Calculating Planck Constant and Work Function in Photoelectric Experiment

  • Thread starter Thread starter georgeh
  • Start date Start date
  • Tags Tags
    Voltage
Click For Summary
SUMMARY

The discussion focuses on calculating the Planck constant and work function in a photoelectric experiment using sodium as the surface material. The stopping potentials of 1.85V at 300nm and 0.820V at 400nm are provided to derive the necessary equations. The correct approach involves using the photon energy equation E_photon = K + φ, where K is the kinetic energy and φ is the work function. Participants emphasize the importance of recognizing the work function in the calculations to accurately determine the Planck constant and cutoff wavelength for sodium.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with the concepts of stopping potential and kinetic energy
  • Knowledge of Planck's equation E = hf
  • Basic algebra for solving simultaneous equations
NEXT STEPS
  • Learn how to derive the work function from experimental data
  • Study the relationship between stopping potential and kinetic energy in photoelectric experiments
  • Explore the significance of cutoff wavelength in photoelectric effect
  • Investigate the implications of Planck's constant in quantum mechanics
USEFUL FOR

Physics students, experimental physicists, and educators interested in the principles of the photoelectric effect and quantum mechanics applications.

georgeh
Messages
67
Reaction score
0
In a photoelecric experiment in which a sodium surface is used, you find a stopping potential of 1.85V for a wavelength of 300nm, and a stopping potential of 0.820V for a wavelength of 400nm. From these data find (a) a value for the Planck constant, (b) the work function for sodium, and (c) the cutoff wavelength for sodium.
I tried solving (a) by stating the following
E=nh*f
h=E/f
h=E/(c/Lambda)
h=e*lamba/c ( i let n = 1)
And we know that K_max=e*delta V, where e is the Elementary charge, and DeltaV is the stopping potential.
I assumed that all the Energy is transffered to K_MAX
thus,
H=e*deltaV*lambda/C
The units end up working out, but the solution is wrong.
Anyone know what am i messing up on?
The rest of the question, i believe i can get once I solve this correctly
 
Physics news on Phys.org
Well you're assumption was incorrect. There is a certain energy called the work function that has to be overcome before an electron can be ejected.

E_{photon} = K + \phi

You can manipulate this equation to determine the equation for the kinetic energy of an ejected electron and it follows the form y=mx+b. The m will be your slope and it will actually be Planck's constant
 
m will be the slope...when x represents the frequency and y represents the KE (stopping potential * e).

Alternatively, using the formula provided by Pengwuino for each of the two events you end up with 2 equations in 2 unknowns (h and \phi). You know how to solve that.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
757
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K