ⎞ ⎜1 − ⎟ = 16 ⎜1 − ⎟ = 2.741 U0 ⎝ U0 ⎠ 41 ⎝ 41 ⎠ 2m(U − E ) 2(9.109 × 10−31 kg)(41 eV − 32 eV)(1.602 × 10−19 J/eV) 1.055 × 10−34 J ⋅ s −1 = 1.536 × 1010 m −1 −9 T = Ae−2κ L = (2.741)e−2(1.536 ×10 m )(0.25 ×10 m) = 2.741e−7.68 = 0.0013 (b) The only change in the mass m, which appears in κ 10 κ= 2m(U − E ) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics κ= 2(1.673 × 10−27 kg)(41 eV − 32 eV)(1.602 × 10−19 J/eV) 1.055 × 10−34 J ⋅ s Then T = Ae−2κ L = (2.741)e−2(6.584 × 10 11 40.38 = 6.584 × 1011 m −1 = 2.741e−392.2 = 10−143 EVALUATE: The more massive proton has a much smaller probability of tunneling than the electron does d 2ψ IDENTIFY: Calculate and insert the result into Eq (40.44) dx SET UP: 2 d −δ x d −δ x e e = −2δ xe−δ x and = (4δ x − 2δ )e−δ x dx dx EXECUTE: Let Eq (40.44) if E = EVALUATE: E = 40.39 m −1 )(0.25 × 10−9 m) 40-11 mk ′ /2 = δ , and so m δ= k ′/m = d 2ψ dψ = (4 x 2δ − 2δ )ψ , and ψ is a solution of = −2 xδψ and dx dx ω ω agrees with Eq (40.46), for n = IDENTIFY and SET UP: The energy levels are given by Eq (40.46), where ω = EXECUTE: ω = k′ m k′ 110 N/m = = 21.0 rad/s m 0.250 kg The ground state energy is given by Eq (40.46): 1 E0 = ω = (1.055 × 10−34 J ⋅ s)(21.0 rad/s) = 1.11 × 10−33 J(1 eV/1.602 × 10−19 J) = 6.93 × 10−15 eV 2 1⎞ ⎛ ⎞ ⎛ En = ⎜ n + ⎟ ω , E( n + 1) = ⎜ n + + ⎟ ω 2⎠ 2⎠ ⎝ ⎝ The energy separation between these adjacent levels is Δ E = En +1 − En = ω = E0 = 2(1.11 × 10−33 J) = 2.22 × 10−33 J = 1.39 × 10−14 eV 40.40 40.41 EVALUATE: These energies are extremely small; quantum effects are not important for this oscillator IDENTIFY: The energy of the absorbed photon must be equal to the energy difference between the two states hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s) = = 0.1433 eV ΔE = ω SET UP and EXECUTE: ΔE = λ 8.65 × 10−6 m ω 0.1433 eV = = 0.0717 eV E0 = 2 EVALUATE: The energy of the photon is not equal to the energy of the ground state, but rather it is the energy difference between the two states IDENTIFY: We can model the molecule as a harmonic oscillator The energy of the photon is equal to the energy difference between the two levels of the oscillator SET UP: The energy of a photon is Eγ = hf = hc/λ , and the energy levels of a harmonic oscillator are 1⎞ ⎛ given by En = ⎜ n + ⎟ 2⎠ ⎝ k′ ⎛ 1⎞ = ⎜ n + ⎟ ω m ⎝ 2⎠ EXECUTE: (a) The photon’s energy is Eγ = hc λ = (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) (b) The transition energy is ΔE = En + − En = ω = we get k ′ = 4π 2c 2m k′ 2π c = , which gives m λ = 0.21 eV k′ Solving for k ′, m 4π (3.00 × 108 m/s) (5.6 × 10−26 kg) = 5,900 N/m (5.8 × 10−6 m) EVALUATE: This would be a rather strong spring in the physics lab = 5.8 ì 106 m â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-12 40.42 Chapter 40 IDENTIFY: The photon energy equals the transition energy for the atom SET UP: According to Eq (40.46), the energy released during the transition between two adjacent levels is twice the ground state energy E3 − E2 = ω = E0 = 11.2 eV EXECUTE: For a photon of energy E , c hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = = 111 nm f E (11.2 eV)(1.60 × 10−19 J/eV) EVALUATE: This photon is in the ultraviolet IDENTIFY and SET UP: Use the energies given in Eq (40.46) to solve for the amplitude A and maximum speed vmax of the oscillator Use these to estimate Δ x and Δpx and compute the uncertainty product E = hf ⇒ λ = 40.43 Δ xΔpx EXECUTE: The total energy of a Newtonian oscillator is given by E = 12 k ′A2 where k′ is the force ( constant and A is the amplitude of the oscillator Set this equal to the energy E = n + 12 level that has quantum number n, where ω = A= k′ , and solve for A: m k ′A2 ( = n + 12 ) ) ω of an excited ω (2n + 1) ω The total energy of the Newtonian oscillator can also be written as E = 12 mvmax Set k′ ( this equal to E = n + 12 ) ω and solve for vmax : mvmax ( = n + 12 ) ω vmax = (2n + 1) ω Thus the m maximum linear momentum of the oscillator is pmax = mvmax = (2n + 1) mω Now A/ represents the uncertainty Δ x in position and that pmax / is the corresponding uncertainty Δ px in momentum Then the uncertainty product is ⎛ (2n + 1) ω ⎞ ⎛ ⎞ (2n + 1) ω m (2n + 1) ω ⎛ ⎞ Δ xΔ px = ⎜⎜ = ⎜ ⎟ = (2n + 1) ⎟⎟ ⎜⎝ (2n + 1) mω ⎟⎠ = ′ k k′ 2 ⎝ω ⎠ ⎝ ⎠ EVALUATE: For n = this gives Δ xΔ px = /2, in agreement with the result derived in Section 40.5 The uncertainty product Δ xΔ px increases with n 40.44 IDENTIFY: Compute the ratio specified in the problem ω k′ SET UP: For n = 0, A = ω= k′ m ⎛ mk ′ ⎞ ω⎞ ⎛ A ⎟⎟ = exp ⎜ − mk ′ ⎟ = e −1 = 0.368 This is consistent with what is = exp ⎜⎜ − k′ ⎠ ⎝ ψ (0) ⎝ ⎠ shown in Figure 40.27 in the textbook EXECUTE: (a) ψ (2 A)2 ⎛ ⎞ mk ′ ω⎞ ⎛ (2 A) ⎟⎟ = exp ⎜ − mk ′ ⎟ = e−4 = 1.83 × 10−2 This figure cannot be read this = exp ⎜⎜ − k′ ⎠ ⎝ ψ (0) ⎝ ⎠ precisely, but the qualitative decrease in amplitude with distance is clear EVALUATE: The wave function decays exponentially as x increases beyond x = A IDENTIFY: We model the atomic vibration in the crystal as a harmonic oscillator ⎞ k′ ⎛ 1⎞ ⎛ SET UP: The energy levels of a harmonic oscillator are given by En = ⎜ n + ⎟ = ⎜ n + ⎟ ω m 2⎠ ⎝ ⎠ ⎝ EXECUTE: (a) The ground state energy of a simple harmonic oscillator is (b) 40.45 ψ ( A) E0 = 1 ω= 2 12.2 N/m k ′ (1.055 × 10−34 J ⋅ s) = = 9.43 × 10−22 J = 5.89 × 10−3 eV m 3.82 × 10−26 kg (b) E4 − E3 = ω = E0 = 0.0118 eV, so λ = hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = 106 μm E 1.88 ì 1021 J â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics 40-13 (c) En +1 − En = ω = E0 = 0.0118 eV EVALUATE: These energy differences are much smaller than those due to electron transitions in the hydrogen atom 40.46 IDENTIFY: For a stationary state, Ψ is time independent SET UP: To calculate Ψ ∗ from Ψ , replace i by −i EXECUTE: For this wave function, Ψ ∗ = ψ 1∗eiω1t + ψ 2∗eiω2t , so Ψ = Ψ ∗Ψ = (ψ 1∗eiω1t + ψ 2∗eiω2t )(ψ 1e−iω1t + ψ 2e−iω2t ) = ψ 1∗ψ + ψ 2∗ψ + ψ 1∗ψ 2ei (ω1 −ω2 )t + ψ 2∗ψ 1ei (ω2 −ω1 )t The frequencies ω1 and ω are given as not being the same, so Ψ is not time-independent, and Ψ is not the wave function for a stationary state EVALUATE: If ω1 = ω , then Ψ is the wave function for a stationary state 40.47 IDENTIFY: We know the wave function of a particle in a box 1 SET UP and EXECUTE: (a) Ψ( x, t ) = ψ 1( x)e− iE1t/ + ψ ( x)e−iE3t/ 2 1 Ψ∗ ( x, t ) = ψ 1( x)e+ iE1t/ + ψ ( x)e+ iE3t/ 2 1⎡ ⎛ [ E − E1 ]t ⎞ ⎤ Ψ ( x, t ) = [ψ 12 + ψ 32 + ψ 1ψ (ei ( E3 − E1 )t / + e−i ( E3 − E1 )t / )] = ⎢ψ 12 + ψ 32 + 2ψ 1ψ cos ⎜ ⎟⎥ 2⎣ ⎝ ⎠⎦ ψ1 = 9π 2 π2 4π 2 ⎛ 3π x ⎞ E E E = and E = , so − = sin ⎜ ⎟ L 2mL2 mL2 2mL2 ⎝ L ⎠ ⎛πx⎞ sin ⎜ ⎟ ψ = L ⎝ L ⎠ Ψ ( x, t ) = ⎡ 2⎛πx⎞ ⎛ π x ⎞ ⎛ 3π x ⎞ ⎛ 4π t ⎞ ⎤ ⎛ 3π x ⎞ ⎢sin ⎜ ⎟ ⎥ At x = L/2, ⎟ + sin ⎜ ⎟ + 2sin ⎜ ⎟ sin ⎜ ⎟ cos ⎜⎜ L ⎢⎣ ⎝ L ⎠ ⎝ L ⎠ ⎝ L ⎠ ⎝ L ⎠ ⎝ mL2 ⎟⎠ ⎥⎦ ⎛πx⎞ ⎛π ⎞ ⎛ 3π x ⎞ ⎛ 3π sin ⎜ ⎟ = sin ⎜ ⎟ = sin ⎜ ⎟ = sin ⎜ ⎝ L ⎠ ⎝2⎠ ⎝ L ⎠ ⎝ (b) ωosc = E3 − E1 = ⎛ 4π t ⎞ ⎤ 2⎡ ⎞ = − Ψ = − ( x , t ) cos ⎢ ⎜⎜ ⎟ ⎟ ⎟⎥ L ⎢⎣ ⎠ ⎝ mL ⎠ ⎥⎦ 4π mL2 EVALUATE: Note that Δ E = ω 40.48 IDENTIFY: Carry out the calculations specified in the problem ∞ −α k ∫0 e SET UP: A standard integral is EXECUTE: (a) B ( k ) = e−α ⇒ kh = α 2 k cos( kx) dk = π − x /4α e 2α B (0) = Bmax = B (kh ) = 2 = e −α kh ⇒ ln(1/2) = −α kh2 ln(2) = wk ∞ (b) ψ ( x) = ∫ e−α 2 k cos kxdk = π 2α (e − x /4α ) ψ ( x) is a maximum when x = − x2 ⇒ h2 = ln(1/2) ⇒ xh = 2α ln = wx 4α 4α h ⎛1 h h ln ⎛ hwk ⎞ ⎞ (d) w p wx = ⎜ = (2ln 2) ln2 ⎟ (2α ln2) = (2ln 2) = ⎟ wx = ⎜ π 2π ⎝ α 2π ⎝ 2π ⎠ ⎠ EVALUATE: The Heisenberg Uncertainty Principle says that ΔxΔpx ≥ /2 If Δx = wx and Δpx = w p , (c) ψ ( xh ) = π when e− xh /4α = 2 then the uncertainty principle says wx w p ≥ /2 So our result is consistent with the uncertainty principle since (2ln 2) > /2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-14 40.49 Chapter 40 ∞ IDENTIFY: Evaluate ψ ( x) = ∫ B( k )cos kx dk for the function B (k ) specified in the problem SET UP: ∫ cos kx dk = x sin kx ∞ k0 ⎛ 0 EXECUTE: (a) ψ ( x) = ∫ B (k )cos kxdk = ∫ k ⎞ sin kx sin k0 x = ⎜ ⎟ cos kxdk = k k x k0 x ⎝ 0⎠ (b) ψ ( x ) has a maximum value at the origin x = ψ ( x0 ) = when k0 x0 = π so x0 = π k0 Thus the width of 2π 2π If k0 = , wx = L B (k ) versus k is graphed in Figure 40.49a The graph of L k0 ψ ( x) versus x is in Figure 40.49b this function wx = x0 = (c) If k0 = π L , wx = L ⎛ hw ⎞ ⎛ 2π ⎞ hwk hk0 EVALUATE: (d) w p wx = ⎜ k ⎟ ⎜ = = h If Δx = wx and Δpx = w p , then the uncertainty ⎟= k0 ⎝ 2π ⎠ ⎝ k0 ⎠ k0 principle states that w p wx ≥ For us, no matter what k0 is, w p wx = h, which is greater than /2 Figure 40.49 40.50 IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation SET UP: The given function is ψ ( x) = Aeikx , and the one-dimensional Schrödinger equation is − d 2ψ ( x) 2m dx + U ( x)ψ ( x) = Eψ ( x) EXECUTE: Start with the given function and take the indicated derivatives: ψ ( x ) = Aeikx 2 dψ ( x ) d 2ψ ( x) d 2ψ ( x) 2 ikx ikx d ψ ( x ) Ai k e Ak e k ( x ) k 2ψ ( x) ψ = Aikeikx = = − = − − = dx 2m dx 2m dx dx Substituting these results into the one-dimensional Schrödinger equation gives 2 k ψ ( x) + U 0ψ ( x) = E ψ ( x) 2m © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics EVALUATE: ψ ( x) = A eikx is a solution to the one-dimensional Schrödinger equation if E − U = or k = 2m ( E − U ) 40-15 2 k 2m (Since U < E was given, k is the square root of a positive quantity.) In terms of the particle’s momentum p: k = p/ , and in terms of the particle’s de Broglie wavelength λ: k = 2π /λ 40.51 IDENTIFY: Let I refer to the region x < and let II refer to the region x > 0, so ψ I ( x) = Aeik1 x + Be− ik1 x dψ I dψ II = at x = dx dx and ψ II ( x) = Ceik2 x Set ψ I (0) = ψ II (0) and d ikx (e ) = ikeikx dx SET UP: dψ I dψ II = at x = gives ik1 A − ik1B = ik2C Solving dx dx ⎛k −k ⎞ ⎛ 2k2 ⎞ this pair of equations for B and C gives B = ⎜ ⎟ A and C = ⎜ ⎟ A + k k ⎝ 2⎠ ⎝ k1 + k2 ⎠ EXECUTE: ψ I (0) = ψ II (0) gives A + B = C EVALUATE: The probability of reflection is R = T= 40.52 C2 A2 = 4k12 (k1 + k2 ) B2 A = (k1 − k2 ) (k1 + k2 ) The probability of transmission is Note that R + T = IDENTIFY: For a particle in a box, En = n2h2 8mL2 SET UP: Δ En = En +1 − En ( n + 1) − n 2n + + This is never larger than it is for n = 1, and R1 = n n n n EVALUATE: (b) Rn approaches zero as n becomes very large In the classical limit there is no EXECUTE: (a) Rn = = = quantization and the spacing of successive levels is vanishingly small compared to the energy levels Therefore, Rn for a particle in a box approaches the classical value as n becomes very large 40.53 IDENTIFY and SET UP: The energy levels are given by Eq (40.31): En = transition and set Δ E = hc/λ , the energy of the photon EXECUTE: (a) Ground level, n = 1, E1 = energy is Δ E = E2 − E1 = This gives hc λ λ = 1.92 × 10 = −5 3h 2 8mL 3h 8mL2 λ= n2h2 8mL2 Calculate Δ E for the h2 4h = = 2, The transition n E First excited level, 8mL2 8mL2 Set the transition energy equal to the energy hc/λ of the emitted photon 8mcL2 8(9.109 × 10−31 kg)(2.998 × 108 m/s)(4.18 × 10−9 m)2 = 3h 3(6.626 × 10−34 J ⋅ s) m = 19.2 μm (b) Second excited level has n = and E3 = − 4h 8mL2 The transition energy is hc 5h 8mcL2 = so λ = = (19.2 μ m) = 11.5 μm 5h 8mL2 8mL2 8mL2 λ 8mL2 EVALUATE: The energy spacing between adjacent levels increases with n, and this corresponds to a shorter wavelength and more energetic photon in part (b) than in part (a) Δ E = E3 − E2 = 9h 9h = 5h © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-16 40.54 Chapter 40 IDENTIFY: The probability of finding the particle between x1 and x2 is SET UP: For the ground state ψ = πx sin sin θ = 12 (1 − cos 2θ ) L L x2 ∫x ψ dx 1 ∫ cos α x dx = α sin α x L/ EXECUTE: (a) L /4 πx L /4 ⎛ 2π x ⎞ 1⎛ L 2π x ⎞ sin dx = ∫ sin ⎜1 − cos ⎟ dx = ⎜ x − ⎟ L ∫0 L L 2⎝ L ⎠ L⎝ 2π L ⎠0 = 1 − , which is 2π about 0.0908 L/ (b) Repeating with limits of L/4 and L/2 gives 1⎛ L 2π x ⎞ 1 sin , about 0.409 ⎜x− ⎟ = + L⎝ 2π L ⎠ L / 4 2π (c) The particle is much likely to be nearer the middle of the box than the edge EVALUATE: (d) The results sum to exactly Since the probability of the particle being anywhere in the box is unity, the probability of the particle being found between x = L/2 and x = L is also This means that the particle is as likely to be between x = and L/2 as it is to be between x = L/2 and x = L (e) These results are consistent with Figure 40.12b in the textbook This figure shows a greater probability near the center of the box It also shows symmetry of ψ 40.55 about the center of the box IDENTIFY: The probability of the particle being between x1 and x2 is x2 ∫x | ψ |2 dx, where ψ is the normalized wave function for the particle ⎛πx⎞ sin ⎜ ⎟ L ⎝ L ⎠ EXECUTE: The probability P of the particle being between x = L/4 and x = 3L/4 is L /4 3L / ⎛ π x ⎞ P=∫ ψ dx = ∫ sin ⎜ ⎟ dx Let y = π x/L; dx = ( L/π ) dy and the integration limits become L/ L L /4 ⎝ L ⎠ π /4 and 3π /4 (a) SET UP: The normalized wave function for the ground state is ψ = 3π /4 P= ⎛ L ⎞ 3π /4 2 ⎡1 ⎤ sin y dy = ⎢ y − sin y ⎥ ⎜ ⎟ π ⎣2 L ⎝ π ⎠ ∫π /4 ⎦π / P= ⎡ 3π π ⎛ 3π ⎞ ⎛ π ⎞ ⎤ − − sin + sin π ⎢⎣ 8 ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎥⎦ P= 1 2⎛π 1 ⎞ 1 − ( −1) + (1) ⎟ = + = 0.818 (Note: The integral formula ∫ sin y dy = y − sin y was used.) π ⎜⎝ 4 ⎠ π (b) SET UP: The normalized wave function for the first excited state is ψ = EXECUTE: P = ∫ L /4 L/ ψ dx = ⎛ 2π x ⎞ sin ⎜ ⎟ L ⎝ L ⎠ L /4 ⎛ 2π x ⎞ sin ⎜ ⎟ dx Let y = 2π x/L; dx = ( L/2π ) dy and the integration L ∫L /4 ⎝ L ⎠ limits become π /2 and 3π /2 P= 3π /2 ⎛ L ⎞ 3π /2 ⎡1 1 ⎛ 3π π ⎞ ⎤ = ⎜ − ⎟ = 0.500 ⎜ ⎟ ∫π / sin y dy = ⎢ y − sin y ⎥ π ⎣2 L ⎝ 2π ⎠ ⎦π /2 π ⎝ 4 ⎠ (c) EVALUATE: These results are consistent with Figure 40.11b in the textbook That figure shows that ψ is more concentrated near the center of the box for the ground state than for the first excited state; this is consistent with the answer to part (a) being larger than the answer to part (b) Also, this figure shows that for the first excited state half the area under ψ curve lies between L/4 and 3L/4, consistent with our answer to part (b) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics 40.56 40-17 IDENTIFY: The probability is ψ dx, with ψ evaluated at the specified value of x SET UP: For the ground state, the normalized wave function is ψ = 2/L sin(π x/L) EXECUTE: (a) (2/L) sin (π /4)dx = dx/L (b) (2/L) sin (π /2)dx = 2dx/L (c) (2 L )sin (3π /4) = dx/L EVALUATE: Our results agree with Figure 40.12b in the textbook ψ is largest at the center of the box, 40.57 at x = L/2 ψ is symmetric about the center of the box, so is the same at x = L/4 as at x = 3L/4 IDENTIFY and SET UP: The normalized wave function for the n = first excited level is ⎛ 2π x ⎞ sin ⎜ ⎟ P = ψ ( x) dx is the probability that the particle will be found in the interval x to x + dx L ⎝ L ⎠ EXECUTE: (a) x = L/4 ψ2 = ⎛ ⎛ 2π ⎞⎛ L ⎞ ⎞ sin ⎜ ⎜ ⎟⎜ ⎟ ⎟ = L ⎝ ⎝ L ⎠⎝ ⎠ ⎠ P = (2/L) dx (b) x = L/2 ψ ( x) = ψ ( x) = ⎛ ⎛ 2π ⎞⎛ L ⎞ ⎞ sin ⎜ ⎜ ⎟⎜ ⎟ ⎟ = L ⎝ ⎝ L ⎠⎝ ⎠ ⎠ ⎛π ⎞ sin ⎜ ⎟ = L ⎝2⎠ L sin(π ) = L P=0 (c) x = 3L/4 ⎛ ⎛ 2π ⎞⎛ 3L ⎞ ⎞ sin ⎜ ⎜ ⎟⎜ ⎟ ⎟ = L ⎝ ⎝ L ⎠⎝ ⎠ ⎠ P = (2/L)dx ψ ( x) = 2 ⎛ 3π ⎞ sin ⎜ ⎟ = − L L ⎝ ⎠ EVALUATE: Our results are consistent with the n = part of Figure 40.12 in the textbook ψ 40.58 is zero at the center of the box and is symmetric about this point IDENTIFY: The impulse applied to a particle equals its change in momentum SET UP: For a particle in a box, the magnitude of its momentum is p = k = nh (Eq 40.29) 2L nπ hn = At x = the initial momentum at the wall is L 2L hn hn ˆ pinitial = − iˆ and the final momentum, after turning around, is pfinal = + i So, 2L 2L hn ˆ hn ˆ ⎛ hn ˆ ⎞ hn i and the final Δp=+ i − ⎜ − i ⎟ = + iˆ At x = L the initial momentum is pinitial = + 2L 2L L ⎝ 2L ⎠ EXECUTE: Δ p = pfinal − pinitial p = k = momentum, after turning around, is pfinal = − 40.59 hn ˆ hn ˆ hn ˆ hn i So, Δ p = − i− i = − iˆ 2L 2L 2L L EVALUATE: The impulse increases with n IDENTIFY: Carry out the calculations that are specified in the problem SET UP: For a free particle, U ( x ) = so Schrödinger’s equation becomes d 2ψ ( x ) dx =− 2m h2 Eψ ( x ) EXECUTE: (a) The graph is given in Figure 40.59 2 dψ ( x ) d 2ψ ( x) 2m κ (b) For x < 0: ψ ( x ) = e +κ x = κ e+κ x = κ 2e+κ x So κ = − E ⇒ E = − dx dx 2m © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-18 Chapter 40 (c) For x > 0: ψ ( x ) = e−κ x d 2ψ ( x) dψ ( x ) 2m − 2κ = −κ e−κ x = κ 2e−κ x So again κ = − E ⇒ E = dx dx 2m Parts (b) and (c) show ψ ( x ) satisfies the Schrödinger’s equation, provided E = − 2κ 2m dψ ( x ) is discontinuous at x = (That is, it is negative for x > and positive for x < 0.) dx Therefore, this ψ is not an acceptable wave function; dψ /dx must be continuous everywhere, except where U → ∞ EVALUATE: (d) Figure 40.59 40.60 IDENTIFY: We start with the penetration distance formula given in the problem SET UP: The given formula is η = 2m(U − E ) EXECUTE: (a) Substitute the given numbers into the formula: 1.055 × 10−34 J ⋅ s = = 7.4 × 10−11 m η= −31 −19 2m(U − E ) 2(9.11 × 10 kg)(20 eV − 13 eV)(1.602 × 10 J/eV) (b) η = 40.61 1.055 × 10−34 J ⋅ s 2(1.67 × 10−27 kg)(30 MeV − 20 MeV)(1.602 × 10−13 J/MeV) = 1.44 × 10−15 m EVALUATE: The penetration depth varies widely depending on the mass and energy of the particle IDENTIFY: Eq (40.38) applies for ≤ x ≤ L Eq (40.40) applies for x < and x > L D = for x < and C = for x > L 2mE d d d κx d −κ x sin kx = k cos kx cos kx = − k sin kx = −κ e −κ x e = κ eκ x e dx dx dx dx EXECUTE: (a) We set the solutions for inside and outside the well equal to each other at the well boundaries, x = and L x = 0: B sin(0) + A = C ⇒ A = C , since we must have D = for x < SET UP: Let k = x = L: B sin 2mE L + A cos 2mE L = + De−κ L since C = for x > L This gives B sin kL + A cos kL = De−κ L , where k = 2mE (b) Requiring continuous derivatives at the boundaries yields dψ x = 0: = kB cos( k ⋅ 0) − kA sin(k ⋅ 0) = kB = κ Cek ⋅ ⇒ kB = κ C dx x = L: kB cos kL − kA sin kL = −κ De−κ L EVALUATE: These boundary conditions allow for B, C, and D to be expressed in terms of an overall normalization constant A © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics 40.62 IDENTIFY: T = Ge−2κ L with G = 16 40-19 2m(U − E ) ⎛T ⎞ E ⎛ E ⎞ , so L = − ln ⎜ ⎟ ⎜1 − ⎟ and κ = 2κ ⎝ G ⎠ U0 ⎝ U0 ⎠ SET UP: E = 5.5 eV, U = 10.0 eV, m = 9.11 × 10−31 kg, and T = 0.0010 2(9.11 × 10−31 kg)(4.5 eV)(1.60 × 10−19 J/eV) EXECUTE: κ = and G = 16 = 1.09 × 1010 m −1 5.5 eV ⎛ 5.5 eV ⎞ ⎜1 − ⎟ = 3.96, 10.0 eV ⎝ 10.0 eV ⎠ ⎛ 0.0010 ⎞ −10 ln ⎜ ⎟ = 3.8 × 10 m = 0.38 nm 2(1.09 × 10 m ) ⎝ 3.96 ⎠ EVALUATE: The energies here are comparable to those of electrons in atoms, and the barrier width we calculated is on the order of the diameter of an atom IDENTIFY and SET UP: When κ L is large, then eκ L is large and e−κ L is small When κ L is small, sinh κ L → κ L Consider both κ L large and κ L small limits so L = − 40.63 (1.054 × 10−34 J ⋅ s) −1 10 ⎡ (U sinh κ L) ⎤ EXECUTE: (a) T = ⎢1 + ⎥ E (U − E ) ⎥⎦ ⎢⎣ −1 sinh κ L = eκ L − e−κ L For κ L 1, sinh κ L → For κ L 1, 16 E (U − E ) + U 02e2κ L → U 02e2κ L T→ 16 E (U − E ) U 02e2κ L (b) κ L = ⎡ eκ L U 02e2κ L ⎤ and T → ⎢1 + ⎥ ⎢⎣ 16 E (U − E ) ⎥⎦ −1 = 16 E (U − E ) 16 E (U − E ) + U 02e2κ L ⎛ E ⎞⎛ E ⎞ −2κ L = 16 ⎜ , which is Eq (40.42) ⎟⎜1 − ⎟e U U 0⎠ ⎝ ⎠⎝ L 2m(U − E ) So κ L when L is large (barrier is wide) or U − E is large (E is small compared to U ) (c) κ = 2m(U − E ) ; κ becomes small as E approaches U For κ small, sinh κ L → κ L and ⎡ U 02κ L2 ⎤ T → ⎢1 + ⎥ ⎣⎢ E (U − E ) ⎦⎥ −1 ⎡ 2U 02 L2m ⎤ Thus T → ⎢1 + ⎥ E ⎦⎥ ⎣⎢ U → E so ⎡ U 2m(U − E ) L2 ⎤ = ⎢1 + ⎥ E (U − E ) ⎦⎥ ⎣⎢ −1 (using the definition of κ ) −1 ⎡ EL2 m ⎤ U 02 → E and T → ⎢1 + ⎥ E ⎦⎥ ⎣⎢ −1 −1 ⎡ ⎛ kL ⎞2 ⎤ But k = , so T → ⎢1 + ⎜ ⎟ ⎥ , as was to be shown ⎢⎣ ⎝ ⎠ ⎥⎦ EVALUATE: When κ L is large Eq (40.41) applies and T is small When E → U , T does not approach unity 40.64 2mE IDENTIFY: Compare the energy E of the oscillator to Eq (40.46) in order to determine n SET UP: At the equilibrium position the potential energy is zero and the kinetic energy equals the total energy © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-20 Chapter 40 EXECUTE: (a) E = mv = [n + (1/2)] ω = [n + (1/2)]hf , and solving for n, 2 mv (1/2)(0.020 kg)(0.360 m/s) − = − = 1.3 × 1030 n= hf (6.63 × 10−34 J ⋅ s)(1.50 Hz) (b) The difference between energies is ω = hf = (6.63 × 10−34 J ⋅ s)(1.50 Hz) = 9.95 × 10−34 J This energy 40.65 is too small to be detected with current technology EVALUATE: This oscillator can be described classically; quantum effects play no measurable role IDENTIFY and SET UP: Calculate the angular frequency ω of the pendulum and apply Eq (40.46) for the energy levels 2π 2π EXECUTE: ω = = = 4π s −1 T 0.500 s 1 The ground-state energy is E0 = ω = (1.055 × 10−34 J ⋅ s)(4π s −1) = 6.63 × 10−34 J 2 E0 = 6.63 × 10−34 J(1 eV/1.602 × 10−19 J) = 4.14 × 10−15 eV 1⎞ ⎛ En = ⎜ n + ⎟ ω 2⎠ ⎝ 1⎞ ⎛ En +1 = ⎜ n + + ⎟ ω 2⎠ ⎝ The energy difference between the adjacent energy levels is Δ E = En +1 − En = ω = E0 = 1.33 × 10−33 J = 8.30 × 10−15 eV 40.66 EVALUATE: These energies are much too small to detect Quantum effects are not important for ordinary size objects IDENTIFY: We model the electrons in the lattice as a particle in a box The energy of the photon is equal to the energy difference between the two energy states in the box n2h2 SET UP: The energy of an electron in the n th level is En = We not know the initial or final 8mL2 levels, but we know they differ by The energy of the photon, hc/λ , is equal to the energy difference between the two states hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) = = EXECUTE: The energy difference between the levels is Δ E = λ 1.649 × 10−7 m 1.206 × 10−18 J Using the formula for the energy levels in a box, this energy difference is equal to 40.67 h2 h2 Δ E = ⎡⎣ n − (n − 1) ⎤⎦ = (2 n − 1) 8mL2 8mL2 ⎞ ⎛ (1.206 × 10−18 J)8(9.11 × 10−31 kg)(0.500 × 10−9 m)2 ⎞ ⎛ Δ E8mL2 + 1⎟ = ⎜ + 1⎟ = Solving for n gives n = ⎜ − 34 ⎜ ⎟ ⎜ ⎟ 2⎝ h (6.626 × 10 J ⋅ s) ⎠ 2⎝ ⎠ The transition is from n = to n = EVALUATE: We know the transition is not from the n = to the n = state because we let n be the higher state and n − the lower state IDENTIFY: At a maximum, the derivative of the probability function is zero 2 mk ′ 2 SET UP and EXECUTE: ψ ( x) = Ce−α x , where α = ψ ( x) = C e−2α x At values of x where ψ ( x) 2 is a maximum, 2 d ψ ( x) d ψ ( x) d ψ ( x) < = and = C ( −2α x)e−2α x = Only dx dx dx © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics solution is x = dx 2 2 d ψ ( x) 2 = C ⎡⎢ −2α e−2α x + 4α x e−2α x ⎤⎥ At x = 0, = C ( −2α ) < 0, so ⎣ ⎦ dx 2 ψ ( x) 40.68 d ψ ( x) 40-21 is a maximum at x = EVALUATE: There is only one maximum, at x = 0, so the probability function peaks only there IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we substitute that wave function into the Schrödinger equation SET UP: The given wave function is ψ 1( x) = A1xe−α − d ψ ( x) 2m dx + 2 x /2 and the Schrödinger equation is k′x ψ ( x) = E ψ ( x) EXECUTE: (a) Start by taking the indicated derivatives: ψ 1( x) = A1xe−α dψ ( x ) = −α x A1e−α dx d 2ψ ( x ) dx d 2ψ 1( x) dx − 2 + A1e−α x /2 2 x /2 =− − A1α x (−α x )e−α 2 x /2 + A1 (−α x)e−α 2 x /2 ⎡ −3α + (α ) x ⎤ ψ 1( x) ⎦ 2m ⎣ d 2ψ ( x) 2m dx + k ′ x2 ψ ( x) = E ψ ( x) Substituting the above result into that equation 2 ⎡ −3α + (α ) x ⎤ ψ ( x) + k ′ x ψ ( x) = E ψ ( x) Since α = mω and ω = k ′ , the ⎣ ⎦ 2m m coefficient of x is − ⎛ mω ⎞ (b) A1 = ⎜ ⎟ ⎝ ⎠ 3/4 2m (α )2 + 2 k′ mω ⎛ mω ⎞ =− = ⎜ ⎟ + 2m ⎝ ⎠ 1/ ⎛4⎞ ⎜ ⎟ ⎝π ⎠ (c) The probability density function ψ is ψ ( x ) = A12 x 2e −α 2 d ψ 1( x) 2 d ψ ( x) d ψ ( x) d ψ ( x) = At x = ± , = α dx dx dx d ψ 1( x) x 2 2 2 2 d ψ 1( x) = A12 xe−α x + A12 x (−α 2 x)e−α x = A12 xe−α x − A12 x3α 2e−α x dx At x = 0, 2 2 At x = 0, ψ = dx Equation (40.44) is − dx x /2 = ⎡ −2α + (α ) x − α ⎤ ψ 1( x) = ⎡ −3α + (α ) x ⎤ ψ 1( x) ⎣ ⎦ ⎣ ⎦ dx gives − x /2 = − A1α 2 xe −α d 2ψ ( x) 2m 2 2 = A12 2e−α 2 + A12 x(−α 2 x)e−α = A12 2e−α 2 − A12 x 2α 2e−α x x 2 x 2 x − A12 2(3 x )α 2e−α − A12 x 2α 2e−α 2 x 2 x − A12 x3α (−α 2 x)e−α + A12 x (α ) e−α 2 x 2 x At x = 0, > So at x = 0, the first derivative is zero and the second derivative is positive Therefore, d ψ 1( x) < So at x = ± , the α α dx first derivative is zero and the second derivative is negative Therefore, the probability density function has maxima at x = ± , corresponding to the classical turning points for n = as found in the previous question α the probability density function has a minimum at x = At x = ± , © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-22 Chapter 40 EVALUATE: ψ ( x) = A1xe−α E= 40.69 2/ x is a solution to Eq (40.44) if − 2m (−3α )ψ 1( x) = E ψ ( x) or 2α ω ω E1 = = corresponds to n = in Equation (40.46) 2m 2 ⎛λ⎞ IDENTIFY: For a standing wave in the box, there must be a node at each wall and n ⎜ ⎟ = L ⎝2⎠ h h SET UP: p = so mv = λ λ EXECUTE: (a) For a standing wave, nλ = L, and En = p (h/λ ) n h = = 2m 2m 8mL2 (b) With L = a0 = 0.5292 × 10−10 m, E1 = 2.15 × 10−17 J = 134 eV EVALUATE: For a hydrogen atom, En is proportional to 1/n so this is a very poor model for a hydrogen 40.70 atom In particular, it gives very inaccurate values for the separations between energy levels IDENTIFY and SET UP: Follow the steps specified in the problem EXECUTE: (a) As with the particle in a box, ψ ( x ) = A sin kx, where A is a constant and k = 2mE/ Unlike the particle in a box, however, k and hence E not have simple forms (b) For x > L, the wave function must have the form of Eq (40.40) For the wave function to remain finite as x → ∞, C = The constant κ = 2m(U − E )/ , as in Eq (40.40) (c) At x = L, A sin kL = De −κ L and kA cos kL = −κ De −κ L Dividing the second of these by the first gives k cot kL = −κ , a transcendental equation that must be solved numerically for different values of the length L and the ratio E/U EVALUATE: When U → ∞, κ → ∞ and 40.71 cos(kL) nπ → ∞ The solutions become k = , n = 1, 2, 3,…, the sin(kL) L same as for a particle in a box IDENTIFY: Require ψ (− L/2) = ψ ( L/2) = p2 λ λ 2m EXECUTE: (a) ψ ( x) = A sin kx and ψ (− L /2) = = ψ (+ L /2) SET UP: k = 2π , p= h and E = 2nπ 2π ⎛ + kL ⎞ + kL ⇒ = A sin ⎜ = nπ ⇒ k = = ⎟⇒ λ L ⎝ ⎠ L h nh p n h (2n) h ⇒ pn = = ⇒ En = = = , where n = 1, 2… λ L n 2m 2mL2 8mL2 (b) ψ ( x) = A cos kx and ψ ( − L /2) = = ψ (+ L /2) ⇒λ = π (2n + 1)π 2π ⎛ kL ⎞ kL ⇒ = A cos ⎜ ⎟ ⇒ = (2n + 1) ⇒ k = = 2 L λ ⎝ ⎠ 2L (2n + 1)h ⇒λ = ⇒ pn = (2n + 1) 2L ⇒ En = (2n + 1) h n = 0, 1, 2… 8mL2 (c) The combination of all the energies in parts (a) and (b) is the same energy levels as given in n2h2 Eq (40.31), where En = 8mL2 EVALUATE: (d) Part (a)’s wave functions are odd, and part (b)’s are even © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Quantum Mechanics 40.72 IDENTIFY and SET UP: Follow the steps specified in the problem p2 h EXECUTE: (a) E = K + U ( x) = + U ( x) ⇒ p = 2m( E − U ( x)) λ = ⇒ λ ( x) = 2m p 40-23 h 2m( E − U ( x)) (b) As U ( x) gets larger (i.e., U ( x) approaches E from below—recall k ≥ 0), E − U ( x ) gets smaller, so λ ( x ) gets larger (c) When E = U ( x ), E − U ( x ) = 0, so λ ( x ) → ∞ b dx b n hn ⇒ ∫ 2m( E − U ( x)) dx = = ∫ 2m( E − U ( x)) dx = a 2 2m( E − U ( x )) h a (e) U ( x) = for < x < L with classical turning points at x = and x = L So, (d) b ∫a b dx b ∫a λ ( x) = ∫a h/ 2m(E − U (x)) dx = ∫ L L 2mEdx = 2mE ∫ dx = 2mE L So, from part (d), hn ⎛ hn ⎞ h2n2 ⇒E= ⎜ ⎟ = 2m ⎝ L ⎠ 8mL2 EVALUATE: (f) Since U ( x ) = in the region between the turning points at x = and x = L, the result is 2mE L = the same as part (e) The height U never enters the calculation WKB is best used with smoothly varying potentials U ( x) 40.73 Perform the calculations specified in the problem SET UP: U ( x) = 12 k ′x DENTIFY: EXECUTE: (a) At the turning points E = (b) nh ⎛ ⎞ 2m ⎜ E − k ′x ⎟dx = To evaluate the integral, we want to get it into a form that matches 2 ⎝ ⎠ + E /k ′ ∫− E /k ′ the standard integral given Letting A2 = ⇒ mk ′ ∫ b a 2E k ′xTP ⇒ xTP = ± k′ 2mE 2E ⎛ ⎞ 2m ⎜ E − k ′x ⎟ = 2mE − mk ′x = mk ′ − x = mk ′ − x2 ′ ′ mk k ⎝ ⎠ 2E 2E 2E ,a=− , and b = + k′ k′ k′ mk ′ A − x dx = 2 ⎡ 2E = mk ′ ⎢ ⎢⎣ k ′ b ⎡ ⎛ x ⎞⎤ 2 ⎢ x A − x + A arcsin ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ A ⎠ ⎥⎦ ⎛ 2E k ′ ⎞⎤ 2E 2E 2E 2E m⎛1⎞ arcsin ⎜ arcsin (1) = E − + ⎟⎟ ⎥ = mk ′ ⎜ ⎟ ⎜ ′ ′ ′ ′ k k k k k′ ⎝ ⎠ ⎝ E k ′ ⎠ ⎥⎦ Using WKB, this is equal to hn m hn k′ h , so E π = Recall ω = , so E = ω n = hω n k′ m 2π ω⎛ ⎞⎞ ⎛ ⎜ recall E = ω ⎜ n + ⎟ ⎟ It ⎝ ⎠⎠ ⎝ underestimates the energy However, our approximation isn’t bad at all! IDENTIFY and SET UP: Perform the calculations specified in the problem E EXECUTE: (a) At the turning points E = A xTP ⇒ xTP = ± A EVALUATE: (c) We are missing the zero-point-energy offset of 40.74 (b) + E /A ∫− E /A 2m( E − A x )dx = 2∫ dy = −2mA dx when x = E /A 2m( E − Ax) dx Let y = 2m( E − Ax) ⇒ E , y = 0, and when x = 0, y = 2mE So A © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 40-24 Chapter 40 E 2∫ A 2m( E − Ax)dx = − 0 3/2 hn (2mE )3/2 Using WKB, this is equal to y1/2dy = − y = ∫ mE 3mA 3mA mA 2 mE 2/3 hn ⎛ 3mAh ⎞ 2/3 ⇒E= (2mE )3/ = ⎜ ⎟ n 3mA 2m ⎝ ⎠ EVALUATE: (c) The difference in energy decreases between successive levels For example: So, 12/3 − 02/3 = 1, 22/3 − 12/3 = 0.59, 33/2 − 23 = 0.49,… • A sharp ∞ step gave ever-increasing level differences (~ n ) • A parabola (~ x ) gave evenly spaced levels (~ n) • Now, a linear potential (~ x) gives ever-decreasing level differences (~ n 2/3 ) Roughly speaking, if the curvature of the potential (~ second derivative) is bigger than that of a parabola, then the level differences will increase If the curvature is less than a parabola, the differences will decrease © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher  ... EXECUTE: An electron in the ground state of hydrogen has an energy of 13. 6 eV, so find the width corresponding to an energy of E1 = 13. 6 eV Solving for L gives L= 40.14 h (6.626 × 10−34 J ⋅ s) = =... −31 −19 2m(U − E ) 2(9.11 × 10 kg)(20 eV − 13 eV)(1.602 × 10 J/eV) (b) η = 40.61 1.055 × 10−34 J ⋅ s 2(1.67 × 10−27 kg)(30 MeV − 20 MeV)(1.602 × 10 13 J/MeV) = 1.44 × 10−15 m EVALUATE: The penetration... kg)(5.0 × 10 eV)(1.602 × 10 J/eV) h (6.626 × 10−34 J ⋅ s) = = 1.66 × 10−10 m 8mE1 8(9.11 × 10−31 kg) (13. 6 eV)(1.602 × 10−19 J/eV) EVALUATE: This width is of the same order of magnitude as the diameter