Determining Planck's Constant With Photoelectric Effect

Click For Summary

Homework Help Overview

The discussion revolves around determining Planck's Constant through an experiment involving the Photoelectric Effect. The original poster is grappling with discrepancies between their calculated values and the expected constant of 6.63e-34, using a graph of Stopping Potential versus the inverse of wavelength (1/λ).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use different equations relating energy, wavelength, and stopping potential to calculate Planck's Constant, noting variations in their results. Some participants suggest plotting frequency against stopping potential instead, while others question the impact of outlier data on the slope of the graph.

Discussion Status

Participants are exploring various methods to analyze the data and are discussing the potential influence of outliers on the results. There is an acknowledgment of the need to adhere to the assignment's requirements while also considering alternative approaches.

Contextual Notes

There is mention of specific constraints related to the assignment, including the requirement to graph Stopping Potential against 1/λ. Additionally, some participants raise concerns about the quality of the data and the types of light sources used in the experiment.

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
1K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 25 ·
Replies
25
Views
5K
  • · 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