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Experimental Value of Planck's Constant?

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Data obtained in a photoelectric experiment:
    Wavelength (nm): 140, 117, 93.3, 70.0
    Maximum Kinetic Energy of Photoelectrons (eV): 2.76, 4.50, 7.19, 11.61

    From the graph, determine an experimental value for Planck's constant (in joules-seconds).

    2. Relevant equations

    Frequency (c/wavelength): 21.43E14, 25.64E14, 32.15E14, 42.85E14
    Maximum Kinetic Energy of Photoelectrons (eV): 2.76, 4.50, 7.19, 11.61

    Conversion factor: 1.6E-19 J / 1 eV

    3. The attempt at a solution

    Plotting Kmax (y-axis) and Frequency (x-axis), generates the
    equation: y = 0.4132x - 6.0936

    h(Planck's constant) = slope * conversion factor * 1 / 10E14 (frequency factor)

    Experimental h value is: 6.6E-34, which is incorrect!

    Please help me find the experimental value of Planck's constant. Thank you!
     
  2. jcsd
  3. May 23, 2009 #2
    It looks like you got it right,Plancks constant is about 6.6 times ten to the power of minus 34 Js
     
  4. May 23, 2009 #3
    It says I'm within 10% of the correct answers when I submitted that answer =(. I'm running out of time, I really need help!
     
  5. May 23, 2009 #4
    10 percent might be in the expected range.Remember that all experiments have errors and uncertainties and you cannot pin anything down to an exact value.Estimate the errors in the experiment you did.What was the uncertainty in your measurement of wavelength etc?
     
  6. May 23, 2009 #5
    I'm sorry, I should have clarified that the data is generated by the hw, and wasn't performed experimentally myself. This is part of a online hw set, and they have already predetermined the correct answer. My answer is "within 10%" of the answer that they have.
     
  7. May 23, 2009 #6
    The answer they provided must have experimental error eg H=6.6etc+or -whatever the percentage error is.Did you get a spread of points on your graph?You should have.Did you draw the line of best fit avoiding any obvious anomolous points?you should have done that also.Did you find the maximum and minimum gradient and find the mean?Ten percent looks pretty good to me but check your points ,how well you drew the line and your calculations.
     
  8. May 23, 2009 #7
    Frequency (c/wavelength): 21.43E14, 25.64E14, 32.15E14, 42.85E14 <= x-axis
    Maximum Kinetic Energy of Photoelectrons (eV): 2.76, 4.50, 7.19, 11.61 <= y-axis

    The above are the data points. I plotted them using Excel and then generated a "linear trendline" with an equation. I am not good with my excel skills, how do I find the percent error? I think this is probably it. It has to do with +/- the precent error. I don't know how I would go about in finding the percent error from that set of points and how do I factor that into the final answer? Thank you so much for the quick reply!
     
  9. May 23, 2009 #8
    So sorry but there you have me stuck.I am useless with excel and computer stuff in general but could you not do it the old fashioned way, by hand.........draw the maximum and minimum gradient work out the average gradient and then work out the percentage difference between the average and the maximum or minimum(which will be the same)
     
  10. May 23, 2009 #9
    I'm sorry but I don't know how to work it out using the minimum and maximum gradient and how to find the percent error. Are you referring to the least squares method? If so, I worked it out to find the slope by hand. I still don't know how to find the percent error and incorporate it into the final answer though. I haven't done statistics in a very long time =(.

    (X) | (Y) | (Xi - Xmean) | (Xi-Xmean)(Yi-Ymean)
    21.43 | 2.76 | 82.6281 | 34.13295
    25.64 | 4.5 | 23.8144 | 9.8332
    32.15 | 7.19 | 2.6569 | 1.10025
    42.85 | 11.61 | 152.0289 | 62.82135

    Summed Total: 261.1283 | 107.88775

    Slope = 107.88775 / 261.1283 = 0.41316 (the same slope I got using Excel's linear trendline)
     
  11. May 23, 2009 #10
    You don't need to be too fussy but you can make a reasoned estimate from your graph.I will illustrate by putting in some numbers.I am assuming you already have the line of best fit:
    1.From your graph draw in what you judge to be the steepest line ie the line with the maximum gradient.Find the gradient and let G max=6
    2.Do the same for the minimum gradient.let G min=4
    3.Work out the average value ie G ave =5(which should be the gradient of the line of best fit)Now you know that the gradient =5+or-1, in other words 5+ or-20%.Present your answer with the estimated percentage error and round it off with the appropriate number of significant figures.With such a small amount of data and no details of the experimental set up a rough estimate is the best you can do.
     
  12. May 23, 2009 #11
    My friend and I tried: 6.0E-34 (way off), 6.4E-34 (within 10%), 6.5E-34 (within 10%), 6.6E-34 (within 10%). Unfortunately, I have to get to work now. I only have 1 last attempt and I'll just guess 6.9E-34 when I return. Thank you for all your help Dadface, but it looks like this one problem I won't be able to get credit for despite knowing the concept.
     
  13. May 23, 2009 #12
    I managed to find some graph paper and I plotted the graph of K.E. against frequency.I found the gradient to be close to 0.4(11.61 divided by42.85-14).Multiplying by 1.6 times ten to the power of minus 19 gives a value for h of 6.4 times ten to the minus 34(this is about 3% away from the published value).All four point on the graph lie close to the line and I estimated the error in finding the gradient to be about 1%,but of course we don't know what the other errors were.It seems that you had it right or nearly right all the time.
    Since all four points lie close to the line you can get the gradient more easily and accurately from the data itself(well I can my eyes are not as good as they used to be).Using the first and last set of data I found the gradient to be 11.61-2.76 divided by42.85-21.43.This gives a value of h of 6.6 times ten to the power of minus thirty four.Anyway good luck with it.
     
    Last edited: May 23, 2009
  14. May 24, 2009 #13
    Omg, you really didn't have to take out a graph sheet and do it for me!! Thank you so much though!! However, because I already tried 6.4E-34 and 6.6E-34, despite it being the APPROPRIATE CORRECT answer, this "online hw" indicate that it was wrong =(. I guessed 6.9E-34, and it was CORRECT! This "online hw" thing is really stupid and pisses me off because understanding the concept here is more important, and the fact that we are limited in our "attempts" at the problem doesn't help us understand the assignment any better. Although, I can't believe the extent that some members of this forum would go to help a physics student with his/her assignment. Thank you again Dadface!!
     
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