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Chapter 36 Applications of the Schrödinger Equation

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Chapter 36 Applications of the Schrödinger Equation CHAPTER 36 Applications of the Schrödinger Equation 1* ∙ True or false: Boundary conditions on the wave function lead to energy quantization. True 2 ∙ Sketch (a) the wave function and (b) the probability distribution for the n = 4 state for the finite square- well potential. (a) The wave function is shown below (b) The probability density is shown below 3 ∙ Sketch (a) the wave function and (b) the probability distribution for the n = 5 state for the finite square- well potential. (a) The wave function is shown below (b) The probability density is shown below Chapter 36 Applications of the Schrödinger Equation 4 ∙∙ Show that the expectation value <x> = ∫ xΨ 2 dx is zero for both the ground and the first excited states of the harmonic oscillator. ∞ 2 The integral ∫ xψ dx = 0 because the integrand is an odd function of x for the ground state as well as any −∞ excited state of the harmonic oscillator. 5* ∙∙ Use the procedure of Example 36-1 to verify that the energy of the first excited state of the harmonic _ ω ω 3 _ 0. (Note: Rather than solve for a again, use the result a = m 0/2h obtained in Example oscillator is E1 = 2 h 36-1.) 2 dψ 2 2 ψ −ax −ax − 2 −ax The wave function is = A1 x e (see Equ. 36-25). Then = A1 e 2 ax A1 e and dx 2ψ d − −ax2 − −ax2 2 3 −ax2 2 3 − −ax2 = 2 axA1 e 4 axA1 e + 4 a x A1 e = (4 a x 6ax) A1 e . We now substitute this into the dx2 _ 2 2 3 2 3 Schrödinger equation. The exponentials and the constant A1 cancel, so (h /2m)(4a x – 6ax) + 1/2mω0 x = E1x. _ 3 _ 2 _ With a = 1/2mω0/h, the terms in x cancel, and solving for the energy E1 we find E1 = 6h a/2m = 3hω0/2 = 3E0. 1/4 6 ∙∙∙ Show that the normalization constant A0 of Equation 36-23 is A0 = (2mω0/h) . ∞ ∞ ∞ 1/4 _ 1/4 ψ 2 _ 1/2 _ 2 π −1/ 2 −s2 = Let A0 = (2mω0/h) = (mω0/πh) . Then ∫ 0 (x) dx = (mω0/πh) ∫ exp[(–mω0/h)x ] dx = ∫ e ds 1 . −∞ −∞ −∞ 7 ∙∙∙ Find the normalization constant A1 for the wave function of the first excited state of the harmonic oscillator, Equation 36-25. ∞ 2 2 −2ax2 = _ 2 2 We require that A1 ∫ x e dx 1 where a = mω0/2h. Let y = 2ax ; then the integral becomes −∞ ∞ 1/4 3 3 3 2 2 π −3/ 2 2 − y 32 a 32 m ω 0π (2a) y e dy = . Consequently, 2 = and = . ∫ 3 / 2 A1 π A1 3 −∞ 2(2a) h 8 ∙∙∙ Find the expectation value <x2> = ∫ x2Ψ 2 dx for the ground state of the harmonic oscillator. Use it to show that the average potential energy equals half the total energy. ∞ ∞ 2 ψ 2 = 2 2 −2ax2 ∫ x dx A0 ∫ x e dx . The integral has already been evaluated in Problem 36-7. Using that result and −∞ −∞ Chapter 36 Applications of the Schrödinger Equation 2 _ 2 2 _ A0 given in Problem 36-6, 〈x 〉 = h/2mω0. The average potential energy of the oscillator is 1/2mω0 〈x 〉 = hω0/4 = E0/2. −ax2 9* ∙∙∙ Verify that Ψ1(x) = A1 xe is the wave function corresponding to the first excited state of a harmonic oscillator by substituting it into the time-independent Schrödinger equation and solving for a and E. From Problem 5 we know that the Schrödinger equation for Ψ1 gives – _ 2 2 3 2 3 (h /2m)(4a x – 6ax) + 1/2mω0 x = E1x. 3 _ If we now set the coefficients of x = 0 and solve for a we find that a = 1/2mω0/h, and using this expression and _ 2 _ solving for the energy E1 we find E1 = 6h a/2m = 3hω0/2 = 3E0. 10 ∙∙∙ Find the expectation value <x2> = ∫ x2 |ψ|2 dx for the first excited state of the harmonic oscillator. ∞ ∞ 2 π 2 ψ 2 = 2 4 −2ax 2 2 2 3 2 ∫∫x 1 (x) dx A1 x e dx . As in Problem 36-7, let 2ax = y . Then <x > = A1 5 . −∞ −∞ 8 32a 32a2 From Problem 36-7, A2 = so 〈x2〉 = 3a/8 = 3πmw /h. 1 π 0 11 ∙∙∙ Classically, the average kinetic energy of the harmonic oscillator equals the average potential energy. We may assume that this is also true for the quantum mechanical harmonic oscillator. Use this condition to determine the expectation value of p2 for the ground state of the harmonic oscillator. 2 2 2 2 2 2 According to the problem statement, 〈p 〉/2m = k〈x 〉/2, from which 〈p 〉 = m ω0 〈x 〉. From Problem 36-7, 2 _ 2 _ 〈x 〉 = h/2mω0. Thus, 〈p 〉 = hω0m/2. 12 ∙∙∙ We know that for the classical harmonic oscillator, pav = 0. It can be shown that for the quantum mechanical harmonic oscillator, <p> = 0. Use the results of Problems 4, 6, and 11 to determine the uncertainty product ∆x ∆p for the ground state of the harmonic oscillator. We associate (∆p)2 and (∆x)2 with the standard deviations as in the statement of Problem 36-41. Then, 2 2 2 2 2 2 2 2 (∆p) = [(p –2ppav – pav )]av and likewise for (∆x) . Since pav = xav = 0, (∆p) = 〈p 〉 and (∆x) = 〈x 〉. Now _ _ from Problems 36-8 and 36-11, (∆p)2(∆x)2 = h2/4 in the ground state of the harmonic oscillator, so ∆p∆x = h/2. 13* ∙∙ A free particle of mass m with wave number k1 is traveling to the right. At x = 0, the potential jumps from zero _ 2 2 to U0 and remains at this value for positive x. (a) If the total energy is E = h k1 /2m = 2U0, what is the wave number k2 in the region x > 0? Express your answer in terms of k1 and in terms of U0. (b) Calculate the reflection coefficient R at the potential step. (c) What is the transmission coefficient T? (d) If one million particles with wave Chapter 36 Applications of the Schrödinger Equation number k1 are incident upon the potential step, how many particles are expected to continue along in the positive x direction? How does this compare with the classical prediction? _ 2 2 _ 2 2 _ (a) We are given that E = h k1 /2m = 2U0. For x > 0, h k2 /2m + U0 = 2U0. So k2 = 2mU 0 /h, whereas _ k1 = 4mU 0 /h. k2 = k1/ 2 . − 2 ( k1 k 2 ) (b) The reflection coefficient is given by Equ. 36-27. So R = 2 = 0.0294. ( k1 + k 2 ) (c) T = 1 – R = 0.971. 5 6 (d) The number of particles that continue beyond the step is N0T = 9.71×10 ; classically, 1×10 would continue to move past the step. 14 ∙∙ Suppose that the potential jumps from zero to –U0 at x = 0 so that the free particle speeds up instead of slowing down. The wave number for the incident particle is again k1, and the total energy is 2U0. (a) What is the wave number for the particle in the region of positive x? (b) Calculate the reflection coefficient R at x = 0. (c) What is the transmission coefficient T? (d) If one million particles with wave number k1 are incident upon the potential step, how many particles are expected to continue along in the positive x direction? How does this compare with the classical prediction? _2 2 _2 2 (a) Proceed as in the preceding problem. Now E = h k2 /2m – U0 = 2U0 = h k1 /2m. Consequently, _2 1/2 k2 = (6mU0/h ) or k 2 = k1 3/2 . (b) Use Equ. 36-27 R = 0.0102 (c) Use Equ. 36-28 T = 0.99 (d) Number of particles transmitted = NT NT = 9.9×105; classically 106 particles are transmitted 15 ∙∙ Work Problem 13 for the case in which the energy of the incident particle is 1.01U0 instead of 2U0. (a) See Problem 36-13; write k1 and k2 _2 1/2 _2 1/2 k1 = (2.02mU0/h ) ; k2 = (0.02mU0/h ) ; k2 = 0.0995k1 2 ()2.02 − 0.02 (b) Use Equ. 36-27 R = 2 = 0.6710 ()2.02 + 0.02 (c) T = 1 – R T = 0.329 5 6 (d) Number of particles transmitted = NT NT = 3.29×10 ; classically 10 particles are transmitted 16 ∙∙ A particle of energy E approaches a step barrier of height U. What should be the ratio E/U so that the 1 reflection coefficient is 2 ? − k 2 2 1 1. Find r = k2/k1 using Equ. 36-27 r = = = 0.1716 k1 2 + 1 2 2 –1 2. Use the result of Problem 36-38 U/E = 1 – r ; E/U = (1 – r ) = 1.03 17* ∙∙ Use Equation 36-29 to calculate the order of magnitude of the probability that a proton will tunnel out of a nucleus in one collision with the nuclear barrier if it has energy 6 MeV below the top of the potential barrier and the barrier thickness is 10-15 m. Chapter 36 Applications of the Schrödinger Equation 2 − _ _ –13 1. Rewrite α of Equ. 36-29 in units of MeV α = 2/mc (U 0 E) hc; hc = 1.974×10 MeV.m –2αa 2 –15 1/2 –13 2. T = e ; mpc = 938 MeV T = exp[–2×10 (2×938×6) /1.974×10 ] = 0.341 18 ∙∙ A 10-eV electron is incident on a potential barrier of height 25 eV and width of 1 nm.
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