Q24.47 P - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

Answer

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

Step 1: Radioactive decay

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

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

Step 2: Calculation

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

- Half-life of  _{${}^{\text{232}}{\text{Th is t}}_{\text{1/2}}{\text{=1.4×10}}^{\text{10}}\text{yr}$}

- The number of decayed ${}^{\text{232}}{\text{Th atoms is 9.5×10}}^{\text{4}}$ atoms

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

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

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.