University Physics with Modern Physics (14th Edition)
14th Edition
ISBN: 9780321973610
Author: Hugh D. Young, Roger A. Freedman
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Question
Chapter 40, Problem 40.65CP
(a)
To determine
The classical turning points for a harmonic oscillator with energy
(b)
To determine
To show: That Energy levels for the harmonic oscillator using
(c)
To determine
How do approximate energy levels
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
A simple harmonic oscillator is in the ground state with A-², where A=1/41/4 jm-1/2 and a = 1/2 m2, What is a (x²) and Ax? (Hint: [ [_xe-x²³ dx=0 and [_x²a-b²³dx==√m/6²³ )
Ax-
μm
m2
um
An electron is confined to move in the xy plane in a rectangle whose dimensions are Lx and Ly. That is, the electron is trapped in a two dimensional potential well having lengths of Lx and Ly. In this situation, the allowed energies of the electron depend on the quant numbers Nx and Ny, the allowed energies are given by
E = H^2/8Me ( Nx^2/ Lx^2 + Ny^2/Ly^2)
i) assuming Lx and Ly =L. Find the energies of the lowest for all energy levels of the electron
ii) construct an energy level diagram for the electron and determine the energy difference between the second exited state and the ground state?
As a 1-dimensional problem, you have Schrodinger's equation, given by:
-h? a2
a
ih
h 4(x, t) =
at
2m Əx² ¥(x,t) + V(x) Þ(x,t)
Suppose for a specific V(x) and certain boundary conditions, the function w, (x, t) is a solution to the above
equation and 42 (x, t) is also a solution. Show that (x, t)
equation, where a, b are complex numbers.
a 41 (x, t) + b w2(x, t) also solves the above
Chapter 40 Solutions
University Physics with Modern Physics (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
- 4. a) Describe the Variational Principle = e-ax? b) The variation method is applied to a harmonic oscillator using a trial function = e Minimize the energy as a function of the variable a. k The solution for energy is E = 24 8a Leave your answer in terms of h, k, and u. Compare your answer to the true energy of a harmonic oscillator, E = v+) ha, where v = 0,1,2, ... and w =arrow_forwardConsider a Face Centered Cubic (FCC) lattice structured Nickel crystal. We are looking to find the surface energy of the new surface that is formed after it is sliced at the (100) plane. a- Find the value of R as function of the lattice constant a. 4R Oa = 2R Oa = 4R Oa = = 2/2R V2 Find the area A11 of (111) surface as function of R. 04R? O16R? O8R? OR? How many atoms lie on the plane (111) within the unit cell? N111 = atoms within the unit cell Find the number of atoms per unit surface area. 2 2 R2 8R2 16R? 4R? Which of the following represents the expression of the surface energt? ON BEPAarrow_forwardA non-isotropic harmonic oscillator has wi = W, W2 w, w2 = ;w, w3 = 2w. Its energy levels up to the fourth excited state level in terms of hw are Select one: О a. 1.75, 2.75, 3.75, 4.75, 5.75 Ob. Оь. 1.75, 2.5, 3.25, 4.0,4.5 с. 1.5,2.5, 3.5, 4.5,5.5 O d. 1.5, 1.75, 2.25, 3.5, 3.75 e. 1.75, 2.25, 2.75, 3.25, 3.75arrow_forward
- The position as a function of time x(t) of a simple harmonic oscillator is given by: x(t) = A cos(wt) %3D where A is the amplitude and w is the angular velocity. a) What is the range of possible values of x permitted for this oscillator? b) Derive the probability density function of p(x) for this oscillator. c) Validate that p(x) is normalized.arrow_forwardSuppose a system contain four identical particles and five energy levels given by the relationship, E;= i x 10-20 J, where i = 0 ,1,2,3,4. If the total energy of the system is Er= 6 €. Find the total number of the microscopic states for the distribution of these particles over the system energy levels keeping the given system conditions. Solution 4 identical particles Energy (10 Joule) Macroscopic state 4 Er= 6 € 3 Levels 1 E(10 ) k 1 2 3 4. 7 N! Wk = no!n;!n2!n3!n4! SK = kglnwkarrow_forwardElectron transfer between redox centers in proteins is controlled by quantum tunneling. We can model the region between two redox centers as an energy barrier which the electron must cross. If the distance between the redox centers is 0.752 nm and the energy of the electron is 0.976 eV lower than the height of the barrier, what is the probability that the electron will successfully cross to the next redox center?arrow_forward
- Atoms vibrate relative to one another in molecules with the bond acting as a spring. Consider the H – CI bond, where the heavy Cl atom forms a stationary anchor for the very light H atom. That is, only the H atom moves, vibrating as a simple harmonic oscillator. (a) Give the equation that describes the allowed vibrational energy levels of the bond. (b) The force constant kf for the H – Cl bond is 516.3 N m'1. Given the mass of H equal to 1.7 x 1027 kg, determine the difference in energy (separation) between adjacent energy levels. (c) Calculate the zero-point energy of this molecular oscillator.arrow_forwardShow that the following function Y(0,9)= sin 0 cos e eiº is the solution of Schrödinger 1 1 equation: sin 0 21 sin 0 00 Y(0,0)= EY (0,9) and find the sin 0 dp? energy, E.arrow_forwardSuppose a system contain four identical particles and five energy levels given by the relationship, E;= i × 10-2º J, where i = 0,1,2 ,3,4. If the total energy of the system is Er= 6 E. Find the total number of the microscopic states for the distribution of these particles over the system energy levels keeping the given system conditions. Solution 4 identical particles Energy (10- Joule) Macroscopic state 4 Er= 6 € 3 Levels 1 E2 E (10-º J) k 1 2 4 5 6 7 N! Wk no! n!n2!n3!n4! Sk = kglnwkarrow_forward
- Impurities in solids can be sometimes described by a particle-in-a-box model. Suppose He is substituted for Xe, and assume a particle-in-a-cubic-box model, the length of whose sides is equal to the atomic diameter of Xe (≈ 2.62 Å). Compute the lowest excitation energy for the He atom’s motion. (This is the energy difference between the ground state and the first excited state.)arrow_forwardSolid metals can be modeled as a set of uncoupled harmonic oscillators of the same frequency with energy levels given by En = ħwn n = 0, 1, 2,... where the zero-point energy (the lowest energy state) of each oscillator has been adjusted to zero for simplicity. In this model, the harmonic oscillators represent the motions of the metal atoms relative to one another. The frequency of these oscillators is low so that ħw = = 224 KB and the system vibrational partition function is given by 3N Z ² = la₁ - (1 1 e-0/T). (a) If the system contains one mole of atoms, find the average energy (in J) of this system at T= 172 K. (You can use = BkB.) T (b) What is the absolute entropy (in J/K) for this system? You can use either the Gibbs expression for S, or the system partition function to make this evaluation (they are equivalent, as your reading assignment indicates).arrow_forwardOne description of the potential energy of a diatomic molecule is given by the Lennard–Jones potential, U = (A)/(r12) - (B)/(r6)where A and B are constants and r is the separation distance between the atoms. For the H2 molecule, take A = 0.124 x 10-120 eV ⋅ m12 and B = 1.488 x 10-60 eV ⋅ m6. Find (a) the separation distance r0 at which the energy of the molecule is a minimum and (b) the energy E required to break up theH2 molecule.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning