Q24.43 P - Chemistry: Molecular Nature Of Matter And Change

Question

A rock contains   _{$\mathbf{\mu }$}mol of ${}^{\mathbf{\text{238}}}\mathbf{\text{U}}\mathbf{\left(}{\mathbf{\text{t}}}_{\mathbf{\text{1/2}}}{\mathbf{\text{=4.5×10}}}^{\mathbf{\text{9}}}\mathbf{\text{yr}}\mathbf{\right)}$   and  110 mol of ${}^{\mathbf{\text{206}}}\mathbf{\text{ Pb}}$ Assuming that all the ${}^{\mathbf{\text{206}}}\mathbf{\text{ Pb}}$ comes from decay of the  , estimate the rock’s age.

 

Step-by-Step Solution

Answer

The rock's age is  _{$2.22\times {10}^9\text{yr}$}

Step 1: Radioactive decay

The radioactive decay follows the first order kinetics. The relation between half-life and rate constant “k” is:

 ${\mathbf{t}}_{1/2}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{693}}{k}$

Step 2: Estimation of the rock’s age

- The half-life of  _{${}^{\text{238}}{\text{U is t}}_{\text{1/2}}{\text{=4.5×10}}^{\text{9}}\text{yr}$}

- Rock now contains ${\text{N}}_{\text{t}}{\text{=270μmol of }}^{\text{238}}\text{U}$ , and  _{${\text{110μmol of }}^{\text{206}}\text{ Pb}$} (which means that ${\text{110μmol of }}^{\text{238}}\text{U}$ decayed into  _{${}^{206} \text{Pb}$} )

Therefore, the rock initially contained

 _{${\text{N}}_{\text{0}}{\text{=270μmol+110μmol}}^{\text{238}}{\text{U=380μmol}}^{\text{238}}\text{U}$}

First, let us calculate the decay constant of  _{${}^{238}\text{U}$}

_{$\begin{array}{c}{\text{t}}_{\text{1/2}}\text{ =}\frac{\text{ln(2)}}{\text{k}}\\ \text{k =}\frac{\text{ln(2)}}{{\text{t}}_{\text{1/2}}}\\ \text{ =}\frac{\text{0.693}}{{\text{4.5×10}}^{\text{9}}\text{yr}}\\ {\text{ =1.54×10}}^{\text{-10}}{\text{yr}}^{\text{-1}}\end{array}$}

Calculation of the rock's age

 _{$\begin{array}{c}\text{ln}\left(\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\right)\text{ =-kt}\\ \text{ t =}\frac{\text{-ln}\left(\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\right)}{\text{k}}\end{array}$}

_{$\begin{array}{l}\text{=}\frac{\text{-ln}\left(\frac{{\text{270μmol}}^{\text{288}}\text{U}}{{\text{380μmol}}^{\text{28U}}}\right)}{{\text{1.54×10}}^{\text{-10}}{\text{yr}}^{\text{-1}}}\\ {\text{=2.22×10}}^{\text{9}}\text{yr}\end{array}$}

Hence the rock's age is  _{$2.22\times {10}^9\text{yr}$}