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Homework Help: Photoelectric effect

  1. Oct 21, 2014 #1
    1. The problem statement, all variables and given/known data
    I am doing the photoelectric effect experiment. And we were trying to verify the relationship between photoelectric current and distance is
    [itex]I \times d^2 = k[/itex], where k is a constant.(At least I believe this to be true.)

    But from the data, I kind of get a [itex]I= k d^{-2} + b [/itex]. And I got a really big b...
    I was using light of [itex]\lambda = 436nm,[/itex] and diaphragm of [itex]\phi=2mm[/itex]. Here is the data.

    i = [310, 198, 131] (10^-10A)
    d^-2 = [1./30**2, 1./35**2, 1./40**2] (cm^-2)
    And I got a relation like[tex]I = 369221.118818 d^{-2} - 101.137960591 [/tex]. I can't understand why is the b like this, big and negative.

    2. Relevant equations

    3. The attempt at a solution

      • I am not alone. My mates have the same problem. And I checked with many of them.(I realize that three points is too small for data...)

      • If distance grows really big, then current should curve. But that should put b higher than origin, while I got a minus sign. Also the line is really really linear.

      • The "apparatus" was OK. Not great, but OK. I measured Planck constant right (to one part in a thousand), and I verified relation between I and \phi (of diaphragm), which has a interception of 0(to one part in a thousand again).

      • My pals and myself can only come up with reasons for the interception to be positive.

    4. Thanks in advance!
    Last edited: Oct 21, 2014
  2. jcsd
  3. Oct 21, 2014 #2


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    What distance do you speak about? Distance of what from what???

  4. Oct 21, 2014 #3
    The distance between the photon emitter (the bulb) and the electron emitter (the metal).

    But strictly speaking, both of them have protective coverings. So I was really measuring the distance between them.
    (There are arrows supposed to imply the exact position of the real emitters.)
  5. Oct 21, 2014 #4


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    So it can be imagined that your measured distance (dm) values are less then the real ones. d=dm+x.

    Then instead of the theoretical intensity I=I0/d2 you measure I(dm). I(dm) = I0/(dm+x)2.
    You can assume that x<<dm.
    You fit a function to the I(dm) data. [tex]I=\frac{I_0}{dm^2(1+\frac{x}{dm})^2}[/tex] If x/dm<<1 the function can be approximated as [tex]I=\frac{I_0}{dm^2}(1-2 \frac{x}{dm})[/tex],
    that is [tex]I=I_0/dm^2 - A[/tex] where A is some average of 2xI0/dm3.

  6. Oct 21, 2014 #5
    Thanks so much!

    This idea came up earlier today. But I didn't find it plausible. For if x << dm, then we get to keep the linearity, but A then must be suffiently small. Note that it takes quite some distance to reach 100e-10 A. One the other hand, if x is only somewhat smaller than dm, then it's hard to keep the linearity.(At least that's what I thought!)

    But I just take the leap and draw the curve. The second guess turns out to be a correct line for the eye. So I find my answer.
  7. Oct 21, 2014 #6


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    I just looked at your data. They are very few and the distances very close to get a really good fit. You should have measured at least at 5 distances, and in a broader range, starting from about 15-20 cm.

  8. Oct 21, 2014 #7
    I'll try to do that today.
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