## models of light

$c=\lambda v$

tonyhu
Posts: 57
Joined: Tue Sep 24, 2019 12:17 am

### models of light

How are the two models -photon and wave model- different? How are they useful in their own way?

John Arambulo 1I
Posts: 58
Joined: Sat Jul 20, 2019 12:15 am

### Re: models of light

The photon (particle) and wave model are different because, in the photon model, the intensity is proportional to the number of photons present at each instant, while in the wave model, the intensity of electromagnetic radiation is proportional to the square of the amplitude of the wave. These show that electromagnetic radiation has wave-particle duality. Electromagnetic radiation behaves like a particle and a wave in different situations. The particle model is useful in understanding the photoelectric effect and can be observed with reflection and refraction. The wave model is useful because reflection, refraction, diffraction, interference, and polarization can be described using waves.

Micah3J
Posts: 100
Joined: Tue Oct 08, 2019 12:16 am

### Re: models of light

When are we able to tell when to use the wavelength model versus the photon model? It is still unclear the relationship between photons and frequency and how that applies to the photoelectric effect

AnnikaMittelhauser4E
Posts: 50
Joined: Fri Aug 30, 2019 12:18 am

### Re: models of light

I'm not sure how to know when to use each model, but my guess is he doesn't expect us to be experts at that. I think understanding the two models and the experiments he talks about (like photoelectric effect experiment) is a good enough place to start. If we understand the models well, we can probably tell which one applies to which situation (like if the experiment involves interference we know it is the wave model because photons could not superimpose and cancel each other out).

As for photons and frequency, the equation is
E=hv which means "energy of individual photon = planck's constant x the frequency of the light."

In the photoelectric experiment, photons of light bombard a metal surface. Each electron attached to the metal requires a fixed amount of energy to "escape" the metal. If a singular photon has sufficient energy, it will "collide" with one electron, and that electron will be released from the metal. If the photon does not have enough energy, no electrons will be released. If the photon has extra energy, the electron will fly off of the metal at a greater speed (light energy becomes kinetic energy). As a higher frequency of light means a higher energy photon, the light must have a sufficiently high frequency to release electrons. Increasing the intensity (aka brightness) of the light would only increase the amount of photons, not the energy of each individual photon. This is unlike the wave model because increasing the intensity of light acting as a wave would increase the amplitude of the wave which would increase its energy. We therefore know that light does not act as a wave in this experiment because increasing the intensity of low frequency light does not release electrons - the energy of the light is not increasing.

Megan Jung 3A
Posts: 50
Joined: Thu Jul 11, 2019 12:17 am

### Re: models of light

From the photoelectric effect:
For waves, the intensity is proportional to the amplitude (size) of the wave so it would have a greater energy. However, this is not the case for the photon or the particle model used for light. In this case, the light model needs a greater frequency to increase its energy in order to remove and electron from a metal's surface. To summarize, in waves the energy is dependent on intensity (amplitude) and for light/photons, energy is dependent on its frequency.

Julia Mazzucato 4D
Posts: 64
Joined: Fri Aug 30, 2019 12:17 am

### Re: models of light

In this course, Dr. Lavelle won't expect us to know when to use each model, but we should be able to look at an experiment that was covered in class and know which model is appropriate to describe the behavior of light. So if a basic photoelectric effect experiment setup is described in a problem, we should be able to know it's a photon model. Same holds true if we are given the De Broglie equation and we should know it's talking about the wave properties.