Interpretation Introduction To determine: The de Broglie wavelength of the auger electron where the electrons are emitted with a kinetic energy of approximately 506 eV. Explanation Given: The kinetic energy of the auger electrons is 506 eV . Explanation: Conversion of eV into Joules can be done as follows: 1 eV = 1.602 × 10 − 19 J Thus: 506 eV = 506 × 1.602 × 10 − 19 J =810 .612 × 10 − 19 J De Broglie equation is written as follows: λ = h m v …… (1) Here, λ is the de Broglie wavelength, h is the Planck’s constant, m is the mass of subatomic particle, and v is its velocity. Kinetic energy of any particle is given as follows: KE = 1 2 m v 2 Here, m is the mass of particle and v is its velocity. Substitute the value of v from the kinetic energy formula to equation (1) as follows: λ = h m ( 2 K E / m ) 1 / 2 = h ( 2 m ⋅ K E ) 1 / 2 The mass of electron is 9

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Chapter 7, Problem 77E

Interpretation Introduction

To determine: The de Broglie wavelength of the auger electron where the electrons are emitted with a kinetic energy of approximately 506 eV.

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Chapter 7, Problem 76E

Chapter 7, Problem 78E

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Chapter 7 Solutions

Chemistry: A Molecular Approach, Books a la Carte Edition; Modified MasteringChemistry with Pearson eText -- ValuePack Access Card -- for Chemistry: A Molecular Approach, 4/e

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