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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?

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  1. anonymous
    • one year ago
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    This is my work so far.

  2. aaronq
    • one year ago
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    What's the ionization energy for the electrons emitted?

  3. aaronq
    • one year ago
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    \(\sf KE=E-\phi =h\nu -I.E.\)

  4. anonymous
    • one year ago
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    Light 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

  5. aaronq
    • one year ago
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    Okay, 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. \(\sf E_{photon }=KE+\phi\rightarrow KE=E_{photon}-\phi\) \( \sf KE=E_{photon}-\phi=h\nu -h\nu_o=h(\nu-\nu_o)\) \(\sf KE=(6.626*10^{-34}J*s)(1.4*10^{15}Hz-9.39*10^{14} Hz)\)

  6. anonymous
    • one year ago
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    Okay, that makes much more sense. Thank you for explaining it to me, I should be able to work it out from here!

  7. aaronq
    • one year ago
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    okay cool! no problem

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