Magnitude of current in photoelelctric effect

[ENgez]

I have been thaught that the magintude of the current generated by photons fired at a metal is unrelated to the frequency of the photons but to the photon beam's srength (photons/sec).

I know from the governing equation: E_{\gamma}=\phi+E_{k_{e^{-}}} that the frequency of the photons determines the amount of kinetic energy given to an electron, and that the kinetic energy grows (the free electron moves faster) as the photon's frequency grows.

The definition of current is electrons/sec through a given cross-section.
that means that the current can grow when:

a) there are more free electrons, achieved by firing more photons (increasing the beams strength)

b) making the electrons faster, by increasing the frequency of the photons.

so the photon's frequency does influence the magnitude of the current. Where am i wrong?

 

[danielakkerma]

You're not mistaken directly, but I would advise you to take a look at this:
http://rugth30.phys.rug.nl/quantummechanics/photoelectric.htm"
Observe that:

The light intensity affects the number of photons and therefore the magnitude of the photoelectric current, but does not affect the cut-off voltage V0...

Does that elucidate matters?
Daniel

 

[ENgez]

well, not exactly, I understand why the light intensity doesn't effect cut off voltage. It is becuase light intensity is measured by the number of photons/sec but their individual energy is dependent only on the frequency.

but this still doesn't explain why the magnitude of the voltage cannot be influenced by the frequency. After all, faster electrons will cross faster, hence more current.

 

[xts]

After all, faster electrons will cross faster, hence more current.

Nope.
Current = number of electrons per second.
Speed of those electrons have no influence on the current.
Like with transportation: people flow is proportional to number of buses leaving the station per hour. It doesn't matter how fast those buses drive on motorway then.

 

[ENgez]

I think i got it. The current in the metal is caused by the electron "holes" left behind by the electrons knocked off the metal. therefore it does not matter how fast the electrons the electrons are flying out, what matters is the number of holes.

more light intenstiy = more photons = more electrons knocked off = more "holes" = more current.

Correct?

 

[xts]

Yes.
You just must remember, that 'light intensity' is not a strictly defined term. You may mean it as 'number of photons per second' - then the current is proportional to the intensity.
Or you may understand it as 'energy transmitted per second'. And in this meaning, number of photons (and thus current) is proportional to number of 'intensity', but also reverse proportional to the frequency of photons.

So - what may sounds paradoxically to you - at the same energy flux, if the photons have higher frequency, the photolectric current is lower.