Explanation Given data: Wavelength of emission line is 656.5 nm. Formula used: Write the expression for energy. E = h c λ (1) Here, E is the energy, h is the Planck’s constant with value 4.14 × 10 − 15 eV ⋅ s , c is the speed of the light 3 × 10 8 m s , and λ is the wavelength. Write the expression for selection rule for emission and absorption. Δ l = ± 1 (2) Here, Δ l is the change in azimuthal or orbital quantum number. Substitute 4.14 × 10 − 15 eV ⋅ s for h , 3 × 10 8 m s for c and 656.5 × 10 − 9 m for λ in equation (1), E = ( 4

Student Workbook for College Physics: A Strategic Approach Volume 1 (Chs. 1-16)

3rd Edition

ISBN: 9780321908865

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 29, Problem 67GP

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Possible transitions that could give the emission.

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Chapter 29, Problem 66GP

Chapter 29, Problem 68GP

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Chapter 29 Solutions

Student Workbook for College Physics: A Strategic Approach Volume 1 (Chs. 1-16)

Ch. 29 - Prob. 1CQ Ch. 29 - Prob. 2CQ Ch. 29 - Prob. 3CQ Ch. 29 - Prob. 4CQ Ch. 29 - Prob. 5CQ Ch. 29 - Prob. 6CQ Ch. 29 - Prob. 7CQ Ch. 29 - Prob. 8CQ Ch. 29 - Prob. 9CQ Ch. 29 - Prob. 10CQ

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