Cutoff Potential vs. Kinetic Energy in Electron Volts: What's the Difference?

  • Thread starter Thread starter garr6120
  • Start date Start date
  • Tags Tags
    Potential
Click For Summary
SUMMARY

The discussion clarifies the distinction between cutoff potential and kinetic energy, emphasizing that while both are related to electron behavior, they are measured in different units. The cutoff potential (V) is defined as the voltage at which the current is zero, while kinetic energy (KE) is calculated using the equation KE=qV, where q is the charge of an electron (1.60 x 10^-19 C). The confusion arises from interpreting negative values in this context; kinetic energy cannot be negative, as it is always a positive quantity derived from the formula KE = 1/2 mv².

PREREQUISITES
  • Understanding of basic physics concepts, particularly electric potential and kinetic energy.
  • Familiarity with the equation KE=qV and its components.
  • Knowledge of electron charge (1.60 x 10^-19 C) and its significance in calculations.
  • Ability to convert between joules and electron volts (1 eV = 1.60 x 10^-19 J).
NEXT STEPS
  • Study the relationship between electric potential and kinetic energy in electron dynamics.
  • Learn about the implications of negative voltage on electron trajectories.
  • Explore the concept of energy conservation in electric fields.
  • Investigate practical applications of cutoff potential in photoelectric experiments.
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism and quantum mechanics, as well as researchers exploring electron behavior in electric fields.

garr6120
Messages
42
Reaction score
0

Homework Statement


Can the cutoff potential (v) be the same as kinetic energy measured in electron volts

Homework Equations

The Attempt at a Solution

 
Physics news on Phys.org
No. electron-volt is a unit of energy; volt is energy only when multiplird by charge.
 
in my class we are doing this physics lab where we get the data, so i have used the equation KE=qV where q is 1.60*10^-19 C which is the charge on an electron, and V is the current at the cutoff potential. So for yellow light i have the values 0.25 J/C. Now I got an answer of 4.0*10^18 J when solving the equation. However, the question also says write your answer in electron volts, I know that one eV is 1.60*10^-19 J/eV, same value as the charge on an electron but different units. anyways i divided 4.0*10^18 J by 1.60*10^-19 J/eV and i got 0.25 eV same as what i had before. I am so confused. This screenshot might make better sense.
 

Attachments

  • Screen Shot 2015-01-07 at 10.07.49 PM.png
    Screen Shot 2015-01-07 at 10.07.49 PM.png
    42.5 KB · Views: 507
"Now I got an answer of 4.0*10^18 J when solving the equation."
You have the charge of the electron wrong. It's 1.6e-19C, not 1.6e19C.

"V is the current at the cutoff potential.". No, V is the cutoff potential.

"i divided 4.0*10^18 J by 1.60*10^-19 J/eV and i got 0.25 eV.
Oh?

Your statement is very confusing on top of that. Try to frame your question precisely. What is it you're solving for?
 
I have made all of my corrections to the charge of an electron and the amount of joules in an electron volt. Now i am trying to solve for the max KE of the electron using the equation KE=qV, where q is the charge on an electron: -1.60*10^-19 C, and V is the cutoff potential when the current is 0. Using this information for green light which has a cutoff potential of 0.25 V. I can now solve for the maximum KE possible, when i do this i get an answer of -4.0*10^-20 J. This question also wants me to convert J to electron volts, since i know that one electron volt is 1.60*10^-19 J/eV, i divided -4.0*10^-20 J by 1.60*10^-19 J/eV to get an answer of -0.25 eV.

Is it possible to get a negative KE value?
 
Instead of just using formulas, try to see what you're really doing. You're firing electrons with an initial k.e. of 0.25 ev at a NEGATIVE plate. So what happens? As you inrease the voltage NEGATIVELY, higher and higher REPULSIVE force is applied to the speeding electrons until the voltage is negative enough to apply just enough repulsive force to prevent the electrons from arriving at the plate. They instead do a 180 in front of the plate and head back to the emitter.
So you have an initial k.e. of 0.25 ev but the electrons LOSE energy as they move to the plate. That energy is qV with q negative and V negative.
Don't just depend on equations to give you the right polarity in any given problem. Think it through.
And no, there is no such thing as negative kinetic energy. k.e. = 1/2 mv2 so even if v is negative, k.e. is positive.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K