Need to understand about the photoelectric effect

In summary, the frequency and intensity of light determines how many electrons are ejected. The wave aspect of light is what determines the color of the light.
  • #1
sssddd
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Hello my question involves what exactly is the frequency and intensity of the light. The E = hf applies to the photon particle aspect of light right? So where is f from, the wave aspect? Also increase f, which increases E makes the electrons eject faster in terms of kinetic energy right, but not increase in numbers. The intensity is what increases the number of electrons ejected but not speed? So is intensity part of the wave aspect? And how is it all related?
 
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  • #2
sssddd said:
Hello my question involves what exactly is the frequency and intensity of the light.

it is what is set by the device emitting it. intensity is the power of radiation crossing a perpendicular surface (to the wave direction) per unit area. power is the time rate of change of the energy passing that surface.

The E=hf applies to the photon particle aspect of light right? So where is f from, the wave aspect?

f is determined by the color of the light. if it's white, there will be lot'sa different frequencies and lot'sa different electron energies.

Also increase f, which increases E makes the electrons eject faster in terms of kinetic energy right, but not increase in numbers.

if the intensity is constant, the number of electrons (per unit time) ejected will have to decrease (but there would be more average energy per electron) if f increases.

The intensity is what increases the number of electrons ejected but not speed?

all other parameters held constant, yes. but if the frequency changes and intensity held constant, the number of electrons emitted per unit time will change (inversely).
 
  • #3
ok sounds like I'm on the right track. Just a little more, if frequency is constant. Increase in intensity the same as increase in power will eject more electrons. How does that work, more energy per length of time is distributed over each electrons, so more electrons ejected right. Now how does intensity connect with the wave aspect.

Also with frequency, could you please clarify a little more about the whole color thing, I thought color was waves and not particle. How does colors tie to the particle aspect. And if intensity is constant, and frequency increases, which is more energy per electron, how does frequency affect the numbers when intensity is constant. You said it decrease?

Oh by the way new question when this is tested in a circuit, why do u need a battery? Does the battery create a voltage that help electrons ejected go from cathode to anode bteween two plates like a capacitor? It will end up like a capacitor right
 
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  • #4
sssddd said:
ok sounds like I'm on the right track. Just a little more, if frequency is constant. Increase in intensity the same as increase in power will eject more electrons. How does that work, more energy per length of time is distributed over each electrons, so more electrons ejected right. Now how does intensity connect with the wave aspect.

increase the number of photons (of the same energy) and more energy is crossing your surface therefore intensity increases.

Also with frequency, could you please clarify a little more about the whole color thing, I thought color was waves and not particle. How does colors tie to the particle aspect.

it's a "wave-particle duality". this is what it is all about. waves with frequencies of [itex]f[/itex] are associated with particles with energy of [itex]E = h f[/itex].

And if intensity is constant, and frequency increases, which is more energy per electron, how does frequency affect the numbers when intensity is constant. You said it decrease?

if the intensity is constant and the frequency increases, the number of photons is less.

Oh by the way new question when this is tested in a circuit, why do u need a battery? Does the battery create a voltage that help electrons ejected go from cathode to anode bteween two plates like a capacitor? It will end up like a capacitor right

you need an adjustable DC voltage source to oppose the electrons emitted and you increase the voltage until the current from emission just barely gets to zero. the electon charge times the voltage is the energy of the electron. that gives you the "E" and the color gives you the "f" and from enough points (different colors, different energies) you can get "h".
 
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1. What is the photoelectric effect?

The photoelectric effect is the phenomenon where electrons are emitted from a material when it is exposed to light of a certain frequency or higher. This was first observed by physicist Heinrich Hertz in 1887.

2. How does the photoelectric effect work?

When light of a certain frequency, known as the threshold frequency, is shone on a metal surface, it transfers enough energy to the electrons in the material to overcome the attractive forces of the metal atoms, causing them to be emitted as photoelectrons.

3. What is the significance of the photoelectric effect?

The photoelectric effect was a crucial experiment in understanding the particle nature of light and the concept of photons. It also led to the development of the field of quantum mechanics and has many practical applications, such as in solar panels and photoelectric cells.

4. What factors affect the photoelectric effect?

The photoelectric effect is affected by the intensity of the light, the frequency of the light, and the material being used. Increasing the intensity or frequency of the light will increase the number of photoelectrons emitted, while using a different material can change the threshold frequency.

5. How does the photoelectric effect support the wave-particle duality of light?

The photoelectric effect provides evidence for the wave-particle duality of light, as it shows that light behaves as both a wave and a particle. The wave-like behavior is demonstrated through the interference and diffraction of light, while the particle-like behavior is seen in the discrete packets of energy (photons) emitted during the photoelectric effect.

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