Photoelectric Effect: Explaining Wavelength Effect on Electron Emission

susan__t
Messages
20
Reaction score
0
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!
 
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
1K