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Photoelectric electrostatic separation

  [vote for,

An electrostatic separator is a well known device for separating metal from non-metals. In a normal electrostatic separator, a mixture of particles is charged with static electricity and each particle retains an amount of charge depending on its composition. Generally speaking, non-metals will retain more charge than metals, and so are more influenced by an electric field.

My idea is to improve the electrostatic separator so it can separate a certain *type* of metal and non-metal, rather than just between metals and non-metals.

The photoelectric effect occurs when when light interacts with matter and electrons are emitted. However, electrons will only be emitted if the matter is exposed to light above a certain threshold frequency. Every material has its own threshold frequency which is known as its work function.

If a mixture of particles is exposed to a particular frequency of light, some particles will emit electrons (and hence form a charge on the surface of those particles), while some will not. This mixture of particles would have to be non-touching so charge doesn't move between particles.

So to make a photoelectric electrostatic separator, a mixture of particles is dropped from a height. As the particles fall they are exposed to light below a certain frequency. Particles with a threshold frequency (i.e. work function) below that frequency become charged and are pulled towards an electric field generator, while the other particles are unaffected.

This could be useful for mining (rare-earth metals perhaps) or recycling garbage.

xaviergisz, Oct 25 2010

Charge induced on an illuminated particle http://www.physicsf...ex.php/t-68739.html
[Wrongfellow, Oct 27 2010]

Energy of a photon http://pvcdrom.pved...UNLIGHT/PENERGY.HTM
[Wrongfellow, Oct 27 2010]

Calutron http://en.wikipedia.org/wiki/Calutron
Prior Art [8th of 7, Oct 27 2010]

Uranium from seawater http://www.ornl.gov...umber=mr20120821-00
[xaviergisz, Aug 26 2012]

Photoelectric Work Function http://www.relativi...ectric_effect.shtml
Scroll to the bottom for a table of threshold wavelengths for various elements [xaviergisz, Oct 13 2013]


       But wouldn't the work function be altered by the present energy level? i.e. I assume that to separate the material is has to be gaseous, which means really hot, wouldn't that base energy level effect the work function and would that be constant across the sample? (+) anyway
MisterQED, Oct 26 2010

       Why would it have to be gaseous? Wouldn't finely powdered solid material work too?
Wrongfellow, Oct 26 2010

       //I assume that to separate the material is has to be gaseous//   

       I had in mind small *solid* particles of about 1mm diameter.   

       Can someone help with rough calculations to see if a practical amount of light could create a charge sufficient to move the particles (in a realistic electric field)?
xaviergisz, Oct 26 2010

       I wondered for a while about how to go about calculating this. A bit of lucky Googling found a thread that suggests a method - see link. So what the heck, let's have a go!   

       The basic idea is that the particle will accumulate charge until electrons no longer gain sufficient energy from the incident photons to escape to infinity. This will happen when the extra energy provided by the photon, after subtracting the work function of the material, is equal to the potential energy required to remove an electron to infinity.   

       Conveniently, the energy required to lift an electron through a potential difference of 1 volt is 1 electron-volt. So the sphere will reach a charge (in volts) equal to the difference between the photon energy and the work function (in eV).   

       Generally, a photon of wavelength w has energy E=hc/w (where h is Planck's constant and c is the speed of light). However for our purposes it's easier to use units of micrometres and electron-volts, and in this case we can use the approximation E = 1.24 / w. (link)   

       Let the work function of the material be W, again in electron-volts. Then the equilibrium voltage of the particle is V = (1.24 / w) - W.   

       To calculate the force, we need to know its charge in coulombs. So we need the capacitor equation: C = Q / V, where C is the capacitance, Q is the charge, and V is the voltage. The "self-capacitance" of a conducting sphere is C = 4 * pi * E0 * R (where E0 is the "electric constant", 8.854x10^-12 farads per metre, and R is the radius of the sphere). Putting these together, we have Q = 1.112 * 10^-10 * R * ((1.24 / w) - W).   

       The force on a charged particle in an electric field F = E * Q where E is the field strength and Q is its charge. So in our application, F = 1.112 * 10^-10 * E * R * ((1.24 / w) - W).   

       Now let's plug in some real numbers.   

       The breakdown field strength of air is about 3 x 10^6 V/m, so let's postulate a field of 10^6 V/m, to stay comfortably below this limit.   

       The work function of iron is 4.5 eV.   

       The wavelength of ultraviolet light is from 0.4 um to 0.01 um. Let's call it 0.1 um to make the numbers simple.   

       You've postulated a particle of 1mm diameter, ie its radius is 5 x 10^-4 m.   

       So the force on our particle is (1.112 * 10^-10) * (10^6) * (5 x 10^-4) * ((1.24 / 0.1) - 4.5)   

       This works out to 4.392 * 10^-7 Newtons, unless I've slipped up.   

       The particle's volume is 4/3 * pi * r^3 = 5.236 * 10^-10 m^3, and its density is 7870 kg/m^3, so its mass is 4.12 * 10^-6 kg.   

       Acceleration a = F / m so our particle accelerates at   

       (4.392 / 4.12) * (10^-7 * 10^6) = 0.106 metres/second^2.   

       If the particle falls for 1 second it will have reached a lateral velocity of 0.106 metres/second, and travelled a distance of 5.3 centimetres horizontally from its injection point.   

       So, this looks kind of workable. Can we tweak the numbers to improve things?   

       The force increases linearly with the particle's radius, but its mass increases with the cube of the radius, so the acceleration (and hence the distance) decreases with the square of the radius. Make the particles 0.1 mm in diameter instead, and you've got 100 times the displacement.   

       Do it in a vacuum instead, and you can crank the static electric field up to higher levels. Again, this affects the displacement linearly.   

       Interestingly, the intensity of the light doesn't affect the equilibrium charge (though I'd expect it will affect the time taken to reach equilibrium; someone else will have to do the BOTEC for this, because I'm rapidly running out of envelopes here.)   

       Of course there's a big ugly gaping hole in this calculation: when considering the electron's escape to infinity, we've completely ignored the presence of the static electric field. In reality, the electrons will be attracted to the positively charged plate, so we need to consider the amount of energy required to lift them to the 'saddle point' where the particle's field exactly cancels the static field. This probably means that these numbers are a load of nonsense.   

       But hey, this is the 'bakery.
Wrongfellow, Oct 27 2010

       It will work perfectly, if you're prepared to shred everything to a molecular level and then do the separation in a vacuum.   

       In which case, you could use a mass spectrometer, or (effectively) a Calutron.   


       // But hey, this is the 'bakery //   

       For which we are truly thankful.
8th of 7, Oct 27 2010

       Thanks for doing the calculations, Wrongfellow.   

       A horizontal accelaration of 1% of gravity should be plenty to separate the particles. It's looking feasible.
xaviergisz, Oct 27 2010

       Of course for a real application you'd need to consider the differential separation between the "want" and "don't want" particles.
Wrongfellow, Oct 28 2010

       I just read that scientists are considering extracting uranium from seawater using absorbent polymers. It got me thinking that photoelectric separation could be used to extract metals disolved in liquid, e.g. uranium in seawater.   

       Basically, water is aerosolised into tiny nanodroplets (e.g. diameter of 100nm) using electrospray techniques. Droplets that contain the desired metal are separated using photoelectric separation, and then the disolved metal concentrate is processed as per usual.   

       Uranium has a concentration of 3 parts ber billion. A seawater nanodroplet with a diameter of 100nm this would contain about 16 million water molecules plus dissolved metals. On average, one droplet in 5000 would contain one uranium atom. Thus this technique could be used to concentrate the seawater containing uranium 5000 times.   

       Obviously this could be used for other dissolved metals.
xaviergisz, Aug 26 2012

       Instead of using the electrostatic charge to move the particles into the collecting basin, could we just use it to identify them? Much lower energy levels would be required compared to physically moving large masses.   

       Currently, a similar method is used to collect alluvial diamonds. The diamond-bearing gravel is dropped in a curtain around a cone-shaped collecting basin. They are fluoresced under black light then a puff of air blasts the diamond into the central collecting basin.
AusCan531, Aug 27 2012

       Continuing on from my last anno, I just realised the process could be done iteratively. The separated water with concentrated uranium would then be diluted with pure water, aerosolised and then photoelectrically separated once again. The process could be repeated until the water contained mainly dissolved uranium (or other metal of choice). The water would then be evaporated away, leaving a uranium rich residue.
xaviergisz, Oct 07 2013

       Gold is meant to have a photoelectric effect threshold of about 250nm (UV).
MaxwellBuchanan, Oct 13 2013

       This idea could also be used for isotope separation. 1. Dissolve an element in acid, 2. aerosolise acid so that each droplet would contain, on average, 1 atom of the element, 3. illuminate aerosol with light that excites one isotope of the element but not the other isotope, 4. apply electric field and collect droplets containing only one of the isotopes.
xaviergisz, Apr 01 2014


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