Physics Help Forum Photoelectric effect error value?

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 Mar 12th 2011, 11:18 AM #1 Junior Member   Join Date: Mar 2011 Posts: 7 Photoelectric effect error value? Hi, In my photoelectric effect experiment, I used stopping voltage Vs frequency to calculate planks constant. my calculation of h = 6.422x10^(-34)j/s which resulted in % error of 2.8%. Now, my highest relative error from the stopping voltage obtained from the multimeter is 0.15%..... How do I justify my value of planks constant is within the 'total experiment error range'? Please.. Thanks Last edited by wolfhound; Mar 12th 2011 at 11:46 AM.
Mar 12th 2011, 12:56 PM   #2

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 Originally Posted by wolfhound Hi, In my photoelectric effect experiment, I used stopping voltage Vs frequency to calculate planks constant. my calculation of h = 6.422x10^(-34)j/s which resulted in % error of 2.8%. Now, my highest relative error from the stopping voltage obtained from the multimeter is 0.15%..... How do I justify my value of planks constant is within the 'total experiment error range'? Please.. Thanks
I presume you are using eVs = hf - (phi) where Vs is the stopping voltage and (phi) is the work function.

I can guess the answer, though I would like to know how you calculated the error in h. Notice that you are multiplying Planck's constant by the frequency, which I assume you measured. So the error in h is going to be affected by the error in both Vs and f.

-Dan
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Mar 12th 2011, 05:14 PM   #3
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 Originally Posted by topsquark I presume you are using eVs = hf - (phi) where Vs is the stopping voltage and (phi) is the work function. I can guess the answer, though I would like to know how you calculated the error in h. Notice that you are multiplying Planck's constant by the frequency, which I assume you measured. So the error in h is going to be affected by the error in both Vs and f. -Dan
Hi yes ,I am using this eVs = hf - (phi)
I was giving the frequencys from a book ,so no error with them.

To calculate the %error in h.. i done (accepted value- experimental value)/(accepted value) x 100

my calculation of h was obtained using slope m from the graph stopping voltage Vs frequency

m=he==> h=m/e

Last edited by wolfhound; Mar 12th 2011 at 05:27 PM.

Mar 12th 2011, 10:16 PM   #4

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 Originally Posted by wolfhound Hi yes ,I am using this eVs = hf - (phi) I was giving the frequencys from a book ,so no error with them. To calculate the %error in h.. i done (accepted value- experimental value)/(accepted value) x 100 my calculation of h was obtained using slope m from the graph stopping voltage Vs frequency m=he==> h=m/e
I presume the work function and Planck's constant were both tabulated as part of a linear regression? Is that here you got the 2.8% number?

-Dan
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Mar 13th 2011, 06:17 AM   #5
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 Originally Posted by topsquark I presume the work function and Planck's constant were both tabulated as part of a linear regression? Is that here you got the 2.8% number? -Dan
I don't understand what you mean by this?

heres how I got 2.8%
I compared my calculated value for h- with the real planks constant from a book.

% error ={6.62x10^(-34) - 6.427 x 10^(-34)/(6.62 x 10^(-34))} x 100 = 2.8%

Mar 13th 2011, 11:15 AM   #6

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 Originally Posted by wolfhound I don't understand what you mean by this? heres how I got 2.8% I compared my calculated value for h- with the real planks constant from a book. % error ={6.62x10^(-34) - 6.427 x 10^(-34)/(6.62 x 10^(-34))} x 100 = 2.8%
Yes, but you said you got your value of h as the slope of a line. I presume you had some kind of data set and graphed it to get the slope?

-Dan
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Mar 13th 2011, 02:44 PM   #7
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 Originally Posted by topsquark Yes, but you said you got your value of h as the slope of a line. I presume you had some kind of data set and graphed it to get the slope? -Dan
Hi, Yes I plotted the stopping voltages against the frequencies of 5 colours of the mercury lamp,
I then obtained a slope m ,used the equation m=h/e => h=me

Mar 13th 2011, 06:09 PM   #8

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 Originally Posted by wolfhound Hi, Yes I plotted the stopping voltages against the frequencies of 5 colours of the mercury lamp, I then obtained a slope m ,used the equation m=h/e => h=me
I guess that's what's confusing me. "h = me" What is m? If m is the slope how did you calculate it? Did you perhaps measure it off the graph? But if that's true why are you multiplying the slope by e? The slope should be h itself...

-Dan

Edit: Wait. I think I understand the factor of e. You plotted the stopping voltages, not eVs. But still, the question remains about how you found the slope?
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Last edited by topsquark; Mar 13th 2011 at 06:11 PM.

Mar 14th 2011, 06:40 AM   #9
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 Originally Posted by topsquark I guess that's what's confusing me. "h = me" What is m? If m is the slope how did you calculate it? Did you perhaps measure it off the graph? But if that's true why are you multiplying the slope by e? The slope should be h itself... -Dan Edit: Wait. I think I understand the factor of e. You plotted the stopping voltages, not eVs. But still, the question remains about how you found the slope?
The slope is not h.
The slope was obtained from excel.
I can't believe you don't know what I am talking about I explained it 3 or 4 times.

Mar 14th 2011, 10:42 AM   #10

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 Originally Posted by wolfhound The slope was obtained from excel.

So. You got the slope using linear regression then. Here is a rundown on how the different variables are computed. The error can be tabulated in two different ways: using the sum of the square of the vertical offsets or by using the correlation coefficient. I am assuming that the cause of the larger percent error has to do directly with the vertical offsets. These "move the fit line" around to find the slope and intercept with the smallest sum of squares error. That's why your standard errors (the +- in the voltage measurements) are lower than the % error in h...The sum of squares is not always the best of the fits to use. (The article I referenced mentions the perpendicular offsets, which I think is a slightly better fit. The downside is that the error is harder to calculate.)

-Dan
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