Question

A laboratory investigation shows that a sample of uranium ore contains $5.37\mathrm{m}\mathrm{g}$ of ${}_{92}^{238}\mathrm{U}$ and $2.52\mathrm{m}\mathrm{g}$of ${}_{82}^{206}\mathrm{P}\mathrm{b}$. Calculate the age of the ore. The half-life of ${}_{92}^{238}\mathrm{U}$is $4.5\times 10^{9}yr$.

Short answer

The age of the ore is calculated as $2.8\times 10^9\mathrm{y}\mathrm{r}\mathrm{s}$.

Step-by-step solution

Concept Introduction

There is a simple formula to calculate the age of the substance:

$\mathrm{t}=\frac{1}{\lambda }\mathrm{l}\mathrm{n}\frac{{\mathrm{n}}_{\mathrm{t}}}{{\mathrm{n}}_0}$

Where$\mathrm{t}$is time,$\lambda$is the decay constant for the radioactive decay,${\mathrm{n}}_0$is the initial concentration of the radioactive substance, and${\mathrm{n}}_{\mathrm{t}}$is the concentration of the radioactive substance over time$\mathrm{t}$.

Calculation for the molar mass

The reaction is as follows:

${}_{92}^{238}\mathrm{U}\to {}_{82}^{206}\mathrm{P}\mathrm{b}$

From the half-life, obtain the value for$\lambda$–

$\begin{array}{c}\lambda =\frac{\mathrm{l}\mathrm{n}(2)}{{\mathrm{t}}_{1/2}}\\ \lambda =\frac{\mathrm{l}\mathrm{n}(2)}{4.5\times 10^9\mathrm{y}}\\ \lambda =1.54\times 10^{-10}{\mathrm{y}}^{-1}\end{array}$

Calculate${\mathrm{n}}_0$by using the mass of rubidium. There is a simple formula to calculate the number of moles from the mass:

$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

Where$\mathrm{M}$is the molar mass.

$\begin{array}{l}{\mathrm{n}}_0=\frac{5.37\times 10^{-3}\mathrm{g}}{238\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}}\\ {\mathrm{n}}_0=2.26\times 10^{-5}\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

Calculation for the age

Lead is formed because of the decomposition of uranium, which means for each mole of lead produced, one mole of uranium needs to be decomposed:

$\begin{array}{l}{\mathrm{n}}_{\mathrm{t}}={\mathrm{n}}_{\mathrm{U}\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}}+{\mathrm{n}}_{Pb}\\ {\mathrm{n}}_{\mathrm{t}}=\frac{5.37\cdot 10^{-3}\mathrm{g}}{238\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}}+\frac{2.52\cdot 10^{-3}\mathrm{g}}{206\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}}\\ {\mathrm{n}}_{\mathrm{t}}=3.48\cdot 10^{-5}\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

Now calculate the age –

$\begin{array}{l}\mathrm{t}=\frac{1}{1.54\times 10^{-11}{\mathrm{y}}^{-1}}\mathrm{l}\mathrm{n}\frac{3.48\times 10^{-5}\mathrm{m}\mathrm{o}\mathrm{l}}{2.26\times 10^{-5}\mathrm{m}\mathrm{o}\mathrm{l}}\\ \mathrm{t}=2.8\times 10^9\mathrm{y}\mathrm{r}\mathrm{s}\end{array}$

Therefore, the value for age is obtained as$2.8\times 10^9\mathrm{y}\mathrm{r}\mathrm{s}$.