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

College Physics: A Strategic Approach Technology Update, Books a la Carte Plus Mastering Physics with Pearson eText -- Access Card Package (3rd Edition)

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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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College Physics: A Strategic Approach Technology Update, Books a la Carte Plus Mastering Physics with Pearson eText -- Access Card Package (3rd Edition)

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