Go Back   Physics Help Forum > College/University Physics Help > Nuclear and Particle Physics

Nuclear and Particle Physics Nuclear and Particle Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Mar 12th 2011, 10:18 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 10:46 AM.
wolfhound is offline   Reply With Quote
Old Mar 12th 2011, 11:56 AM   #2
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,387
Originally Posted by wolfhound View Post
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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.
topsquark is offline   Reply With Quote
Old Mar 12th 2011, 04:14 PM   #3
Junior Member
 
Join Date: Mar 2011
Posts: 7
Originally Posted by topsquark View Post
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 04:27 PM.
wolfhound is offline   Reply With Quote
Old Mar 12th 2011, 09:16 PM   #4
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,387
Originally Posted by wolfhound View Post
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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.
topsquark is offline   Reply With Quote
Old Mar 13th 2011, 05:17 AM   #5
Junior Member
 
Join Date: Mar 2011
Posts: 7
Originally Posted by topsquark View Post
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%
wolfhound is offline   Reply With Quote
Old Mar 13th 2011, 10:15 AM   #6
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,387
Originally Posted by wolfhound View Post
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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.
topsquark is offline   Reply With Quote
Old Mar 13th 2011, 01:44 PM   #7
Junior Member
 
Join Date: Mar 2011
Posts: 7
Originally Posted by topsquark View Post
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
wolfhound is offline   Reply With Quote
Old Mar 13th 2011, 05:09 PM   #8
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,387
Originally Posted by wolfhound View Post
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?
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

Last edited by topsquark; Mar 13th 2011 at 05:11 PM.
topsquark is offline   Reply With Quote
Old Mar 14th 2011, 05:40 AM   #9
Junior Member
 
Join Date: Mar 2011
Posts: 7
Originally Posted by topsquark View Post
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.
wolfhound is offline   Reply With Quote
Old Mar 14th 2011, 09:42 AM   #10
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,387
Originally Posted by wolfhound View Post
The slope was obtained from excel.
That was the part you didn't tell me. I asked about this in post #4 in this thread.

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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.
topsquark is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > Nuclear and Particle Physics

Tags
effect, error, photoelectric



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Quaries about photoelectric effect spixmacaw Quantum Physics 1 Nov 3rd 2012 10:38 AM
PhotoElectric effect Steve George Quantum Physics 3 May 16th 2012 04:27 PM
Photoelectric Effect. Akshay Quantum Physics 3 Sep 11th 2009 10:43 AM
maximum photocurrent ejected by electrons in photoelectric effect mola Nuclear and Particle Physics 1 Jun 13th 2009 12:07 PM
Photoelectric effect werehk Light and Optics 1 Aug 17th 2008 05:21 PM


Facebook Twitter Google+ RSS Feed