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Work function and wavelength

  • Thread starter kingwinner
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  • #1
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1) The maximum kinetic energy of photoelectrons is 2.8eV. When the wavelength of the light is increased by 50%, the maximum energy decreases to 1.1 eV. What are the "work function" of the cathode and the "initial wavelength"?
K.E.initial=K.E.max=2.8eV=4.48x10^-19J
lambda.final=1.5*lambda.initial
Final energy = 1.1 eV (<---is this purely kinetic energy? why?)

K.E.max = hc/(lambda.initial) - work function
But now I have 2 unknowns: work function and lambda.initial, what can I do to solve for both?

I am stuck here...

Does anyone have any idea or insight? Any help is greatly appreciated.
 

Answers and Replies

  • #2
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Can someone please help me? I am sure that there are a lot of genius here...
 
  • #3
Dick
Science Advisor
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K.E.max = hc/(lambda.initial) - work function
Make that:

K.E.max.initial = hc/(lambda.initial) - work function

Then how about

K.E.max.final = hc/(lambda.final) - work function

as a second equation?
 
  • #4
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Thanks for your help! I got it!
 

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