Photoelectric Effect: Explaining Wavelength Effect on Electron Emission

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The discussion clarifies the relationship between wavelength, frequency, and the photoelectric effect, emphasizing that the energy of a photon is inversely proportional to its wavelength. While classical theory suggested that the amplitude of light waves determined energy, the photoelectric effect demonstrates that frequency is the key factor in photon energy, with E=hf being central to this understanding. Electrons are only emitted when photons exceed a certain energy threshold, known as the work function, regardless of the number of lower-energy photons. Intensity relates to the number of photons impacting a surface, affecting the rate of electron emission once the cutoff frequency is surpassed. Overall, the interplay between frequency, energy, and intensity is crucial in explaining electron emission in the photoelectric effect.
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I was reading about the photoelectric effect and it described that in the classical description of light, the changing of wavelength would effect the rate at which electrons were emitted. This theory was later proved wrong when Einstein introduced the idea of photons. However this theory goes on to say that the energy of a photon is inversely proportional to its wavelength, and therefore the wavelength does effect the rate at which electrons are emitted from metals. I'm confused to why wavelength can be used in both theories of the photoelectric effect. Please explain! thank you!
 
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It's not exactly the wavelength per se, but the frequency of the light waves. But that's ok since frequency and wavelength are related by c, speed of light. You have E=hf for a single photon. Classical theory says that energy of light waves is determined by the amplitude of the wave and not its frequency. So classically, frequency (and hence wavelength) does not matter whereas amplitude does. The photoelectric effect shows that the frequency determines the quantised energy of photons.
 
As I understand it, the energy of the electrons is proportional to the wavelength, E=hf, and the rate of electron flow is proportional to the intensity of the light. Both ideas you mentioned aren't contradictory
 
The important thing about the photoelectric effect (which lead to quantum mechanics) is that you need a certain amount of energy in a single photon to eject an electron - you cannot make up for lower energyby using lots of them.
The energy of a photon depends on wavelength (or frequency) the number of photons depends on the intensity.

The classical theroy said that the total energy mattered, so you could use lots of long wavelength photons to make up for their lower energy - in a real experiment there is no emmission until a certain cuttoff frequency.
 
thank you very much that makes so much more sense
 
Frequency = energy of a single photon = the energy an electron needs + get when he is existed.

What this means is that if the work function is 1ev and you have a photon hitting the substance with 2ev then an electron will be released with 1ev of kinetic energy.

Intensity= the rate of which photons hit a surface.

If they don't have the right energy (frequency) then no electron will be existed , if they are above the frequency cutoff then the energy generated will be in direct relation to the intensity because that will mean more electrons are existed per unit of time and area.
 
It may be shown from the equations of electromagnetism, by James Clerk Maxwell in the 1860’s, that the speed of light in the vacuum of free space is related to electric permittivity (ϵ) and magnetic permeability (μ) by the equation: c=1/√( μ ϵ ) . This value is a constant for the vacuum of free space and is independent of the motion of the observer. It was this fact, in part, that led Albert Einstein to Special Relativity.
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