- #1

Zeynaz

- 29

- 0

- Homework Statement
- The intensity of the light that falls onto the cathode is 6.0 W/m2. The area of the cathode of the photocell is 3.5 cm^2

b) Calculate the percentage of the photons incident on the cathode that release an electron.

c) what happens to the energy of the photons that do not release an electron

- Relevant Equations
- E=hf

V=IR

P=IV

I=n(q-electron)

The full questions is in the picture. I already solved a) and found 5.6E14 electrons per second

For b) i first found the power of the light but just multiplying the intensity with the area: (6.0 W/m2)(3.5E-4 m^2) = 0.0021 W

Then I tried to use the voltage from the graph but i am not sure which value i should use because i don't know the voltage that corresponds to the intensity given.

So could you show me a way that can help me work it out?

After finding the number of electrons that the light releases i would calculate the percetage by dividing the value in question a) by the value i find and multiply it with 100.

(The correct answer should be 13% but i don't know how to get there)

For c) i said that because the energy of the photons are not enough to eject the electrons on the plate are a reflected back. (But I am not sure if that's a correct statement)