Photoelectric effect and what happens to the electron

[MarkFarrell82]

Hi everyone,
I've been reading bits about the photoelectric effect, about how when you shine a light on a strip of metal its spits out electrons etc. What do they mean by spits out electrons? Is it that it actually spits the electrons out? If so where do they go?

The book I'm reading uses the example of automatic supermarket doors. The only way i can make sense of it is that the photons hit the electons in the metal and that sets off a current within the metal. Therefore the photons are creating a kind of voltage to make the current flow. If someone disturbs the beam of light it stops the photons hitting the electrons and so cuts the current. Once the current stops it triggers the doors to open.

Is that right or am I way off the mark? I don't get the spitting of electrons. I can understand it if it means it pushes the electrons along within the metal creating a current but not if its literally spitting the electrons out which is what the statement implies to me.

I do admit I'm no genius and new to trying to understand the world we live in so it doesn't take much to confuse me and make me ask lots of potentially stupid and straight forward questions.

Hope someone can help lift the mist for me.

Mark

 

[xts]

Is it that it actually spits the electrons out? If so where do they go?

Out of metal - to surrounding space. Usually you put such piece of metal in a vacuum bulb, then you have another electrode (on +V potential) collecting the electrons kicked out - you may measure the electric current appears as you lit one electrode.

Therefore the photons are creating a kind of voltage to make the current flow.

Exactly! Actually it is more convinient to support the current flow helping it with external voltage, but, of course, if you don't connect it such, the other electrode would gain negative charge. Such device makes (pretty inefiicient though) solar cell.

 

[MarkFarrell82]

Thanks for that. What happens to the metal or the atom in the metal that loses the electron? Does the electron get replaced so that it can repeat the effect again? If so, where does it come from? Do you know of any animations for this effect on the internet? I've looked around but I must be looking for the wrong thing..
Thanks again
Mark

 

[xts]

Thanks for that. What happens to the metal or the atom in the metal that loses the electron? Does the electron get replaced ...

In metals electrons are not fixed at their mother atoms. If a large piece of metal loses one, two or million electrons, nothing dramatic happens. They'll be replaced by electrons transported over wire as soon as you close the circuit

Do you know of any animations for this effect on the internet?

C'mon! Science is not taught through animations, but through written texts. I may recommend you a treatise in Latin (joking... but in 100 years old German maybe?), but not animations!

 

[sophiecentaur]

When an electron gains enough energy from a photon of light, it can leave the surface of the metal - leaving the metal with a positive charge. But this positively charged piece of metal will attract an electron to become neutral again. Often, if 'left to itself' the photo electron will just fall back to the metal surface it came from. If there is a positively charged electrode nearby, the electron will go there instead and a sensitive miliameter placed between the metal and the electrode will detect a small current as electrons will flow through the meter to make up the deficit.

 

[MarkFarrell82]

Thanks both of you for your help..

 

[magpie5]

Energy perspective:
The electron is bound (held) to the metal by a force (think of water atoms held to the water) If a photon interracts with an electron ( the electron absorbs the photon) the electron will ecape the binding (energy) force of the metal only if the energy of the photon is greater or equal to the binding energy of the metal.
Imagine water heated, if a water molecule gains enough energy from being heated, then it can escape the liquid phase and move into the gas phase. ie it escapes the liquid
Another examle, we are held or bound to the Earth by a gravitational force, If we can get a rocket with enough energy to escape the gravitational binding energy than we can escape.

 

[jewbinson]

Say the detector is a 1x1 square 100 units (of distance) away from the point of impact of the light to the strip of metal.

Then the way I look at it is that the electrons are emitted from the strip of metal at an equal rate in all directions, so the detector detects number of electrons actually emitted from the surface can be calculated (approx.) by:

(no. of electrons detected)[(4*pi/3)*(100^3)].

Is this the right way of thinking, or do the electrons favour going in a particular direction?

 

[sophiecentaur]

This could be complicated. You need to bear in mind that there is a whole range of velocities for the ejected electrons. The Einstein equation only refers to the maximum KE of the photo electrons. Some will barely stagger away from the surface and fall back easily.
The strip will become charged and the electric field around it won't be uniform (the sharp edges will have a stronger, attractive, field around them). More electrons will end up escaping perpendicularly from the centre of the strip because the field over the flat surface would be less than at the edges.
If it's insulated, the charging collector plate would reach a maximum (negative) potential such that photoelectrons are eventually repelled from it.

If the strip and collector are connected via a low resistance wire, in a big loop going round behind both, then I think you are probably right that electrons could leave at all angles. However, I have a feeling that the distribution of energies of the ejected electrons could depend upon the angle of ejection, with the maximum energies being normal to the surface. My argument is a bit intuitive but I suggest that the electrons emerging from slightly below the surface could have 'further to go' through the top layers of metal (analogous to a higher work function?) and so would have a lower mean KE. This could mean fewer leaving in directions away from the normal

 

[MarkFarrell82]

Sorry to take this back a little bit chaps but just a basic one to help visualise what's happening. As I understand it, a photon collides with an electron and passes on some of its energy. This is enough energy to dislodge the electron from the atom in the metal because its' only loosly bound. This is happening all along the metal strip which is in the path of the photon source so there are elctrons flying everywhere but at the same time the electrons are either falling back to the metal or being replaced by another electron from somewhere else looking for a home.

Further back in the post I asked if it was the photons hitting the electrons creating the the voltage. Coming back to that, is it when the electrons are dislodge that there becomes a potential difference allowing the current to flow? And not the photons pushing the electrons like a battery would?

And also, in the example of the supermarket doors, where are the sources of photons and the metal strip and how are the located with respect to each other? I've been through some automatic doors recently and had a look for any related equipment and all I can see is a little black box above the door. I must admit, I've not fully investigated for fear of people thinking I have some automatic door fetish..

Thanks again
Mark

 

[ZapperZ]

In a standard photoelectric effect, the cathode (in this case, the metal) is grounded. So each time an electron leaves, another one comes into replenish the cathode so that it remains neutral.

If this is not done, then eventually, you'll have a charging effect and at some point, no more electrons will be emitted since the cathode has become too positively charged. We encounter this scenario some time when we do photoemission studies on semiconductors.

Zz.

 

[xts]

is it when the electrons are dislodge that there becomes a potential difference allowing the current to flow? And not the photons pushing the electrons like a battery would?

Yes, but that is very unefficient solar battery.
There is a demonstration often shown for 1st year physics students, or even in high schools:
You have a photoelectric photocell (vacuum bulb with two electrodes, one of them larger, flat, covered with a metal of good photoelectric properties), connected to voltmeter or microampmeter. As you lit the cathode with strong lamp, you may measure the voltage of 0.5V or so, or current of microampers.

in the example of the supermarket doors,

No one uses such photoelectric cells nowadays for such purposes. For technical reasons (and price) semiconductor photodiodes or photoresistors are used instead. But the idea is the same: if the photocell is lit, it allows current to flow, while in dark the current stops to flow, and the current is then amplified and used to switch on the machanism opening the door (or letting water go in the automatic toilet)

 

[sophiecentaur]

All the photon does is to give the metal lattice some energy. This energy may cause it to eject an electron away from the metal surface. Many electrons will be caused to move internally (you don't get a photo electron for every photon that arrives) and end up as heat. There is a minimum energy required to get an electron to leave the surface (the work function). If an arriving photon has more than this, then the electron will have the excess energy in the form of Kinetic Energy. Many electrons require more than the Work Function energy to be ejected - so there will be a distribution of kinetic energies of the photo electrons, with a maximum - given by photon energy minus work function (this is the Einstein equation)

You cannot be correct about your supermarket door system. Light shining on a metal strip in air cannot produce photoelectrons. To get photo-emission from a metal surface using visible light, you need a vacuum and a very clean (not oxidised) surface. The whole thing needs to be in a vacuum or else the photoelectrons will hit an air molecule and can't be detected. There are very few metals which actually work for optical frequencies (Potassium is the favourite for demos). You CAN get UV to produce photoelectrons from a clean zinc surface in air and this can be detected by charging a zinc plate and watching it discharge faster when you shine UV on it..
Light detectors are normally (solid state) photo diodes or photo transistors which work on a similar but different principle from the conventional photoelectric effect. That "strip'" you have seen is not photosensitive in itself. It could possibly be reflective and the "source" you are seeing could contain a source (lamp / diode laser) which sends out a beam of light and a detector which senses the reflected beam. When you stand in the way, the beam is broken and you have been 'spotted'.

 

[MarkFarrell82]

Thanks again guys. I love this forum. I'm learning loads and if not finding out other things to look at. Could be on here all day at the minute