Determining Planck's Constant With Photoelectric Effect

Click For Summary
SUMMARY

The discussion focuses on determining Planck's Constant through the Photoelectric Effect experiment, specifically using the equations E = hc/λ and e(V + Φ) = hc/λ. The participant encountered discrepancies in their calculated values for Planck's Constant, yielding results of 7.76e-34, 3.88e-34, and 5.72e-34, while the expected value is 6.63e-34. The participant suspects that outlier data may be skewing their slope calculations in Excel, which is critical for accurate results.

PREREQUISITES
  • Understanding of the Photoelectric Effect and its equations
  • Proficiency in using Excel for data analysis and graphing
  • Knowledge of wavelength (λ) and stopping potential (V) relationships
  • Familiarity with the concept of outlier data in experimental physics
NEXT STEPS
  • Learn how to accurately identify and handle outlier data in experimental results
  • Explore advanced graphing techniques in Excel for better slope determination
  • Research the implications of using different light sources, such as LEDs versus mercury lamps, in experiments
  • Study the derivation and application of the equation h = 2Veλ/c in the context of the Photoelectric Effect
USEFUL FOR

Students and educators in physics, particularly those involved in experimental physics and the study of quantum mechanics, will benefit from this discussion.

MASmith
Messages
3
Reaction score
0

Homework Statement


I'm trying to determine Planck's Constant through an experiment with the Photoelectric Effect, however, the equations I'm given and the data I've collected are not getting me to the 6.63e-34 that I need to be at. I'm graphing Stopping Potential (V) vs 1/λ and then using the slope to find Planck's Constant. The experiment is designed so that stopping potential equals the work function. I don't know whether the trendline's slope that Excel plotted for me is wrong because of outlier data or whether my manipulation of equations is wrong.

Homework Equations


E = hc/λ
e(V + Φ) = hc/λ
e(V + Φ) = hc/ λ --> 2eV = hc/λ if stopping potential = work function --> h = 2Veλ/c
V = hc/eλ (what's given on assignment) --> h = Veλ/c

The Attempt at a Solution


Using h = 2Veλ/c, where Vλ is the slope of the line, I get an average value for Planck's Constant of 7.76e-34. Using h = Veλ/c, I get an average value of Planck's Constant of 3.88e-34. Also, if I disregard my stopping potential reading for λ = 470 nm, I get a value for Planck's constant that is probably the most accurate, 5.72e-34.
 
Physics news on Phys.org
Calcualate freq. By given wavelengths. Plot a graph between freq. And stopping potential then slop of the graph will give you h/e multiply it with charge it will give value of Planck constant.
 
If I multiply that by 2 it gives me the most accurate result so far, but the assignment specifically asks for a graph of V vs 1/λ.
 
MASmith said:
If I multiply that by 2 it gives me the most accurate result so far, but the assignment specifically asks for a graph of V vs 1/λ.
Then plot graph between V and 1/λ but the slop will give you hc/e.
 
So my problem is most likely outlier data that is skewing the slope of the line then.
 
Ye
MASmith said:
So my problem is most likely outlier data that is skewing the slope of the line then.
Yeah try plotting graph by urself by adjusting the slope to get required value..
 
Show your data. Typically you have only a few frequencies of spectral lines to work with. (Or did you use LEDs instead of a mercury lamp?)
Low-frequency light may be polluted with some of the other frequencies to give a tail out to high lambda. Just inserting a factor of 2 out of nowhere isn't physics.

Finding a much lower h isn't uncommon (see http://www.jcon999.com/Physics/PlanckLab.htm )
 
Oh, and: Hello MA, welcome to PF :smile: !
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
844
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K