Light energy (eV) 2.12 2.13 2.14 2.15 2.16 Electron emitted? Part A no no Electron KE (eV) yes yes yes *Note that 1 eV (electron volt) = 1.60 × 10-¹⁹ J. What is the threshold frequency 0 of cesium? Express your answer numerically in hertz. - 0 0.01 0.02
Light energy (eV) 2.12 2.13 2.14 2.15 2.16 Electron emitted? Part A no no Electron KE (eV) yes yes yes *Note that 1 eV (electron volt) = 1.60 × 10-¹⁹ J. What is the threshold frequency 0 of cesium? Express your answer numerically in hertz. - 0 0.01 0.02
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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Transcribed Image Text: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.
The equation for this phenomenon is
KE = hv - hvo
where KE is the kinetic energy of the emitted
electron, h = 6.63 × 10-³4 J·s is Planck's
constant, is the frequency of the light, and vis
the threshold frequency of the metal.
Also, since E = hv, the equation can also be
written as
KE=E-O
Here are some data collected on a sample of cesium exposed to various energies of light.
Light energy
Electron emitted?
Electron KE
(eV)
(eV)
2.12
2.13
2.14
2.15
2.16
no
Part A
no
yes
yes
yes
*Note that 1 eV (electron volt) = 1.60 × 10-¹⁹ J.
What is the threshold frequency o of cesium?
Express your answer numerically in hertz.
0
0.01
0.02
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