Help on photoelectric effect (1 Viewer)

[noobynewby]

This question is from success one, but i dont understand the answers, can someone please explain them to me? It is part c) i do not understand

During an experiment, a beam of ultraviolet light, of wavelength 200nm, is incident on a target metal sheet. The UV light releases photoelectrons from the metal. The maximum kinetic energy of the photoelectron was measured.

a) calculate the energy of each photo in the ultraviolet light
b) explain why the measured maximum kinetic energy of the photoelectrons would be less than the answer you calculated in a)

c) If a second beam of ultraviolet light, of equal intensity to the original ultraviolet beam, but of a shorter wavelength, was used in the experiment, compare the photoelectrons released by the original beam with those produced when the second ultraviolet source is used.

The answer from the book is:

the frequency of light has increased and this means that the energy of the photon is larger. The new beam will liberate photoelectrons with a greater maximum kinetic energy, but because the intensity of the beam is equal to the original, the total energy per unit area is unchanged and, because each photon has a higher individual energy, there will be less photoelectrons being liberated from the surface than was the case with the original UV source.

Why is there less photoelectrons being emitted, i dont get that part

Thanks

 

[Sastrawan]

Yeah, that seems wrong to me. Could you say which exam it's from, because we might be able to cross-check with the HSC marking guidelines...

 

[noobynewby]

Yeah, that seems wrong to me. Could you say which exam it's from, because we might be able to cross-check with the HSC marking guidelines...

its actually from one of the questions that success one set, not an actual HSC question

 

[Sastrawan]

Damn it, that means the only thing we have to go on is Success One's word...

This is going to be tricky. I learned, in this course, then when dealing with the photoelectric effect, that the intensity of the incident is proportional to the number of photons. With this logic, if you had the same intensity of photons with increased frequency (hence, increased energy), as you do here, then the total energy imparted would be greater.

However, looking back over my notes, and Wikipedia's over 9000 definitions of the word "intensity", it seems as though it means something different, or at least I was thinking of something different when I wrote those notes. It seems that this is the case: "For constant frequency/energy, if you increase the number of photons, you increase the intensity". What is happening here is by increasing the number of photons, you increase the density of photons, which seems to be proportional to intensity (and incidentally, to the current produced).

So perhaps "intensity" means "total energy per unit area". I think this fits with the observations with stopping potential and all that jazz.

Thus, when SuccessOne says "increase the energy of EACH photon, but maintain the intensity of the whole beam", it means, increase the energy of each photon, but maintain the total energy per square millimetre of the incident beam. To do that, you need to reduce the number of photons per unit area. So there will be fewer photoelectrons emitted per unit area [because of that awkward stipulation "maintain the same intensity"].

Is that somewhat coherent? I'm not very happy about this either, but there's an explanation that, at least in the first few minutes of the new day, seems to work alright. But why not ask some more people and see how they put it?

 

[noobynewby]

Damn it, that means the only thing we have to go on is Success One's word...

This is going to be tricky. I learned, in this course, then when dealing with the photoelectric effect, that the intensity of the incident is proportional to the number of photons. With this logic, if you had the same intensity of photons with increased frequency (hence, increased energy), as you do here, then the total energy imparted would be greater.

However, looking back over my notes, and Wikipedia's over 9000 definitions of the word "intensity", it seems as though it means something different, or at least I was thinking of something different when I wrote those notes. It seems that this is the case: "For constant frequency/energy, if you increase the number of photons, you increase the intensity". What is happening here is by increasing the number of photons, you increase the density of photons, which seems to be proportional to intensity (and incidentally, to the current produced).

So perhaps "intensity" means "total energy per unit area". I think this fits with the observations with stopping potential and all that jazz.

Thus, when SuccessOne says "increase the energy of EACH photon, but maintain the intensity of the whole beam", it means, increase the energy of each photon, but maintain the total energy per square millimetre of the incident beam. To do that, you need to reduce the number of photons per unit area. So there will be fewer photoelectrons emitted per unit area [because of that awkward stipulation "maintain the same intensity"].

Is that somewhat coherent? I'm not very happy about this either, but there's an explanation that, at least in the first few minutes of the new day, seems to work alright. But why not ask some more people and see how they put it?

Oh i think i get it now, i think its just the way i misinterpreted the question xD

Thanks