24.104 CP - Chemistry: Molecular Nature Of Matter And Change

Question

${}^{238}U\left(t_{1/2}=4.5\times {10}^{9}yr\right)$ begins a decay series that ultimately forms 206Pb . The scene below depicts the relative number of ${}^{238}U$  atoms (red) and ${}^{206} Pb$   atoms (green) in a mineral. If all the Pb comes from ${}^{238}U$ , calculate the age of the sample.

Step-by-Step Solution

Answer

Age of the sample is  $3.3\times {10}^9$ years.

Step 1: Half-life

The half-life of a radioactive substance is defined as the amount of time required for the decay of half the initial amount of a substance. Half-life is denoted by  $t_{1/2}$.

Radioactive decay is a first-order reaction. The expression for half-life is given below.

     $t_{1/2}=\frac{0.693}{k}$                

Where, k is the decay constant.

Step 2: Calculation of decay constant from half-life.

The rate constant, k, is calculated as follows:

 $$\begin{gathered} t_{1/2}=\frac{\ln 2}{k}k \\ =\frac{\ln 2}{t_{1/2}} \\ =\frac{0.693}{\left(4.5\times {10}^{9}yr\right)} \\ =1.54\times {10}^{-10}y{r}^{-1} \end{gathered}$$

Step 3: Calculate the age of the sample

Considering the given information:

 

No. of  ${}^{206} Pb$ in the sample (green) =9

No. of  ${}^{238}U$ in the sample (red)=6 .

 decays to form ${}^{206} Pb$.

So, the initial amount of  ${}^{238}U$N= 9+6=15 .

The following equation can be used to calculate the sample's age:

 $$\begin{gathered} t=\frac{1}{k}\ln \frac{N_0}{N_t} \\ =\frac{1}{\left(1.54\times {10}^{-10}y{r}^{-1}\right)}\ln \frac{\left(15\right)}{\left(9\right)} \\ =3.3\times {10}^{9}years \end{gathered}$$

Therefore, the age of the sample is  $3.3\times {10}^9$ years.