91Ó°ÊÓ

The half-life of a chemical reaction is the time it takes for a particular reactant's concentration to reach of its initial concentration (i.e., the time taken for the reactant concentration to reach half of its initial value).It is commonly represented in seconds and is denoted by the sign ' t_1/2'.

In this problem, we need to determine the amount of ²³⁸U first.

$$\begin{gathered} Mass=0.023g^{206}Pb\left[\frac{1mo{l}^{206}Pb}{206g^{206}Pb}\right]\left[\frac{1mo{l}^{238}U}{1mo{l}^{206}Pb}\right]\left[\frac{238g^{238}U}{1mo{l}^{238}U}\right] \\ Mass=0.0266g^{238}U \end{gathered}$$

Original Amount of ²³⁸U :

$(0.065g+0.0266g{)}^{238}U=0.0916g^{238}U$

We need to identify the constant (k) using the formula for half-life.

$$\begin{gathered} t_{\frac{1}{2}}=\frac{\ln 2}{k} \\ k=\frac{\ln 2}{t_{\frac{1}{2}}} \\ k=\frac{\ln 2}{t_{\frac{1}{2}}} \\ =\frac{\ln 2}{4.5x{10}^{9}yr} \\ =\frac{1.540327x{10}^{-10}}{yr} \end{gathered}$$

Using the expression for finding the number of nuclei remaining, we can solve for the value of t.

$\ln \frac{N_0}{N_t}=kt$

$\ln \frac{0.0916g}{0.065g}=\frac{1.540327×{10}^{-10}×t}{yr}$

$t=2.2271x{10}^{9}years$