# Homework Help: Effects of frequency on rate of emission of photoelectrons

1. Apr 13, 2014

### coconut62

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.

2. Apr 13, 2014

### suvendu

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).

3. Apr 13, 2014

### coconut62

But why does the book say rate of emission decreases when frequency increases?

4. Apr 13, 2014

### haruspex

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.
Agreed
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.

5. Apr 13, 2014

### suvendu

Haruspex, thanks for the note.But I stick to my point. Intensity depends only on number of incoming photon, not frequency.

6. Apr 13, 2014

### haruspex

So what units would you use for intensity?

7. Apr 13, 2014

### coconut62

Shouldn't the photon rate be constant? I thought it is just the <energy> of the photons that is increasing.

8. Apr 14, 2014

### haruspex

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.

9. Apr 14, 2014

### suvendu

For any radiation, Intensity is no. of photons incident.

10. Apr 14, 2014

### suvendu

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. Apr 14, 2014

### haruspex

http://en.wikipedia.org/wiki/Light_intensity lists these:
• 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. Feb 20, 2017

### AaronBae

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.