Photoelectric effect threshold

AI Thread Summary
The discussion revolves around calculating the threshold wavelength for a surface illuminated by light of different wavelengths, where the stopping potentials are given. Participants explore the use of Einstein's equation relating kinetic energy to photon energy and the work function, noting that the work function should be assumed to be non-zero. The thickness of the surface is deemed irrelevant for the calculations at hand, as it is considered suitable for the experiment. There is uncertainty regarding the correct answer provided in the textbook. The conversation highlights the need for clarity on the role of the work function in the context of the photoelectric effect.
Nairabhi
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When a cm thick surface is illuminated with light of wavelength lambda,the stopping potential is V. When the same surface is illuminated by light of wavelength 2lambda , the stopping potential is V/3.The threshold wavelength for the surface is ?

Homework Equations

The Attempt at a Solution

 
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Did you try to do this problem yourself?
 
Matterwave said:
Did you try to do this problem yourself?
Yes i tried but i am unable to solve it.
 
Nairabhi said:
Yes i tried but i am unable to solve it.

Hi Nairabhi. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

What equations do you have that you'll use here?
 
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NascentOxygen said:
Hi Nairabhi. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

What equations do you have that you'll use here?
See according to Einstein equation KE(Max)= Energy of photons - work function of surface, but since work function is not given i have assumed it
to be 0. And energy of photons is equal to hc/lamda and this energy is equal to KE(Max) or the stopping potential so using these equations and i am trying to find a relation between them and find the threshold wavelength .However i don't understand what role does the thickness of surface plays which is 1cm ?
 
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You should assume work function is non-zero, and is fixed at some value.

That material thickness may simply mean "of suitable thickness" for our purposes.

Does the textbook give the correct answer?
 
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