# Photoelectric current equation

1. Sep 16, 2011

### logearav

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 16, 2011

### danielakkerma

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

3. Sep 16, 2011

### logearav

Mr. Daniel,
I got your relation U = qV but i dont understand your line " All of the potential energy was converted to the particle's KE"
Why the work done is equated to KE, here?

4. Sep 16, 2011

### danielakkerma

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

5. Sep 17, 2011

### logearav

Thanks Mr. Daniel. So, when potential energy to kinetic energy, no particle will move. Am I right? Does this hold true in all situations?

6. Sep 17, 2011

### danielakkerma

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