Photoelectric Effect: Intensity & Frequency Effects

[Ukitake Jyuushirou]

hello

from my understanding of the photo electric effect, the minimum workfunction must be met before electrons are emitted prior to that, intensity doesn't play a part. if the frequency is increased to juz above the work function and stays constant there and i increase the intensity shouldn't the current detected be constant and the number of electrons emitted per second is the same?

 

[daniel_i_l]

If the photons hitting the metal have enough energy to emit photons then the bigger the intensity the bigger the current. Cause more intensity means that more photons are hitting the metal. That means that more electrons are emitted: I = q/t

 

[quicknote]

Hello, thank you for your question. The photoelectric effect is a phenomenon where electrons are emitted from a material when it is exposed to electromagnetic radiation, such as light. The minimum work function refers to the amount of energy needed to remove an electron from the material. If the frequency of the radiation is below this minimum, no electrons will be emitted regardless of the intensity. However, if the frequency is above the minimum, increasing the intensity of the radiation will result in an increase in the number of electrons emitted per second, but the current detected may not necessarily be constant. This is because the intensity of the radiation affects the number of photons hitting the material, which in turn affects the number of electrons that can be emitted. So, while the number of electrons emitted per second may increase with intensity, the current detected may not be constant due to other factors such as the efficiency of the detector. I hope this helps clarify the relationship between intensity and frequency in the photoelectric effect.