Question

A rock that contains${\mathbf{\text{3.1×10}}}^{\mathbf{\text{-15}}}$ mol of${}^{\mathbf{\text{232}}}\mathbf{\text{Th}}$ ( ${\mathbf{\text{t}}}_{\mathbf{\text{1/2}}}{\mathbf{\text{=1.4×10}}}^{\mathbf{\text{10}}}\mathbf{\text{yr}}$) has ${\mathit{\text{9.5×10}}}^{\mathit{\text{4}}}$fission tracks, each track representing the fission of one atom of ${}^{\mathbf{\text{232}}}\mathbf{\text{Th}}$. How old is the rock?

Short answer

The age of the rock is$3.78\times {10}^6$ years

Step-by-step solution

Radioactive decay

Radioactive decay follows the first order kinetics. The expression for rate constant is shown below:

$\mathbf{k}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{303}}{t}\mathbf{ln}\frac{\left[{\mathbf{N}}_t\right]}{\left[{\mathbf{N}}_0\right]}$

Calculation

- A rock contains_{${\text{N}}_{\text{t}}{\text{=3.1×10}}^{\text{-15}}{\text{mol}}^{\text{232}}\text{Th}$}

- Half-life of_{${}^{\text{232}}{\text{Thist}}_{\text{1/2}}{\text{=1.4×10}}^{\text{10}}\text{yr}$}

- The number of decayed${}^{\text{232}}{\text{Thatomsis9.5×10}}^{\text{4}}$atoms

- The number of$232$Th atoms in a rock is

_{$\begin{array}{c}{\text{N}}_{\text{t}}{\text{=3.1×10}}^{\text{-15}}\text{mol×}\frac{{\text{6.022×10}}^{\text{23}}\text{atoms}}{\text{1.0mol}}\\ {\text{=1.86682×10}}^{\text{9}}\text{atoms}\end{array}$}

First, let us calculate the initial amount of${}^{232}\text{Th}\left({\text{N}}_0\right)$

_{$\begin{array}{c}{\text{N}}_{\text{0}}{\text{=N}}_{\text{t}}{\text{+9.5×10}}^{\text{4}}\text{atoms}\\ {\text{=1.86682×10}}^{\text{9}}{\text{atoms+9.5×10}}^{\text{4}}\text{atoms}\\ {\text{=1.866915×10}}^{\text{9}}\text{atoms}\end{array}$}

Now, let us calculate the decay constant

$\begin{array}{c}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{ln(2)}}{\text{k}}\\ \text{k=}\frac{\text{ln(2)}}{{\text{t}}_{\text{1/2}}}\\ \text{=}\frac{\text{0.693}}{{\text{1.4×10}}^{\text{10}}\text{yr}}\\ {\text{=4.95×10}}^{\text{-11}}{\text{yr}}^{\text{-1}}\end{array}$

Therefore, the age of the rock is (t)

_{$\begin{array}{c}\text{ln}\left(\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\right)\text{=-kt}\\ \text{t=}\frac{\text{-ln}\left(\frac{{\text{N}}_{\text{N}}}{{\text{N}}_{\text{0}}}\right)}{\text{k}}\\ \text{=}\frac{\text{-ln}}{{\text{4.95×10}}^{\text{-11}}{\text{yr}}^{\text{-1}}}\\ {\text{=3.78×10}}^{\text{6}}\text{yr}\end{array}$}

Hence,the age of the rock is_{$3.78\times {10}^6$} years.