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posted by  J.F on 10/30/2007 11:54:59 AM  |  status: Live  

Exam on Friday! I don't want to fail!

Course Textbook Chapter Problem
General Chemistry Chemistry: The Central Science (10th) by Brown, Lemay, Bursten N/A N/A
Question Details:

Cesium metal is used in the photocells of automatic door openers. The minimum energy needed to knock an electron out of an atom of cesium is 3.43 x 10-19 J.
A) What is the maximum wavelength of light (in nm) that can be used to operate an automatic door opener with a cesium photocell?
B)What colors of light could be used?
C) What is this phenomenon known as and how does it work?

I got  2.34 x 10-8 for question A but I think I'm wrong. Can someone walk me through this one? Thanks for your time and help! I'll take any help I can get at this point.

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posted by io2chip on 10/30/2007 1:16:05 PM  |  status: Live
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J.F's comment:
"That really helps! Thanks!"
Response Details:
A) For electromagnetic radiation of wavelength λ, the energy of each photon is given by the expression:
      Eproton = hc/λ
  h = Plank's constant = 6.626 * 10-34 J.s
  c = speed of light = 2.9979 * 108 m/s
  E = 3.43 * 10-19 J (given)
 
   λ = hc/E
      = [(6.626 * 10-34 J.s)(2.9979 * 108 m/s)] / (3.43 * 10-19 J ) = 5.7913 * 10-7 m
   5.7913 * 10-7 m * (10-9nm/ 1m) = 5.7913 * 10-16 nm
 
B) Since 5.7913 * 10-16 nm is outside the spectrum, the light WHITE could be used.
C) I can not help you with this. Sorry.
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