The first rule of microanalysis

 There are lots of rules in microanalysis. In the sort that I practice, elemental and isotopic in-situ mass spectrometry, most of these are more what you’d call guidelines than actual “rules.” But there is one rule that is not so malleable; the first rule. And that is this:

Don’t run out of atoms.

Figure 1. The protagonist in this figure has insufficient iron atoms for his proposed procedure.

In ordinary life, we think of atoms a something so gobsmackingly small that they are essentially innumerable in any visible object (like a rock).  And for things you can pick up and throw at crows, that is generally correct.  In fact, even for things that need a handlens or a low power optical microscope to see, there are still a lot of atoms there.  But the combination of small scales (particularly nanoscales), low concentration of trace elements, and high required analytical precisions can all multiply rather quickly to leave one atomically short handed.  Fortunately, atom counting is fairly straight forward.

How many atoms do you have?

A good ballpark number of atoms in crystalline solids is 100 atoms per cubic nanometer.  For the mineral zircon (the Mick Jagger of mineralogical microanalysis), the actual number is about 92.  It is higher in densely packed phases like corundum (119), and lower in loosely packed ones like sanidine (72). Of course, the great thing about cubic nanometers is that you get one BILLION of them per cubic micron.  So a smallish SHRIMP spot (e.g. 15x10x1µm, or 111 µm³) would have about 10¹³ atoms in it, while a large (160x160x50µm, or 1 million µm³) laser ICPMS spot would have 10¹⁷ atoms.

These are heaps of atoms.  

How many atoms could we possibly need?

 This depends on three things.  The concentration of the analyte, the precision of the measurement which is desired, and the detection efficiency of the instrument. Since the presence or absence of individual ions of interest in an analytical volume is a poisson process, the minimum number of atoms needed in the least abundant species measured is the square of the inverse of the desired precision.  So 1% precision  requires 10,000 of the least abundant atom.  A permil requires a million, and 0.1 permil requires 10⁸.

For isotopic ratios, the abundance of the least abundant isotope must be multiplied by the chemical abundance of the element of interest in the sample. And finally, the detection efficiency of the analytical equipment must be considered.

So, the atoms you have are:

~100 x # of cubic nanometers

The atoms you need are: 1/ (required precision^2 * lowest isotopic abundance x volumetric concentration * detection efficiency)

Detection efficiency cam be further broken down into useful yield * dwell time.

Some examples:

A small (100µm³) SHRIMP spot for 3 permil ²⁰⁷Pb/²⁰⁶Pb ratios in a 1.5 billion year old zircon with 50ppm (atomic) ²⁰⁶Pb.

You need:

1/ (0.003^2 (precision) * 0.1 (²⁰⁷Pb/²⁰⁶Pb ratio)  * 0.00005 (²⁰⁶Pb concentration) * 0.02 (Pb useful yield) * 0.15 (dwell time on ²⁰⁷Pb) = 7.4E12

You have:

100 (atom/nm3) * 10⁹ (nm³ per µm³) * 100 (analytical volume- see above) = 1E13

So there are enough atoms, but only barely. All of a sudden, ten trillion atoms seems rather stingy instead of extravagant.

If you want a 0.1 permil ⁴⁶Ca/⁴⁰Ca ratio in a  multi-collector SHRIMP spot of the same size on calcite,

You need:

1/ (0.0001^2 (precision) * 0.00004 (⁴⁶Ca abundance)  * 0.2 (Ca concentration) * 0.1 (estimated useful yield) *0.9 (estimated multicollector dwell time).

1.4E14 atoms required.  

You have:

100 (atom/nm3) * 10⁹ (nm³ per µm³) * 100 (analytical volume in µm³- see above) = 1E13

So you need 14 times more atoms than the measly ten trillion atoms the small spot contains.

Insufficient atoms available.

Even though the target is a major element with great ion yield, the high precision required, combined with the low isotopic abundance of ⁴⁶Ca, means that a larger volume is needed.

Alternatively, a lower analytical precision could be desired, or a more abundant isotope could be targeted. Or, you could choose the option which every instrumentalist wants the inquiring scientist to propose:

“Just use a bigger spot”

A 2000 µm³ Spot is about the largest analytical volume that most sane ion probers would generally use. So although a patient man could do that with SIMS (It’s hard to work with SIMS craters of more than a couple thousand cubic microns), there’s always a shark willing to sell laser beams to the impatient scientist.  Would LA-MC-ICPMS would work?  While the volume of the big laser spot above is sufficient, the order-of-magnitude lower useful yield needs to be considered.  So you need 1.4E15 atoms.  Luckily, a the 1 million µm³ blast hole yields 10¹⁷ atoms.  So this problem can be solved with a bigger laser. And really, who doesn't want a solution like that?