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Photoelectric Effect and atoms

  1. Jun 23, 2008 #1
    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?

    I'm a bit confused here... when it says an atom can continually absorb light, does this mean that you can keep feeding it energy until it escapes from the metal? I'm assuming you can do this...

    okay so the atom is 0.1 m away from the light source, so if you draw straight lines from the point source to the top and bottom of the atom, you get an isoceles triangle, two of the lengths which are (very close to) 0.1 m and the base which is just the diametre of the atom, 2 angstroms. Now, since the angle theta, which we define to be the vertex angle of this isoceles triangle, is extremely small, we can approximate it by saying

    0.1 * sin theta = 2 e - 10
    0.1 * theta = 2 e - 10
    theta = 2 e - 9 rad

    Now, this means that the atom receives

    2e-9 / (2 pi) * 25 joules/sec of energy.

    But a total of 4 eV is needed, so we have that the total time taken would be 2.01 e -11 sec.
     
  2. jcsd
  3. Jun 23, 2008 #2

    mgb_phys

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    Strange question.
    Since it's a point source I think you should assume that the light goes into a sphere.
    You can work out the J/s/m^2 at the target and so the photon/s/m^2.

    The photo-electric effect says that you need 1 photon of the correct wavelength to emit an electron - unless you are going to do a QM calculation of the electron I can't see how you get the time - except just 'c' and 0,1m distance?
     
  4. Jun 23, 2008 #3
    Well, see, the thing is that one photon doesn't have enough energy to release the electron. So you need another photon (it is really strange, because this would never happen in reality... the question is hypothetical).
     
  5. Jun 23, 2008 #4
    [tex]4*\pi*r^{2}[/tex] is the surface area of a sphere. If the plate is a certain distance away, you can simply assume the the source is emitting isotropically. Divide the surface area of the plate by the spherical surface area (r= the distance from the plate). This will tell you what percentage of your photons are hitting the surface.

    You can determine the energy departed by assuming the photon is completely absorbed (hence photoelectric absorption). The energy of a photon is [tex]E=h\nu[/tex] and [tex]c=\lambda\nu[/tex]. You probably need the density of the plate and some other things, but this should be enough to get you going.
     
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