TSFX 2010 exam Question

[tridol]

In AoS 2, Question 3:
"Which of the lights had the greater intensity: Blue, Ultraviolet or Unable to determine?"
There is a graph for the question of photocurrent versus voltage, for blue and ultraviolet light. Blue light has a smaller stopping voltage and a larger maximum photocurrent than ultraviolet.
At first my answer was Blue, as it had a larger photocurrent, yet the answer states:
Unable to determine. High frequency light has photons of higher energy, so less photons are needed to provide an equivalent intensity of radiation. Less photons would result in a lower photocurrent. Therefore, because Blue has a higher photocurrent doesn't necessarily imply it has higher intensity.

I'm having a little trouble getting my head around why this is so, my textbook as far as i know does not touch on the differences in photocurrent for lights of different frequency, only that photocurrent is proportional to intensity for a light of same frequency. I assumed that since each photon releases one photoelectron from the metal, that blue must have a larger intensity. How does photocurrent relate to number of electrons emitted?

Please help

[Lasercookie]

For a given metal and frequency of incident radiation, the rate at which photoelectrons are ejected is directly proportional to the intensity of the incident light.Going by that, What TSFX said is correct. Very few questions seemed to explicitly question it like this one though.

If you look closely at the textbook (I checked and couldn't find any counterexamples), they all followed that same rule. If you can find one, then please post it here, I'd like to see it. Also take note that the I in the graphs you see stands for current, not intensity.

Well current is moving charge (electrons), so if you have more charge (more electrons) you'd have greater current. So more electrons emitted means greater photocurrent.

So that TSFX graph showed that UV had lesser photocurrent than that of the blue light. To have more current, you'd need more electrons being emitted. One photon can only eject one electron - energy is required to eject an electron. Energy is dependent on frequency (E=hf) and not intensity. Because of this we cannot compare UV and Blue light. I hope that makes sense.

However, if the frequencies are the same, then we know that intensity is comparable.

[tridol]

So what you are implying is that if both light sources have equal intensity (and assuming work function is low enough), blue light of lower energy will have more photons and so a larger photocurrent than UV light (which would have larger Ek of electrons in the vacuumed tube)? I may be confused in the meaning of Light intensity, my current definition is that intensity is the number of photons that pass a point per second. Is this correct, or is it only the Power per square metre (Wm^-2)?

[Lasercookie]

Well first off, this is outside the course, we are talking about electrons other than the 'most energetic' here.

Intensity is the number of photons in the beam. More electrons emitted will result in greater photocurrent. If the frequency and metal are the same, then the only variable is intensity (the number of photons). This results in more photons that can eject electrons and therefore higher photo current.

However, if we start talking about different levels of energies:

You know how the electrons are located in different energy levels. If you have fixed intensity, then you have the same number of photons. The energy in each photon remains the same. So if you had an electron that's deep in the atom, that would require more energy to knock out than the valence electrons. Light with lower energy might not have the energy to knock this out. Light with higher energy will be able to reach that electron and knock it out.

I think the TSFX graph is wrong? UV has a lower wavelength than blue light, so it would have more energy. Since it has more energy, more of the deeper electrons can be knocked out and contribute to the photocurrent.

I found a few sources that back up what I'm saying, but this explanation was the clearest and direct (and didn't go off into a tangent about probability theory).

Not every photon emits an electron, even if the photons have enough energy to emit electrons.  If a photon is absorbed by an electron with binding energy greater than the photon energy, the electron will not be released.  Photons with higher energies are more likely to release electrons because a greater proportion of the electrons in the metal have binding energy less than the photon energy.  Therefore, as you increase the frequency, the number of emitted electrons (and therefore the current) will increase until all photons are emitting electrons.  Note that this behavior is different from the simplified model used by many textbooks, in which every photon with frequency greater than the threshold frequency releases an electron, so the current is constant above the threshold frequency

[tridol]

The graph definitely shows that UV light has a lower photocurrent, so it could be in error. If this is the case then: with the same intensity, both light sources have the same number of photons that strike the metal. Yet since UV light has higher energy, a larger proportion of photons will release an electron and so would result in a larger photocurrent?

[Lasercookie]

The graph definitely shows that UV light has a lower photocurrent, so it could be in error. If this is the case then: with the same intensity, both light sources have the same number of photons that strike the metal. Yet since UV light has higher energy, a larger proportion of photons will release an electron and so would result in a larger photocurrent?

Yeah, UV light has more energy, so the photons have the sufficient energy to remove more electrons. How large the difference would be, I'm not sure.

[tridol]

Sweet, thanks for clearing this up