Light Intensity and Aperture Variable for Planck's Constant

Click For Summary
Using a Planck's constant kit, the discussion revolves around verifying Planck's constant by varying light frequency and aperture size. The confusion lies in how to relate the data from light intensity to Planck's constant, as light intensity does not appear in Einstein's photoelectric effect formula. Instead, the focus should be on measuring the stopping voltage across a photoelectric tube and applying the formula hf = W + qV, where the slope of the resulting graph can yield Planck's constant. Proper rearrangement of the formula is necessary to facilitate graphing, with the stopping voltage on the y-axis and the frequency on the x-axis. Ultimately, the goal is to achieve a linear relationship that confirms the value of Planck's constant within experimental error.
Procrastinate
Messages
155
Reaction score
0
I used a Planck's constant kit and varied the frequency of light in order to verify Planck's constant.

However I have also used apertures to vary the surface area and I am quite confused as to what to do with this data to relate it back to Planck's constant.

The data is as seen in in the attachment. I'm sorry they are neat and tidy, they are currently just rough drafts at the moment. I had to leave out the blue light because it wouldn't let me go over 300kb.

Is there anyway i can use light intensity as a variable in order to verify Planck's constant? Or any other thing similar to this.
 

Attachments

Physics news on Phys.org
I am not familiar with your "Plank's constant kit"; it would be useful to know what it is. Anyway, you must be shining light on a photoelectric tube and measuring the electric potential needed to stop the current from flowing - your "backing voltage"? The next step is to compare your data with Einstein's formula for the photoelectric effect (graph it so the formula for the line of best fit matches Einstein's formula). Use the form of the formula where the kinetic energy of the electron is replaced by qV, the electric energy that is absorbed in stopping it. You will notice that light intensity is not in the formula, so you cannot find Planck's constant using intensity. But you can find it by fitting your data to the formula. Look up "photoelectric effect" in Wikipedia if you don't have it handy in your textbook.
 
http://www.scientrific.com.au/PDFs/ap2341-001.pdf

This is basically the instruction manual for what it does. Einstein's formula

I looked it up on Wikipedia and found that Einstein's formula was

5ec67b75d0ee8b9f636b86efdb77f952.png


However, I'm not sure how to apply it with the data I have currently.
 
Last edited by a moderator:
The kinetic energy of the electrons emitted from the photo tube's cathode can be determined by putting an electric potential across the tube. The Ek is equal to the electrical energy absorbed, qV when V is adjusted to just the voltage necessary to stop the electrons so no current flows. So you have
hf = W + qV where W is the work function of the cathode (energy needed to push one electron out)

The trick is to rearrange this formula so it represents a graph where it is convenient to find the h value you are looking for. It would be nice if h was the slope on the graph. Recall that in the math formula for a straight line, y = mx + b, m is the slope. Also nice if the manipulated variable (the thing you set in the experiment) is on the right. And the responding variable - stopping voltage - is on the left. Once you have this sorted out, you will know what to graph. If your formula was y = mx + b, you would put x on the horizontal axis and y on the vertical axis. According to Einstein's theory, you should get a straight line to within experimental error and if all goes well the slope will be Planck's constant to within experimental error. It would be nice if you had an error estimate on your measured values so you get error bars on the graph. Then you can tell if it is a straight line to within experimental error.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 3 ·
Replies
3
Views
649
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
978
Replies
1
Views
857
  • · Replies 5 ·
Replies
5
Views
2K
Replies
15
Views
2K
Replies
5
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
2K