Why Is the Kinetic Energy Equation Equated to eV in Lenard's Experiment?

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SUMMARY

In Lenard's experiment, the equation mv²/2 = eV is used to equate the kinetic energy of photoelectrons to the potential energy in an electric field. Here, e represents the charge of the electron, m is the mass of the photoelectron, and V is the applied voltage. The potential energy (U = qV) is converted to kinetic energy (KE) when electrons are accelerated by the electric field. This relationship is crucial for measuring the maximum kinetic energy of emitted electrons, known as Kmax, and is foundational for understanding the photoelectric effect.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with kinetic and potential energy concepts
  • Knowledge of electric fields and voltage
  • Basic grasp of electron charge and mass
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  • Study the principles of the photoelectric effect in detail
  • Learn about the derivation of the stopping voltage in photoelectric experiments
  • Explore the concept of Compton scattering and its implications
  • Investigate the relationship between frequency and photoelectron emission threshold (ν0)
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Homework Statement



In Lenard's experiment to determine e/m for photoelectrons, he puts forwards this equation
mv2/2 = eV,
where e is the charge, m is the mass of photoelectron, and V is the potential applied.
Why the kinetic energy equation is equated to eV? Thanks in advance

Homework Equations





The Attempt at a Solution


 
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Hi...
eV is the potential energy of a charged particle in an Electric field.
U = qV irrespective of the Photoelectric effect, and is true for any such system.
In your case, a particle is accelerated/stopped by an electric field; All of the potential energy was converted to the particle's kinetic energy or vice-versa, depending on the case.
Daniel
 
Mr. Daniel,
I got your relation U = qV but i don't understand your line " All of the potential energy was converted to the particle's KE"
Could u please elaborate more?
Why the work done is equated to KE, here?
 
Certainly,
In the scheme for the photoelectric effect, one measures, on top of other things, the voltage needed to thwart the current, this is known as the "Stopping voltage".
In other words, what sort of an electric field must be created(i.e, what's the voltage difference necessary), in order to stop the electrons ejected from the metal being irradiated.
So, all of the kinetic energy the electrons had upon leaving the metal is negated by the work done by the Electric field, and the energy deposited in this process is qV in this case.
The voltage is an alternative way, and a mandatory one, of estimating K_{max}, or the energy with which electrons are being emitted. This gave way to the first quantitative assessments of the Photoelectric effect, as done, as you mentioned, by Lenard.

Daniel
 
Thanks Mr. Daniel. So, when potential energy to kinetic energy, no particle will move. Am I right? Does this hold true in all situations?
 
Hi there,
Yes, when the voltage is applied, it is done for the explicit purpose of stopping the charges.
This only works however, naturally, when electrons are actually emitted; Therefore, the photoelectric experiments also require one to derive the minimal frequency, \nu_0 needed to start the reaction while the voltage is first set to 0.
In the broader sense, you can look into effects created in the energy spectrum exceeding Photoelectric ranges; Compton scattering is an example, whereby the energy of the incidient photon is large enough for a secondary photon to recoil, and then, there would naturally be no effect on it by the electric field.
Daniel
 

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