Explanation Given Info: The transition is from a p state to an s state. The original wavelength is 575.050 nm , the number of lines after splitting is 3 and the distance between the adjacent lines is 0.0462 nm . Write the equation for the energy of the transition. E = h c λ (1) Here, E is the energy of the transition h is the Planck’s constant c is the speed of light λ is the wavelength The wavelength difference between the adjacent levels is very small and thus the energy is also very small. So we can use the differentials. Differentiate equation (1). d E = − h c λ 2 d λ | d E | = h c λ 2 d λ Δ E = h c Δ λ λ 2 (2) Write the equation for the energy difference between the adjacent lines. Δ E = μ B B (3) Here, μ B is the Bohr magneton B is the value of the magnetic field Put equation (3) in (2). μ B B = h c Δ λ λ 2 Rewrite the equation for B . B = h c Δ λ μ B λ 2 The value h in electron volts is 4.136 × 10 − 15 eV ⋅ s , the value of c is 3.00 × 10 8 m/s and the value of μ B is 5.788 × 10 − 5 eV/T

University Physics with Modern Physics (14th Edition)

14th Edition

ISBN: 9780321973610

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Chapter 41, Problem 41.55P

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The strength of the magnetic field.

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Chapter 41, Problem 41.54P

Chapter 41, Problem 41.56P

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Chapter 41 Solutions

University Physics with Modern Physics (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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The strength of the magnetic field.

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Chapter 41 Solutions