Max Photo Current Calculation Using Platinum's Work Function

[xago]

1. Platinum has a very high work function of w_o = 6.35 eV.What is the maximal photo current I_o that can be achieved with an irradiation power of
P = 1.27 W ?

Homework Equations

hf = w_o

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.

 

[nasu]

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...

 

[thomate1]

For the voltage, I'm thinking of using V = hf/q. So I'm thinking of using the given work function as the energy and then solving for V, but I'm not sure if that's the right approach. Any suggestions?I would approach this problem by first understanding the relationship between the work function and the maximum photo current. The work function represents the minimum energy required for an electron to escape from the surface of a material. In other words, it is the energy needed to overcome the attractive forces of the material and become a free electron.

In the context of this problem, the work function of platinum tells us that it requires 6.35 eV of energy for an electron to escape from its surface. This energy can come from an external source, such as light. When light with a frequency (f) corresponding to 6.35 eV is incident on the platinum surface, it can provide the necessary energy to release electrons from the metal and create a current.

Now, to find the maximum photo current (Io) that can be achieved with an irradiation power of P = 1.27 W, we can use the formula Io = P/hf, where h is Planck's constant. This formula tells us that the maximum photo current is directly proportional to the irradiation power and inversely proportional to the frequency of the incident light.

Substituting the given values, we get Io = (1.27 W)/(6.63 x 10^-34 J s x 6.35 eV) = 3.29 x 10^19 electrons per second.

In conclusion, the maximal photo current that can be achieved with an irradiation power of P = 1.27 W is 3.29 x 10^19 electrons per second. This value can be increased by increasing the irradiation power or using light with a higher frequency.