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Photoelectric effect and determining the planckconstant

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
    In an experiment with the photoelectric effect, the kinetic energy of the expelled electrons was measured. Draw a graph using this data and find the planckconstant.

    Data:
    f(1014)hz : 5, 7, 9, 11
    Ek(10-18)J: 0.07,0.19,0.3,0.43

    2. Relevant equations
    well, Ek + W = hf where Ek is the kinetic energy of the electron, W= the energy needed to liberate it and hf the energy of the photon hitting an atom.


    3. The attempt at a solution

    As far as I know, one can simplify the above formula to Ek=h(f-fw) where fw is the frequency needed to liberate the electron in the first place. To determine planck's constant I thought about ΔEk/Δ(f-fw) but I'm a bit confused since Ek isn't proportional to f, but to (f-fw) I'm not sure if that's going to make any difference tho..

    So, am I on the right track?

    thx 4 all help in advance.
     
  2. jcsd
  3. Oct 12, 2011 #2
    Did you draw the graph? That could be helpful.

    Anyway, look at the equation you wrote:
    [itex]K=h\nu-\phi=h\nu-h\nu_f[/itex]
    What's the relationship between the frequency [itex]\nu[/itex] and the energy?
     
  4. Oct 12, 2011 #3
    v=K/h + vf

    vf remains constant while K increases, but when v increases by a factor of 10, this doesn't mean that K will increase by a factor of 10...
     
    Last edited: Oct 12, 2011
  5. Oct 12, 2011 #4
    Electron/Positron Impact & Photon Energies

    BTW, can you help me with another thing? "An electron and a positron are moving against each other. at the point of impact both have the velocity 0.6c

    Two gamma photons are sent out. What is each of theirs energy?"

    Well, the sum of momentum equals zero in both cases and thus the momentum of each of the photons must equal [lorentzfactor]*9.11*(10^-31)*1.8*10^8 = 0.615*10^-13. But the book says I'm wrong with the momentum. Please help with this...

    You'd better move the second question and write it in a new topic.

    Returning to the photoelectric problem,
    [itex]K=h\nu-h\nu_f[/itex] is in the same form as [itex]y=mx+q[/itex], where m=h.

    What does this suggest you?
     
    Last edited by a moderator: Oct 12, 2011
  6. Oct 13, 2011 #5
    ah right. thanks, i forgot about that. so h equals the growth rate
     
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