## Light Intensity

$c=\lambda v$

mia alexander 2K
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Joined: Wed Sep 30, 2020 10:02 pm

### Light Intensity

I just wanted some clarification on what light intensity is. In Dr. Lavelle's focus topic video on the photoelectric effect he said that increasing light intensity is the wave's amplitude as well as the number of photons. Is this correct or did I misunderstand what he was saying?

Sophia Hu 1A
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### Re: Light Intensity

Yes, intensity is related to amplitude. An increase in intensity is synonymous to an increase in amplitude. An increase in intensity also does increase the number of photons ejected. However, you have to remember increasing the intensity does NOT increase the energy of these photons. Therefore, if the light does not have sufficient energy to overcome the threshold energy, then increasing the intensity still leads to no electrons ejected. Only when the light meets or overcomes this minimum energy (through shortening the wavelength/increasing the frequency of the light), then electrons are ejected, and then increasing intensity will cause more ejections. The reason for this is because light has both wave and particle properties.

Hope this helps!

Posts: 164
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### Re: Light Intensity

Hi! Increasing the light intensity means increasing the amplitude of the waves. This does not increase the amount of electrons emitted while increasing the number of photons would increase the number of electrons emitted. That is why Dr. Lavelle told us to think about it as photons and not waves.
Hope this helps

Kamille Kibria 2A
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### Re: Light Intensity

Yes, you're right. because light can act as both a wave or a particle, there are only certain circumstances where the energy of light will be enough to eject electrons from a given metal, and that is if it meets a threshold frequency. simply increasing the "intensity" of light will not always result in emitted electrons because it is related to the amplitude and # of photons like you said. increasing the intensity of light means that the amplitude is increased, which doesn't necessarily have an effect on the frequency/energy of light. also increasing the intensity means increasing the # of photons, but the energy of the photons does not change. hope that helps :)

Alexa Pham 1D
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Joined: Wed Sep 30, 2020 10:03 pm

### Re: Light Intensity

Yes you are correct. It just depends on whether you're looking at the wave model of light or the quantum one. If we look at the wave model of light, then the intensity is the wave's amplitude. If we look at the quantum description of it, then the intensity of light is proportional to the number of photons.

Sandy Lin 1L
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Joined: Wed Sep 30, 2020 9:48 pm

### Re: Light Intensity

Yes, that is correct. Since the interaction between photons and electrons is one on one, increasing the intensity will not affect whether or not the electrons would be ejected as only the number of photons is increased. In order for electrons to be ejected, it depends on the energy of the individual photons having greater energy or energy equivalent to the binding energy of the electron.

Malakai Espinosa 3E
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### Re: Light Intensity

Not only does this concept apply to light waves, but waves in general! You could apply this to sound for example. If you increase the amplitude of the sound waves, the intensity/volume of the sound you hear will be perceived as louder.

Inderpal Singh 2L
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### Re: Light Intensity

Yes, that is correct. Since the interaction between photons and electrons is one on one, increasing the intensity will not affect whether or not the electrons would be ejected as only the number of photons is increased. In order for electrons to be ejected, it depends on the energy of the individual photons having greater energy or energy equivalent to the binding energy of the electron.

Remember, photons and electrons have a relationship where one photon will interact with one electron. By increasing the intensity, all you will do is increase the number of photons being released. Since we are talking about Quantum mechanics, we have to focus on the energy of a singular photon.