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Among the stocked items in the maintenance supply shop I work are thoriated tungsten inert gas (TIG) welding rods. Remembering what I read several decades ago, thorium is radioactive. So, I was curious, just how much radioactivity is each of these TIG rods producing?

Below is my own personal research transcribed from a LaTeX document I created. (I was practicing math, physics, and TeX all at the same time.) :-)

My questions are:

  1. Does the value I calculated (564.5 Bq) appear accurate? (i.e. correct math and physics calculations)
  2. Since this only shows the radioactivity of the thorium, and its decay chain has ten decay products, then is the total radioactivity ten times as much?
  3. Six of the ten decays create an alpha particle. Does this mean that the TIG rod is "producing" six $^{4}$He nuclei per decay event? (Imagining rod being "filled" with helium...)

From my LaTeX document:

The TIG rod is a cylinder 2.4 mm in diameter and 175 mm long. It is made of 2% thorium dioxide (ThO$_{2}$) and the rest is tungsten.

Assumptions made:

  • The proportions of tungsten and thorium dioxide are by volume, not mass.
  • Thorium is comprised of a single isotope, $^{232}$Th, with a standard atomic weight of 232.0377 u.
  • Oxygen has a standard atomic weight of 15.9995 u. This is based on the natural terrestrial occurrence of the three stable isotopes of oxygen.
  • Thorium's radioactive half-life ($t_{1/2}$) is 1.405 $\times$ 10$^{10}$ years.
  • Tungsten and oxygen are stable--all radioactivity is from the thorium.

NOTE: Modern accurate measurements indicate that all five naturally-occurring isotopes of tungsten are very mildly radioactive, with each having a $t_{1/2}$ > 10$^{18}$ years. This can be ignored for the purposes of this discussion.

  • Thorium dioxide's density ($\rho$) is 10.00 g/cm$^{3}$.
  • Avogadro's constant ($N_{A}$) is 6.02214 $\times$ 10$^{23}$ mol$^{-1}$
  • There are 3.15576 $\times$ 10$^{7}$ seconds in a year.

First, calculate the volume of the rod. $V_{cylinder} = \pi r^{2}h$.

$$ V_{rod} = \pi \times (1.2 \text{ mm})^{2} \times 175 \text{ mm} = 791.6835 \text{ mm}^{3} = 0.7916835 \text{ cm}^{3}$$ Next, find the fractional volume of the ThO$_{2}$.

$$ V_{ThO_{2}} = 2\% \text{ of } V_{rod} = 0.02 \times 0.7916835 \text{ cm}^{3} = 1.583367 \times 10^{-2} \text{ cm}^{3} $$ Find the mass of the ThO$_{2}$.

$$ m_{ThO_{2}} = \rho_{ThO_{2}} \times V_{ThO_{2}} = 10.00 \dfrac{\text{g}}{\text{cm}^{3}} \times 1.583367 \times 10^{-2} \text{ cm}^{3} = 0.1583367\text{ g} $$ Calculate the mass fraction of thorium in ThO$_{2}$. $$ M_{Th} = \dfrac{u_{Th}}{u_{Th} + 2 u_{O}} = \dfrac{232.0377\text{ u}} {232.0377\text{ u} + 2 \times 15.9995\text{ u}} = 0.8788 ~M_{ThO_{2}} $$

Find the mass of the thorium given its fraction of the mass of ThO$_{2}$. $$ m_{Th} = 0.8788 \times 0.1583367\text{ g} = 0.1391463\text{ g} $$ Calculate the number of moles of Th. This is calculated by dividing the mass by its standard atomic weight. $$ n_{Th} = \dfrac{m_{Th}}{u_{Th}} = \dfrac{0.1391463\text{ g}}{232.0377\text{ u}} = 5.99671 \times 10^{-4}\text{ mol} $$ Multiply moles by Avogadro's Constant to find the number of thorium atoms in the rod. $$ N_{Th} = (5.99671 \times 10^{-4}\text{ mol}) \times (6.02214 \times 10^{23}\text{ mol}^{-1}) = 3.6113035 \times 10^{20} $$ Convert thorium's half life from years to seconds. $$ t_{1/2} = (1.405 \times 10^{10} \text{ yr}) \times \left(3.15576 \times 10^{7} \dfrac{\text{s}}{\text{yr}}\right) = 4.43395977 \times 10^{17}\text{ s} $$ The activity (in Bequerels) is calculated by multiplying the number of thorium atoms by the natural logarithm of 2 divided by its half-life. $$ A_{Th} = 3.6113035 \times 10^{20} \times \dfrac{0.69314718}{4.43395977 \times 10^{17}\text{ s}} = 564.544 \text{ Bq} $$ Thus, there are about 565 radioactive decays of $^{232}$Th each second in the thoriated TIG rod. For reference, the specific activity of pure thorium is 4075 $\dfrac{\text{Bq}}{\text{g}}$.

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    $\begingroup$ it's probably 2% by weight, not volume $\endgroup$ – Manu de Hanoi Sep 22 '18 at 10:23
  • $\begingroup$ what happens if you just put it on a geiger counter? $\endgroup$ – wlexxx Mar 25 at 11:07
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Answer to #1:

Assuming the rod has 0.1391463 g of Th, and the specific activity of Th is 4075 $\dfrac{\text{Bq}}{\text{g}}$, then

$$ 0.1391463 \text{ g} \times 4075 ~\dfrac{\text{Bq}}{\text{g}} = 567 \text{ Bq} $$

Pretty close to my calculated value.

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  • $\begingroup$ P.S. I added the specific activity to the question part just today (not in my original research paper). Link1 Link2 $\endgroup$ – pr1268 Sep 1 '16 at 18:39

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