Photo electric effect review liboff ch2

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Homework Statement



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?

Homework Equations



My attempt at an answer. I used the 100 cm placement away from the source for the radius and put that into 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.


The Attempt at a Solution



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.
 
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5000 A = 500 nm => 2.5 eV. Which is lower than the work function. Is the photoeffect possible?
 
The problem statement says to "Assume that the atom can continually absorb light." It's not meant to be realistic.
 
Ah. I missed that statement completely. Then the proposed solution seems rational.
 
Yes this is the problem. The problem wants you to discount the reality of the situation and assume that the atom would continually absorb energy. So at what point would it emit an electron. Once it got to the work function level or some time after.
 
There is no more useful information in the problem. All you can do is compute the lower bound for time.
 
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