Question

In a certain rock, the ratio of lead atoms to uranium atoms is 0.300. Assume that uranium has a half-life of$\mathbf{4}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathit{y}$and that the rock had no lead atoms when it formed. How old is the rock?

Short answer

The rock is $1.69\times {10}^{9}y$old.

Step-by-step solution

Given data

Ratio of lead atoms to uranium atoms,$\frac{N_{Pb}}{N_U}=0.300$

Half-life of uranium, $T_{1/2}=4.47\times {10}^9\mathrm{y}$

Understanding the concept of radioactive dating  

Radioactive dating is a method used by geologists to calculate the age of a sample by knowing the radioactive substance in the sample with its given mass or abundance. Using the exponential function for calculating the number of undecayed nuclei, we can calculate the age of the rock as it decays with the decay of the radioactive nuclides.

Formulae:

The undecayed sample after a given time,$N=N_0{e}^{-\lambda t}................\left(1\right)$

The disintegration constant,$\lambda =\frac{In2}{T_{1/2}}............\left(2\right)$

Where, $T_{1/2}$ is the half-life of the substance.

Calculation of the age of the rock

Using equation (1), the amount of uranium atoms and lead atoms present in the rock at time is given as follows:

$$\begin{gathered} N_U=N_0{e}^{-\lambda t} \\ {N}_{Pb}=N_0-N_U \\ =N_0-N_0{e}^{-\lambda t} \\ =N_0\left(1-e^{-\lambda t}\right) \end{gathered}$$

Now, the ratio of the value of the lead atoms to the uranium atoms can be given as follows that gives the age of the rock as:

$$\begin{gathered} \frac{N_{Pb}}{N_U}=\frac{{\mathrm{N}}_0\left(1-{\mathrm{e}}^{-\mathrm{\lambda t}}\right)}{{\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}} \\ 0.3=\frac{\left(1-{\mathrm{e}}^{-\mathrm{\lambda t}}\right)}{{\mathrm{e}}^{-\mathrm{\lambda t}}} \\ {e}^{\lambda t}-=0.3 \\ {e}^{\lambda t}=1.3 \\ \lambda t=\mathrm{In}\left(1.3\right) \\ \left(\frac{\mathrm{In}2}{{\mathrm{T}}_{1/2}}\right)\mathrm{t}=\mathrm{In}\left(1.3\right)\left(\because \mathrm{from}\mathrm{equation}\left(2\right)\right) \\ t=\frac{\left(4.47\times {10}^9\mathrm{y}\right)\mathrm{In}\left(1.3\right)}{\mathrm{In}2} \\ =1.69\times {10}^9\mathrm{y} \end{gathered}$$

Hence, the age of the rock is $1.69\times {10}^9\mathrm{y}$.