Use linear regression to find Planck’s constant

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member 731016
Homework Statement
Please see below
Relevant Equations
E = hf
I am trying to find Planck's constant using Excel given the data:

Frequency [Hz]Photon Energy [J]
7.5E+14​
4.90E-19​
6.7E+14​
4.50E-19​
6E+14​
4.00E-19​
5.5E+14​
3.60E-19​
5E+14​
3.30E-19​
4.6E+14​
3.00E-19​
4.3E+14​
2.80E-19​
4E+14​
2.65E-19​
3.75E+14​
2.50E-19​
I am using Linear regression and I have set the intercept to zero since photon energy is directly proportional to frequency from Planck's Law

My graph is,
1678592068711.png

However, why is the linear regression giving an incorrect value for h? According to the data Planck's Constant is ##7 \times 10^{-34} Js ## but it meant to be ##6.63 \times 10^{-34} Js##

Many thanks!
 
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Callumnc1 said:
However, why is the linear regression giving an incorrect value for h? According to the data Planck's Constant is ##7 \times 10^{-34} Js ## but it meant to be ##6.63 \times 10^{-34} Js##

Many thanks!
You can tell the spreadsheet software how many sig figs to show in the formula.
 
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haruspex said:
You can tell the spreadsheet software how many sig figs to show in the formula.
Thank you for your reply @haruspex!

I will try 3 sig figs just like the Planck's constant then.

Many thanks!
 
Callumnc1 said:
Thank you for your reply @haruspex!

I will try 3 sig figs just like the Planck's constant then.

Many thanks!
Thank you @haruspex!

I have done that and got:
1678596242696.png

That is probably the most accurate experimental value for h I'm going to get.

Thank you!
 
Callumnc1 said:
Thank you @haruspex!

I have done that and got:
View attachment 323496
That is probably the most accurate experimental value for h I'm going to get.

Thank you!
Sorry that is 4 sig fig. To 3 sig fig it is 6.60.

Many thanks!
 
Depending on the experiment I would say that 7 is a very good result, because it is less that 6% greater than the value in tables :)
 
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malawi_glenn said:
Depending on the experiment I would say that 7 is a very good result, because it is less that 6% greater than the value in tables :)
Thank you for your reply @malawi_glenn !
 
How did you collect these data? It is unlikely that you measured the energy and frequency independently. If they are obtained from a photoelectric effect experiment, the honest treatment would be to plot frequency vs. photoelectron energy (or stopping voltage) and extract both the work function and Planck's as the intercept and slope of the regression line. If, as you may have done here, you pre-calculate the work function and then force a zero intercept, you are biasing the data. The linear regression algorithm in Excel should have the freedom to trade slope for intercept.
 
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kuruman said:
How did you collect these data? It is unlikely that you measured the energy and frequency independently. If they are obtained from a photoelectric effect experiment, the honest treatment would be to plot frequency vs. photoelectron energy (or stopping voltage) and extract both the work function and Planck's as the intercept and slope of the regression line. If, as you may have done here, you pre-calculate the work function and then force a zero intercept, you are biasing the data. The linear regression algorithm in Excel should have the freedom to trade slope for intercept.
Thank you for your reply @kuruman!

I did not collect that data. I think the professor found it online. We were taught to set the intercept equal to zero in Excel since we are not that advanced yet. They may teach us that stuff you are saying in their second- and third-year experimental physics courses.

But thank you mentioning those points, that is helpful to keep in mind.

Many thanks!
 
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Callumnc1 said:
Thank you for your reply @kuruman!

I did not collect that data. I think the professor found it online. We were taught to set the intercept equal to zero in Excel since we are not that advanced yet. They may teach us that stuff you are saying in their second- and third-year experimental physics courses.

But thank you mentioning those points, that is helpful to keep in mind.

Many thanks!
That's fine then. Thanks for the clarification.
 
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