Essential University Physics, 3e (Wolfson) Chapter 35 Quantum Mechanics 35.1 Conceptual Questions 1) The square of the wave function of a particle, |ψ(x)|2, gives the probability of finding the particle at the point x. A) True B) False Answer: B Var: 1 2) The smallest kinetic energy that an electron in a box (an infinite well) can have is zero. A) True B) False Answer: B Var: 1 3) The wave function for a particle must be normalizable because A) the particle's momentum must be conserved. B) the particle's angular momentum must be conserved. C) the particle's charge must be conserved. D) the particle must be somewhere. E) the particle cannot be in two places at the same time. Answer: D Var: 1 4) A set of five possible wave functions is given below, where L is a positive real number. ψ1(x) = Ae-x, for all x ψ2(x) = A cos x, for all x ψ3(x) = ψ4(x) = ψ5(x) = Which of the five possible wave functions are normalizable? (There may be more than one correct choice.) A) ψ1(x) B) ψ2(x) C) ψ3(x) D) ψ4(x) E) ψ5(x) Answer: C, D Var: 1 00 mm? A) 140 B) 1400 C) 450 D) 45 Answer: A Var: 1 2) The wave function for an electron that is confined to x ≥ 0 nm is ψ(x) = (a) What must be the value of b? (b) What is the probability of finding the electron in a 0.10 mm-wide strip centered at x = 0.17 Var: 1 .0023 Var: 1 3) The wave function for an electron that is confined to x ≥ 0 nm is ψ(x) = (a) What must be the value of A? (b) What is the probability of finding the electron in the interval 1.2 Problems 1) The probability density for an electron that has passed through an experimental apparatus is shown in the figure.15 nm ≤ x ≤ 1.010 nm-wide region centered at x = 1. then it is impossible for the atom to have zero kinetic energy. If 4100 electrons pass through the apparatus.0 nm? Answer: (a) 0.93 (nm)-1/2 (b) 0.5) If an atom in a crystal is acted upon by a restoring force that is directly proportional to the distance of the atom from its equilibrium position in the crystal. what is the expected number that will land in a 0. A) True B) False Answer: A Var: 1 35.56 (nm)-1/2 (b) 0.84 nm? Answer: (a) 0. 4) Find the value of A to normalize the wave function ψ(x) = . What . is given by ψ(x) = sin is the probability of finding the particle between x = 0 and x = L/3? A) 0. which is in the ground state. E) 1/ Answer: E Var: 1 5) Find the value of A to normalize the wave function ψ(x) = . The potential height of the walls of the box is infinite. with 0 ≤ x ≤ L.24 D) 0.28 Answer: A Var: 1 . A) B) C) D) E) Answer: D Var: 1 6) A particle is confined to a one-dimensional box (an infinite well) on the x-axis between x = 0 and x = L.22 C) 0.20 B) 0. A) 1/L B) C) 1/L2 D) 1.26 E) 0. The normalized wave function of the particle. What is the ground state energy of the electron? (h = 6.6 nm wide. mel = 9.626 × 10-34 J ∙ s. 1 eV = 1. with 0 ≤ x ≤ L.11 × 10-31 kg.60 × 10-19) A) 880 nm B) 750 nm C) 610 nm D) 1000 nm E) 1100 nm Answer: A Var: 50+ 10) An electron is in an infinite square well that is 2.60 × 10-19) A) 0.0076 eV E) 0. What is the smallest value of the state quantum number n for which the energy level exceeds 100 eV? (h = 6.9 nm wide. The electron makes a transition from the to the state.0057 eV C) 0.0 nm wide. The potential height of the walls of the box is infinite.60 × 10-19) A) 43 B) 44 C) 45 D) 42 E) 41 Answer: A Var: 50+ . is given by ψ(x) = sin . which is in the ground state.11 × 10-31 kg. 1 eV = 1.626 × 10-34 J ∙ s. what is the wavelength of the emitted photon?(h = 6. What is the maximum probability per unit length of finding the particle? A) 1/ B) C) 2/ D) 1/L E) 2/L Answer: E Var: 1 8) An electron is in an infinite square well (a box) that is 8.0066 eV D) 0.0048 eV B) 0. 1 eV = 1. The normalized wave function of the particle.11 × 10-31 kg.0085 eV Answer: A Var: 1 9) An electron is in an infinite square well (a box) that is 2. mel = 9. mel = 9.626 × 10-34 J ∙ s.7) A particle is confined to a one-dimensional box (an infinite well) on the x-axis between x = 0 and x = L. h = 6.60 × 10-19 J) A) 1.11 × 10-31 kg) (a) What is the width of this well? (b) What wavelength photon would be required to excite the electron from its original level to the next higher one? Answer: (a) 0.626 × 10-34J ∙ s. the normalized wave function of the electron is given by: ψ(x) = sin (2πx/L). mel = 9. 1 eV = 1.10 eV B) 0.00 × 108 m/s.026 eV Answer: A Var: 50+ 12) An electron is in the ground state of an infinite well (a box) where its energy is 5.225 nm (b) 23.0 eV D) 15. mel = 9.11) An electron is bound in an infinite square-well potential (a box) on the x-axis. mel = 9. If an identical particle is confined to a similar region with fixed length L/6.0 eV E) 20. What is the energy of the electron in this state?(h = 6.25 eV B) 2.626 × 10-34 J ∙ s.9 nm.10 nm.9 nm Var: 1 . or infinite well) with fixed length L is E0.60 × 10-19) A) 0. find the wavelength of the emitted photon. Answer: 36 Var: 9 14) An electron is bound in an infinite well (a box) of width 0.11 × 10-31 kg.0 eV Answer: D Var: 1 13) The lowest energy level of a particle confined to a one-dimensional region of space (a box. In the next higher level.00 eV.052 eV C) 0.50 eV C) 10.2 nm Var: 1 15) An electron in an infinite potential well (a box) makes a transition from the n = 3 level to the ground state and in so doing emits a photon of wavelength 20.00 nm to In its present state. (c = 3.13 eV D) 0.078 eV E) 0. h = 6.11 × 10-31 kg) Answer: 2.626 × 10-34 J ∙ s. The width of the well is L and the well extends from x = 0.00 × 108 m/s. If the electron is initially in the n = 8 state and falls to the n = 7 state. (c = 3. what would its energy be? (1 eV = 1. what is the energy of the lowest energy level that the particles have in common? Express your answer in terms of E0. 0 × 10-18 J.16) You want to confine an electron in a box (an infinite well) so that its ground state energy is 5. mel = 9.5 µm B) 11 µm C) 2.3 × D) zero Answer: A J Var: 50+ 18) You want to have an electron in an energy level where its speed is no more than 66 m/s.11 × 10-31 kg) A) 1.068 × 10-18 J and 1.18 nm Answer: A Var: 50+ 17) A 10.4 µm Answer: A Var: 1 19) An electron is confined in a one-dimensional box (an infinite well). mel = 9.1 nm D) 2.3-cm-long box (an infinite well). If we model the ball as a point particle.626 × 10-34 J ∙ s.93 nm C) 1. What is the length of the smallest box (an infinite well) in which you can do this? (h = 6.626 × 10-34 J ∙ s) A) 8. Two adjacent allowed energies of the electron are 1.2 × J C) 1.0-g bouncy ball is confined in a 8.0 × J B) 3. What is the length of the box? (h = 6.3 nm Answer: A Var: 1 .11 × 10-31 kg) A) 5.626 × 10-34 J ∙ s.22 nm C) 0. what is the minimum kinetic energy of the ball? (h = 6.11 × 10-31 kg) A) 0. mel = 9.9 nm B) 0.626 × 10-34 J ∙ s.15 nm D) 0. What should be the length of the box? (h = 6.8 µm D) 1.11 nm B) 0.352 × 10-18 J. 626 × 10-34J ∙ s.07 μm Answer: A Var: 50+ 21) One fairly crude method of determining the size of a molecule is to treat the molecule as an infinite square well (a box) with an electron trapped inside.626 × 10-34J ∙ s. mel = 9.11 × 10-31 kg) A) 6.00 × 108 m/s.45 nm Answer: A Var: 1 22) The wave function of an electron in a rigid box (infinite well) is shown in the figure.12 nm C) 1.11 × 10-31 kg) A) 6.91 μm D) 7.00 × 108 m/s. (c = 3. If the electron energy 98.5 eV Answer: A Var: 50+ .49 μm B) 5.33 nm B) 1.9 eV D) 24.11 × 10-31 kg) A) 1.92 eV C) 10.45 μm C) 5. h = 6. what is the energy of the electron's ground state?( mel = 9.0 eV. what is the width of the molecule? (c = 3. mel = 9. If the photon emitted during the n = 2 to n = 1 transition has wavelength 1940 nm. h = 6.20) An electron is trapped in an infinite square well (a box) of width Find the wavelength of photons emitted when the electron drops from the n = 5 state to the n = 1 state in this system.21 nm D) 1.13 eV B) 3. and to measure the wavelengths of emitted photons. 019 eV C) 0.0 eV Answer: C Var: 1 .3 × 1013 rad/s. h = 1.500 MeV D) 4.50 MeV Answer: A Var: 50+ 24) A lithium atom.015 eV B) 0.75 × 10-26 kg.027 eV E) 0.23) A particle confined in a rigid one-dimensional box (an infinite well) of length 17. vibrates with simple harmonic motion in a crystal lattice. mass 1.023 eV D) 0. The mass of a nickel atom is 9.055 × 10-34 J ∙ s.5 MeV. What is the value of the ground state energy? (1 eV = 1. h = 6.0 eV D) 20.14 × 10-2 eV (b) 2.60 × 10 J) (a) What is the ground state energy of this system.60 × 10-19 J) A) 1.5 MeV C) 0.50 MeV B) 13. 1 eV = 1.00 eV. What is the difference in energy between adjacent vibrational energy levels of nickel? (h = 6. in eV? (b) What is the wavelength of the photon that could excite this system from the ground state to the first excited state? Answer: (a) 2.91 × 10-5 m Var: 50+ 25) The atoms in a nickel crystal vibrate as harmonic oscillators with an angular frequency of 2.626 × 10 J ∙ s. where the effective force constant of the forces on the atom is k = (c = 3.00 × 8 -34 -34 -19 10 m/s. What is the energy of the next higher level? A) 7.626 × 10-34 J ∙ s.50 eV B) 10.60 × 10-19 J) A) 0.17 × 10-26 kg. 1 eV = 1.055 × 10 J ∙ s.0 eV E) 50.0 fm has an energy level and an adjacent energy level En+1 = 37.0 eV C) 15. h = 1.031 eV Answer: A Var: 1 26) The lowest energy level of a certain quantum harmonic oscillator is 5. 055 × 10-34 J ∙ s.27) Calculate the ground state energy of a harmonic oscillator with a classical frequency of 3.95 × 1015 rad/s Answer: A Var: 50+ 29) Find the wavelength of the photon emitted during the transition from the second EXCITED state to the ground state in a harmonic oscillator with a classical frequency of (c = 3.626 × 10-34 J ∙ s) A) 4. What is the wavelength of the emission if the net force on the electron behaves as though it has a spring constant of (mel = 9.626 × 10-34 J ∙ s) A) 290 nm B) 150 nm C) 190 nm D) 580 nm Answer: A Var: 50+ . h = 6. h = 1. h = 1. 1 eV = 1.00 × 108 m/s.11 × 10-31 kg.60 × 10-19 J. h = 1.31 × 1015 rad/s B) 2. A photon is emitted when the electron undergoes a 3→1 quantum jump.26 μm C) 2.626 × 10-34 J ∙ s) A) 7.03 μm B) 2.62 eV B) 15.08 × 1016 rad/s C) 4.71 eV Answer: A Var: 1 28) The energy of a particle in the second EXCITED state of a harmonic oscillator potential is What is the classical angular frequency of oscillation of this particle? (1 eV = 1. h = 6. (1 eV = 1.24 μm Answer: A Var: 50+ 30) An electron is confined in a harmonic oscillator potential well. c = 3.00 × 108 m/s.60 × 10-19 J.60 × 10-19 J.055 × 10-34 J ∙ s. 1 eV = 1. h = 1.97 × 1015 rad/s D) 6.2 eV C) 11.98 μm D) 5.68 × 1015 Hz.055 × 10-34 J ∙ s.626 × 10-34 J ∙ s) A) 3.4 eV D) 5. h = 6. h = 6.055 × 10-34 J ∙ s.60 × 1019 J. 11 × 10-31 kg.00 × 108 m/s. What is the longest wavelength of light that the electron can absorb if the net force on the electron behaves as though it has a spring constant of 74 N/m? (mel = 9.626 × 10-34 J ∙ s) A) 210 nm B) 200 nm C) 220 nm D) 230 nm Answer: A Var: 1 .31) An electron is confined in a harmonic oscillator potential well.60 × 10-19 J. c = 3. 1 eV = 1. h = 1. h = 6.055 × 10-34 J ∙ s.