Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term
5th Edition
ISBN: 9781133422013
Author: Raymond A. Serway; John W. Jewett
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Question
Chapter 28, Problem 42P
To determine
The probability that the particle is located between
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
A quantum particle in an infinitely deep square well has a wave function given by ψ2(x) = √2/L sin (2πx/L)for 0 ≤ x ≤ L and zero otherwise. (a) Determine the expectation value of x. (b) Determine the probability of finding the particle near 1/2 L by calculating the probability that the particle lies in the range 0.490L ≤ x ≤ 0.510L. (c) What If? Determine the probability of finding the particle near 1/4L bycalculating the probability that the particle lies in the range 0.240L ≤ x ≤ 0.260L. (d) Argue that the result of part (a)does not contradict the results of parts (b) and (c).
The wave function for a quantum particle is
a
4(x)
π (x² + a²)
for a > 0 and -*
An electron is trapped in a region between two infinitely high energy barriers. In the region between the barriers the potential energy of the electron is zero. The normalized wave function of the electron in the region between the walls is ψ(x) = Asin(bx), where A=0.5nm1/2 and b=1.18nm-1. What is the probability to find the electron between x = 0.99nm and x = 1.01nm.
Chapter 28 Solutions
Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term
Ch. 28.1 - Prob. 28.1QQCh. 28.2 - Prob. 28.2QQCh. 28.2 - Prob. 28.3QQCh. 28.2 - Prob. 28.4QQCh. 28.5 - Prob. 28.5QQCh. 28.5 - Prob. 28.6QQCh. 28.6 - Prob. 28.7QQCh. 28.10 - Prob. 28.8QQCh. 28.10 - Prob. 28.9QQCh. 28.13 - Prob. 28.10QQ
Ch. 28 - Prob. 1OQCh. 28 - Prob. 2OQCh. 28 - Prob. 3OQCh. 28 - Prob. 4OQCh. 28 - Prob. 5OQCh. 28 - Prob. 6OQCh. 28 - Prob. 7OQCh. 28 - Prob. 8OQCh. 28 - Prob. 9OQCh. 28 - Prob. 10OQCh. 28 - Prob. 11OQCh. 28 - Prob. 12OQCh. 28 - Prob. 13OQCh. 28 - Prob. 14OQCh. 28 - Prob. 15OQCh. 28 - Prob. 16OQCh. 28 - Prob. 17OQCh. 28 - Prob. 18OQCh. 28 - Prob. 1CQCh. 28 - Prob. 2CQCh. 28 - Prob. 3CQCh. 28 - Prob. 4CQCh. 28 - Prob. 5CQCh. 28 - Prob. 6CQCh. 28 - Prob. 7CQCh. 28 - Prob. 8CQCh. 28 - Prob. 9CQCh. 28 - Prob. 10CQCh. 28 - Prob. 11CQCh. 28 - Prob. 12CQCh. 28 - Prob. 13CQCh. 28 - Prob. 14CQCh. 28 - Prob. 15CQCh. 28 - Prob. 16CQCh. 28 - Prob. 17CQCh. 28 - Prob. 18CQCh. 28 - Prob. 19CQCh. 28 - Prob. 20CQCh. 28 - Prob. 1PCh. 28 - Prob. 2PCh. 28 - Prob. 3PCh. 28 - Prob. 4PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 13PCh. 28 - Prob. 14PCh. 28 - Prob. 15PCh. 28 - Prob. 16PCh. 28 - Prob. 17PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Prob. 20PCh. 28 - Prob. 21PCh. 28 - Prob. 22PCh. 28 - Prob. 23PCh. 28 - Prob. 24PCh. 28 - Prob. 25PCh. 28 - Prob. 26PCh. 28 - Prob. 27PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 31PCh. 28 - Prob. 32PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - Prob. 37PCh. 28 - Prob. 38PCh. 28 - Prob. 39PCh. 28 - Prob. 40PCh. 28 - Prob. 41PCh. 28 - Prob. 42PCh. 28 - Prob. 43PCh. 28 - Prob. 44PCh. 28 - Prob. 45PCh. 28 - Prob. 46PCh. 28 - Prob. 47PCh. 28 - Prob. 48PCh. 28 - Prob. 49PCh. 28 - Prob. 50PCh. 28 - Prob. 51PCh. 28 - Prob. 52PCh. 28 - Prob. 53PCh. 28 - Prob. 54PCh. 28 - Prob. 55PCh. 28 - Prob. 56PCh. 28 - Prob. 57PCh. 28 - Prob. 58PCh. 28 - Prob. 59PCh. 28 - Prob. 60PCh. 28 - Prob. 61PCh. 28 - Prob. 62PCh. 28 - Prob. 63PCh. 28 - Prob. 64PCh. 28 - Prob. 65PCh. 28 - Prob. 66PCh. 28 - Prob. 67PCh. 28 - Prob. 68PCh. 28 - Prob. 69PCh. 28 - Prob. 70PCh. 28 - Prob. 71PCh. 28 - Prob. 72PCh. 28 - Prob. 73PCh. 28 - Prob. 74P
Knowledge Booster
Similar questions
- For a particle in a one-dimensional box, calculate the probability of the particle to exists between the length of 0.30L and 0.70L if n = 5.arrow_forwardWhat is the probability of the particle that in the box with a length of 2 nm is between x = 0.2 and x = 1.0 nm? Ѱ=√(2/L)*sin(nπx/L)arrow_forwardIf the particle in the box in the second excited state(i.e. n=3), what is the probability P that it is between x=L/2 and x=L/3 ?arrow_forward
- For a "particle in a box" of length, L, the wavelength for the nth level is given by An 2L %3D 2п and the wave function is n(x) = A sin (x) = A sin (x). The energy levels are пп %3D n?h? given by En : %3D 8mL2 lPn(x)|2 is the probability of finding the particle at position x in the box. Since the particle must be somewhere in the box, the integral of this function over the length of the box must be equal to 1. This is the normalization condition and ensuring that this is the case is called “normalizing" the wave function. Find the value of A the amplitude of the wave function, that normalizes it. Write the normalized wave function for the nth state of the particle in a box.arrow_forwardAn 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 two quantum numbers nx and ny and are given by E = h2/8me (nx2/Lx2 + ny2/Ly2)Using this information, we wish to find the wavelength of a photon needed to excite the electron from the ground state to the second excited state, assuming Lx = Ly = L. (a) Using the assumption on the lengths, write an expression for the allowed energies of the electron in terms of the quantumnumbers nx and ny. (b) What values of nx and ny correspond to the ground state? (c) Find the energy of the ground state. (d) What are the possible values of nx and ny for the first excited state, that is, the next-highest state in terms of energy? (e) What are the possible values of nx and ny for thesecond excited state?…arrow_forwardA particle of mass m is moving in an infinite 1D quantum well of width L. y,(x) = J? sinx. sin nAx L (a) How much energy must be given to the particle so it can transition from the ground state to the second excited state? (b) If the particle is in the first excited state, what is the probability of finding the particle between x = and x = ;? 2.arrow_forward
- (a) Find the normalization constant A for a wave function made up of the two lowest states of a quantum particle in a box extending from x= 0 to x = L: x) = A sin + 4 sin L. (b) A particle is described in the space -aSxs a by the wave function (x) = A cos + B sin 2a a Determine the relationship between the values of A and B required for normalization.arrow_forwardConsider a particle trapped in a 1D box with zero potential energy with walls at x = o and x = L. The general wavefunction solutions for this problem with quantum number, n, are: V,6) = sin ) 4n(x) = The corresponding energy (level) for each wavefunction solution is: n²h? En 8mL? a) What is the probability of finding the particle between x = L/4 and x = 3L/4 when the particle is in quantum state n = 1, 2 and 3. You can use calculator or a numerical program to do the integral. For people who want to try doing the integral by hand, the following identity will be helpful: sin²(x) = (1 – cos (2x))/2.arrow_forwardConsider an electron in a one-dimensional box of length L= 6 Å. The wavefunction for the particle is given as follows: Pn(x) = where n is the quantum number. Sketch the 2 and |Þ2|². Calculate the probability of finding electron in the first half of the box at n=2 level.arrow_forward
- An electron confined to a one-dimensional box of length is in its third energy state. What is the probability of finding the electron in the region x = 0 to x = {/4?|arrow_forwardA hypothetical one dimensional quantum particle has a normalised wave function given by (x) = ax - iß, where a and 3 are real constants and i = √-1. What is the most likely. x-position, II(x), for the particle to be found at? 0 11(x) == ○ II(2) = 0 ○ II(r) = 2/ O II(z) = 011(r) = ± √ +√ 13 aarrow_forwardA quantum system has a ground state with energy E0 = 0 meV and a 7-fold degenerate excited state with energy E1 = 78 meV. Calculate the probability of finding the system in its ground state when it is at T = 300 K.arrow_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 LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
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 with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning