Explanation 1) Concept: We can equate the centripetal force and Coulomb force between the muon and the H e + nucleus to find the equation for square of velocity. By using quantization condition, we can find the radius. By substituting these values in the equation for total energy, we can prove the given equation. 2) Formula: i) F = - m v 2 r ii) F = - k Z e 2 r 2 iii) L = n ℏ = m v r iv) K = 1 2 m v 2 v) U = - k Z e 2 r 3) Given: i) Charge of muon is - e ii) Mass of muon is m = 207 m e = 207 × 9.11 × 10 - 31 k g 4) Calculations: Suppose the muon of mass m is orbiting in circular uniform motion around the H e + nucleus with velocity v .It, thus, experiences a centripetal force given by F = - m v 2 r This force must be balanced by the Coulomb force between the muon and the H e + nucleus, which is given as F = - k Z e 2 r 2 Where, k = 1 / 4 π ε 0 Therefore, - m v 2 r = - k Z e 2 r 2 v 2 = k Z e 2 m r … … … … … … … … … … … … … … … … … … … … … … ( 1 ) Since the magnitude of the angular momentum is quantized as L = n ℏ where ℏ = h / 2 π And as L = m v r , we get, m v r = n ℏ Rearranging this equation for v , v = n ℏ m r Substituting this in equation ( 1 ) , n ℏ m r 2 = k Z e 2 m r On simplifying, r = n 2 ℏ 2 k Z e 2 m … … … … … … … … … … … … … … … … … … … … … … ( 2 ) Now, the kinetic energy of muon is K = 1 2 m v 2

11th Edition

ISBN: 9781119573913

Author: Halliday

Publisher: WILEY

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Chapter 39, Problem 68P

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The energy levels of the system are given by

E=-11.3 keVn2

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Chapter 39, Problem 67P

Chapter 39, Problem 69P

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Chapter 39 Solutions

FUNDAMENTALS OF PHYSICS V.1

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Ch. 39 - Prob. 11Q Ch. 39 - Prob. 12Q Ch. 39 - Prob. 13Q Ch. 39 - Prob. 14Q Ch. 39 - Prob. 15Q Ch. 39 - Prob. 1P Ch. 39 - Prob. 2P Ch. 39 - Prob. 3P Ch. 39 - Prob. 4P Ch. 39 - Prob. 5P Ch. 39 - Prob. 6P Ch. 39 - Prob. 7P Ch. 39 - Prob. 8P Ch. 39 - Prob. 9P Ch. 39 - Prob. 10P Ch. 39 - Prob. 11P Ch. 39 - Prob. 12P Ch. 39 - Prob. 13P Ch. 39 - Prob. 14P Ch. 39 - Prob. 15P Ch. 39 - Prob. 16P Ch. 39 - Prob. 17P Ch. 39 - Prob. 18P Ch. 39 - Prob. 19P Ch. 39 - Prob. 20P Ch. 39 - Prob. 21P Ch. 39 - Prob. 22P Ch. 39 - Prob. 23P Ch. 39 - Prob. 24P Ch. 39 - Prob. 25P Ch. 39 - Prob. 26P Ch. 39 - Prob. 27P Ch. 39 - Prob. 28P Ch. 39 - Prob. 29P Ch. 39 - Prob. 30P Ch. 39 - Prob. 31P Ch. 39 - Prob. 32P Ch. 39 - Prob. 33P Ch. 39 - Prob. 34P Ch. 39 - Prob. 35P Ch. 39 - Prob. 36P Ch. 39 - Prob. 37P Ch. 39 - Prob. 38P Ch. 39 - Prob. 39P Ch. 39 - Prob. 40P Ch. 39 - Prob. 41P Ch. 39 - Prob. 42P Ch. 39 - Prob. 43P Ch. 39 - Prob. 44P Ch. 39 - Prob. 45P Ch. 39 - Prob. 46P Ch. 39 - Prob. 47P Ch. 39 - Prob. 48P Ch. 39 - Prob. 49P Ch. 39 - Prob. 50P Ch. 39 - Prob. 51P Ch. 39 - Prob. 52P Ch. 39 - Prob. 53P Ch. 39 - Prob. 54P Ch. 39 - Prob. 55P Ch. 39 - Prob. 56P Ch. 39 - Prob. 57P Ch. 39 - Prob. 58P Ch. 39 - Prob. 59P Ch. 39 - Prob. 60P Ch. 39 - Prob. 61P Ch. 39 - Prob. 62P Ch. 39 - Prob. 63P Ch. 39 - Prob. 64P Ch. 39 - A diatomic gas molcculc consistsof two atoms of...Ch. 39 - Prob. 66P Ch. 39 - Prob. 67P Ch. 39 - Prob. 68P Ch. 39 - Prob. 69P Ch. 39 - Prob. 70P Ch. 39 - An old model of a hydrogen atom has the charge e...Ch. 39 - Prob. 72P Ch. 39 - Prob. 73P

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The energy levels of the system are given by E = - 11.3 k e V n 2

Solution Summary: The author explains that the energy levels of the system are given by E=-11.3keVn2.

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To show:

The energy levels of the system are given by

E=-11.3 keVn2

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Chapter 39 Solutions

The energy levels of the system are given by E = - 11.3 k e V n 2

Solution Summary: The author explains that the energy levels of the system are given by E=-11.3keVn2.

Concept explainers

To show: The energy levels of the system are given by E=-11.3 keVn2

Want to see the full answer?

Chapter 39 Solutions

To show:

The energy levels of the system are given by

E=-11.3 keVn2