# (a) Interpretation: The maximum kinetic energy of an ejected photoelectron from cesium when irradiated by 520 nm light should be calculated. Concept introduction: Data given: Longest wavelength of radiation needed to eject photoelectrons from cesium = λ min = 660 nm Formulae to use: Energy involved in photoelectric effect is expressed as, E λ = KE m + ω E λ = energy of the incident radiation KE m = kinetic energy of the ejected electrons ω = work function (minimum energy needed to eject the electrons) E = h ν Where, E = energy of light h = Plank’s constant = 6.6254 × 10 − 34 Js ν = frequency of light For air or vacuum, ν = c λ Where, c = velocity of light in vacuum (3.00 × 10 8 m/s ) λ = wavelength of light ### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213 ### Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

#### Solutions

Chapter 6, Problem 6.9QAP
Interpretation Introduction

## (a)Interpretation:The maximum kinetic energy of an ejected photoelectron from cesium when irradiated by 520 nm light should be calculated.Concept introduction:Data given:Longest wavelength of radiation needed to eject photoelectrons from cesium = λmin=660 nmFormulae to use:Energy involved in photoelectric effect is expressed as,Eλ = KEm+ωEλ= energy of the incident radiationKEm = kinetic energy of the ejected electronsω= work function (minimum energy needed to eject the electrons)E = hνWhere,E = energy of lighth = Plank’s constant = 6.6254×10−34 Jsν = frequency of lightFor air or vacuum,ν=cλWhere,c = velocity of light in vacuum (3.00 ×108 m/s )λ = wavelength of light

Interpretation Introduction

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