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Photoelectric effect and a metal plate

  1. Feb 5, 2014 #1
    First of all, Hi we haven't even learned the E = ((hc)/λ) in the physics class but we have this question on our test-exam and I hope that it's fairly easy:

    1. The problem statement, all variables and given/known data

    A metal plate is shone upon with a monochromatic light. When the wave-length is 550 nm a photo-electric effect is observed where the fastest photo electrons have the energy of 0.45 eV. When the light have 410 nm wave-length the quickest photo electrons get the energy 1.26 eV. Decide the planck constant with help of these values.

    First λ = 550 x 10^9 m
    First E = 0.45 eV
    Second λ = 410 x 10^9 m
    Second E = 1.26 eV

    2. Relevant equations

    E = h*v
    λν = c

    3. The attempt at a solution

    Ok, so I start out with, I'm not sure if I even is to use the conversion of eV to J so I can find the constant but it's the only way I can see that we can do this:

    0.45 x (1.60217657 × 10^-19) = E

    c/λ= v

    (3*10^8)/(550 x 10^9) = v

    E/v = h

    (0.45 x (1.60217657 × 10^-19))/((3*10^8)/(550 x 10^9)) = h

    1.3217957e-16 Js = h

    Which doesn't look right. But I guess this has to do with photoelectric effect which haven't learned anything about yet, and that I should use the two values and wavelengths in co-operation in some form to get the Plancks constant.

    Anybody who could make me see what to do?

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 5, 2014 #2

    rude man

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    Homework Helper
    Gold Member

    You need an additional parameter, to wit, the work function of the metal.

    Basically, E = hf - W
    where W is the work function of the metal, f is the frequency of the impinging light, and E is the maximum observed kinetic energy of the emitted electrons. E cannot be negative.

    So you have 2 equations and 2 unknowns: h and W. Solve for h.
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