Max Photo Current Calculation Using Platinum's Work Function

AI Thread Summary
Platinum's work function is 6.35 eV, and the discussion revolves around calculating the maximum photo current (Io) achievable with an irradiation power of 1.27 W. The relationship between energy and power is highlighted, with the equation hf = wo indicating that the energy of each photon must exceed the work function to free electrons. The challenge lies in determining the voltage to use in the equation P = VI, as the current depends on the number of electrons liberated per second. The available power indicates the total energy per second, which is crucial for calculating the maximum current. Understanding these relationships is essential for solving the problem effectively.
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1. Platinum has a very high work function of wo = 6.35 eV.What is the maximal photo current Io that can be achieved with an irradiation power of
P = 1.27 W ?




Homework Equations


hf = wo


The Attempt at a Solution


Well, Since the only information they give here is the work function, then i assume that it is related to the voltage. I'm trying to use P=VI to find I but it's the voltage part that's confusing me.
 
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The current is charge per unit time so it will depend on how many electrons can you break free in one second.
For each one you need the energy wo.
The given power tells you how much energy is available per second...
 
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