Little Confused, PhotoElectric Effect?

[EIRE2003]

When white light is radiated on a metal surface such as a zinc plate, do any of the zinc atoms emit any electrons?
I know that only the high freq photons that are above the threshold freq such as ultra-violet do, but what confuses me now that I am thinking about it is that, of course white light is composed of the 7 colours of light, each of with a certain freq, in which UV light is the highest energy of them, so why can't e- be emitted from the surface of a metal when white light is shone upon a metal plate?

 

[ZapperZ]

When white light is radiated on a metal surface such as a zinc plate, do any of the zinc atoms emit any electrons?
I know that only the high freq photons that are above the threshold freq such as ultra-violet do, but what confuses me now that I am thinking about it is that, of course white light is composed of the 7 colours of light, each of with a certain freq, in which UV light is the highest energy of them, so why can't e- be emitted from the surface of a metal when white light is shone upon a metal plate?

"White light", if it's the way you mean by the combination of the "7 colors" does NOT include UV. UV is not one of the colors that you typically remember as part of a rainbow, is it?

VIBGYOR... no "U" here.

Zz.

 

[EIRE2003]

sorry not ultra violet but violet light.

And I mean white light is made up of the combination of the 7 colours each of with increasing frequencies ROYGBIV where violet is the highest.

 

[ZapperZ]

sorry not ultra violet but violet light.

And I mean white light is made up of the combination of the 7 colours each of with increasing frequencies ROYGBIV where violet is the highest.

If the work function is lower than the photon energy of any of the "spectrum" of light in that white light, then yes, there will be electrons emitted. However, take note that most metals (including zinc) forms an insulating oxide layer if it is exposed to air, which can raise the necessary energy to make the transition. That is why most metals in a photoelectric effect experiment is kept and done in low vacuum. So you just don't get photoelectrons that easily off some old zinc that you find.

Zz.

 

[radjan]

The photoelectric effect is x-rays or gamma rays interact with materials...the incident photon emitted by an x-ray or gamma ray literally knocks the material's e- out of one of the inner shells...usually the K or L shell and that wondering e- is totally absorped. Please note that the incident photon must be slighty greater than that of the materials binding energy. So i really don't think that is possible for UV light to have enough energy to knock zinc's electrons out of orbit.
I am really not an expert but I do think an x-ray photon or a higher energy photon is required in order to get a photoelectric effect :D

 

[Pieter Kuiper]

Most metals (including zinc) forms an insulating oxide layer if it is exposed to air, which can raise the necessary energy to make the transition. That is why most metals in a photoelectric effect experiment is kept and done in low vacuum. So you just don't get photoelectrons that easily off some old zinc that you find.

Hallwachs did the experiment in air with white light from a magnesium flame:
http://www.acolyte.co.uk/origins/herz.html

You just need to polish the piece of zinc with abrasive paper.

 

[ZapperZ]

The photoelectric effect is x-rays or gamma rays interact with materials...the incident photon emitted by an x-ray or gamma ray literally knocks the material's e- out of one of the inner shells...usually the K or L shell and that wondering e- is totally absorped. Please note that the incident photon must be slighty greater than that of the materials binding energy. So i really don't think that is possible for UV light to have enough energy to knock zinc's electrons out of orbit.
I am really not an expert but I do think an x-ray photon or a higher energy photon is required in order to get a photoelectric effect :D

No, the photoelectric effect is NOT just x-ray and gamma rays "interacting with materials". If this is true, those undergraduate labs in photoelectric effect using mercury arc lamp would not work! I use 3.3 eV and 5 eV photons to generate photoelectrons in a linear accelerator. Those energies are certainly NOT anywhere near x-rays and gamma rays spectrum.

Secondly, in a typical photoemission experiment, the excitation is NOT from the core level as you have indicated, but from the conducton band of the metal. The conduction band is NOT part of any particular atom within the material due to overlapping of the valence shell. The work function is measured from the top of the conduction band- the Fermi energy. This is all that matters to get photoelectrons, NOT the inner core shells.

Zz.

 

[EIRE2003]

The Principle quantum number describes the energy levels of the atomic orbitals and the distance from the nucleus, the shell nearest the nucleus has the lowest energy e-, with increasing energies the higher the orbits.
So if a photon of a particular energy if fired at an atom, will it emit an e- of the same energy? Or is some of the energy lost during the process and e- of less energies, ie, e- in low energy orbits released?

 

[ZapperZ]

The Principle quantum number describes the energy levels of the atomic orbitals and the distance from the nucleus, the shell nearest the nucleus has the lowest energy e-, with increasing energies the higher the orbits.
So if a photon of a particular energy if fired at an atom, will it emit an e- of the same energy? Or is some of the energy lost during the process and e- of less energies, ie, e- in low energy orbits released?

First of all, let's be clear that this is not your typical photoelectric effect, because photoelectric effect is seldom done on "isolated atoms". It is done 99% of the time on solids.

Secondly, if you take the example of a hydrogenic atom, for example, the lowest level corresponds to an electron having the largest binding energy to the atom. Now, this is technically the "lowest" energy in the sense that it's energy is "maximum negative"! But don't confuse this to be a "small" amount of absolute energy required to ionize this atom. It isn't!

If a photon is fired to a ground state hydrogen atom, then only if the photon has an energy of 13.6 eV will it barely able to produce an electron. The electron will have ZERO kinetic energy, because all of the photon's energy was used to overcome the -13.6 eV binding energy of the 1S1 energy state.

What I just described to you is highly simplistic. The experiment to do just this isn't as simple.

Zz.

 

[EIRE2003]

First of all, let's be clear that this is not your typical photoelectric effect, because photoelectric effect is seldom done on "isolated atoms". It is done 99% of the time on solids.

I was talking about the atomic level of solids.

I get you now!

 

[CharlesP]

In case you never noticed the photoelectric effect is happening to nearly every piece of metal that you can see. Ordinary light is sufficient to send electrons flying. A vacuum helps if you want to send them somewhere. Many common gas discharge lights like neon and florescent will start their arcs faster if there is already light present. That is because there are photoelectrons loose in the medium. Such neon lamps cost $0.50 and require 120 VAC. You can play with one. The effect is more pronounced as they age.