answer 10min The Photoelectric EffectElectrons are emitted from the surface of a metal when it's exposed to light.This is called the photoelectric effect. Each metal has a certain thresholdfrequency of light, below which nothing happens. Right at this thresholdfrequency, an electron is emitted. Above this frequency, the electron isemitted and the extra energy is transferred to the electron.Here are some data collected on a sample of cesium exposed to various energies of light.Light energy(eV)Electron KE(eV)Electron emitted?The equation for this phenomenon is3.87noKE= hv – hv3.88nowhere KEis the kinetic energy of the emitted electron,h= 6.63 x 10 34 J.s is Planck's constant, v is the frequency of the light,3.89yes3.90yes0.01and vo is the threshold frequency of the metal.3.91yes0.02Also, since E== hv, the equation can also be written as*Note that 1 eV (electron volt) = 1.60 x 10–19 J.КЕ-Е-Фwhere E is the energy of the light and ø is the binding energy of theelectron in the metal.Part AWhat is the threshold frequency vo of cesium?Express your answer numerically in hertz.View Available Hint(s)ν ΑΣφ?Vo =Hz

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What is the threshold frequency of cesium?

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

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Question

answer 10min

The Photoelectric Effect
Electrons are emitted from the surface of a metal when it's exposed to light.
This is called the photoelectric effect. Each metal has a certain 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.
Here are some data collected on a sample of cesium exposed to various energies of light.
Light energy
(eV)
Electron KE
(eV)
Electron emitted?
The equation for this phenomenon is
3.87
no
KE= hv – hv
3.88
no
where KEis the kinetic energy of the emitted electron,
h= 6.63 x 10 34 J.s is Planck's constant, v is the frequency of the light,
3.89
yes
3.90
yes
0.01
and vo is the threshold frequency of the metal.
3.91
yes
0.02
Also, since E== hv, the equation can also be written as
*Note that 1 eV (electron volt) = 1.60 x 10–19 J.
КЕ-Е-Ф
where E is the energy of the light and ø is the binding energy of the
electron in the metal.
Part A
What is the threshold frequency vo of cesium?
Express your answer numerically in hertz.
View Available Hint(s)
ν ΑΣφ
?
Vo =
Hz

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