Part CWhat is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.10 x 1015HZ ?Express your answer in joules to three significant figures.• View Available Hint(s)Ην ΑΣφ?X•10"KE=JSubmitPrevious Answers I RevElectrons are emitted from the surface of a metal when it's exposed tolight. This is called the photoelectric effect. Each metal has a certainthreshold frequency of light, below which nothing happens. Right atthis threshold frequency, an electron is emitted. Above this frequency,the electron is emitted and the extra energy is transferred to theelectron.Here are some data collected on a sample of cesium exposed to various energies of light.Light energy(eV)Electron KE(eV)Electron emitted?3.87noThe equation for this phenomenon is3.88noKE= hv – hvo3.89yeswhere 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 the3.90yes0.01light, and vy is the threshold frequency of the metal.3.91yes0.02*Note that 1 eV (electron volt) =1.60 × 10 1º J.Also, since E = hv, the equation can also be written asKE=E-6Part Awhere E is the energy of the light and ø is the binding energy of theelectron in the metal.

Kinetic energy of cesium E = hv Where h = plank's constant = 6.626 ×10-34 J s v = frequency of…

Part C What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.10 x 1015HZ ? Express your answer in joules to three significant figures. • View Available Hint(s) Ην ΑΣφ ? x" X IXI X•10" КЕ- J Submit Previous Answers

Part C What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.10 x 1015HZ ? Express your answer in joules to three significant figures. • View Available Hint(s) Ην ΑΣφ ? x" X IXI X•10" КЕ- J Submit Previous Answers

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Question

Part C
What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.10 x 1015HZ ?
Express your answer in joules to three significant figures.
• View Available Hint(s)
Ην ΑΣφ
?
X•10"
KE=
J
Submit
Previous Answers

I Rev
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?
3.87
no
The equation for this phenomenon is
3.88
no
KE= hv – hvo
3.89
yes
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
3.90
yes
0.01
light, and vy is the threshold frequency of the metal.
3.91
yes
0.02
*Note that 1 eV (electron volt) =1.60 × 10 1º J.
Also, since E = hv, the equation can also be written as
KE=E-6
Part A
where E is the energy of the light and ø is the binding energy of the
electron in the metal.

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