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  • 21-09-2021
  • Chemistry
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What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.20x10^15Hz?

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Аноним Аноним
  • 21-09-2021

v=Frequency=1.2×10^15Hz

Now

[tex]\\ \sf{:}\!\!\implies E=hv[/tex]

h is Planck's constant

[tex]\\ \sf{:}\!\!\implies E=6.626\times 10^{-34}Js\times 1.2\times 10^{15}s^{-1}[/tex]

[tex]\\ \sf{:}\!\!\implies E=7.95\times 10^{-19}J[/tex]

[tex]\\ \sf{:}\!\!\implies E=79.5\times 10^{-18}J[/tex]

[tex]\\ \sf{:}\!\!\implies E=79.5am[/tex]

  • am means atto meters
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