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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.

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: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter5: Electron Configurations And The Periodic Table
Section: Chapter Questions
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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.

It is not 9.17×10−19 J or 12.589 x 10-19 J

By using the following formula, you can calculate the kinetic energy of the emitted electron:

KE=E−ϕ=hν−hν0

where h=6.63×10−34 J⋅s is Planck's constant, ν=1.90×1015Hz is the given frequency, and ν0=9.39×1014 Hz is your answer from Part A. The threshold frequency ν0 of cesium is 9.39×1014 Hz

Electrons are emitted from the surface of a metal What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of when it's exposed to light. This is called the photoelectric effect. Each metal has a certain frequency 1.90 × 1015 Hz ? Express your answer in joules to three significant figures. threshold frequency of light, below which nothing happens. Right at this threshold frequency, an electron is emitted. Above this frequency, the electron is emitted and the extra energy is transferred to the electron. v View Available Hint(s) Hint 1. How to approach this problem The equation for this phenomenon is By using the following formula, you can calculate the kinetic energy of the emitted electron: KE= hv – hvo KE= E – ¢ = hv – hvo where KE is the kinetic energy of the emitted electron, h = 6.63 × 10¬34 J.s is Planck's constant, v is the frequency of the light, and vọ is the threshold frequency of the metal. where h = 6.63 × 10 34 J ·s is Planck's constant, v = 1.90 × 105HZ is the given frequency, and vo = 9.39 × 1014 Hz is your answer from Part A. Also, since E = hv, the equation can also be written as ? KE = E – o 19 where E is the energy of the light and o is the binding energy of the electron in the metal. KE = 12.6 • 10¯ J Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining
Transcribed Image Text:Electrons are emitted from the surface of a metal What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of when it's exposed to light. This is called the photoelectric effect. Each metal has a certain frequency 1.90 × 1015 Hz ? Express your answer in joules to three significant figures. threshold frequency of light, below which nothing happens. Right at this threshold frequency, an electron is emitted. Above this frequency, the electron is emitted and the extra energy is transferred to the electron. v View Available Hint(s) Hint 1. How to approach this problem The equation for this phenomenon is By using the following formula, you can calculate the kinetic energy of the emitted electron: KE= hv – hvo KE= E – ¢ = hv – hvo where KE is the kinetic energy of the emitted electron, h = 6.63 × 10¬34 J.s is Planck's constant, v is the frequency of the light, and vọ is the threshold frequency of the metal. where h = 6.63 × 10 34 J ·s is Planck's constant, v = 1.90 × 105HZ is the given frequency, and vo = 9.39 × 1014 Hz is your answer from Part A. Also, since E = hv, the equation can also be written as ? KE = E – o 19 where E is the energy of the light and o is the binding energy of the electron in the metal. KE = 12.6 • 10¯ J Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining
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