Photoelectric current VS frequency

[ImranParukZA]

Hello All, I have a problem that has been troubling me and I have been unable to find an answer on Google. It is said that 1 photon = 1 photoelectron ejected, Therefore the photoelectric current produced is directly proportional to the intensity of light used. It is also said that the frequency of the photons does not effect the photoelectric current produced. I disagree with this. If one source of light is used and an experiment of two different frequencies is done with a constant intensity of light (Frequencies above threshold frequencies of the metal), the second frequency is twice as high as the first. The kinetic energy of the photoelectrons produced by the higher frequency light used will be greater than the first, therefore in a vacuum it will travel from the cathode to the anode in a shorter time and since the photoelectric current is the amount of charge passing a point in a particular time, the amount of charge is constant due to the constant intensity while the time is shorter therefore a greater current is produced.

My teacher tells me I am incorrect by stating this and I am sure she is correct but she has no evidence to tell me why. Could someone please tell me where I am wrong.

Thanks.

 

[Jasso]

Since number of photons per time interval is only dependent on the intensity, then only intensity affects the current.

Higher frequencies may increase the energy of the ejected photoelectrons and make it cross a distance faster, but the time between each successive photoelectron remains the same because the time between each successive photon impact remains the same for the same intensity. It does not increase the total number of photoelectrons per photon.

 

[ImranParukZA]

Since number of photons per time interval is only dependent on the intensity, then only intensity affects the current.

Higher frequencies may increase the energy of the ejected photoelectrons and make it cross a distance faster, but the time between each successive photoelectron remains the same because the time between each successive photon impact remains the same for the same intensity. It does not increase the total number of photoelectrons per photon.

Thanks for the reply. I do understand that the number of photo electrons emitted do rely on the number of photons that hit the metal. 1 photoelectron emitted = 1 photon. I'm not sure if I phrased my question correctly...

Say for instance 10 photoelectrons are emitted in that experiment of two frequencies, The higher frequency would emmit the electrons at a higher velocity and cross the gap in a shorter time and since the current is directly inversely proprtional to time, it would increase the current would it not ?

If you look at this little applet from this website phet.colorado.edu/en/simulation/photoelectric. It seems to state the same thing, when you increase the frequency of light and keep the intensity constant the current increases.

Please just bear with me, my understanding of this section is very limited.

 

[mfb]

Say for instance 10 photoelectrons are emitted in that experiment of two frequencies, The higher frequency would emmit the electrons at a higher velocity and cross the gap in a shorter time and since the current is directly inversely proprtional to time, it would increase the current would it not ?

If the electrons have a higher velocity, there are fewer electrons flying at the same time. Simple example: Imagine a gap of 10m, and a person shooting 1 object per second across this gap. With a velocity of 1m/s, the objects have a spacing of 1m, there are always 10 objects in the gap at the same time, and the receiver will get 1object/s. With a velocity of 2m/s, the spacing is 2m, there are 5 objects in the gap, and the receiver will get 1object/s.The rate of incoming electrons is proportional to the rate of outgoing electrons, which is (roughly) proportional to the rate of incoming photons.

 

[sophiecentaur]

Say for instance 10 photoelectrons are emitted in that experiment of two frequencies, The higher frequency would emmit the electrons at a higher velocity and cross the gap in a shorter time and since the current is directly inversely proprtional to time, it would increase the current would it not ?

The current only depends upon the number of photons emitted per second.How long they take to get across the gap is not relevant. (Think of cars on a motorway, once the traffic has settled down at the lead on and lead off sections.)

However, the Einstein Photoelectric equation tells you the Maximum KE of photoelectrons for a given photon energy. Φ, the work function represents the minimum energy required to shift an electron from the surface. There will be a statistical distribution (Fermi?) of the actual energies required for 'potentially' photoelectrons so I should expect there to be a value of f, just near the threshold, where some but not all photons actually manage to shift an electron. As f increases, more of the photons would be expected to produce the effect until all the photons are successful, so I would expect to see some sort of curvature in the theoretical straight line, right where it intersects the x axis. I have tried to Google for a reference but not found one. Can anyone confirm or comment on this? Perhaps the effect is masked by experimental uncertainties.

Also, I believe a static DC voltage can affect the threshold frequency and I wouldn't be surprised if heating the metal target would make a difference too.

 

[mfb]

@sophiecentaur: I think you are right. However, the typical scale for those thermal fluctuations at room temperature is ~25meV, which is small compared to the typical photon energy of >1eV.
In addition, the energy of the free electrons can influence the fraction which reaches the readout, if some voltage is applied between metal and this readout.

But those are experimental issues - you can cool the metal and modify the geometry of the setup to reduce their effect.

 

[sophiecentaur]

@sophiecentaur:
But those are experimental issues - you can cool the metal and modify the geometry of the setup to reduce their effect.

Oh yes. But I was after exaggerating the effect!