Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
bartleby

Concept explainers

Question
Book Icon
Chapter 28, Problem 50P

(a)

To determine

The quantum number of the marble.

(a)

Expert Solution
Check Mark

Answer to Problem 50P

The quantum number of the marble is 6×1028.

Explanation of Solution

The mass of the marble is 10 g, its speed is 2cms1 and the length of the box is 10 cm.

Write the expression for momentum of a particle in a box.

pn=nh2L                                                            (I)

Here, pn is the momentum of the marble, n is the quantum number, h is the Planck’s constant and L is the length of the box.

Rearranging (I) for n

n=2pnLh                                                            (II)

Write the classical expression for momentum.

pn=mv                                                           (III)

Here, m is the mass of the marble and v is its speed.

Substituting (III) in (II)

n=2mvLh                                                            (IV)

Substituting 10 g for m, 2cms1 for v, 6.626×1034 Js for h and 10 cm for L in (IV) to find n

n=2×10 g×2cms1×6.626×1034 kgm2s110 cm=2×10×103 kg×2cms1×6.626×1034 kgm2s110 cm=6×1028

Thus, the quantum number of the marble is 6×1028.

(b)

To determine

The reason why the quantization of the marble isn’t observable.

(b)

Expert Solution
Check Mark

Answer to Problem 50P

Because of the small energy difference between two quantized levels of the marble, the quantization isn’t observable.

Explanation of Solution

Write the expression for energy of the quantized level.

En=n2h28mL2                                                            (V)

Here, En is the energy of the level n

Substituting n+1 for n in (I)

En+1=(n+1)2h28mL2                                                            (VI)

Here, En+1 is the energy of the level n+1

Subtracting (I) from (II)

ΔE=(n+1)2h28mL2n2h28mL2ΔE=h28mL2[(n+1)2n2]

Here, ΔE is the energy difference.

Thus, the energy difference is

ΔE=h28mL2(2n+1)                                                           (VII)

Substituting 10 g for m, 2cms1 for v, 6.626×1034 Js for h, 10 cm for L, 6×1028 for n in (VII) to find ΔE

 ΔE=(2(6×1028)+1)×(6.626×1034 Js)28×10 g×(10 cm)2=(12×1028+1)×(6.626×1034 Js)28×10×103 g×(10×102 cm)2=7×1935 J

The energy difference is 7×1935 J which is very small. Thus, they cannot be observed.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 28 Solutions

Physics

Ch. 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. 1MCQCh. 28 - Prob. 2MCQCh. 28 - Prob. 3MCQCh. 28 - Prob. 4MCQCh. 28 - Prob. 5MCQCh. 28 - Prob. 6MCQCh. 28 - Prob. 7MCQCh. 28 - Prob. 8MCQCh. 28 - Prob. 9MCQCh. 28 - Prob. 10MCQCh. 28 - Prob. 1PCh. 28 - Prob. 2PCh. 28 - Prob. 3PCh. 28 - Prob. 4PCh. 28 - Prob. 5PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Prob. 13PCh. 28 - Prob. 15PCh. 28 - Prob. 14PCh. 28 - Prob. 17PCh. 28 - Prob. 16PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Prob. 20PCh. 28 - Prob. 21PCh. 28 - Prob. 23PCh. 28 - Prob. 22PCh. 28 - Prob. 25PCh. 28 - Prob. 24PCh. 28 - Prob. 26PCh. 28 - Prob. 27PCh. 28 - Prob. 28PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 32PCh. 28 - Prob. 31PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - Prob. 37PCh. 28 - Prob. 39PCh. 28 - Prob. 41PCh. 28 - Prob. 40PCh. 28 - Prob. 38PCh. 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. 65PCh. 28 - Prob. 64PCh. 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. 74PCh. 28 - Prob. 75PCh. 28 - Prob. 76PCh. 28 - Prob. 77PCh. 28 - Prob. 79PCh. 28 - Prob. 78PCh. 28 - Prob. 80PCh. 28 - Prob. 81PCh. 28 - Prob. 82PCh. 28 - Prob. 83PCh. 28 - Prob. 84P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON