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:
- Does the value I calculated (564.5 Bq) appear accurate? (i.e. correct math and physics calculations)
- 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?
- 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 $\dfrac{\text{Bq}}{\text{g}}$.
Best Answer
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.