Lecture 5

Moderators: Chem_Mod, Chem_Admin

Kayko Lee 1C
Posts: 117
Joined: Wed Sep 30, 2020 9:39 pm

Lecture 5

Postby Kayko Lee 1C » Mon Oct 12, 2020 12:04 pm

In the video "Unless E(Photon) ≥ E(energy remove e-) then electrons are not emitted even for high intensity light" was mentioned. Does high intensity light mean wavelength? As in even though you increase the frequency of the wavelength electrons aren't emitted as long as it isn't greater than or equal to the energy to remove electrons?

Ria Nawathe 1C
Posts: 108
Joined: Wed Sep 30, 2020 9:39 pm
Been upvoted: 3 times

Re: Lecture 5

Postby Ria Nawathe 1C » Mon Oct 12, 2020 12:10 pm

According to the particle model of light, the intensity of light is proportional to the number of photons, which translates into how bright the light is. Intensity is not related to wavelength.

EnricoArambulo3H
Posts: 112
Joined: Wed Sep 30, 2020 9:33 pm
Been upvoted: 4 times

Re: Lecture 5

Postby EnricoArambulo3H » Mon Oct 12, 2020 12:11 pm

High intensity light means the high amounts of photons within the light source. If the frequency increases, so does the energy. Even if you increase the intensity of light, even though they have low energy, they won't be able to remove the electron. Hope this helped!

Margaret Wang 2D
Posts: 100
Joined: Wed Sep 30, 2020 9:58 pm

Re: Lecture 5

Postby Margaret Wang 2D » Mon Oct 12, 2020 12:45 pm

Intensity refers to the number of protons emitted from the light source. For example, if you have 1 photon emitted from the source and hitting the surface per second, that is less intense than if 5 photons were hitting the surface per second (the numbers are usually way higher than this obviously). Intensity and wavelength are unrelated characteristics of light, meaning that changing intensity does not directly affect wavelength and vice versa.

905290504
Posts: 133
Joined: Wed Sep 30, 2020 9:44 pm
Been upvoted: 1 time

Re: Lecture 5

Postby 905290504 » Mon Oct 12, 2020 1:44 pm

the intensity of a light ray is proportional to the number of photons which then confers brightness. what we care about with the photoelectric effect is the individual energy of each photon, not how many photons there are. to increase the energy of the individual photons, we increase the frequency of the wave and therefore decrease the wavelength. once the frequency is high enough that each photon has an energy greater than or equal to the threshold energy, then the photon can eject an electron from the metal.

Jay Solanki 3A
Posts: 137
Joined: Wed Sep 30, 2020 9:59 pm
Been upvoted: 1 time

Re: Lecture 5

Postby Jay Solanki 3A » Mon Oct 12, 2020 2:14 pm

Hello all!

So I think as I was watching the lecture 5 this morning I made an important distinction. Increasing the intensity of the light implies that you are increasing its amplitude, which means that there is more light, but it has the same wavelength and frequency since amplitude is not correlated to either of those properties, and therefore has the same energy. This implies that more photons were being directed towards the electrons, but they could not eject them because their FREQUENCY was too low. To make the frequency higher you have to change the light source completely, and increasing the frequency would increase the energy to allow the electrons to be ejected. This provides evidence of the one photon-one electron model. In other words, each of the photons in the higher intensity light could not eject the electron because their energies cannot be combined, and the energy to eject the electrons must come from a single photon. Hope this helps!

Jay Solanki


Return to “Photoelectric Effect”

Who is online

Users browsing this forum: No registered users and 4 guests