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

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Homework Help Overview

The discussion revolves around the relationship between cutoff potential and kinetic energy, particularly in the context of electron volts. Participants explore the definitions and conversions between these concepts in a physics lab setting.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definitions of electron volts and cutoff potential, questioning how they relate to each other. There are attempts to apply the equation KE=qV, with varying interpretations of the variables involved. Some participants express confusion about the calculations and the implications of negative values for kinetic energy.

Discussion Status

There is an ongoing exploration of the concepts, with some participants providing corrections and clarifications regarding the charge of the electron and the interpretation of voltage. The discussion includes questions about the validity of negative kinetic energy and the physical implications of the setup being described.

Contextual Notes

Participants are working within the constraints of a physics lab assignment, which includes specific values for cutoff potentials and the requirement to express energy in electron volts. There is a noted confusion regarding the application of formulas and the physical meaning behind the calculations.

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Homework Statement


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

Homework Equations

The Attempt at a Solution

 
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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.
 

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"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.
 

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