# How do you calculate the time to emission of an electron from a metal given the incident radiation?

Here's the question:

A monochromatic point source of light radiates 25 W at a wavelength of 5000 angstroms. A plate of metal is placed 100 cm from the source. Atoms in the metal have a radius of 1 angstrom. Assume that the atom can continually absorb light. The work function of the metal is 4 eV. How long is it before an electron is emitted from the metal?

My attempt at an answer. I used the 100 cm placement away from the source for the radius and put that in to $4\pi r^2$ to get the total surface area. Divided the energy of the source by the total surface area. Multiplied that by the square of the atom spacing to get how much energy per sec was falling on that atom. I then used the work function of 4eV divided by the energy falling on that area per sec to find out how long it takes to reach the work function energy level.

My only issue is, is that the time they are looking for or is there some other concept I am missing. I know using the normal photo electric effect theory it is hf - work function= energy of emitted electron. If I stop timing at the point I reach the work function value then in theory would I not be emitting an electron with 0 energy. (tech impossible).

If its absorbing continuously would I add the time to allow for one more wave cycle above that of the energy and time needed to get to the work function.

-
I don't understand the question: 500nm light has an energy of 2.48eV and will not eject any photoelectrons if the work function is 4eV. – John Rennie Sep 7 '12 at 8:20
Its because it wants you to imagine it "continuously" absorbs light energy. – aaron burns Sep 8 '12 at 16:58
I believe the key is to treat the atom like a blackbody radiator. When the energy is low you won't get electrons emitted but you will get photons reemitted. When a photon of high enough energy is emitted then an electron will be emitted. I say this because I have the same book and the section just before it deals with blackbody radiation. – cspirou Mar 6 '13 at 1:27
Your calculation seems to be along what the author of the question seemed to want. However, this way you will get much greater value than the real interval according to experiments. This is because this reasoning is based on a flawed idea that atom gets its energy by absorbing energy that passes through its cross-section according to the prescribed 25 W. In reality intensity of radiation may be much higher due to interactions of the atoms in the metal and presence of background radiation. See also Marty Green's answer. – Ján Lalinský Jan 30 '14 at 12:39

## 5 Answers

You use the photoelectric effect to calculate the amplitude and probabilities that a photon enters an atom. They said to you the surface and the energy, so now calculate the probability.

-
This doesn't really contribute to the question in any way. – udiboy1209 Sep 2 '13 at 10:53

Since the wave function of an electron in the conduction band can occupy the entire volume of the metal plate, the right way to calculate the expected time to eject an electron would be to take the power density of the classical light and multiply it by the area of the whole plate. The cross-section of a single atom is irrelelevant, although people have used it for years to debunk the wave theory of light.

-

I think the question is trying to get you to think about the differences between what you might think under the classical and intuitive model of the photoelectric effect and what actually happens.

Einstein won the Nobel Prize in 1921 for his work in precisely this area: he found, among other things, that no matter what the intensity, as long as there is enough energy to overcome the work function (i.e., as long as the frequency of incident light is greater than the cutoff frequency), photoelectrons will be emitted immediately.

-

I think it requires a single atom to absorb 2 photons at the same time(upto the frequency of the light particle) to emit the electron. Can you calculate this probability from the data given? Now this is the only reasonable conclusion I can make because otherwise as some one pointed out there is no emission of photo electrons.

-

Remember that you are dealing with a quantum phenomenon. As you said, you know that the quantum theory of the photoelectric effect tells you $hf - W = E_e$. So think about that: what does the term $hf$ represent the energy of? How is the energy of the incident radiation distributed? Is it possible for an electron in a metal to continuously absorb radiation?

Note that this is kind of a trick question, in the sense that the way it is worded may be a little misleading as to what sort of answer you are supposed to be getting.

-
The question wants you to assume the atom is continually absorbing contrary to the know mechanics of the situation. – aaron burns Sep 6 '12 at 18:16
Oh, then it would help to specify that explicitly in the question. And it shouldn't have the quantum-mechanics tag on it. So, is your last paragraph in the question the main thing you'd like to ask? – David Z Sep 6 '12 at 18:30
the first part paragraph is the question and it is out of a quantum mech book. liboff chapter 2. I don't want to argue with you but you edited my question and the first paragraph says ...absorbs continuously. Thats the reason I was having trouble. – aaron burns Sep 6 '12 at 19:26
Oh, I see. Sorry, I missed that part when I read it. Anyway, what I was getting at in my last comment is that it's not exactly clear to me what you are asking. I can see what the question you are given is, but I'm not sure what exactly you want to ask about it - in other words, what exactly is confusing you about how this problem is to be solved? I'm trying to figure this out so I can edit my answer accordingly. – David Z Sep 6 '12 at 20:02
Im confused because if the work function is 4eV and I find the time it takes to get one atom to 4eV then when the electron is emitted it will have 0 energy because of hf-work function=energy of emitted electron. I think it just wants me to find the time until the work function value is reached but it does not clearly ask for that. – aaron burns Sep 6 '12 at 22:18