University Physics with Modern Physics, Volume 2 (Chs. 21-37); Mastering Physics with Pearson eText -- ValuePack Access Card (14th Edition)
14th Edition
ISBN: 9780134265414
Author: Hugh D. Young, Roger A. Freedman
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Question
Chapter 40, Problem 40.6DQ
To determine
To explain: Why the values
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Could the function ψ(x)=Asin(π*x/L) for 0≤x≤L and ψ(x)=0 elsewhere (where A and L are real and positive constants) possibly be a valid quanton wavefunction under the right circumstances? Why or why not? (Hint: Sketching a graph of this wavefunction and its absolute square might prove helpful.)
Calculate the probability that an electron will be found (a) between x = 0.1 and 0.2 nm, (b) between 4.9 and 5.2 nm in a box of length L = 10 nm when its wavefunction is Ψ = (2/L)1/2 sin(2πx/L). Treat the wavefunction as a constant in the small region of interest and interpret δV as δx.
An electron in a one-dimensional region of length L is described by the wavefunction ψn(x) = sin(nπx/L), where n = 1, 2, …, in the range x = 0 to x = L; outside this range the wavefunction is zero. The orthogonality of these wavefunctions is confirmed by considering the integralI= ∫0L sin(nπx/L)sin(mπx/L)dx(a) Use the identity sinAsinB = 1/2{cos(A-B)-cos(A+B)} to rewrite the integrand as a sum of two terms. (b) Consider the case n = 2, m = 1, and make separate sketch graphs of the two terms identified in (a) in the range x = 0 to x = L. (c) Make use of the properties of the cosine function to argue that the area enclosed between the curves and the x axis is zero in both cases, and hence that the integral is zero. (d) Generalize the argument for the case of arbitrary n and m (n ≠ m).
Chapter 40 Solutions
University Physics with Modern Physics, Volume 2 (Chs. 21-37); Mastering Physics with Pearson eText -- ValuePack Access Card (14th Edition)
Ch. 40.1 - Does a wave packet given by Eq. (40.19) represent...Ch. 40.2 - Prob. 40.2TYUCh. 40.3 - Prob. 40.3TYUCh. 40.4 - Prob. 40.4TYUCh. 40.5 - Prob. 40.5TYUCh. 40.6 - Prob. 40.6TYUCh. 40 - Prob. 40.1DQCh. 40 - Prob. 40.2DQCh. 40 - Prob. 40.3DQCh. 40 - Prob. 40.4DQ
Ch. 40 - If a panicle is in a stationary state, does that...Ch. 40 - Prob. 40.6DQCh. 40 - Prob. 40.7DQCh. 40 - Prob. 40.8DQCh. 40 - Prob. 40.9DQCh. 40 - Prob. 40.10DQCh. 40 - Prob. 40.11DQCh. 40 - Prob. 40.12DQCh. 40 - Prob. 40.13DQCh. 40 - Prob. 40.14DQCh. 40 - Prob. 40.15DQCh. 40 - Prob. 40.16DQCh. 40 - Prob. 40.17DQCh. 40 - Prob. 40.18DQCh. 40 - Prob. 40.19DQCh. 40 - Prob. 40.20DQCh. 40 - Prob. 40.21DQCh. 40 - Prob. 40.22DQCh. 40 - Prob. 40.23DQCh. 40 - Prob. 40.24DQCh. 40 - Prob. 40.25DQCh. 40 - Prob. 40.26DQCh. 40 - Prob. 40.27DQCh. 40 - Prob. 40.1ECh. 40 - Prob. 40.2ECh. 40 - Prob. 40.3ECh. 40 - Prob. 40.4ECh. 40 - Prob. 40.5ECh. 40 - Prob. 40.6ECh. 40 - Prob. 40.7ECh. 40 - Prob. 40.8ECh. 40 - Prob. 40.9ECh. 40 - Prob. 40.10ECh. 40 - Prob. 40.11ECh. 40 - Prob. 40.12ECh. 40 - Prob. 40.13ECh. 40 - Prob. 40.14ECh. 40 - Prob. 40.15ECh. 40 - Prob. 40.16ECh. 40 - Prob. 40.17ECh. 40 - Prob. 40.18ECh. 40 - Prob. 40.19ECh. 40 - Prob. 40.20ECh. 40 - Prob. 40.21ECh. 40 - Prob. 40.22ECh. 40 - Prob. 40.23ECh. 40 - Prob. 40.24ECh. 40 - Prob. 40.25ECh. 40 - Prob. 40.26ECh. 40 - Prob. 40.27ECh. 40 - Prob. 40.28ECh. 40 - Prob. 40.29ECh. 40 - Prob. 40.30ECh. 40 - Prob. 40.31ECh. 40 - Prob. 40.32ECh. 40 - Prob. 40.33ECh. 40 - Prob. 40.34ECh. 40 - Prob. 40.35ECh. 40 - Prob. 40.36ECh. 40 - Prob. 40.37ECh. 40 - Prob. 40.38ECh. 40 - Prob. 40.39ECh. 40 - Prob. 40.40ECh. 40 - Prob. 40.41ECh. 40 - Prob. 40.42PCh. 40 - Prob. 40.43PCh. 40 - Prob. 40.44PCh. 40 - Prob. 40.45PCh. 40 - Prob. 40.46PCh. 40 - Prob. 40.47PCh. 40 - Prob. 40.48PCh. 40 - Prob. 40.49PCh. 40 - Prob. 40.50PCh. 40 - Prob. 40.51PCh. 40 - Prob. 40.52PCh. 40 - Prob. 40.53PCh. 40 - Prob. 40.54PCh. 40 - Prob. 40.55PCh. 40 - Prob. 40.56PCh. 40 - Prob. 40.57PCh. 40 - Prob. 40.58PCh. 40 - Prob. 40.59PCh. 40 - Prob. 40.60PCh. 40 - Prob. 40.61PCh. 40 - Prob. 40.62PCh. 40 - Prob. 40.63PCh. 40 - Prob. 40.64CPCh. 40 - Prob. 40.65CPCh. 40 - Prob. 40.66CPCh. 40 - Prob. 40.67PPCh. 40 - Prob. 40.68PPCh. 40 - Prob. 40.69PPCh. 40 - Prob. 40.70PP
Knowledge Booster
Similar questions
- Can we simultaneously measure position and energy of a quantum oscillator? Why? Why not?arrow_forward(a) Let n = a + ib be a complex number, where a and b are real (positive or negative) numbers. Show that the product nn* is always a positive real number. (b) Let m = c + id be another complex number. Show that |nm| = |n| |m|.arrow_forwardConsider a macroscopic object of mass 90 grams confined to move between two rigid walls separated by 2 m. What is the minimum speed of the object? What should the quantum number n be if the object is moving with a speed 1 ms-1? What is the separation of the energy levels of the object moving with that speed?arrow_forward
- Harmonic oscillator eigenstates have the general form 1 μω ,1/4 μω AG)(√(-) n ħ In this formula, which part determines the number of nodes in the harmonic oscillator state? = y (x) 1 √(™ ћn 2"n! Holev 1/4 μω 1 2"n! exp(-1022²) 2ħ μω ħ 2"n! exp μω χ 2ħ 2arrow_forwardWhen an electron trapped in a one-dimensional box transitions from its n= 2 state to its n= 1 state, a photon with a wavelength of 9 nm is emitted. What is the length of the box (in nm)? What If? If electrons in the box also occupied the n= 3 state, what other wavelengths of light (in nm) could possibly be emitted? Enter the shorter wavelength first.arrow_forwardA particle of mass 1.60 x 10-28 kg is confined to a one-dimensional box of length 1.90 x 10-10 m. For n = 1, answer the following. (a) What is the wavelength (in m) of the wave function for the particle? m (b) What is its ground-state energy (in eV)? eV (c) What If? Suppose there is a second box. What would be the length L (in m) for this box if the energy for a particle in the n = 5 state of this box is the same as the ground-state energy found for the first box in part (b)? m (d) What would be the wavelength (in m) of the wave function for the particle in that case? marrow_forward
- The wavefunction for v =1 for a simple harmonic oscillator is Ψ = (2)1/2 ( α3/π)1/4 x exp (-αx2/2) Find the values of x such that ψ* ψ is a maximum.Hint: Differentiate dψ*ψ/dx and set the result equal to zero and solve for the value of x.arrow_forward. An electron in a long organic molecule used in a dye laser behave approximately like a particle in a box with width 4.18 nm. (a) What is the λ of the proton emitted when the electron undergoes a transition from the first excited level to the ground level. (b) What is the λ of the proton emitted when electron undergoes a transition from the second excited level to the first excited level?arrow_forwardThe treatment of electrons in atoms must be a quantum treatment, but classical physics still works for baseballs. Where is the dividing line? Suppose we consider a spherical virus, with a diameter of 30 nm, constrained to exist in a long, narrow cell of length 1.0 μm. If we treat the virus as a particle in a box, what is the lowest energy level? Is a quantum treatment necessary for the motion of the virus?arrow_forward
- Going from -(h_bar^2/2m) (d^2/dx^2) ψ to the momentum operator squared (1/2m) p_hat^2, how is the negative sign lost? I must be missing something fundamental since it looks to me like momentum operator ->. p_hat^2 = (-ih_bar d/dx)(-ih_bar d/dx)=+i^2 h_bar^2 (d^2/dx^2)= -h_bar^2 (d^2/dx^2) ? Thank you!arrow_forwardAn electron is trapped in an infinitely deep one- dimensional well of width 0.285 nm. Initially, the electron occupies the n = 4 state. (a) Suppose the electron jumps to the ground state with the accompanying emission of a photon. What is the energy of the photon? (b) Find the energies of other photons that might be emitted if the electron takes other paths between the n = 4 state and the ground state.arrow_forwardA particle is confined in a box of length L as shown in the figure. If the potential is treated as a perturbation, including the first order correction, the ground state energy is (a) E = ħ²π² 2mL² + V (b) E = ħ²π² Vo 2mL² ħ²π² Vo ħ²π² Vo (c) E = + (d) E = + 2mL² 4 2mL² L/2arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax