Physical Chemistry 2 - Work function - HELP NEEDED

AI Thread Summary
The discussion focuses on solving a problem related to the photoelectric effect, specifically calculating the number of electrons ejected, their energy, and velocity from a laser pulse. The work function of rubidium (Rb) is given as 2.09 eV, and the relevant equations provided include kinetic energy (K.E = hv - work function), velocity (v = speed of light/wavelength), and energy (E = nhv). Participants encourage the poster to apply these equations to find the solution and to share any difficulties encountered. The thread emphasizes the importance of understanding the photoelectric effect in this context.
gunner05
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I really need help with the following question,

If the work function of Rb is 2.09 eV, what is the number of electrons ejected, their energy in eV, and their velocity, from a microjoule pulse of 500nm laser light?
 
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This is a standard application of the photoelectric effect - what have you tried so far? What equations do you have that may be useful?
 
I have K.E = hv-(work function)
and v=speed of light/wavelength
E = nhv
 
t!m said:
This is a standard application of the photoelectric effect - what have you tried so far? What equations do you have that may be useful?

I have K.E = hv-(work function)
and v=speed of light/wavelength
E = nhv
 
You have pretty much everything you need then, just plug away. If you get stuck, show us where.
 
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