Q43P 聽A particle of mass聽m聽in a on... [FREE SOLUTION]

91影视

University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 6: Q43P (page 1356)

A particle of massmin a one-dimensional box has the following wave function in the regionx= 0 tox=L:
$蠄 (x, t) = \frac{1}{\sqrt{2}} 蠄_{1} (x) e^{- i E_{1} t / h} + \frac{1}{\sqrt{2}} 蠄_{3} (x) e^{- i E_{3} t / h}$

Here $蠄_{1} (x)$ and $蠄_{3} (x)$ are the normalized stationary-state wave functions for the
n = 1 and n = 3 levels, and E1 and E3 are the energies of these levels. The wave function is zero for x < 0 and for x > L. (a) Find the value of the probability distribution function at x = L/2 as a function of time. (b) Find the angular frequency at which the probability distribution function oscillates.

Short Answer

Expert verified

a) The value of the probability distribution function at x = L/2 is $| 蠄 (x, t) |^{2} = \frac{2}{L} [1 - \cos (\frac{4 蟺^{2} 鈩� t}{{mL}^{2}})]$

b) The angular frequency at which the probability distribution function oscillates is $\frac{4 蟺^{2} 鈩�}{{mL}^{2}}$

Step by step solution

(a) Determination of the value of the probability distribution function at x = L/2.

Given,

$蠄 (x, t) = \frac{1}{\sqrt{2}} 蠄_{1} (x) e^{- i 鈥� E_{1} t / 鈩�} + \frac{1}{\sqrt{2}} 蠄_{3} (x) e^{- i 鈥� E_{3} t / 鈩�}$

And,

$蠄 (x, t) = \frac{1}{\sqrt{2}} 蠄_{1} (x) e^{- i 鈥� E_{1} t / 鈩�} + \frac{1}{\sqrt{2}} 蠄_{3} (x) e^{- i 鈥� E_{3} t / 鈩�}$ ........(i)

${|蠄 (x, t)|}^{2} = \frac{1}{2} {蠄^{2}}_{1} + {蠄^{2}}_{3} + 蠄_{1} 蠄_{3} (e^{- i 鈥� (E_{3} - E_{1}) t / 鈩�} + e^{- i 鈥� (E_{3} - E_{1}) t / 鈩�})$

The expression of wave function of particle in a box is,

$蠄_{1} = \sqrt{\frac{2}{L}} \sin (\frac{蟺虫}{L}) 鈥� and 鈥� 蠄_{3} = \sqrt{\frac{2}{L}} \sin \frac{3 蟺虫}{L}$

And, the energy of a particle in a box is,

$E = \frac{n^{2} 蟺^{2} {鈩�}^{2}}{2 {mL}^{2}}$

Now, for level n = 3 and n = 1, the energy is,

role="math" localid="1664005923581" $E = \frac{9 蟺^{2} 鈩�}{2 {mL}^{2}} and E = \frac{蟺^{2} {h^{2}}^{}}{2 {mL}^{2}}$

So,

$E_{3} - E_{1} = \frac{4 蟺^{2} h^{2}}{2 {mL}^{2}}$

Substitute all the values in equation (i) and solve for the probability distribution,

$| 蠄 (x, t) |^{2} = \frac{2}{L} [1 - \cos (\frac{4 蟺^{2} 鈩� t}{{mL}^{2}})]$

(b) Determination of the angular frequency at which the probability distribution function oscillates.

The angular oscillation frequency is,

${^{(i)}}_{osc} = \frac{鈭� E}{h} = \frac{E_{3} - E_{1}}{h} = \frac{4 蟺^{2} h^{}}{{mL}^{2}}$

Thus, the oscillation frequency is $\frac{4 蟺^{2} h^{}}{{mL}^{2}}$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

(a) Determination of the value of the probability distribution function at x = L/2.

(b) Determination of the angular frequency at which the probability distribution function oscillates.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Mechanics and Materials

Oscillations

Electricity

Physical Quantities And Units

Fluids

Thermodynamics

Study anywhere. Anytime. Across all devices.