Question on photoelectric effect and this particular experim

[CJit]

question is based on the video above (1.5 mins).

The main question is this, if photoelectrons are released based on the frequency of light that hits it, why is it that there is a large difference on the multimeter between when there is no filter, and when the blue filter is used (small wavelength=higher frequency). Shouldn't they be the same or almost the same? As the blue wavelength IS the smaller wavelength OF the light?

Anyone familiar with such experiment? It says in the video that the multimeter reads the energy of photoelectrons released by the light. Is this the kinetic energy of the electron?

**My assumption
If the multimeter reads the energy (in joules) of the photoelectrons, then amplitude should not be the affecting factor. Hence, the frequency/wavelength is the only factor, which leads to my question, shouldn't the readings of blue filter and no filter be close to each other? As blue is the smaller wavelength of the light

 

[Simon Bridge]

Welcome to PF;

question is based on the video above (1.5 mins).

The main question is this, if photoelectrons are released based on the frequency of light that hits it, why is it that there is a large difference on the multimeter between when there is no filter, and when the blue filter is used (small wavelength=higher frequency). Shouldn't they be the same or almost the same? As the blue wavelength IS the smaller wavelength OF the light?

You are only seeing the effect of the narrow range of blue wavelengths that the filter allows - while the unfiltered light also contains much higher energy photons such as higher energies of blue, violet and ultraviolet. The blue filter used simply did not select the shortest wavelength possible.

Anyone familiar with such experiment?

This is a classic experiment in physics so everyone has to study it several times in order to become a physicist. So that's a yes.

It says in the video that the multimeter reads the energy of photoelectrons released by the light. Is this the kinetic energy of the electron?

This is basically correct - it takes a specific minimum energy just to get the electron out of the metal (called the "work function") any additional energy provided by the photon is converted to kinetic energy. The dial is a representation of the kinetic energy: they didn't show you units because this is a qualitative demonstration only - they only want you to get the idea without going into details.

**My assumption
If the multimeter reads the energy (in joules) of the photoelectrons, then amplitude should not be the affecting factor. Hence, the frequency/wavelength is the only factor, which leads to my question, shouldn't the readings of blue filter and no filter be close to each other? As blue is the smaller wavelength of the light

There is no reason why the reading with and without the blue filter should be close to each other since visible blue is not even close to the highest energy of light.

 

[Let'sthink]

You have not told whether the current increases with filter or without filter. Please specify that.

 

[Simon Bridge]

You have not told whether the current increases with filter or without filter. Please specify that.

video does not say that either... it is not a standard setup.
The energy of the electrons is the stopping potential.
It is unclear what the meter in the vid directly measures.

 

[Let'sthink]

current represents the number of electrons coming out per second and stopping potential represents the kinetic energy of the fastest electron. i think in a standard experiment there is one microammeter and other voltmeter.

 

[CJit]

You are only seeing the effect of the narrow range of blue wavelengths that the filter allows - while the unfiltered light also contains much higher energy photons such as higher energies of blue, violet and ultraviolet.

Thank you for the clear explanation, everything makes sense. One thing tho, why would there be UV? Shouldn't it only consists of wavelength inside the visible light spectrum, as it is what its exposed onto the specimen?

 

[ZapperZ]

Thank you for the clear explanation, everything makes sense. One thing tho, why would there be UV? Shouldn't it only consists of wavelength inside the visible light spectrum, as it is what its exposed onto the specimen?

But why shouldn't there be UV? Why does a light source have to somehow confined itself only to a range that humans can see?

It is the nature of the light source, something you can't dictate. You can control what you let through by using filters, but you can't tell an atomic gas "You must only emit light in this range!". And since there are UV range in the source, the sample can't help but be exposed to it.

Just to reiterate, even WITHOUT such UV source, when you shine on it without a filter, you're getting more photons hitting the sample. As long as these photons have energies above the work function (even if it isn't blue), it will be another photon that can liberate an electron. So more photons, more photoelectrons, more current.

Zz.

 

[Simon Bridge]

I cannot speak to why the particular light source used in the video had a UV component - but, in general, incandescent bulbs, like regular household bulbs, do emmit in the UV... as well as in shades of blue that were not passed by the blue filter. Engineering a light source to be restricted to visible wavelengths is tricky.

Also, as Zz points out, there are other processes that can add to the results.
A proper photoelectric effect experiment is much more carefully set up to eliminate these extra effects.