Explanation Given Info: The potential energy function for each of the electron is U ( x ) = 1 2 k ′ x 2 where k ′ is a positive constant and the mass of the electron is m . The potential U ( x ) = 1 2 k ′ x 2 is that of simple harmonic oscillator. Treat the problem quantum mechanically. Write the equation for the energy of a state by using the energy expression for quantum mechanical harmonic oscillator. E n = ℏ ω ( n + 1 2 ) (1) Here, E n is the energy of a particular state ℏ is h 2 π where h is the Planck’s constant ω is the angular frequency of the oscillator n is the quantum number of the harmonic oscillator Electrons obey the exclusion principle. This permits to put two electrons; one for each spin magnetic quantum number m s = ± 1 2 , for every value of n . So each quantum state is defined by the ordered pair of quantum numbers ( n , m s ) . Place two electrons in each energy level. Write the equation for the lowest energy of the system of 2 N electrons. E = 2 ( ∑ n = 0 N − 1 E n ) Put equation (1) in the above summation. E = 2 ( ∑ n = 0 N − 1 ℏ ω ( n + 1 2 ) ) Solve the above summation

University Physics with Modern Physics, Volume 1 (Chs. 1-20) and Mastering Physics with Pearson eText & ValuePack Access Card (14th Edition)

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ISBN: 9780134209586

Author: Hugh D. Young

Publisher: PEARSON

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Question

Chapter 41, Problem 41.68CP

To determine

To show: The minimum energy of the system of 2N electrons is ℏN2k′/m by using exclusion principle.

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Chapter 41, Problem 41.67P

Chapter 41, Problem 41.69CP

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Chapter 41 Solutions

University Physics with Modern Physics, Volume 1 (Chs. 1-20) and Mastering Physics with Pearson eText & ValuePack Access Card (14th Edition)

Ch. 41.1 - Prob. 41.1TYU Ch. 41.2 - Prob. 41.2TYU Ch. 41.3 - Prob. 41.3TYU Ch. 41.4 - In this section we assumed that the magnetic field...Ch. 41.5 - In which of the following situations is the...Ch. 41.6 - Prob. 41.6TYU Ch. 41.7 - Prob. 41.7TYU Ch. 41.8 - Prob. 41.8TYU Ch. 41 - Prob. 41.1DQ Ch. 41 - Prob. 41.2DQ

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