# A Pound of Electrons?

I propose a Fermi Problem.

Over the lifetime of an average light bulb, what is the total mass of all the electrons that have flowed through?

Work on that if you have an idea how to proceed, or just take your best plausible guess. Remember it's a Fermi problem, so we're looking for estimates rather than detailed calculations. My suggested solution method under the link.

Ok, here's my rough shot. Each electron will transfer energy to the bulb roughly equal to the energy associated with the potential of the electric field produced by the voltage at the plug. For wall current, that's about 120 volts. That times the electron charge gives you, oh, about 2e-17 joules per electron. A good old-fashioned energy-eating incandescent might burn 75 watts, where a watt is of course one joule per second.

Divide the energy per time divided by energy per electron gives time per electron. Doing the division, I get about 4e18 electrons per second. Electrons are very, very light. They have a mass of around 9e-31 kilograms, which multiplied by our electron flow rate gives 3.5e-12 kilograms per second. That's trillionths of a kilogram, so not a whole lot.

But bulbs last a while. Maybe 1000 hours? Multiply that by the mass rate (remembering the unit conversions), and I get about 13ish milligrams. Not big, but still macroscopic. It's about the mass of a very tiny pebble.

To be honest this was a little surprising to me. I's have guessed it would be smaller. But having the numbers tell you something unexpected is sometimes even more interesting,

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I like Fermi problems, but I'm not a scientist (I do math), and didn't know enough to even get started on this one.

I often do the piano tuners in NYC problem with my math classes, to get them thinking about estimating.

I see two approaches which get two different answers, one of which is close to yours, one quite different.

Approach 1. P=VI. We know P (I assumed a 100 W light bulb) and V (I used 110 V), giving I just under 1 A (round it up for simplicity). 1 A = 1 C/s = 6*10^18 electrons/s, over a lifetime of ~1000 hours, which is 3.6*10^6 s. That gives you roughly 2.2*10^25 electrons at 9*10^-31 kg/electron for a total of ~10 milligrams.

Approach 2. Remember that household current is AC, not DC, so the net total must perforce average to 0: for each electron that passes through one way, another passes through the opposite way.

Which approach you use depends on whether you are interested in omnidirectional flux or net flux.

By Eric Lund (not verified) on 15 Oct 2009 #permalink

I never imagined I could say anything useful about a Fermi problem, but I think it's safe to eliminate Eric's "average to O" answer, on the ground (ouch) that we got a lot of light and heat out of the bulb while it operated, and can't have gotten that from zero electrons. OR if we have, patent it, quick ....

I concur with Eric. The drift velocity of electrons in metals around 500K is millimeters per second. Under an AC electromotive force, at 50-60 Hz, your typical electron won't get to move the length of the filament before the force changes sign. The electrons that do all the work during the life of the bulb are the ones that were in there when you bought it.

Awesome... I figured about 100 Watts over 100 Volts for 1000 hours to get about 10 milligrams too.

Using an approach very similar to Eric's 'Approach 1', I got around 24 milligrams. I used slightly different constants: 900 hour lifetime, 75 Watt bulb, 6.242*10^18 electrons/Amp*s.

All in all, not too shabby...all answers within an order of magnitude.

I concur with Joe. No electrons flow through the filament, unless you're running it off DC. They just jiggle back and forward in place by a small fraction of a millimetre.

Hank, you're confusing electrons with energy. there's no reason you need to actually add electrons to do work. One example: electric toothbrushes are often charge by induction -no electrons actually flow into the toothbrush while it's charging, and none actually flow out when it's discharging. In a perfect electric toothbrush charged by induction, no electrons actually flow *through* the brush, the same ones just move back and forth due to external forces. It's the same with a lightbulb.

Power stations don't really send us electrons, they just push the electrons we have back and forth at 60 Hz.

I lean to the "approximately zero" side of things.

Now, ask yourself, what weight of conduction electrons are present in the two 2 metre copper wires that form the cord that plugs this lamp into the wall?

Also, your house isn't connected to the high tension mains by continuous metal wires. Power enters through a transformer located nearby, so your house wiring forms a closed loop. What weight of conduction electrons did you buy with the house?

With AC, you're not going to get any through-flow unless the frequency is low enough for a charge carrier to move from one end of the path to the other. I would use the P=VI method Eric used for DC through an incandescent.

Hank: What you are overlooking is that while the averages of V and I must vanish in an AC circuit, the average of their product in general does not. It depends on the relative phase of the two. If you have two sine waves in phase (which is approximately true of V and I in household current), their product is sin^2(ω*t), which averages to 1/2. If they happen to be 90 degrees out of phase, the product is sin(ω*t) * cos(ω*t) = 0.5 * sin(2*ω*t), the average of which does vanish.

By Eric Lund (not verified) on 15 Oct 2009 #permalink

Seemed like a complicated way to calculate it. Wouldn't it be faster to calculate it from the amp flow? A 110 watt bulb draws about one amp making everything simple.

I'm not a scientist or a mathematician, but I would have assumed electrons don't flow as water through a hose, but vibrate as the waves on a guitar string, thus making the answer 0. If this is incorrect, please let me know so that I may reteach my kids.

I vote for zero.

But it depends on what you mean by "through." When you say an airplane flies through the air you don't mean that it goes in one end of the air and out the other end, you mean that it stays within the limits of the medium. You wouldn't say that zero airplanes fly through the air just because none of them ever leave the air.

The "zero" votes have a perfectly correct point, but the problem is less interesting that way. We can combine the best of both worlds to say that from an AC perspective, think of a cut right at the center of the filament and ask how many electrons pass through that cut going in either direction. A drift velocity argument should get you to roughly the same neighborhood.

In round numbers... The bulb draws IV = 75 W or 0.625 amps. 1 amp-second is a coulomb, 96,500 coulombs are a mole of electrons. A 75 W soft white lasts 750 hrs (on the package), 2.7 million seconds, 17.5 moles. Electron molar mass is 5.5x10^(-4) g, hence 10 mg overall. Given 750/1000 scaling, Matt was on the money. (Given 60 Hz AC no charge flowed - it jiggled in place.)

Never send a physicist to do a chemist's job. That is why the parity Eotvos experiment is so vital: no physicist could think of it, so that is where the answer hides. The universe is devoted to irony - ask any supernova.

http://www.mazepath.com/uncleal/erotor1.jpg
Somebody should look.

At first I would have said that about 1/120 sec.'s worth of electrons flow through the filament, over and over, but then, is it "the same electrons" each cycle? You might say that due to randomness, there's a distribution of electrons in the filament and wires around it, that through random walks eventually get a chance at the filament. But this is already making a mess of the quantum version of the situation, and imagining classical electrons so you can do statistical calculations probably just gives you meaningless numbers.

By the way, Matt, it's funny to see you use physics to solve what's a first-year textbook problem in electronics.

By Steve Witham (not verified) on 15 Oct 2009 #permalink

Matt, this is alternating current. The net charge over time is very, very near zero.

OK, so we count both ways.

Wait! If we count both ways, that implies that we have to count each electron as it passes some arbitrary point, even if all it does is wander back and forth thanks to thermal noise. Which means that you can't restrict your analysis to the large induced currents when the light is on, you also have to count the smaller thermal currents (at very low potential, please note) when it's off.

Do the math -- I think you'll find that the total thermal charge is quite large in a metal, especially when it's sneaking up on a thousand degrees Kelvin.

By D. C. Sessions (not verified) on 15 Oct 2009 #permalink

If you don't like zero as the answer, why not look at it the following way?

Every gram of tungsten wire contains 0.2 mg of electrons. During the lifetime of the bulb, every one of those electrons will have flowed through the filament, back and forth many times. So multiply the number of grams of mass in the filament by 0.2 mg to get the total mass of electrons.

This neglects the fact that some electrons in the filament will exchange with electrons in the wires that the filament is connected to, but so what? How can we calculate that number?

This should please the people who like zero as the answer while still giving a mass of electrons that meets the alternate definition of "through" that doesn't require electrons to exit the filament.

The problems gets MORE interesting when we consider AC.

Electrons flow back and forth at 60 Hz that means they change direction every 16.7 milliseconds.

The speed of electrons flowing in an insulated copper wire is approximately 66% the speed of light (according to the ever-reliable internets): 66% of 299,792,458 meters/sec is 199,861,639 meters/sec.

Therefore the electrons move about 333,769 meters in each direction. That's about 200 miles.

I'm thinking that most people are closer than 200 miles to their power plant.

Now I get stuck; what happens when electrons return to the power plant? My understanding stops here.

But in any case, it seems that it is NOT the same electrons going through the lightbulb each time.

By Farley Gwazda (not verified) on 15 Oct 2009 #permalink

Farley: Wrong velocity. You're taking the speed of light in a medium -- how fast the electrons experience other electrons -- as the speed of the electrons. You might as well say that Led Zeppelin fans banged their heads together at 1100 feet per second. Imagine the cranial fractures!

By Nathan Myers (not verified) on 15 Oct 2009 #permalink

The speed of electrons flowing in an insulated copper wire is approximately 66% the speed of light

No, that's the speed of electromagnetic waves in the space around the wire.

Electrons in the wire drift at a much lower rate.

By D. C. Sessions (not verified) on 15 Oct 2009 #permalink

For AC the net flux of electrons across an arbitrary cross sectional surface is ~ zero.

For DC:

Power = VI

Let V ~ 100 Volt
Let P ~ 100 Watt

I ~ 1 Amp = 1 C/s ~ 6E18 electrons/sec

Lifetime over 3 years (note choice of 3 in terms of order of magnitude for Fermi Problems) at ~ 3 hours a day
=> LT ~ 1000 hr ~ 4E6 sec

#e ~ 2E25

Mass ~ (9E-31 kg/electron)(2E25 electron)~ 10 mg.

Given an old fashioned LP record, spinning at it's usual speed, how many grooves are traveled during a 3 min 33 second song?

the mass of a very tiny pebble

... which, of course, weighs the same as a duck.

By Nathan Myers (not verified) on 15 Oct 2009 #permalink

0 is what you get when you +1 for each electron passing in one direction across the "cut" and -1 for the other direction. However, I interpreted Matt's followup question as allowing only a +1 for either direction, not counting each electron more than once.

@Sue

The great thing about Fermi problems is that you don't need to be a scientist to enjoy them, or to even do them!
And, there are no strict rules when it comes to solving a Fermi problem.

Some "purists" will say that you are only allowed to use what you already know to estimate in a Fermi problem.
Rubbish I say. That actually is contrary to the spirit of a Fermi problem, which is actually to explore your world, and as a filter to what may be absurd.

The Fermi problem allows one to expand their knowledge, and to apply critical thinking, not to test ones ability to be a warehouse of facts.

Yes it is fun to see how far you can go with what you already know or can guess, but once you run up on the wall of your knowledge, then hit a book or two to find out a fact that can help you!

I say: almost nothing. The current through a light bulb is AC, so the electrons don't really flow through - they just sit there waggling back and forth. It's the same old bunch of electrons each time, so we are just looking at the mass of the electrons in a light-bulb filament.

It's one of those "As I was going to St Ives" problems.

Okay, a variant question.

Put 10 mg of excess electrons inside the glass of a light bulb. How much energy does the bulb contain? What is the internal pressure?

Given an old fashioned LP record, spinning at it's usual speed, how many grooves are traveled during a 3 min 33 second song?

One, of course. :)

Do we count only those which have flowed through or can we include those which have tunneled through?

Ok, another variant on the question: What is the wieght of the electrons that energized the beta and alpah waves that were formed during the thought processes Matt generated to conceive and write the initial question?

Tow things we would need to know: a) how long Matt was thinking about the idea, and b), how hard his brain worked during the analytical and communication process.

Any care to work this out?

By layperson (not verified) on 16 Oct 2009 #permalink

Wish I had seen this right away and beat Eric to that answer, since my initial response was "zero".

It is a rather remarkable accident of nomenclature that power plants actually transmit power (which adds up to energy over time) rather than a "current" of electrons to us.

By CCPhysicist (not verified) on 16 Oct 2009 #permalink

I assume that the light and heat from the bulb come from the electrons being slowed down by collisions in the filament. Isn't there some mean free path related to the resistance of the filament and the push of the voltage that would let us estimate how many electrons have to slam into something to light the bulb over its lifetime? This might get us an interesting number if we look at a cross section of the filament with a thickness on the order of the mean atomic separation in the alloy. As for an answer, I'm going to guess about 10 mg.

Of course you are allowed to do stuff to solve Fermi problems. Look at one of Fermi's most famous solutions. He had to help build and detonate an atomic bomb, and he had to tear up a piece of paper to estimate the bomb's yield based on the paths of the paper bits in the wind of the explosion.

I wonder how much weight escapes a light bulb from usage?

The "zero" argument is irrelevant to this problem since the light bulb doesn't care whether it's being energized by AC or DC current. In either case, an average voltage of 120 volts produces a characteristic current in the filament and dissipates a characteristic number of watts. Ergo, AC or DC, the problem doesn't change.

One groove.

"Its usual speed".

Why can't people learn to use apostrophes correctly? It isn't rocket science.

By Vince Whirlwind (not verified) on 18 Oct 2009 #permalink

It is also an immense pet peeve of mine. You can rest assured that if I ever do it, it's a crossed-neurons typo.

By Matt Springer (not verified) on 18 Oct 2009 #permalink