* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chapter 6 Quantum Mechanics in One Dimension. Home
Aharonov–Bohm effect wikipedia , lookup
Probability amplitude wikipedia , lookup
Franck–Condon principle wikipedia , lookup
Wheeler's delayed choice experiment wikipedia , lookup
Copenhagen interpretation wikipedia , lookup
Scalar field theory wikipedia , lookup
Molecular Hamiltonian wikipedia , lookup
Quantum key distribution wikipedia , lookup
Interpretations of quantum mechanics wikipedia , lookup
X-ray fluorescence wikipedia , lookup
Quantum teleportation wikipedia , lookup
Path integral formulation wikipedia , lookup
Casimir effect wikipedia , lookup
Elementary particle wikipedia , lookup
Atomic theory wikipedia , lookup
EPR paradox wikipedia , lookup
Double-slit experiment wikipedia , lookup
Quantum electrodynamics wikipedia , lookup
Quantum state wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
History of quantum field theory wikipedia , lookup
Coherent states wikipedia , lookup
Hydrogen atom wikipedia , lookup
Hidden variable theory wikipedia , lookup
Bohr–Einstein debates wikipedia , lookup
Renormalization group wikipedia , lookup
Renormalization wikipedia , lookup
Electron scattering wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Canonical quantization wikipedia , lookup
Matter wave wikipedia , lookup
Wave–particle duality wikipedia , lookup
Particle in a box wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Chapter 6 Quantum Mechanics in One Dimension. Home Work Solutions 6.1 Problem 6.9 (In the text book) The nuclear potential that binds protons and neutrons in the nucleus of an atom is often approximated by a square well. Imagine a proton confined in an infinite square well of length 10−5 nm, a typical nuclear diameter. Calculate the wavelength and energy associated with the photon that is emitted when the proton undergoes a transition from the first excited state (n = 2) to the ground state (n = 1). In what region of the electromagnetic spectrum does this wavelength belong? Solution The energy levels of a particle in an infinite square well width L are given by: En = n2 h2 8mL2 n = 1, 2, 3, · · · where m is the mass the particle. The energy given off in the transition is: ∆E = E2 − E1 = 3h2 3(hc)2 = 8mp L2 8mp c2 L2 where mp is the mass of the proton, ∆E = 3 × (1.24 × 103 )2 = 6.145 M eV 8 × 938.3 × 106 × (10−5 )2 2 CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. HOME WORK SOLUTIONS this energy will be taken up by a photon, with a wave length λ given by: hc 1.24 × 10−3 = = 2.018 × 10−4 nm ∆E 6.145 This is the gamma-ray region of the electromagnetic spectrum λ= Physics 205:Modern Physics I, Chapter 6 Fall 2004 Ahmed H. Hussein 6.2. PROBLEM 6.12 (IN THE TEXT BOOK) 6.2 3 Problem 6.12 (In the text book) A ruby laser emits light of wavelength 694.3 nm. If this light is due to transitions from the n = 2 state to the n = 1 state of an electron in a box, find the width of the box. Solution In a box, an electron is allowed to have the following energies: n2 h2 8mL2 and the transition energy is related to the emitted photon wavelength by En = ∆E = h2 c2 hc (22 − 12 ) = 2 2 8mc L λ we can then find the length of the box from; r r 3hcλ 3 × 1.24 × 103 × 694.3 = = 0.795 nm L= 8mc2 8 × 511 × 103 Physics 205:Modern Physics I, Chapter 6 Fall 2004 Ahmed H. Hussein 4 CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. HOME WORK SOLUTIONS 6.3 Problem 6.21 (In the text book) Sketch the wavefunction ψ(x) and the probability density |ψ(x)|2 for the n = 4 state of a particle in a finite potential well. Solution Canceled Physics 205:Modern Physics I, Chapter 6 Fall 2004 Ahmed H. Hussein 6.4. PROBLEM 6.25 (IN THE TEXT BOOK) 6.4 5 Problem 6.25 (In the text book) Show that the oscillator energies in Equation 6.29 correspond to the classical amplitudes r (2n + 1)~ An = mω Solution The total energy of a classical particle oscillating under the effect of a restoring force is: 1 1 E = mv 2 + kx2 2 2 where k is the force constant, m is the mass of the particle, v is particle’s velocity, and x is the position of the particle relative to the equilibrium position. At the extreme points x = ±A the total energy becomes totally potential, i.e. 1 E = kA2 2 with k = mω 2 , then 1 E = mω 2 A2 2 If the energy is quantized as in the quantum oscillator, then 1 1 E = n+ ~ω = mω 2 A2 2 2 for a particular n the amplitude is given by: r An = Physics 205:Modern Physics I, Chapter 6 (2n + 1)~ mω Fall 2004 Ahmed H. Hussein 6 CHAPTER 6. QUANTUM MECHANICS IN ONE DIMENSION. HOME WORK SOLUTIONS 6.5 Problem 6.33 (In the text book) (a) What value do you expect for < px > for the quantum oscillator? Support your answer with a symmetry argument rather than a calculation. (b) Energy principles for the quantum oscillator can be used to relate < p2x > to < x2 >. Use this relation, along with the value of < x2 > from Problem 32, to find < p2x > for the oscillator ground state. (c) Evaluate ∆px , using the results of (a) and (b). Solution (a) In quantum oscillator, the re is no preference for the direction of the motion, so the particle will spend equal times moving to the left of the center and to the right. As a result the average value of the momentum < px > = 0. (b) We can find the < p2x > from the energy principles by writing < E >: <E>= p2x 2m 2 p + < U (x) > = x + < U (x) > 2m In a quantum oscillator, the energy is sharp and in the ground state the energy is: 1 < E >= E◦ = ~ω 2 Using U (x) = 12 mω 2 x2 and take the value of < x2 > = ~/2mω (see solution of problem 6.32 in Selected Solutions of chapter 6 on the web) we get: 1 1 < U (x) > = mω 2 < x2 > = ~ω 2 4 Now, < p2x > becomes: < p2x 1 1 1 > = 2m(< E > − < U (x) >) = 2m(E◦ − ~ω) = 2m( ~ω− ~ω) = 2m 4 2 4 ~ω 4 1 = m~ω 2 (c) ∆px is given by; r ∆px = p Physics 205:Modern Physics I, Chapter 6 < p2x > − < px >2 = Fall 2004 1 m~ω − 0 = 2 r m~ω 2 Ahmed H. Hussein