What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.90×1015Hz?Express your answer in joules to three significant figures. I Review | Constants | Periodic TableElectrons are emitted from the surface of a metalE = n4.23 × 1019 J-7.4.70×10¬´mwhen it's exposed to light. This is called thephotoelectric effect. Each metal has a certainthreshold frequency of light, below which nothinghappens. Right at this threshold frequency, anelectron is emitted. Above this frequency, theelectron is emitted and the extra energy isPart Ctransferred to the electron.What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays ofThe equation for this phenomenon isfrequency 1.90 × 1015 Hz ?ΚΕhv-hvoExpress your answer in joules to three significant figures.where KE is the kinetic energy of the emittedelectron, h = 6.63 × 10¬34 J.s is Planck'sconstant, v is the frequency of the light, and vọ isthe threshold frequency of the metal.• View Available Hint(s)ΑΣΦ?Also, since E= hv, the equation can also bewritten asKE = 5.76JKE= E – ¢where E is the energy of the light and o is thebinding energy of the electron in the metal.SubmitPrevious AnswersX Incorrect; Try Again; 2 attempts remaining

The frequency of the UV rays, (ν) = 1.90×1015Hz The threshold frequency of the cesium metal, (ν0) =…

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What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.90×1015Hz? Express your answer in joules to three significant figures.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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