Explanation Given Info: The energies for the electron in the K, L and M shells of the tungsten atom are − 69 , 500 eV, − 12 , 000 eV, − 2200 eV respectively. The energies of the x rays will be equal to the energy difference between the shells. Write the equation for the wavelength of the x ray produced. λ = h c Δ E (1) Here, λ is the wavelength of the x ray h is the Planck’s constant c is the speed of light Δ E is the energy difference between the shells A K α x ray is produced in a L → K transition. Write the equation for Δ E of L → K transition. Δ E = − 12 , 000 eV − ( − 69 , 500 eV ) = 57 , 500 eV The value of Planck’s constant in eV ⋅ s is 4.136 × 10 − 15 eV ⋅ s . Substitute 4.136 × 10 − 15 eV ⋅ s for h , 3.00 × 10 8 m/s for c and 57 , 500 eV for Δ E in equation (1) to find the wavelength of the K α x ray of tungsten. λ = ( 4.136 × 10 − 15 eV ⋅ s ) ( 3

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 41.38E

To determine

The wavelengths of the K_α and K_β x rays of tungsten.

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Chapter 41, Problem 41.37E

Chapter 41, Problem 41.39P

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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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