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anonymous
 one year ago
What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.4×1015Hz?
anonymous
 one year ago
What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.4×1015Hz?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is my work so far.

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2What's the ionization energy for the electrons emitted?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2\(\sf KE=E\phi =h\nu I.E.\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Light energy (eV) Electron emitted? Electron KE (eV) 3.87 no — 3.88 no — 3.89 yes 0 3.90 yes 0.01 3.91 yes 0.02 Here's the data table I was given for Cesium. I found the threshold frequency for Cesium to be 9.39*10^14 Hz. I'm just not sure how to solve this problem and where I'm going wrong, @aaronq

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2Okay, the threshold frequency is also called the work function, \(\phi\). The energy of a photon is used to remove the electron, the leftover is the kinetic energy. \(\sf E_{photon }=KE+\phi\rightarrow KE=E_{photon}\phi\) \( \sf KE=E_{photon}\phi=h\nu h\nu_o=h(\nu\nu_o)\) \(\sf KE=(6.626*10^{34}J*s)(1.4*10^{15}Hz9.39*10^{14} Hz)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, that makes much more sense. Thank you for explaining it to me, I should be able to work it out from here!
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