Question

Question: The isotope ${}^{\mathbf{232}}\mathbf{Th}$decays to ${}^{\mathbf{208}}\mathbf{Pb}$by the emission of six alpha particles, with a half-life of localid="1660822741483" $\mathbf{1}\mathbf{.}\mathbf{39}\mathit{x}{\mathbf{10}}^{\mathbf{10}}$years. Analysis of of ocean sediment shows it to contain localid="1660822800077" $\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{mg}$of ${}^{\mathbf{232}}\mathbf{Th}$and $\mathbf{4}\mathbf{.}\mathbf{9}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{cm}}^{\mathbf{3}}$of gaseous helium at ${\mathbf{0}}^{\mathbf{o}}\mathbf{C}$and atmospheric pressure. Estimate the age of the sediment assuming no loss or gain of thorium or helium from the sediment since its formation and assuming that helium arose entirely from the decay of thorium.

Short answer

The age of the sediment is $2.26\mathrm{x}{10}^7\mathrm{years}$.

Step-by-step solution

Decay constant of Th232

The decay constant of can be calculated from the half-life of with the help of the following expression:

$\begin{array}{l}{\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ \mathrm{k}=\frac{0.693}{1.39\times {10}^{10}}{\mathrm{years}}^{-1}\\ \mathrm{k}=0.498\times {10}^{-10}{\mathrm{years}}^{-1}\end{array}$

Number of moles of Th232 present

From the ideal gas law, we have to calculate the number of moles of He present in $4.9x{10}^{-3}{\mathrm{cm}}^3$.

role="math" localid="1660823379540" $$\begin{gathered} \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}} \\ \mathrm{n}=\frac{1\mathrm{atm}\times 4.9\times {10}^{-6}\mathrm{L}}{0.0826\mathrm{L}\mathrm{atm}/\mathrm{mol}\hspace{0.17em}\mathrm{K}\times 273\mathrm{K}} \\ \mathrm{n}=2.1\times {10}^{-7}\mathrm{mol} \end{gathered}$$

Since one mole of ${}^{232}\mathrm{Th}$decays to form 6 moles of He,

Number of moles of ${}^{232}\mathrm{Th}$decayed$=\frac{2.1\mathrm{x}{10}^{-7}}{6}\mathrm{mol}=0.35\mathrm{x}{10}^{-7}\mathrm{mol}$

Number of moles of ${}^{232}\mathrm{Th}$remaining data-custom-editor="chemistry" $=\frac{7.4\mathrm{x}{10}^{-3}\mathrm{g}}{232\mathrm{g}/\mathrm{mol}}=3.1\mathrm{x}{10}^{-5}\mathrm{mol}$

Total number of moles of${}^{232}\mathrm{Th}$$=0.35\times {10}^{-7}\mathrm{mol}+3.1\times {10}^{-5}\mathrm{mol}=3.1035\times {10}^{-5}\mathrm{mol}$

Age of Sediment

The age of sediment is calculated as follows:

$$\begin{gathered} \mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{{\mathrm{N}}_0}{{\mathrm{N}}_{\mathrm{t}}} \\ \mathrm{t}=\frac{2.303}{0.498\times {10}^{-10}{\mathrm{years}}^{-1}}\times \log \frac{3.1035\times {10}^{-5}}{3.1\times {10}^{-5}} \\ \mathrm{t}=2.26\times {10}^7\mathrm{years} \end{gathered}$$

Therefore, the age of the sediment is $2.26\times {10}^7\mathrm{years}$.