Is there any niobium-92 left?

What primordial radioactive nuclides (isotopes) still exist in the Earth’s crust? To figure that out, we might need to know the original abundances of various nuclides in the matter that now makes up the Earth’s crust. I haven’t found good data on that. But it’s pretty easy to calculate what the abundances would have to have been.

I’ll show a way to do the calculations, using a spreadsheet. You can check my work, if you know at least a little about spreadsheets. I used LibreOffice (which you can get for free), but other spreadsheets applications like Excel are similar. Note that I’m not an expert spreadsheet user, and I’m not going to try to do this the best way, or to make it pretty.

Here’s the end result:

niobium-92

(Click here for a full-size image.)

Here’s how to make it. First, define a few “variables” to use in later formulas:

  • Avogadro’s number: The number of atomic mass units (u) in a gram.
  • The Earth’s crust’s mass. I couldn’t find a simple answer to this, but I don’t need it to be very precise. It’s “less than 1% of the Earth’s volume”, so I’m going with 0.5% of its mass, or 3.0×10^25 g.
  • The age of the Earth: I’ll use 4.54 billion years.

Now, make a table of the nuclides that might be of interest. Of the thousands of nuclides that exist, only a few have a half life that is in the interesting range (somewhat less than the age of the Earth). You can get them from a list of “nuclides by half life”, such as this one: <https://en.wikipedia.org/wiki/List_of_radioactive_isotopes_by_half-life>.

The first three columns of the table (Element, Isotope mass number, Half life) must be entered manually. The remaining columns are calculated by simple formulas.

Number of half lives elapsed

The age of the Earth, divided by the half life. In cell D8, type the formula “=$B$5/C8”. Then copy the formula to the remaining cells in column D. One way to do that is to click on the cell to select it, then drag the lower-right corner of the selection box down over the rest of the cells in the column.

Fraction remaining

The reciprocal of (2 to the power of the number of half lives elapsed). E8: “=1/(2^D8)”.

Number of original atoms needed

The number of atoms you’d have to start with, for there to be just one left today. The reciprocal of the previous column. F8: “=2^D8”.

Original mass needed (u)

The mass of the atoms counted in the previous column, in atomic mass units. G8: “=F8*B8”.

Original mass needed (g)

The previous column, converted to grams. H8: “=G8/$B$3”.

Abundance in Earth’s crust needed

The previous column, divided by the mass of the Earth’s crust. I8: “=H8/$B$4”.

If the number in the last column is greater than 1, then we would need to have started with more than one Earth in order for there to be any atoms remaining. The numbers get big fast. For example, we would have had to start with over a billion Earths made of solid uranium-236, for there to be one atom of it left today.

There are just three intermediate nuclides, which might still exist but be hard to detect: samarium-146, plutonium-244, and niobium-92.

Samarium-146 and plutonium-244 are sometimes considered to be extinct, but they definitely still exist on Earth in tiny amounts, and I believe that both of them have been successfully detected.

Niobium-92 is the strange one. The required initial abundance I calculated is in the ballpark of what the actual initial abundance probably was, which means there could literally be just one or two atoms of primordial niobium-92 left in the Earth’s crust. Or zero. Or a few hundred. But no more than that. It’s definitely not detectable.

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