Effects of frequency on rate of emission of photoelectrons

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Homework Help Overview

The discussion revolves around the effects of frequency on the rate of emission of photoelectrons when light of constant intensity is incident on a metal surface. Participants are exploring concepts related to the photoelectric effect, including the relationship between photon frequency, intensity, and electron emission rates.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are attempting to reconcile the relationship between photon frequency and the rate of electron emission, questioning why an increase in frequency leads to a decrease in the number of emitted electrons if intensity is held constant. Some are discussing the particle nature of light and its implications for photon interactions with electrons.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's interpretations of intensity and photon rates. There is an ongoing exploration of the definitions and implications of intensity in different contexts, but no consensus has been reached regarding the relationship between frequency and emission rates.

Contextual Notes

Participants are navigating various definitions of intensity and its dependence on photon rate and frequency, indicating potential confusion stemming from differing contexts in physics.

coconut62
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Light of constant intensity is incident on a metal surface, causing electrons to be emitted. State and explain why the rate of emission of electrons changes as the frequency of the incident light is increased.

My answer: As frequency increases, energy of photons also increases, so more energy is available per electron, so more electrons can escape and rate of emission increases

Book's answer: at higher frequency, fewer photons (per second) for same intensity, so rate of emission decreases. (Allow argument based on photoelectric efficiency)

I totally don't understand what the book is saying. Why fewer photons at higher frequency? And what is photoelectric efficiency?

Someone please explain to me, thanks.
 
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Rate of emission depends on intensity of incident ligh only. And the energy of the ejected electron depends on the frequency of photons. It all can be explained if you consider particle nature of light : light consists of packets. Each packet interacts with a single electron. So if you increase intensity of incident light,that is no. of photons or packets in light, then no of outcoming electrons increase(i.e. rate of emission).
 
But why does the book say rate of emission decreases when frequency increases?
 
suvendu said:
Rate of emission depends on intensity of incident ligh only.
That's not quite right. The rate of emission depends on rate of incoming photons (assuming they are energetic enough). Above a threshold, it does not depend on the frequency. The intensity is the power; it is proportional to both photon rate and photon frequency.
And the energy of the ejected electron depends on the frequency of photons. It all can be explained if you consider particle nature of light : light consists of packets. Each packet interacts with a single electron.
Agreed
So if you increase intensity of incident light,that is no. of photons or packets in light,
That's only the same thing if the frequency is held constant.

coconut62, the book is saying that if you increase the frequency but keep the intensity constant then that must mean you have reduced the photon rate, so the emission rate drops.
 
Haruspex, thanks for the note.But I stick to my point. Intensity depends only on number of incoming photon, not frequency.
 
suvendu said:
Haruspex, thanks for the note.But I stick to my point. Intensity depends only on number of incoming photon, not frequency.
So what units would you use for intensity?
 
haruspex said:
coconut62, the book is saying that if you increase the frequency but keep the intensity constant then that must mean you have reduced the photon rate, so the emission rate drops.

Shouldn't the photon rate be constant? I thought it is just the <energy> of the photons that is increasing.
 
coconut62 said:
Shouldn't the photon rate be constant? I thought it is just the <energy> of the photons that is increasing.
Intensity is power per unit area. Power is energy per photon x photon rate. Energy per photon is proportional to its frequency. If the intensity stays constant over an area but the energy per photon increases then the photon rate must decrease.
 
For any radiation, Intensity is no. of photons incident.
 
  • #10
The word 'Intensity' is used in a lot of cases differently. As like in Astro the way it is defined is different from what is used in optics..!
 
  • #11
suvendu said:
For any radiation, Intensity is no. of photons incident.
The word 'Intensity' is used in a lot of cases differently. As like in Astro the way it is defined is different from what is used in optics..!

http://en.wikipedia.org/wiki/Light_intensity lists these:
  • Radiant intensity, a radiometric quantity measured in watts per steradian (W/sr)
  • Luminous intensity, a photometric quantity measured in lumens per steradian (lm/sr), or candela (cd)
  • Irradiance, a radiometric quantity, measured in watts per meter squared (W/m2)
  • Intensity (physics), the name for irradiance used in other branches of physics (W/m2)
  • Radiance, commonly called "intensity" in astronomy and astrophysics (W·sr-1·m-2)
They are all some version of either power per unit area or power per steradian.
Luminous intensity is a little different in that the weighting given to a photon is based on the eye's sensitivity to it, not its intrinsic energy. But none of them merely count photons.
 
  • #12
Intensity = Power / Area
Power = Energy per photon * photon per time(Photon rate) (Which will just simplify to Energy per time which is power)

Thus,
Intensity = ( E/γ * γ/t ) / Area
When intensity is held constant and frequency increases,
E increases as E=hf
but the intensity should remain constant, so
γ/t decreases.
 

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