What’s the smallest number of atoms that I can see?

It depends what you mean by see. Single air molecules scatter light (that’s why the sky glows) so with a dark background and an absurdly intense light source you would presumably be able to visually detect a single atom suspended in a vacuum.

But that doesn’t really feel like seeing to me. The question I’m going to answer is: what is the smallest number of atoms that I can quickly assemble using the stuff in my flat, that I can see with my unaided eye by ordinary reflection in typical room lighting?

I’m sure I could look this up somewhere but there’s no fun in that.

pencil dot
The dot is where the two pencil lines would meet. The ruler scale is 100ths of an inch.

My assemblage of atoms was a tiny pencil dot made on white printer paper. There it is, on the right. The dot was definitely visible but so small that I needed to draw marks nearby so that I didn’t lose it.

I estimate that the number of atoms in that minute mark was about 1013, with an uncertainty of at least a factor of 10 in both directions.

In other words, 10 million million, very roughly.

That’s a lot. We talk about atoms very casually, drawing diagrams of chemical structures and so on, and it’s easy to forget how exceedingly tiny they are. It’s useful to do experiments like this one now and again to remind ourselves that atoms really are small beyond our comprehension.

The experiment

I made the dot by rubbing the end of a propelling pencil to a point and then touching the point lightly against a sheet of white paper.

To estimate the thickness of the layer of pencil lead, I held the pencil perpendicular to some paper and scribbled until the lead had a flat end to it. I then adjusted it so that, as far as I could tell, 1 mm of lead protruded from the pencil. Then, using normal pencil pressure, I drew lines 10cm long until the exposed millimetre of lead had all worn away. I took care to hold the pencil perpendicular to the paper so that the lines I drew were the full width of the lead. I could draw 520 such lines with the millimetre of lead, a total of 52 metres of line.

Calculation 1: volume and mass of the dot

I used the thickness (don’t confuse this with the width) of the lines described in the previous paragraph as a proxy for the thickness of the dot (and in doing so introduced probably the biggest uncertainty in the whole procedure). Propelling pencils leads appear to come in sizes of 0.5 mm, 0.7 mm and 0.9 mm and larger. Holding mine against a ruler showed that it was clearly a 0.7 mm lead. The volume of the initial protruding cylinder of lead was therefore

π × (0.35 mm)² = 0.385 mm³ or 3.85 × 10-10

If the line lines I drew were uniformly 0.7 mm wide (and that’s quite a big if – tilting the pencil will make them narrower) then I can equate the volume of the lines and the volume of the lead cylinder thus:

3.85 × 10-10 m³ = (52 m) × (0.7 × 10-3 m) × t

where t is the average thickness of the layer of lead on the paper in metres. This gives us

t = 1.06 × 10-8 m

It certainly doesn’t deserve 3 significant figures but I’m going to leave more reasonable rounding to the end.

To measure the area of my dot, I took a photograph of it next to the finest scale on my ruler, which is 100ths of an inch (see earlier). Things aren’t made any easier by the non-roundness of the dot, but if I were to say that the dot was 1/300 of an inch in each direction, I don’t think I’d be too far wide of the mark. That makes its area

(1/300 in)² × (25.4 × 10-3 m in-1)² = 7.17 × 10-9

(the 25.4  × 10-3 being the conversion from inches into metres). Using our estimate for the thickness of the pencil layer above, this makes the volume of the dot 7.60 × 10-17 m3.

Next, we need to know the density of the pencil lead. If it was a clay brick, its density would be 2400 kg m-3, and if it was pure graphite its density would be in the range 2090-2230 kg m-3, so it’s a reasonable guess that the density of the graphite/clay mix is about 2300 kg m-3.

So using the volume calculated earlier, the mass of my pencil dot is about

(2300 kg m-3) × (7.60 × 10-17 m3)= 1.75 × 10-13 kg

Calculation 2: how many atoms in a kilogram of pencil lead?

From the Cumberland Pencil Company, cited here, I infer that an HB pencil lead is roughly 50% clay and 50% graphite. They don’t say whether that’s before or after firing (the clay will lose water on firing). I’m going to assume that it’s after firing, but given all the other uncertainties in this calculation, I don’t think it’ll matter much if I’m wrong.

Clay is variable in composition, but a typical constituent of fired clay appears to be various minerals or combinations of minerals of overall composition Al2Si2O7. The relative “molecular” mass of such a compound/mixture will be 220. The relative atomic mass of carbon (in the graphite) is 12, so a 50:50 (by mass) mix of clay and graphite will need about 18 atoms of carbon for every unit of Al2Si2O7, giving a total of 29 atoms per unit of this mixture/compound, and a relative “molecular” mass of 440.

440 g of pencil lead is therefore one mole of pencil lead, and with 29 atoms per elementary entity of this mole, it will contain about 29 × 6.02 × 1023 = 1.75 × 1025 atoms; this is about 3.97 × 1025 atoms per kilogram. (6.02 × 1023 is Avogadro’s number: the number of elementary entities in a mole of a substance)

Calculation 3: number of atoms in the dot

From calculation 1, we know that the dot weighs 1.75 × 10-13 kg, and therefore with 3.97 × 1025 atoms per kilogram, the number of atoms in the dot is

(1.75 × 10-13 kg) × (3.97 × 1025 kg-1) = 6.95 × 1012

Rounded more reasonably, this is 1013 atoms in the pencil dot.

With the uncertainties in the size of the dot and the composition of the lead, I wouldn’t want to quote the answer any more precisely than this.

Checking

We’ve done quite a few steps here. Can we check that this answer looks about right?

Suppose the dot were pure graphite. Its mass would be (2150 kg m-3) × (7.17 × 10-17 m³) kg, which is 1.54 × 10-11 moles and hence about 9.2 × 1012 atoms in the dot. As carbon atoms are smaller than aluminium or silicon atoms, it’s not surprising that this number is a little bit bigger than the unrounded number of atoms calculated in the dot.

Now suppose that it was pure aluminium, with density 2700 kg m-3. Its mass would be (2700 kg m-3)× (7.2 × 10-17 m-3) = 1.94 × 10-13 kg which is 7.46 × 10-12 moles and hence 4.5 × 1012 atoms in the dot. As aluminium atoms are larger than carbon atoms, it’s not surprising that this number is a little bit smaller than the unrounded number of atoms calculated in the dot.

So our clay mineral calculations look at least plausible. The bit I’m really worried about is the thickness of the dot. Making a mark with a pointed lead and drawing a line with a flat end of the lead are likely to involve different pressures and hence different mark thicknesses. My feeling is that I’m most likely to have underestimated the thickness of the dot, and hence the number of atoms in it.

If I suddenly became weightless, what would happen to me?

Suppose that you were standing perfectly still, and gravity suddenly stopped operating on your body. What would happen? Nothing much, you might think, apart from a queasy feeling of weightlessness. After all, an object won’t start to move unless a force acts on it, and no force is acting on your body.

nogravity1
Diagram of the Earth, looking straight down on the north pole.

However, when you stand “still”, in fact you’re travelling in a very large circle at rather high speed, as the Earth turns on its axis and carries you with it. Newton’s 1st Law tells us that things travel in straight lines unless a sideways force acts upon them. The force that keeps tugging you to make you travel in a circle rather than in a straight line is gravity. This means that the moment gravity stops acting on you, you’ll start moving along a straight line (the red line in the diagram) while the ground continues to move in a circular path underneath you (the blue line).

The consequence is that you’ll lose contact with the Earth and float upwards, serenely or otherwise. At least, that’s what it will look like to earthbound observers. But what’s really happening is that the ground is accelerating downwards away from you as it moves on its curved path. Try to remember that as you watch your footprints receding beneath you.

I wondered how fast this would all happen. The answer is: remarkably quickly. I give the geometry later, for those who are interested, but here are some example results for a person standing in Edinburgh, on a latitude of 56° N. Just for now, we’ll pretend that there isn’t any air.

  • After 1 second your feet will be 5 millimetres off the ground.
  • After 10 seconds you’ll be 53 centimetres off the ground.
  • After a minute, you’ll be 19 metres up
  • After an hour you’ll be at an altitude of over 68 km (though, being half frozen to death by now, you may be losing interest).

The lower your latitude, the quicker your ascent. At the equator, you’ll rise about three times as fast as in Edinburgh, and at the poles, you won’t lose contact with the ground at all.

Why on earth should I be interested in a situation that is contrived and physically impossible? It’s because it brings home the fact that each of us is constantly moving along a curved path as the Earth rotates. On the equator, it takes only 10 seconds for our trajectory to deviate from a straight line by 1.7 metres (during which time we’ve travelled 465 metres).

Why did I pretend that there isn’t any air? Because the presence of air muddies the waters by adding another force: the upward force of our buoyancy in the air.  At low altitudes this force isn’t negligible: it slightly more than doubles the first three figures above. As you rise further and the air gets thinner, it matters less and less. I left it out because I wanted to make the effect of the Earth’s rotation clear.

The question that I’ve just answered is a trimmed-down version of a question that my friend Malcolm and I occupied ourselves with once when we were on a rather long and boring tramp along a glen at the end of a camping trip in the Cairngorms in Scotland. The question we asked then was: what would we observe if gravity suddenly stopped operating altogether? I may return to that subject in a future post.

The geometry

Let the centre of the Earth be at O, the origin, and let the Earth’s radius be r and its angular velocity about its own axis be \omega. You are standing at latitude \phi and are therefore a distance r \cos \phi from the Earth’s axis. Your linear velocity as you stand still on the rotating Earth will be \omega r \cos \phi, tangential to the Earth’s surface.

nogravity3Suppose that gravity stops acting on you at time zero, when you are at point Q. With no gravity acting on you, will now travel in a straight line tangential to the Earth’s surface. The diagram shows the situation from a suitable vantage point, looking sideways on to your direction of travel. We are not looking down on the north pole.

After a time t, you will have travelled a distance \omega t r \cos \phi, to point P.

Your altitude is the distance PR, where R is the point on the Earth’s surface for which P is directly overhead.  R lies on OP, the line from P to the centre of the Earth.  The length of OP is given directly by Pythagoras’ Theorem in triangle OPQ: it’s \sqrt{(\omega t r \cos \phi)^2 + r^2}. As OP=PR+r, the altitude of point P is OP - r. So

PR = ((\omega t r \cos \phi)^2 + r^2)^{1/2} - r

All we need now is \omega = 2\pi/86400 \text{ s}^{-1}, because the Earth does one full rotation in 86400 seconds, and r=6.4\times10^6 \text{ m}, because that’s how big the Earth is. We can now choose \phi and calculate PR for any value of t.

 

nogravity4We can check this answer in two ways. Firstly, we can use the very useful intersecting chords theorem to calculate the distance marked h in the diagram on the right. For small values of t, where h \ll r, then h should be approximately equal to the your altitude PR. For values of t of 1, 10, or 60 seconds, h and PR agree to 4 significant figures. As we expect, as t increases, the agreement gets less good: h and PR differ by about 2% after 1 hour.

The second check is to differentiate the expression for PR twice with respect to time. The first differentation gives us an expression for the rate of change of PR with respect to time, that is, your rate of gain of altitude:

\frac{dPR}{dt} = \frac{\omega^2rt}{  (1+\omega^2t^2)^{1/2}}

(Note: to keep things clear, I’ve omitted \cos \phi, which merely accompanies r everywhere and doesn’t change the conclusions.) Where t=0, and your motion is purely tangential, this expression for your speed away from the ground should be zero, and where t is very large (\omega t \gg 1), and your motion is purely radial, your speed away from the ground should be \omega r. Both are true.

The second differentiation gives us an expression for your upward radial acceleration:

\frac{d^2PR}{dt^2} = \frac{\omega^2r}{(1+\omega^2t^2)^{3/2}}

As it’s not you accelerating, but the ground that is accelerating away from you as it continues on its circular path, this expression for your upwards acceleration should, where t=0 and your path is tangential to the surface, become the same as that for the centripetal acceleration of the ground, \omega^2 r, which it does. In addition, when \omega t r \gg r, and your path is almost radial, the expression for your acceleration should approach zero, which it does.