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Ionizing Cesium Atom

  1. Oct 9, 2008 #1
    Cesium is often used in "electric eyes" for self-opening doors in an application of the photoelectric effect. The amount of energy required to ionize (remove an electron from ) a cesium atom is 3.89 electron volts (1 eV = 1.60 x 10-19 J). Show by calculation whether a beam of yellow light with wavelength 5230Å would ionize a cesium atom.

    I converted wavelength: 5230*1x10^-10 = 5.23*10^-7m

    Then I calculated the energy by using hc/[tex]\lambda[/tex] :
    (6.62*10^-34 * 3*10^8)/(5.23*10^-7) = 3.797*10^-19 J

    I also calculated 3.89eV*(1.6*10^-19) = 6.224*10^-19 J energy required to ionize Cesium

    How can I tell the beam of yellow light ionizes the Cesium atom?
  2. jcsd
  3. Oct 9, 2008 #2


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    What value of planck's constant did you use? Remember to include units. Keep track of your units!
  4. Oct 9, 2008 #3
    h = 6.62*10^-34 J.s
  5. Oct 9, 2008 #4
    I think since 6.224*10^-19 J is required to ionize cesium and there is 3.797*10^-19 J of energy in one photon of yellow light, then you can prove it does ionize because 6.224*10^-19 is less than 3.797*10^-19 J, therefore there is enough energy to ionize cesium. Hope that's right.
  6. Oct 10, 2008 #5


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    Look at what you just stated...

    My earlier comment was directed at using the value of planck's constant in eV to make the calculation simpler for you...

    h = 4.14 X 10-15 eV s
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