Here are some data collected on a sample of cesium exposed to various energies of light. Electrons are emitted from the surface of a metal Light energy (eV) Electron emitted? Electron KE 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. (eV) 2.12 no 2.13 no 2.14 yes 2.15 yes 0.01 The equation for this phenomenon is 2.16 yes 0.02 KE= hv – hvo *Note that 1 eV (electron volt) = 1.60 × 10¬19 J. where KEİS 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 vo is the threshold frequency of the metal. Part A Also, since E = hv, the equation can also be written as What is the threshold frequency vo of cesium? КЕ-Е-Ф Express your answer numerically in hertz. • View Available Hint(s) where E is the energy of the light and o is the binding energy of the electron in the metal. ? Vo = Hz Submit

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Here are some data collected on a sample of cesium exposed to various energies of light.
Electrons are emitted from the surface of a metal
Light energy
(eV)
Electron emitted?
Electron KE
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.
(eV)
2.12
no
2.13
no
2.14
yes
2.15
yes
0.01
The equation for this phenomenon is
2.16
yes
0.02
KE= hv – hvo
*Note that 1 eV (electron volt) = 1.60 × 10¬19 J.
where KEİS 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 vo is the
threshold frequency of the metal.
Part A
Also, since E = hv, the equation can also be written
as
What is the threshold frequency vo of cesium?
КЕ-Е-Ф
Express your answer numerically in hertz.
• View Available Hint(s)
where E is the energy of the light and o is the
binding energy of the electron in the metal.
?
Vo =
Hz
Submit
Transcribed Image Text:Here are some data collected on a sample of cesium exposed to various energies of light. Electrons are emitted from the surface of a metal Light energy (eV) Electron emitted? Electron KE 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. (eV) 2.12 no 2.13 no 2.14 yes 2.15 yes 0.01 The equation for this phenomenon is 2.16 yes 0.02 KE= hv – hvo *Note that 1 eV (electron volt) = 1.60 × 10¬19 J. where KEİS 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 vo is the threshold frequency of the metal. Part A Also, since E = hv, the equation can also be written as What is the threshold frequency vo of cesium? КЕ-Е-Ф Express your answer numerically in hertz. • View Available Hint(s) where E is the energy of the light and o is the binding energy of the electron in the metal. ? Vo = Hz Submit
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