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.

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 10^{13}, 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} m³

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} m²

(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} m^{3}.

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} m^{3})= 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 Al_{2}Si_{2}O_{7}. 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 Al_{2}Si_{2}O_{7}, 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 × 10^{23} = 1.75 × 10^{25 }atoms; this is about 3.97 × 10^{25} atoms per kilogram. (6.02 × 10^{23} 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 × 10^{25} atoms per kilogram, the number of atoms in the dot is

(1.75 × 10^{-13} kg) × (3.97 × 10^{25} kg^{-1}) = 6.95 × 10^{12}

Rounded more reasonably, this is 10^{13} 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 × 10^{12} 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 × 10^{12} 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.

So how many atoms do I carry around with my 200 pound body?

I’ve done this rather quickly, but I make it about 8 x 10

^{27}atoms or 8,000,000,000,000,000,000,000,000,000 if you prefer. Of these, the most numerous are hydrogen atoms (62% by number). This is because there’s a lot of water (H_{2}0) in your body, and many of the other molecules have a lot of hydrogen atoms in them. After that it’s oxygen (26% by number, all that water again), and then carbon (10% by number; carbon chains form the backbone of organic molecules). Nitrogen staggers over the finishing line at 1.5% by number, with the rest nowhere.Wow, very wordy! But a great kid with the experiment and the results!!

I know what you mean about the length of the post. It’s a difficult balance. People recycle a lot of science stuff on the internet without checking it (and it’s not always right), and I wanted to show that I’ve thought about this from the bottom up. I also want to show people how they might approach similar calculation questions that they might have.

To counter this, in all my posts I aim to have the “punchline” as early as I can. (In this case, the smallest visible number of atoms). I leave the justification until afterwards. This way, people can easily get the key message without having to go through all the gory details.